TITLE: Arbitrarily large finite irreducible matrix groups in odd dimension? QUESTION [10 upvotes]: I consider a finite irreducible matrix group $\Gamma\subseteq\mathrm{GL}(\Bbb R^d)$. I am interested in the maximal size of $\Gamma$ depending on $d$. But this question makes only sense if there is an upper limit. In even dimension there is no such limit. This is easiest seen in dimension $d=2$, where we have the cyclic groups or dihedral groups of arbitrarily large size. More generally, in dimension $d=2n$ we can consier the $n$-th cartesian power of a regular $k$-gon $P_k$: $$\overbrace{P_k\times \cdots\times P_k}^{\text{n times}}.$$ Its symmetry group is irreducible and gets arbitrarily large with $k\to\infty$. Question: What about odd dimensions? Can there be arbitrarily large finite irreducible matrix groups in dimension $d=2n+1$? For example, in dimension $d=3$ we have the arbitrarily large symmetry groups of prisms and antiprisms, which are reducible. The largest irreducible group is probably the symmetry group of the icosahedron. I have the feeling that in sufficiently large odd dimensions, the largest such group is the reflection group $B_d$. REPLY [13 votes]: Indeed, in odd dimension it's bounded. Indeed, let $\Gamma$ be such a matrix group. By Jordan's theorem, it has a normal abelian subgroup $\Lambda$ of index $\le c_d$. (An explicit bound for $d\ge 71$ is $c_d=(d+1)!$, by work of Collins and Weisfeiler, see Breuillard - An exposition of Jordan's original proof of his theorem on finite subgroups of $\operatorname{GL}_n(\mathbb C)$.) If $\Lambda$ acts diagonalizably, then it has cardinal $\le 2^d$ and hence $\Gamma$ has cardinal $\le 2^dc_d$. Otherwise, $\Lambda$ has blocks of size 2. By irreducibility, the sum of blocks of size two being invariant, it equals $\mathbf{R}^d$. Hence $d$ is even. REPLY [8 votes]: View your group as a subgroup $\Gamma$ of ${\rm GL}(d,\mathbb{C})$ by extending scalars. If it remains irreducible as such , then it has an Abelian normal subgroup $A$ with the index $[\Gamma :A]$ bounded in terms of $d$ (by a Theorem of Jordan). The rank of $A$ is clearly at most $d$, so we only need to consider the exponent of $A$. Let $\chi$ be the character afforded by $\Gamma$. By Clifford's Theorem $\chi$ decomposes on restriction to $a$ as a sum $e(\lambda_{1} + \lambda_{2} + \ldots + \lambda_{t})$, where $et = d$ and each $\lambda_{i}$ is a linear character of $A$, with $\Gamma$ transitively permuting the $\lambda_{i}$. Note that $e$ and $t$ are both odd. However, since $\chi$ is real-valued on restriction to $A$, we see that $\lambda_{i}$ and $\overline{\lambda_{i}}$ occur with equal multiplicity in ${\rm Res}^{\Gamma}_{A}(\chi)$ for each $i$. Since $t$ is odd, at least one $\lambda_{i}$ is real-valued. But $[A:{\rm ker}\lambda_{i}]$ is independent of $i$ by the transitive action of $\Gamma$, and $A/{\rm ker} \lambda_{i}$ is cyclic for each $i$. Hence $[A: {\rm ker} \lambda_{i}] = 2$ for each $i$. Thus $A$ is an elementary Abelian $2$-group in the case that $\Gamma$ reamins irreducible as a complex linear group, and $|\Gamma|$ is bounded in terms of $d$ in that case. In this case, it is true for large enough $d$ that the maximum possible order is attained by the group $(\mathbb{Z}/2\mathbb{Z}) \wr S_{d}$, though this requires theorems of B. Weisfeiler and M.J. Collins, which require the Classification of finite simple groups. More generally, if the character $\chi$ afforded by $\Gamma$ is a sum of real-valued irreducible complex characters, we get a bound in terms of $d$ for $|\Gamma|$ by the same argument (applied to each irreducible summand). Otherwise, (since we are given that $\Gamma$ is irreducible as a real linear group) we may write $\chi = \mu + \overline{\mu}$ for some non-real irreducible complex character $\mu$, contrary to the fact that $d$ is odd.<|endoftext|> TITLE: Does every geometric progression contain a small remainder modulo a large prime? QUESTION [10 upvotes]: The exact question I am interested in is the following. Fix a small $\varepsilon\in(0,1)$ and an integer $q\ge 2$ (you may assume that $q$ is prime if it helps though I believe it shouldn't matter too much). For a large prime $P$ and an integer $a\in\mathbb Z$, define $G(a,P)=\{aq^m\mod P: m=0,1,2,\dots\}$ where the remainders are taken in the range $(-P/2, P/2)$ (i.e., with the minimal possible absolute value). Is it true that for all primes $P$ outside of a set of density at most $\varepsilon$ (in any sense of the word "density" that is subadditive), $G(a,P)$ contains a remainder in the range $(-\varepsilon P,\varepsilon P)$ for every choice of $a\in \mathbb Z$? However I'll be also interested in any nontrivial results in the same direction even if they fall somewhat short of a complete answer (be it affirmative or negative). REPLY [3 votes]: $\newcommand{\F}{\mathbb F}$ $\newcommand{\eps}{\varepsilon}$ (As reqested by the OP, and to address @Mark Lewko's comments, here is the argument showing that the statement is true for the primes satisfying a certain condition; the missing counterpart is to prove that almost all primes satisfy the condition in question.) Claim. Suppose that $p$ is a prime, $H<\F_p^\times$, and $a\in\F_p^\times$. If $|H|>(2\eps)^{-1}\sqrt p\log p$, then the coset $aH$ has a non-empty intersection with the interval $I:=(-\eps p,\eps p)\subset\F_p$. Proof. Let $H^\perp<\widehat{\F_p^\times}$; that is, $H^\perp$ is the subgroup of those multiplicative characters of $\F_p$ with $H$ in their kernel. Assuming for a contradiction that $aH\cap I=\varnothing$, we have $$ \sum_{z\in\F_p^\times} \Big(\sum_{\chi\in H^\perp}\chi(a^{-1}z)\Big)\ \Big(\sum_{g\in I} \sum_{\psi\in\widehat{\F_p}} \psi(z-g)\Big) = 0 $$ where $\psi$ runs over all additive characters of $\F_p$. The contribution of the principal character $\psi=1$ is $$ |I| \sum_{z\in\F_p^\times} \sum_{\chi\in H^\perp}\chi(a^{-1}z) = |I|p; $$ therefore, changing the order of summation and separating the terms with $\psi=1$, we get $$ |I|p \le \sum_{\psi\ne 1} \Big| \sum_{g\in I} \psi(-g)\Big| \sum_{\chi\in H^\perp} \Big|\sum_{z\in\F_p^\times} \chi(a^{-1}z)\chi(z) \Big|. $$ The sum over $z$ is a Gauss sum; as such, it does not exceed $\sqrt p$ in the absolute value. This gives $$ |I|p \le |H^\perp|\,\sqrt p \sum_{\psi\ne 1} \Big| \sum_{g\in I} \psi(-g)\Big|. $$ The outer sum in the right-hand side can be written explicitly as $$ \sum_{u=1}^{p-1} \Big| \sum_{g=-\eps p}^{\eps p} e^{2\pi i ug/p}\Big|, $$ which easily yields the (well-known) upper bound $p\log p$ for the whole sum. As a result, $$ |I|p \le |H^\perp| p^{3/2}\log p $$ and the assertion follows in view of $|H^\perp|=(p-1)/|H|$.<|endoftext|> TITLE: Virtually large groups of small rank (related to 3-manifolds) QUESTION [7 upvotes]: Edited 25.05.21: the assumptions of the question were incorrect, but as the discussion may be helpful for future MOnauts, I'll strike my mistakes and add clearly marked explanations afterwards. I am looking for a reason why a 3-manifold group $G$ that is virtually $\mathbb{Z}\times F$, $F$ being either non-cyclic free or a surface group, does not admit a presentation on two generators. These are the fundamental groups of closed 3-manifolds with $\mathbb{H}^2\times\mathbb{R}$ geometry (Added: thanks @HJRW for pointing out that the strike-through case above corresponds to a non-empty boundary), and it turns out that all other geometries admit examples with fundamental group of rank two, with notable highlight of euclidean geometry where all fundamental groups are virtually $\mathbb{Z}^3$ (and rank two example being the Hantzche–Wendt/Fibonacci manifold). Thus the 3-manifold groups admit examples of virtually high rank groups being nonetheless of small rank themselves. Of course it is well known that a free group on two generators is virtually of arbitrarily high rank. However, by Boileau & Zieschang, Theorem 1.1, the rank of $\mathbb{H}^2\times\mathbb{R}$ manifolds depends on the genus of the base surface and number of singular fibers of the Seifert fibration (and is at least 3), so apparently being virtually $\mathbb{Z}\times F$ forces the group to be of at least the same rank. Added: this is my initial confusion - I assumed that the base orbifold of a $\mathbb{H}^2\times\mathbb{R}$ manifold must have genus at least 2, but this is not true. In fact, following the Wikipedia's conventions for Seifert spaces, $\{-1,(o_1,0);(5,1),(5,2),(5,2)\}$ is a $\mathbb{H}^2\times\mathbb{R}$ manifold Seifert-fibering over a shpere, which in particular fits into Theorem 1.1, case ii) of the cited paper (just don't let the initial $g>0$ mislead you) and is indeed of both rank and genus 2. I thank again @HJRW for their comments which got me on the right track eventually. This of course makes the question that followed invalid. What is the cause that this subgroup bounds the rank of the ambient group from below and, say, free groups or abelian free $\mathbb{Z}^3$ do not? I would be happy if there is a geometric 3-dimensional reason in play here, but would be grateful for refreshing my general group theory as well. REPLY [5 votes]: The question stems from a misinterpretation of Theorem 1.1 in the paper by Boileau and Zieschang. Theorem 1.1 excludes a fair number of cases, in particular, it does not apply to (totally oriented) closed Seifert manifolds with 3 singular fibers and base of genus 0. Some of these excluded Seifert manifolds provide counter-examples to your claim about rank $\ge 3$. For instance, take the exterior $N$ of a $(p,q)$-torus knot which is nontrivial and not the trefoil. The genus of this knot is $$ g=\frac{(p-1)(q-1)}{2}\ge 2 $$ (because I excluded the trefoil which has genus 1). The manifold $N$ is a surface bundle over the circle whose fiber $F$ is the once-punctured surface of genus $g$. The monodromy of this fibration is a finite order (actually, the order is $pq$) homeomorphism $h: F\to F$. Thus, if we collapse the boundary of $F$ to point, we obtain a closed surface $S$ of genus $g$ and $h$ will project to a finite order homeomorphism $f: S\to S$. The mapping torus $M=M_f$ is a Seifert manifold of type ${\mathbb H}^2\times {\mathbb R}$ obtained by a Dehn filling of the boundary of $N$. The base of the Seifert fibration will have three singular points and genus 0: Two of the singular fibers come from $N$ and one comes from the solid torus attached to $\partial N$ as the result of our Dehn filling. (It is a general fact that the mapping torus of a finite order homeomorphism of a hyperbolic surface is a Seifert manifold of type ${\mathbb H}^2\times {\mathbb R}$.) Since the group $\pi_1(N)$ is 2-generated, the quotient group $\pi_1(M)$ is also 2-generated.<|endoftext|> TITLE: How do we describe the right adjoint? QUESTION [6 upvotes]: I am interested in the category-theoretic description of trees (and operads?) and have started a course of study that will allow me to engage with these two (1, 2) manuscripts of Joachim Kock. An essential prerequisite to the early portions of the manuscript involves an understanding of a pair of adjoint functors. The left adjoint is the change of basis functor $g^{\ast}: \mathbf{Set}_{/A} \rightarrow \mathbf{Set}_{/B}$ associated with the change of basis between sets $g: B \rightarrow A$. I find it very easy to reason about the image of a bundle under the change of basis functor. This functor takes a bundle $f: X \rightarrow A$ to the pullback of $f$ by $g$, which I can easily interpret in $\mathbf{Set}_{/B}$ as the fibered product with the canonical projection $h: X \times_{A} B \rightarrow B$ given by $h: (x, b) \mapsto b$. Kock utilizes the right adjoint, which I have read is called the dependent product: $$ (g^* \dashv \prod_g) \colon \mathbf{Set}_{/B} \stackrel{\overset{g^* }{\leftarrow}}{\underset{\prod_g}{\to}} \mathbf{Set}_{/A} \,. $$ I am trying to understand this functor, but am finding it very difficult. Are there any set-theoretic descriptions of the image of a bundle $f: X \rightarrow B$ under $\prod_g$ that would help me in this context? Although I've only had a little exposure to it, the process of re-describing adjoint functors in the "internal language of their categories" (I hesitate to use this phrase here-- I have seen it used in the literature of toposes and do not know its formal definition-- perhaps I'll just call this process reifying?) has proven very difficult for me. Are there any mental algorithms that offer any help? Are there a set of useful exercises I can undertake to develop the skill? Are there theorems I can study that will provide insight into the process? It has struck me in the course of mulling over the problem that I would also like to know whether category-theorists even bother with reifying their constructions. Does this process have a name? Is it done often? REPLY [7 votes]: It's worth thinking about the simplest example, namely when $A$ is a single point. In this case, $g^*$ is the product functor $-\times B$, and its right adjoint is the set of sections: it sends $f\colon X\to B$ to the set $\Gamma(B,X)$ of sections $s\colon B\to X$ of $f$. This is vaguely similar behaviour to how the internal hom in a cartesian closed category is right adjoint to taking a product. In particular, if $X = Y\times B = g^*Y$, then $\Pi_g f = Y^B$. The space of sections is the same thing as the product of all the fibres of $f$, so one can think of the general dependent product this way, now just parameterised by the elements of $A$. Given a point $a\colon *\to A$, which is an object of $\mathbf{Set}/A$, we can calculate $$ \mathbf{Set}/A(a,\Pi_g f) \simeq \mathbf{Set}/B(B\times_A *,f) = \mathbf{Set}/B(g^{-1}(a),f) $$ Now the latter is the set $\Gamma(g^{-1}(a),X)$ of sections of $f$ over $g^{-1}(a)\subset B$. But $\mathbf{Set}/A(a,\Pi_g f)$ is the fibre of $\Pi_g f \to A$ over $a$, so $$ \Pi_g f = \coprod_{a\in A} \Gamma(g^{-1}(a),X) \to A. $$ A similar mental picture works, to some extent, in categories that behave like the category of sets, except that you shouldn't think of $\Pi_g f$ as a disjoint union, but something more like a continuous family of spaces of sections. Note howoever, how I used the defining property that it was a right adjoint to get a handle on the construction. In general, one can use generalised elements in place of $1\to A$ to construct the thing, at least as a presheaf, and then knowing the dependent product exists means this is representable.<|endoftext|> TITLE: What is the direct proof of the recurrence relation about lattice path enumeration given by Bizley? QUESTION [6 upvotes]: Let $k$ be a nonnegative integer and let $m,n$ be coprime positive integers. Let $\phi_k$ be the number of lattice paths from $(0,0)$ to $(km,kn)$ with steps $(0,1)$ and $(1,0)$ that are never rising above the line $my=nx$. A path having this property will be called a $\phi$-path. Then, $\phi_k$ satisfies the recurrence relation $$ k(m+n)\phi_k = \sum_{j=1}^{k}\binom{j(m+n)}{jm}\phi_{k-j} $$ for all $k \in \mathbb{Z}^+$, as shown by Bizley (1954). Bizley has stated that “these relations can be deduced directly by general reasoning from the geometrical properties of the paths”. However, I could not manage to obtain a combinatorial proof of this theorem. Question: What is the direct proof of the recurrence relation mentioned above? My first thought about this relation was that the left-hand side of the equation counts the number of the cyclical permutations of all $\phi$-paths from $(0,0)$ to $(km,kn)$. In his paper, Bizley defines the highest point of a lattice path as “a lattice point $X$ on the path such that the line of gradient $\frac{n}{m}$ through $X$ cuts the y-axis at a value of $y$ not less than that corresponding to any other lattice point of the path”. (It is important to note that the first point $(0,0)$ is regarded as not belonging to the path.) Thus, the number of the cyclical permutations of all $\phi$-paths may be expressed as the sum of $t$ times the number of all the lattice paths with exactly $t$ highest points for all $t=1,2,\ldots,k$. However, apparently the right-hand side of the equation has nothing to do with the number of the lattice paths with a specified number of highest points. I am afraid I am missing something obvious about the geometrical properties of the $\phi$-paths and I would be so glad if anyone can provide a combinatorial proof or trick that I could not manage to see. Thanks for your attention in advance. REPLY [3 votes]: The trick is to add $\phi_k$ to both sides of the equation, and interpret the left hand side as counting paths with a prepended horizontal step, and one of its steps marked. Then, make the marked step the first step of a path from $(-1, 0)$ to $(km, kn)$. Let $j$ be minimal such that this path hits $(jm, jn)$ and stays below $my = nx$ after that. Then the steps before the meeting point, excluding the final horizontal step, form an arbitrary path counted by the binomial coefficient in the summand with index $j$, and the remaining steps form a path counted by $\phi_{k-j}$.<|endoftext|> TITLE: Cutting a Julia set into infinitely many pieces at finitely many points QUESTION [6 upvotes]: Let $f\colon \widehat{\mathbb{C}}\to \widehat{\mathbb{C}}$ be a rational function of degree two or greater whose Julia set $J_f$ is connected. If $S\subseteq J_f$ is a finite set of periodic points, is it possible that the complement $J_f\setminus S$ has infinitely many connected components? I am particularly interested in the case where $f$ is hyperbolic. REPLY [9 votes]: For a polynomial, this is equivalent to asking whether there can be infinitely many external rays landing at a single point. This could happen only if the function has a Cremer point (i.e., a non-linearisable irrationally indifferent periodic point). If the Cremer point is accessible from the complement, then any external ray landing at it would have to be non-periodic by the so-called "snail lemma", which means that then there would indeed be infinitely many rays landing at the same point. However, I believe that it is still an open question whether or not this actually happens. Since hyperbolic maps have no Cremer points, what you ask about is impossible in the hyperbolic case. I think that, similarly, for hyperbolic rational maps it is impossible, and probably more generally whenever the Julia set is locally connected.<|endoftext|> TITLE: Non-measurable sets on groups from translation invariance QUESTION [9 upvotes]: The most well-known construction of a non-measurable set is the Vitali set. The idea behind Vitali sets is to split up the space (such as $[0,1]$) into equal-sized copies (guaranteed by translation invariance), by looking at something like $\mathbb{R}/\mathbb{Q}$. This same idea is used in "Visualizing a Nonmeasurable Set" to construct non-measurable sets on the torus. Another construction I know about is on the probability space of infinitely many coin tosses $\{ 0, 1 \}^\mathbb{N}$. In this case, instead of modding out by $\mathbb{Q}$, you can mod out by "switching the outcome of finitely many coins". This approach is taken in these probability lecture notes. All of these constructions seem closely related: each time, we have a way to decompose our set into "translation-invariant" copies. My question is how these sorts of constructions of non-measurable sets generalize. In the Wikipedia article on Haar measure I read: Unless $G$ is a discrete group, it is impossible to define a countably additive left-invariant regular measure on all subsets of $G$, assuming the axiom of choice, according to the theory of non-measurable sets. This seems very close to an answer to my question, but the Wikipedia article doesn't elaborate here. So, how does the construction of non-measurable sets on a non-discrete group $G$ work? Is the general intuition similar to the examples I've described here? What happens for discrete groups (could this be related to amenable groups, which I know you can put measures on)? REPLY [4 votes]: The proof for the reals can be generalized to any non-discrete locally compact group $G$. We let $K \subset G$ be any compact set with positive Haar measure $\lambda(K) > 0$ (e.g., $K = [0, 1]$ when $G = \mathbb R$), and we let $\Lambda < G$ be any subgroup such that $\Lambda \cap KK^{-1}$ is countably infinite (e.g., $\Lambda = \mathbb Q$ when $G = \mathbb R$). We define an equivalence relation on $G$ by the $\Lambda$-orbits coming from left multiplication and we let $V \subset K$ be a set containing exactly one representative of each equivalence class that intersects non-trivially with $\Lambda K$. We then have $K \subset (\Lambda \cap K K^{-1}) V \subset K K^{-1} K$. The first inclusion here follows from the fact that if $k \in K$, then we may find $t \in \Lambda$ such that $tk \in V \subset K$. It then follows that $t = (tk) k^{-1} \in K K^{-1}$ and hence $k \in (\Lambda \cap K K^{-1})V$. If we were able to extend the Haar measure to a countably additive left-invariant measure defined on all subsets of $G$, then we would have $\lambda(( \Lambda \cap K K^{-1} ) V) = \sum_{t \in \Lambda \cap K K^{-1}} \lambda(V) \in \{ 0, \infty \}$, but this would then contradict the inequalities $$ 0 < \lambda(K) \leq \lambda( ( \Lambda \cap K K^{-1} ) V ) \leq \lambda(K K^{-1} K) < \infty. $$<|endoftext|> TITLE: how mathematicians do research? QUESTION [9 upvotes]: I am a first-year physics graduate student with a deep interest in mathematics. I am specifically interested in algebraic geometry and algebraic topology, and I would like to employ advanced mathematics in my physics research. So far, I haven't chosen an advisor. However, I have been trying to do research on my own without any guidance for a few years and I failed miserably. I do not have any good idea how the best mathematicians do research. I will try my best to summarize one of the problems I am currently facing hoping that someone might help: I never start by stating some open research problem. I start with the goal of understanding some well-known result or subject. In particular, suppose I want to understand the known ideas about the quantum dynamics of $\mathcal{N}=2$ Super Yang-Mills theory. I try to derive everything starting from scratch e.g. by deriving the SUSY algebra, Lagrangian, and so forth. I try to do this in my own way. I pretend that these results are unknown, and I try to re-derive them. However, it seems I'm not able to go so far in the subject. I'm not sure what the reason is. I believe that $\mathcal{N}=2$ supersymmetric theory is connected to some ideas in mathematics, such as algebraic geometry and supergeometry, so I try to motivate the study of these mathematical subjects using supersymmetric dynamics. My question is: is this how mathematicians do research? is this approach good? I find that recent papers are impenetrable because they assume I have a lot of background knowledge/ did a lot of computations that I did not do. So I have to study the background. However, when I decide to study the background, it takes me a lot of time because it seems I'm trying to re-invent the wheel, re-deriving everything on my own. I hope my question is not too vague. Any help would be greatly appreciated. REPLY [2 votes]: To answer your original question, you could try reading these books: J. Hadamard The Psychology of Invention in the Mathematical Field (1954) J.E. Littlewood, "The Mathematician's Art of Work" in Littlewood's miscellany (revised edition, 1986) I. Stewart Letters to a Young Mathematician (2006) C. Villani Birth of a Theorem (2015) However, I suspect that your real question is along the lines of "what's the most effective way of learning algebraic geometry and algebraic topology, so I can use them as tools in my research?" The resources at https://mathvault.ca/ may be useful for this purpose.<|endoftext|> TITLE: What did Paul Cohen mean by saying that generic sets of natural numbers have "no asymptotic density?" QUESTION [12 upvotes]: In Paul Cohen's original 1963 paper on forcing, The independence of the Continuum Hypothesis, published in PNAS, he gives his general proof sketch of how he intends to create a model of ZFC that doesn't support CH: Start with a countable model of ZFC. Within the model, there are the model's cardinals $\aleph_0^*, \aleph_1^*, \aleph_2^*,$ etc, which are different from the cardinals in the ambient theory $\aleph_0, \aleph_1, $ etc. Cohen then proceeds building an ordinal-length sequence of subsets of the true $\aleph_0$ which is of length $\aleph_2^*$, and uses this to build a new model in which CH doesn't hold. Of these subsets, Cohen writes: "Only those properties which are true in a "uniform" manner for "generic" subsets of $\omega$ in $\mathcal{M}$ [the first model] shall be true for the $a_\delta$ [the ordinal-length sequence of subsets] in $\mathcal{N}$ [the new model]. For example, each $a_\delta$ contains infinitely many primes, has no asymptotic density, etc." I interpret "no asymptotic density" as meaning the natural density doesn't converge to anything as $N \to \infty$. I am not sure what Cohen means with the language "generic," but if the most straightforward interpretation is that it means "almost all," then this doesn't seem like a true statement for "generic subsets" of $\Bbb N$, as almost all have natural density equal to 0.5. If identify any such subset with its characteristic function from $\mathbb N \to \{0, 1\}$, then you have an infinite length binary string, and it would seem that almost all of those have 50% 0's and 50% 1's, similarly to how almost all real numbers have 50% of their bits equal to 0 and 50% equal to 1 in their binary expansion. Does anyone have clarification on this? Is this a simple error or is there a different interpretation here of a "generic" subset? REPLY [6 votes]: To compliment the comments and Nate Eldredge's answer, here's a useful lemma for showing that a given property $P$ of infinite binary sequences is comeager: Lemma: Consider the game $G_P$ where players $1$ and $2$ alternate playing finite binary strings for $\omega$-many moves, and then player $2$ wins iff the concatenation of those strings has property $P$. Then the following are equivalent: Player $2$ has a winning strategy in $G_P$. The set of $f$ with property $P$ is comeager in $2^\omega$. So for example: The set of sequences with infinitely many $0$s is comeager: just have player $2$ always play "$0$." The set of sequences with arbitrarily long strings of $1$s is comeager: just have player $2$ play a length-$i$ string of $1$s on their $i$th move. And, as desired: The set of sequences with undefined asymptotic density is comeager: have player $2$ alternately play "really long" strings of $0$s or of $1$s. Note that here we have to take the opponent's behavior into account in a meaningful way, since the right notion of "really long" depends on the total length of the sequence built so far, not just how many moves have been played. In my experience it's much easier to think in terms of winning strategies than in terms of (co)meagerness per se. Note that while I've phrased the lemma for the specific context of $2^\omega$, it holds much more generally (and so we can use the same intuition when working with wildly different forcings).<|endoftext|> TITLE: Constant Gaussian curvature disks QUESTION [7 upvotes]: This question has also been posted on MSE, but maybe here is the right place to post it. Is it true that if $D$ is a Riemannian $2$-disk having constant Gaussian curvature equal to $1$ and whose boundary has constant geodesic curvature, then $D$ is isometric to some geodesic ball of the unit sphere $\mathbb{S}^2 \subset \mathbb{R}^3$? I strongly suspect so, but I couldn't find a reasonable argument. REPLY [5 votes]: This is an addendum to the proofs by Anton and Deane, completing the missing part of the argument. Lemma. Let $D$ be the closed unit disk and $f: D\to S^2$ an immersion such that $f(\partial D)$ is a circle $C$ in $S^2$. Then $f$ is 1-1. Proof. Let $J: S^2\to S^2$ denote the reflection in $C$. Double $D$ cross its boundary to obtain the 2-sphere $\Sigma$ and let $j: \Sigma\to\Sigma$ denote the reflection in $\partial D$. Then extend $f$ to a local homeomorphism $F: \Sigma\to S^2$ by $$ F(j(z))=J f(z). $$ Since $\Sigma$ is compact, $F$ is a covering map, hence, a homeomorphism. Thus, $f$ is 1-1. qed Edit. As for the existence of the isometric immersion $\iota$ in Anton's answer (above, $\iota=f$), it is an application of Riemann's theorem (the local form of the Killing–Hopf theorem): It shows that for every $z\in D$ there exists neighborhood $U\subset D$ and an isometric embedding $U\to S^2$. Joe Wolf in his book attributes the local result to Riemann and he is probably right, but it is likely (since it is about surfaces) that Gauss already knew how to prove this. Since $D$ is simply-connected, these local isometries can be combined to produce a globally-defined isometric immersion, see for instance, the answer to this question.<|endoftext|> TITLE: Two inequalities in $C^*$ algebras QUESTION [9 upvotes]: Under what conditions on a $C^*$ algebra $A$ we have the following inequality: $$x^*a^*ax+a^*x^*xa\leq x^*x+a^*x^*ax+x^*a^*xa\;\;\; \forall x,a\in A$$ The second identity which I am looking for is the following: Does the following inequality imply that the algebra is commutative: $$xx^*\leq k x^*x\;\;\forall x\in A$$ for some positive real $k$? What about if we assume that $k$ is a fixed positive elemnt of the algebra rather than a positive scalar? The second question is motivated by the fact that every Banach algebra which satisfies $|ab|\leq k|ba|,\;\;\forall a,b$ is necessarily a commutative algebra. The proof is based on Liouville theorem. But in our case the conjucation arising from $*$ operation destroy holomorphicity. REPLY [7 votes]: The second condition also implies that $A$ is commutative. If $A$ is not commutative then it has an irreducible representation on some Hilbert space $H$ of dimension at least $2$. Find unit vectors $v,w \in H$ with $\langle v, w\rangle = 0$. By Kadison transitivity there exists $x \in A$ with $xv = 0$ and $xw = v$. Then $\langle x^*xv, v\rangle = \|xv\|^2 = 0$ but $\langle xx^*v, v\rangle = \|x^*v\|^2 \neq 0$ because $\langle x^*v, w\rangle = \langle v, xw\rangle = 1$. So $xx^* \leq kx^*x$ is impossible. (The second part of the question doesn't really make sense because if $k \in A$ then $kx^*x$ will not be positive in general. You could ask about $xx^* \leq x^*kx$, but this would imply $xx^* \leq \|k\|x^*x$ and therefore $A$ must be commutative by the scalar case.)<|endoftext|> TITLE: Are there "typical" formal systems that have mutual consistency proofs? How long a chain of these can we build? QUESTION [5 upvotes]: Sufficiently powerful theories (Peano arithmetic, ZFC, and so on — this question came from thinking about Coq) can't prove their own consistency. However, are there cases of two theories, $A$ and $B$, where $A$ proves $B$ is consistent and $B$ proves $A$ is consistent? (To make up a potential example, "Peano arithmetic proves ZFC is consistent, and ZFC proves Peano arithmetic is consistent".) If so, are there long chains of these sorts of proofs we can build, so that, if any of $k$ theories were inconsistent, all of them would be? (The context here is idle curiosity about whether we can get in-practice reassurances about our theories by noting that many separate systems would need to have "bugs" at once.) REPLY [5 votes]: No, this cannot happen, although it's a little bit trickier than one might expect to prove this! First, a miniature result: Suppose $T,S$ are computably axiomatizable theories in the language of arithmetic, each containing the theory $\mathsf{I\Sigma_1}$, with $T\vdash Con(S)$ and $S\vdash Con(T)$. Then $T$ and $S$ are inconsistent. If you haven't seen $\mathsf{I\Sigma_1}$ before, the only points you need to know are that it is finitely axiomatizable, strong enough for Godel's theorems to be applicable, and self-provably $\Sigma_1$-complete. Note that neither of the better-known arithmetics $\mathsf{Q}$ or $\mathsf{PA}$ will suffice: $\mathsf{Q}$ doesn't prove its own $\Sigma_1$-completeness since it lacks induction, and $\mathsf{PA}$ isn't finitely axiomatizable. PROOF. It will be enough (by symmetry) to show that $T$ is inconsistent. Since $\mathsf{I\Sigma_1}$ is finitely axiomatizable and proves its own $\Sigma_1$-completeness, we have that $T$ proves "$S$ is $\Sigma_1$-complete:" just verify in $T$ an $S$-proof of any single sentence axiomatizing $\mathsf{I\Sigma_1}$. Consequently, $T$ proves the sentence $\neg Con(T)\rightarrow [S\vdash (\neg Con(T))]$. On the other hand, since $S\vdash Con(T)$ and $T$ is $\Sigma_1$-complete we have that $T$ proves $S\vdash Con(T)$. Putting this together with the above paragraph, we get a $T$-proof of "If $T$ is inconsistent, then $S$ proves $Con(T)\wedge\neg Con(T)$" - that is, a $T$-proof of $\neg Con(T)\rightarrow\neg Con(S)$. But since $T\vdash Con(S)$, this gives a $T$-proof of $Con(T)$ - so $T$ is inconsistent. The above can be improved, however. First there's the issue of generalizing beyond $n=2$. This isn't very interesting though, since it's clear how to proceed: simply iterate the above idea by applying "provable $\Sigma_1$-completeness" over and over again. More interestingly there's the language issue: $\mathsf{ZFC}$ for example is not a theory in the language of arithmetic, so the above result doesn't immediately apply to it. This can be handled via the notion of an interpretation. Basically, a theory $A$ interprets a theory $B$ if there is some tuple of formulas $\Phi_A$ in the language of $A$ such that for each sentence $\varphi\in B$, the theory $A$ proves that the structure defined by $\Phi_A$ satisfies $\varphi$. (Think about how $\mathsf{ZFC}$ implements arithmetic via the finite ordinals, for example.) Via interpretations, we can generalize the argument above to arbitrary languages. Combined with the generalization past $n=2$ above, this gives the stronger result: Suppose $T_1,...,T_n$ are computably axiomatizable theories, each of which interprets $I\Sigma_1$, such that $T_1\vdash Con(T_2)$, $T_2\vdash Con(T_3)$, ..., $T_n\vdash Con(T_1)$. Then each $T_i$ is inconsistent. The most difficult part here is being precise about what "$Con(-)$" should mean in each of the relevant languages (basically, we just "work along interpretations"). The final improvement to be made is with respect to the base theory. We can replace $\mathsf{I\Sigma_1}$ with substantially weaker theories without changing the argument, but this doesn't get us all the way to $\mathsf{Q}$. So - dropping back to a more manageable level of generality along the other axes - we're left with a natural question: Can there be two computably axiomatizable consistent theories $T,S$ in the language of arithmetic containing $\mathsf{Q}$ such that $T\vdash Con(S)$ and $S\vdash Con(T)$? As Emil Jerabek comments below, the answer is still negative. However, at this point I'm not familiar with the relevant methods, so I can't say anything meaningful. REPLY [4 votes]: No, it is not possible for two sufficiently strong, consistent theories to prove each other's consistency. Here's a sketch of the proof: Suppose A and B are sufficiently strong and prove each other's consistency. Then A proves every true $\Sigma^0_1$ sentence. (This needs only that $A$ extends Robinson's Q.) Furthermore, Peano arithmetic PA proves this fact, and therefore so does B (being sufficiently strong). Apply that to the negation of an arbitrary $\Pi^0_1$ sentence $\pi$, and you find that B proves that "if $\pi$ is consistent with A then $\pi$ is true." Consider the particular $\Pi^0_1$ sentence $\pi$ expressing "B is consistent" and remember our assumption that A proves this $\pi$. The fact that A proves $\pi$ is a $\Sigma^0_1$ statement, so it's provable in B. That lets us simplify the conclusion of the previous paragraph to: B proves "if A is consistent then B is consistent." But by assumption, B proves "A is consistent" and therefore B also proves "B is consistent." But now Gödel's theorem tells us that B is inconsistent. Of course, a symmetrical argument gives that A is also inconsistent. Alternatively, we can argue that the inconsistency of B, being a $\Sigma^0_1$ truth, is provable in A. But so is "B is consistent" by assumption, and thus we have an inconsistency in A. REPLY [2 votes]: Would you be happy with equiconsistency? That is, $A$ being consistent would imply $B$ consistent, and vice versa? There are lots of cases like this! I guess to make things rigorous, one would have to say how one is supposed to count the different theories, but presumably mutually contradictory theories should count as different, or theories that are proper extensions (as in, they prove strictly more theorems). Under this notion, you can say that ZF, ZFC, ZFC+CH, ZFC+(not CH) are all equiconsistent, for example. There are many more examples that can be added to this list, via models constructed by forcing. Set theorists measure the different classes of such theories via the large cardinal hierarchy.<|endoftext|> TITLE: Vopěnka's principle and contravariant full embeddings between module categories QUESTION [19 upvotes]: I was recently reminded about this old question on math.stackexchange. Let $\operatorname{Mod}R$ be the category of (right) modules for a ring $R$. The questioner mistakenly thought that the Freyd-Mitchell embedding theorem implied that for every ring $R$ there was another ring $S$ and a full embedding $(\operatorname{Mod}R)^{\text{op}}\to\operatorname{Mod}S$, and asked for an explicit description of such an $S$. I realized (see this answer) that general facts about locally presentable categories imply that, if Vopěnka's principle is true, then there is no such embedding (for nonzero $R$). Is there a proof that doesn't assume Vopěnka's principle? Or could it be that this fact genuinely depends on very strong large cardinal assumptions? REPLY [11 votes]: Assuming (M) (= there is only a set of measurable cardinals), the category $\bf{Vec}$ of vector spaces (over every field) has a small dense subcategory. This is an old result of Isbell (see also https://arxiv.org/pdf/1812.10649.pdf). Hence $\bf{Vec}$$^{\text{op}}$ is boundable, i.e., it can be fully embedded to a category of algebras (I do not know whether to a category of modules). Hence the set-theoretical strength of "no opposite category of modules is boundable" lies between $\neg$(M) and VP. Moreover, the existence of a full embedding of $\bf{Ab}^{op}$ to $\bf{Mod}$ $R$ implies WVP (weak Vopěnka's principle) is false. Indeed, the accessible category of graphs with monomorphisms can be fully embedded to the category $\bf{Gra}$ of graphs. A. J. Przezdziecki https://arxiv.org/pdf/1104.5689.pdf constructed an embedding $G:\bf{Gra}\to\bf{Ab}$ such that $\bf{Ab}$$(GX,GY)$ is the free abelian group on $\bf{Gra}$$(X,Y)$. This yields $R$-modules $A_i$ indexed by ordinals such that the only morphisms $A_i\to A_j$, $i TITLE: Idempotent Laurent polynomials (in noncommuting variables) QUESTION [7 upvotes]: Let $K$ be a field and $R=K\langle X_1,\dots,X_n,X_1^{-1},\dots,X_n^{-1}\rangle$ the Laurent polynomial ring in $n$ noncommuting variables. Can $R$ have idempotents distinct from $0$ and $1$? REPLY [6 votes]: Here's a self-contained proof (which is certainly Higman's proof), following Fedor Petrov's answer. Let $G$ be a locally indicable group (= every nontrivial f.g. subgroup has $\mathbf{Z}$ as quotient). The $KG$ has no nontrivial idempotent. Indeed, suppose $u^2=u$ in $KG$ with $u\neq 0,1$. Then passing to the subgroup generated by $\mathrm{Supp}(u)$, we can suppose that the support of $u$ generates $G$, and that $G$ is finitely generated. Clearly $G\neq 1$. Then fix a surjective homomorphism $G\to\mathbf{Z}$. Push forward $u$ to $K[\mathbf{Z}]$ to get an idempotent, whose support generates $\mathbf{Z}$. But $K[\mathbf{Z}]=K[t^{\pm 1}]$ has no nontrivial idempotent, contradiction.<|endoftext|> TITLE: Are the class numbers of $\mathbb{Q}(\cos(2\pi / m))$ $O(m^n)$ for some fixed $n$? QUESTION [9 upvotes]: Question: Are the class numbers of $\mathbb{Q}(\cos(\frac{2\pi}m))$ $O(m^n)$ for some fixed $n$? Evidences (e.g. a recent paper) showing that the question above is open are also OK. Remark: If such $n$ exists, then $n\geq2$. By the paper unit groups and class numbers of real cyclic octic fields by Yuan-Yuan Shen, there exists infinitely many cyclic octic fields with conductor $32b$ and class numbers at least $c(\epsilon)b^{3-\epsilon}$ for every $\epsilon>0$. As every number field extension $E/F$ satisfies $h(F)$ divides $[E:F]h(E)$, the class numbers of such $\mathbb{Q}(\cos(\frac{2\pi}{32b}))$s are at least $c(\epsilon)b^{2-\epsilon}$. A key feature of Yuan-Yuan Shen's octic fields is that they have regulators of size $\log^6(b)$. If such an infinite family of number fields (i.e. real abelian with poly-logarithmic regulators) exists for every fixed degree, the answer for this question will be "no". REPLY [4 votes]: The answer is no. By Proposition 2 of Gary Cornell and Michael I. Rosen 's paper The -rank of the real class group of cyclotomic fields (paraphased): Let $L/\mathbb Q$ be an abelian -extension of the rationals with Galois group $G$. Assume the inertia group of $2$ is cyclic. If $G$ is the direct sum of $t$ cyclic groups and $s$ finite primes of $\mathbb Q$ ramify in $L$ then the -rank of the class group of $L / \mathbb Q \geq \frac{t(t-1)}{2}-s-e$, where $e=0$ for $$ odd. Let $A$ be the first $n$ primes with the property that $x^3-2$ has three roots in $\mathbb F_a$ ($a\in A$). By Chebotarev's density theorem, the largest element of $A$ is $O(n \log n)$. Let $L_a$ be the cubic field with conductor $a$. Such cubic fields must exist, as the property of $a$ ensures that $1$ has three cubic roots in $\mathbb F_a$, so $a=1 \text{ mod }3$. Notice that $2$ splits completely in $L_a$. Let $L$ be the compositium of all $L_a$ s. Then $L/\mathbb Q$ is an abelian $3$-extension with Galois group $C_3^n$, and $2$ splits completely (so the inertia group of $2$ is trivial). There are only $n$ primes that ramify in $L$, so the size of the class group is at least $\exp (c_1n^2)$ for some $c_1>0$. Also the conductor is bounded above by $\exp(c_2n\log^2 n)$ for some $c_2$, so the extension $\mathbb Q (2 \cos (2\pi / \prod_{a\in A} a))/L$ has degree at most $\exp(c_2n\log^2 n)$, and thus the class number of $\mathbb Q (2 \cos (2\pi / \prod_{a\in A} a))/L$ is at least $\exp(c_1n^2-c_2n\log^2 n)$. This is asymptotically larger than $\exp(c_2n\log^2 n)^m$ for any $m$.<|endoftext|> TITLE: Subwords of the infinite Fibonacci word QUESTION [12 upvotes]: Let $W = 01001010010010 \ldots$ be the infinite Fibonacci word, A003849 in the OEIS. Let $B(m)$ be the set of $m+1$ subwords of $W$ that have length $m$, and for each such subword $u$, let $p(u)$ be the sequence of positions in $W$ where $u$ begins. Then $p(u)$ is a composite of the famous Wythoff sequences $A=$A000201 and $B=$A001950. Can someone figure out (or cite a reference) exactly which composites represent the subwords in $B(m)$? Here's how it looks for $m=4$: \begin{array}{|c|c|c|c|} \text{subword, } u & \text{positions, } p(u) & \text{composite} & \text{OEIS} \\ 0100 & 1,6,9,14,\ldots & AAA & A134859 \\ 1001 & 2,7,10,15,\ldots & BA & A035336 \\ 0010 & 3,8,11,16,\ldots & AB & A003623 \\ 0101 & 4,12,17,25,\ldots & AAB & A134860 \\ 1010 & 5,13,18,26,\ldots & BB & A101864 \end{array} (Links: A134859, A035336, A003623, A134860, A101864) (Note that for every $m$, the difference sequence of every $p(u)$ consists of Fibonacci numbers.) Following Sam Hopkins's note, here's a definition. Start with $0$ and apply the substitutions $0 \rightarrow 01$ and $1 \rightarrow 0$ repeatedly, like this: $$0,01,010,01001,01001010,0100101001001,\ldots.$$ The limiting word is A003849, one of several called the infinite Fibonacci word, but this one is regarded as the standard form, according to the Crossrefs section of A014675. Writing those words as $w_0,w_1,w_2,\ldots$, respectively, note that $w_n$ is, for $n \geq 2$, the concatenation indicated by $w_n=w_{n-1}w_{n-2}$, so that the length of $w_n$ is a Fibonacci number. Some more background: suppose that $w$ is a word in $B(m)$. Then at least one of the words $w0$ and $w1$ must be in $B(m+1)$. However, there is only one $w$ in $B(m)$ such that both $w0$ and $w1$ are in $B(m+1)$. Such a "splitter" turns out to be simply a reversal of an initial word of $W$, so that the first few splitters are $,0,10,010,0010,10010,\ldots$. The corresponding Wythoff composites are $$A,B,AA,AB,BA,AAA,BB,AAB,ABA,BAA,AAAA,ABB,\ldots$$ So, if someone can tell specifically how to generate this sequence, the problem will be solved. REPLY [3 votes]: Such a "splitter" turns out to be simply a reversal of an initial word of $W$, so that the first few splitters are $,0,10,010,0010,10010,\ldots$. The corresponding Wythoff composites are $$A,B,AA,AB,BA,AAA,BB,AAB,ABA,BAA,AAAA,ABB,\ldots$$ So, if someone can tell specifically how to generate this sequence, the problem will be solved. I construct some parallel families of finite sequences in "generations", where the full sequence is obtained by concatenating all of the generations in order. Let $W$ denote the infinite Fibonacci word. Let $P_0 = [w_0]$, $P_1 = [w_1, w_1 w_0]$, $P_{k+2} = w_{k+2} P_k + w_{k+2} P_{k+1}$. It's easy to show by induction that the lengths of the words in $P = P_0 + P_1 + P_2 + \cdots$ are $1, 2, 3, \ldots$ and that the words themselves are all prefixes of $W$. (A useful lemma to show the latter is that $w_n w_{n-1} \cdots w_0$ is a prefix of $w_{n+2}$). Let $r_k = \textrm{reverse}(w_k)$; equivalently $r_0 = 0$, $r_1 = 10$, $r_{k+2} = r_k r_{k+1}$. (Note that the first symbol of $r_i$ is $i \bmod 2$). Then the family $R_k$ is defined inductively as $R_0 = [r_0]$, $R_1 = [r_1, r_0 r_1]$, $R_{k+2} = R_k r_{k+2} + R_{k+1} r_{k+2}$. By construction, the $i$th word in $R_k$ is the reverse of the $i$th word in $P_k$, and therefore $R = R_0 + R_1 + R_2 + \cdots$ is the sequence of the (non-empty) splitters. A point of interest is that the last word in $R_k$ is $r_0 r_1 \cdots r_k$, and this is a prefix of $W$ (see e.g. proposition 17 of Factorizations of the Fibonacci Infinite Word, Fici, J. Int. Seq. article 15.9.3). But it's also the reverse of a prefix, and therefore is a palindrome. Then we can work backwards along the generation and say that the $i$th last word in $R_k$ can be found at position $i$ in $W$. The main sequence is $G_0 = [A]$, $G_1 = [B, AA]$, $G_{k+2} = G_k B + G_{k+1} A$. Clearly every (non-empty) Wythoff composite occurs exactly once in the sequence $G = G_0 + G_1 + G_2 + \cdots$. It will be useful to note that if $A$ has weight $1$ and $B$ has weight $2$ then every composite in $G_k$ has weight $k+1$. (In fact, $G_k$ contains precisely the composites of weight $k+1$ in lexicographic order of their reversals, but we won't use this). Assuming the claim made in the question that compositions of $A$ and $B$ exactly index the splitters (which I don't know how to prove), we can show that the $i$th element of $G_k$ exactly indexes the $i$th element of $R_k$, so that $G$ is the desired sequence. Observe that since $A$ and $B$ partition the domain of a Wythoff composite $S$, $SA$ and $SB$ partition its range. If the longest word indexed by $S$ is the splitter $s \in R_k$, then the construction of the generations of $R$ tells us that $sr_{k+1}$ and $sr_{k+2}$ are both splitters whose common prefix is precisely $s$. Take the inductive hypothesis: The elements of $G_k$ exactly index the corresponding elements of $R_k$ for every $0 \le k \le n$. This can be checked for $n = 1$. Suppose it holds for a given $n \ge 1$. Then for any given splitter $s \in R_{n-1}$ with corresponding composite $S \in G_{n-1}$, the splitter $sr_n$ is indexed by $SA$, and the splitter $sr_{n+1}$ must therefore be indexed by $SBT$ for some $T$. But $T$ must be empty as otherwise there is no splitter indexed by $SB$, contradicting our assumption. Now, for the $i$th last $s' \in R_n$ with corresponding composite $S' \in G_n$ we note that $s'r_{n+1} \in R_{n+1}$ and occurs at position $i$ in $W$. Our inductive hypothesis suffices to show that $s'$ does not occur at any position $j < i$ in $W$, since otherwise it would be indexed by multiple composites in $R_n$, requiring one to be a prefix of another, which is incompatible with the observation about their weights. Therefore $i$ is the first occurrence of both $s'$ and $s'r_{n+1}$, and $s'r_{n+1}$ must be indexed by $S'A^m$ for some $m > 0$. But in fact $m = 1$ since otherwise there is no splitter indexed by $S'A$, contradicting our assumption. As a by-product, we obtain an effective algorithm to determine the composite which corresponds to any subword of $W$ without explicitly calculating anything more than Fibonacci numbers. Let $s$ be the subword of interest, and initialise $i = 0$. While $s$ is non-empty, if its first symbol is $i \bmod 2$ then emit $A$, delete (up to) $f_{i+2}$ symbols from the start of $s$, and increment $i$; otherwise emit $B$, delete (up to) $f_{i+3}$ symbols from the start of $s$, and increment $i$ twice. And as a side-note, I observe that A341258 describes a sequence of words beginning $0$, $1$, $00$, $01$, $10$, $000$, $11$, $001$, $010$, $100$, $000$, $011$, $101$, $0001$, $110$, $0010$, $0100$, $1000$, $00000$, $111$ which appears to correspond to $G$ under the substitution $0 \to A, 1 \to B$ and was submitted by Clark Kimberling in March. I assume that this is no coincidence and that you have already solved the problem by a different route.<|endoftext|> TITLE: Do we need full choice to "efficiently" use (sub)bases? QUESTION [14 upvotes]: This question was previously asked and bountied at MSE without success. Suppose $(X,\tau)$ is a topological space, $B$ is a base for $\tau$, and $U\in \tau$ is an open set. Consider the following two strategies for writing $U$ as a union of elements of $B$: We have $U=\bigcup\{V\in B: V\subseteq U\}$. For each $u\in U$ pick some $V_u\in B$ with $u\in V_u\subseteq U$; then $U=\bigcup\{V_u: u\in U\}$. The first strategy has the advantage of not requiring the axiom of choice. If we pay attention to the number of basic opens required, however, it is noticeably inefficient: the first strategy might involve as many as $2^{\vert U\vert}$-many basic open sets, while the second involves at most $\vert U\vert$-many. It's not hard to show that in fact this drop in efficiency is unavoidable: it is consistent with $\mathsf{ZF}$ that there is a space $(X,\tau)$, a base $B$ for $\tau$, and an open set $U\in\tau$ such that there is no map $f:U\rightarrow B$ with $\bigcup_{u\in U}f(u)=U$. I'm interested in the exact strength of the corresponding efficiency principle, as well as its "subbase" variation: Over $\mathsf{ZF}$, are either of the following statements equivalent to $\mathsf{AC}$? For every topological space $(X,\tau)$, every base $B$ for $\tau$, and every $U\in\tau$, there is some $f:U\rightarrow B$ with $\bigcup_{u\in U}f(u)=U$. For every topological space $(X,\tau)$, every subbase $B$ for $\tau$, and every $U\in\tau$, there is some $f:U\rightarrow [B]^{<\omega}$ with $\bigcup_{u\in U}(\bigcap f(u))=U$. (Above, "$[A]^{<\omega}$" denotes the set of finite subsets of $A$. So the subbase version of the principle is saying that we can write $U$ as the union of $U$-many finite intersections of subbase elements.) EDIT: I do not require that $u\in f(u)$ (resp. $u\in\bigcap f(u)$) in the principles above, although that is a very natural requirement to include. REPLY [10 votes]: EDIT: this answer addresses a variant of this question in which we require that $u\in f(u)$, respectively $u\in\bigcap f(u)$. Unfortunately it doesn't seem to say anything about the question as stated. The subbase version implies AC. Let $X_i,i\in I$ be a family of nonempty sets. We may assume they are pairwise disjoint and are disjoint from $I$, and that $|X_i|>1$ for all $i$. Define a topology on $I\cup\bigcup_{i\in I}X_i$ by taking a subbasis $B$ consisting of sets $\{i,x\}$ for all $x\in X_i$. Then $I$ is an open subset in this topology: indeed, for any $i\in I$ and distinct $x,y\in X_i$ we have $\{i\}=\{i,x\}\cap\{i,y\}$. Now any function $f:I\to[B]^{<\omega}$ such that $I=\bigcup_i(\bigcap f(i))$ and $i\in f(i)$ must satisfy that $f(i)$ consists of (finitely many) sets of the form $\{i,x\}$, and so $\bigcup f(i)\setminus\{i\}$ is a nonempty finite subset of $X_i$. We conclude that axiom of multiple choice holds, and hence so does axiom of choice. Update: the basis version also implies (multiple) choice. Take the same set $X=I\cup\bigcup_{i\in I}X_i$ as above, but this time take the following basis of a topology: we take $B$ to consist of sets of the form $\{i\}\cup A$, where $A$ is a proper subset of $X_i$ whose complement in $X_i$ is finite. Any two elements of $B$ are either disjoint or their intersection is in $B$ as well. As $\bigcup B=X$, this implies $B$ is a basis of a topology. In particular $X$ itself is a union of elements of $B$. Let $f:X\to B$ be a function such that $x\in f(x)$ for all $x$. Then $X_i\setminus f(i)$ is a nonempty finite subset of $X_i$, giving multiple choice.<|endoftext|> TITLE: Sum of $\sin$ when angles shrink by $1/n$ QUESTION [6 upvotes]: There are many identities known like $$\sum_{k=0}^{n-1} \sin (k \cdot \theta + \varphi) = \frac{\sin\left(n \cdot \frac{\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} \cdot \sin \left(\frac{2 \varphi + (n-1)\cdot \theta}{2} \right)$$ However, in such situations the angles add up and are equidistant. Instead, I pose myself the question whether there is a simplifying formula for $$\sum_{k=1}^{n} k \cdot \sin \left(\frac{x}{k} \right),$$ particularly, if $x$ is much larger than $n$. REPLY [3 votes]: If you rewrite everything using complex exponentials, then it's clear why a sum such as $$\sum_{k=0}^{n-1}\sin(k\theta+\phi)$$ has a nice closed formula, since it's just two geometric sums $$ e^{i\phi}\sum_{k=1}^{n-1} e^{\pm i\theta k}. $$ And if you want $\sum_{k=0}^{n-1}k\sin(k\theta+\phi)$, it's similar, there are nice formulas for $\sum_{k=1}^{n-1}kT^k$. On the other hand, the sum you're asking about looks like $$ \sum_{k=1}^n k\sin\left(\frac{x}{k}\right) = \frac12\sum_{k=1}^n \Bigl(ke^{\pi i x/k} + ke^{-\pi i x/k}\Bigr). $$ I don't think that there's a nice closed formula in general for $$ \sum_{k=1}^n kT^{1/k},\quad\text{nor for}\quad \sum_{k=1}^n T^{1/k}. $$ However, as Carlo indicated, one can often get a good approximation using an integral.<|endoftext|> TITLE: Bezout’s identity for analytic functions of several variables QUESTION [7 upvotes]: In (single-variable) complex analysis, given analytic functions $f$ and $g$ with no common zeros, one can find analytic functions $u$ and $v$ such that $uf+vg=1$. I’d like to know if the same holds in several variables; as a simple case, specifically, Let $f,g\colon\mathbb{D}^2\to\mathbb{C}$ be analytic (in the bi-disc $\mathbb{D}^2$) with no common zeros. Does there exist analytic functions $u,v\colon \mathbb{D}^2\to\mathbb{C}$ such that $uf+vg=1$? REPLY [8 votes]: Make an open cover $D^2=\cup_j(U_j\cup V_j)$, for example, by polydisks such that $f$ has zeros only in $U_j$ and $g$ has no zeros in $U_j$. This is possible since zeros of $f$ and $g$ are disjoint. Solve the 1st Cousin problem with Cousin data $-1/(fg)$ in $U_j$ and $0$ in $V_j$. The solution is a meromorphic function $\phi$ such that $\phi+1/(fg)$ is holomorphic in $U_j$ and $\phi$ is holomorphic in $V_j$. Let $v:=-f\phi$. Then $-v+1/g$ is holomorphic and divisible by $f$ in $U_j$, and thus $v$ is also holomorphic in $U_j$ since $1/g$ is holomorphic in $U_j$. So $v$ is holomorphic everywhere. Now $-v+1/g$ is divisible by $f$ also in $V_j$ since $f$ has no zeros in $V_j$. Then since $-vg+1$ is holomorphic and divisible by $f$, then $u:=(1-vg)/f$ is holomorphic and $uf+vg=1$ as required. In modern texts they refer to H. Cartan's theorems A and B, but the case of polydisk of dimension 2 this was in the original paper of Cousin. REPLY [8 votes]: For the record, let me translate Alexandre Eremenko's answer in modern terms (after all, this is why sheaf theory was invented). The hypothesis implies an exact sequence $$0\rightarrow \mathscr{O}_{\mathbb{D}}\xrightarrow{\ (-v,u)\ } \mathscr{O}_{\mathbb{D}}^2\xrightarrow{\ (u,v)\ } \mathscr{O}_{\mathbb{D}}\rightarrow 0$$hence, using $H^1(\mathbb{D}^2,\mathscr{O})=0$ (Cartan's theorem B), the map $(u,v):H^0(\mathbb{D}^2,\mathscr{O})^2\rightarrow H^0(\mathbb{D}^2,\mathscr{O})$ is surjective.<|endoftext|> TITLE: Grassmannian cluster algebra of infinite type has no trees in its mutation class QUESTION [8 upvotes]: The question is why the statement in the title is true (is it?). To elaborate, recall that Grassmannian cluster algebra, according to Scott`s paper Grassmannians and Cluster Algebras, is the cluster structure on $\mathbb{C}[\operatorname{Gr}(k,n)]$, corresponding to the following quiver, called $\Gamma_{k,n}$: In case $k=2$ this is a chain of length $n-3$, so the corresponding cluster algebra is of type $\mathsf{A}_{n-3}$. It is also not hard to show that for $k=3$ and $n=6,7,8$ a suitable sequence of mutations transforms $\Gamma_{k,n}$ into a tree, which happens to be a Dynkin diagram of type $\mathsf{D}_4$, $\mathsf{E}_6$ or $\mathsf{E}_8$, respectively. In all other cases this cluster algebras is of infinite type. Now I ran some computer experiments and it seems that in the infinite case the mutation class of $\Gamma_{k,n}$ contains no trees at all. Q1: Does this experimental observation hold in all cases? If so, why? Q2: Can one at least find some nice representative in each mutation class? Say, having the smallest number of (unoriented) cycles. REPLY [3 votes]: Regarding question Q2, one can go a little bit further and describe simple diagrams with few edges for some more cases. Let us talk about $Gr(p, p+q)$, so that there is a symmetry between $p$ and $q$. Assume moreover that $p\geq 2$ and $q\geq 2$. For $p=2$ and $2\leq q$, one can mutate to the tree $A_{q-1}$. For $p=3$ and $2\leq q\leq 7$, one can mutate to $A_2, D_4$, $E_6$, $E_8$, $E^{(1,1)}_8$ and $K_{12}$. for $p=4$ and $2\leq q\leq 5$, one can mutate to $A_3, E_6$, $E^{(1,1)}_7$ and $W_{12}$. The symbols $E^{(1,1)}_8$ and $E^{(1,1)}_7$ stand for quivers related to elliptic singularities. These quivers are obtained from a tree by replacing the unique triple point by a double edge surrounded by three oriented triangles. The symbols $K_{12}$ and $W_{12}$ stands for quivers related to two of the 14 unimodal singularities of Arnold. Unoriented graphs describing these singularities can be found in many references about singularity theory. For a picture, see page 3 of arXiv:0708.0210. The quivers are almost trees, and only have one double edge and three oriented triangles.<|endoftext|> TITLE: What is the simplest proof that the density of primes goes to zero? QUESTION [51 upvotes]: By density of primes, I mean the proportion of integers between $1$ and $x$ which are prime. The prime number theorem says that this is asymptotically $1/\log(x)$. I want something much weaker, namely that the proportion just goes to zero, at whatever rate. And I want the easiest proof possible. The simplest proof I know uses estimates involving the binomial coefficient $\binom{2n}{n}$, but the argument still feels a bit involved. Does anyone know an even simpler proof that the density of primes goes to zero? REPLY [6 votes]: In my opinion, the simplest way to establish that $$\lim_{n \to +\infty} \frac{\pi(n)}{n}=0$$ is via the elementary inequality $$ \prod_{p \leq n} p \leq 4^{n-1} \qquad \mbox{(*)}$$ which holds for every $n \in \mathbb{Z}^{+}$. In 1939, Erdös and Kalmár found a proof of this inequality "which comes out of THE BOOK" (cf. P. Erdös, Ramanujan and I. In: K. Alladi (ed.), Number Theory - Madras 1987. Lecture Notes in Mathematics, vol. 1395. Springer Verlag, p. 2.). Clearly enough, as an immediate consequence of the Erdös-Kalmár ineq. we have that $$ \pi(n) < (2\log 4+1)\frac{n}{\log n}$$ for every integer $n>2$ and whence the result on the natural density of the primes.<|endoftext|> TITLE: Indeterminacy locus of meromorphic maps of rigid analytic spaces QUESTION [7 upvotes]: Setup. Let $k$ be an algebraically closed field of characteristic zero. Let $X/k$ be a normal variety, and let $Y/k$ be a proper variety. It is well-known that the indeterminacy locus of a rational map $X \dashrightarrow Y$ has codimension at least $2$ (see Lemma 3.2 of here for a proof). The complex analytic variant of this statement where rational map is replaced by meromorphic map was proved by Remmert; see e.g., Theorem 2.5 of here for the statement. Question. Is there an analogous result for rigid analytic/Berkovich/adic spaces? More precisely, if I have a meromorphic map from a normal rigid analytic space to a proper rigid analytic space, does the indeterminacy locus have codimension at least $2$? I am hopeful that such a result has been proved somewhere as there are a large quantity of articles proving non-Archimedean variants of results of Remmert and studying meromorphic maps; however I have not been able to find such a reference. I would appreciate any references for this question or any examples providing a negative answer to the question. Thanks! REPLY [2 votes]: Since this answer might be of use to other people in the future, I wanted to add an answer to this question. Brian Conrad provided me with a proof of this fact, which is written up Section 3 of arXiv 2105.04352.<|endoftext|> TITLE: Free extension of algebra for an operad QUESTION [7 upvotes]: I fix $C$ a symmetric monoidal model category (with a cofibrant unit if it helps). I'm assuming that it is closed, or at least that the tensor product commutes to colimits in each variable. If $X$ is a monoid in $C$, and $A \to B$ is a map in $C$, I'm calling free extension of $X$ by $A \to B$ (along some map $A \to X$) the monoid object $X \coprod_{F(A)} F(B)$ where $F(A)$ and $F(B)$ are the free monoids on $A$ and $B$ and the pushout is a pushout of monoid object (using some arbitrary map $A \to X$). It can be shown that if $A \to B$ is a (trivial) cofibration and $X$ is a cofibrant as an object of $C$, then the map $X \to X \coprod_{F(A)} F(B)$ is again a (trivial) cofibration in $C$. The reason for this is that the map $X \to X \coprod_{F(A)} F(B)$ can be written as an $\omega$-composition of maps that are obtained by taking iterated "pushout-product" of $A \to B$ with itself and tensor product with $X$ (so in particular, are all cofibrations/trivial cofibrations as soon as $A \to B$ is). What I wonder about is whether this can be generalized when "monoid" is replaced by $\mathcal{O}$-algebras for a general operad $\mathcal{O}$. My guess after some computation is that something like the above can be done when $\mathcal{O}$ is $\Sigma$-cofibrant (i.e. cofibrant for the projective model structure on collections) and when $X$ is cofibrant for the projective model structure on $\mathcal{O}(1)$-object (assuming $X$ cofibrant in $C$ do not seem enough in general) But I feel this might have been treated somewhere or that there might be other set of conditions under which something like this can be done. REPLY [5 votes]: This result is true and is due to Spitzweck, Berger–Moerdijk, Fresse, and Elmendorf–Mandell. A complete set of references can be found around Proposition 5.7 in the paper https://arxiv.org/abs/1410.5675.<|endoftext|> TITLE: Conditions equivalent to finiteness QUESTION [12 upvotes]: We've all probably come across some conditions that naturally imply finiteness, or are equivalent to it. For ZFC examples: A set $X$ can be ordered in such a way that the ordering is well-founded and reverse well-founded iff $X$ is finite. The discrete topology on $X$ is compact iff $X$ is finite (this can probably be strengthened topologists). A set $X$ is finite iff it has no countably infinite subsets. A set $X$ is finite iff every self-injection is a surjection. (The last two 'Dedekind finiteness' conditions fail if we remove choice, I believe) What other conditions are equivalent to being finite in various background theories? More ZFC examples are definitely welcome, but I'm also interested if working in weaker/stronger background theories can make fewer/more conditions equivalent to finiteness, or perhaps break an iff in one direction. REPLY [3 votes]: Here are some notions of "finiteness" which make sense in lots of categories, and which characterize the finite sets when the category is $Set$. I view this as an addendum to the notions discussed in Hanul Jeon's answer. A salient point is that notions of finiteness which agree in $Set$ may diverge when considering more general categories, even when those more general categories are things like toposes, i.e. "generalized categories of sets". A set $X$ is finite if and only if it is Noetherian, i.e. it satisfies the ascending chain condition for subobjects. A set $X$ is finite if and only if it is Artinian, i.e. it satisfies the descending chain condition for subobjects. A set $X$ is finite if and only if it is Hopfian, i.e. every surjection $X \to X$ is a bijection. A set $X$ is finite if and only if it is co-Hopfian, i.e. every injection $X \to X$ is a bijection. In the 1-topos literature, the term "Dedekind-finite" is used to mean co-Hopfian. A set $X$ is finite if and only if it is finitely-presentable, i.e. for every chain $Y_0 \to Y_1 \to \dots$ and every function $f: X \to \varinjlim Y_n$, there is some $m$ such that $f$ factors through $Y_m \to \varinjlim Y_n$. Variants of (5) include: considering all diagrams of $Y_n$'s indexed by any directed poset, or by any filtered category. In the $\infty$-categorical literature, the term "compact" is used instead of "finitely-presentable". If we require the transition maps between the $Y_n$'s to be injections, we get the notion of a finitely-generated object. A set is finite if and only it is quasicompact in the sense that the top element of its powerset lattice is finitely-presentable (equivalently, finitely-generated) as an element of that lattice. The nlab says "compact" instead of "quasicompact" here, but I think in the original literature "quasicompact" is used -- if I recall correctly, the nlab has a general convention to never use the term "quasicompact", insisting instead that "compact" never implies "Hausdorff". A set $X$ is finite if and only if its coherent, i.e. it is "stably quasicompact" in the sense defined at the link. (7) and (8) come from topos theory. There are corresponding notions in $\infty$-topos theory. A set is finite if and only if it lies in the closure of $\{ \{\emptyset\}\}$ under pushouts and intial objects. Here are some other fun ones: A set $X$ is finite if and only if all linear orders on $X$ are isomorphic. A set $X$ is finite if and only if there exists a first-order theory $T$ such that $T$ has a unique model $M$ up to isomorphism, and moreover $M$ may be taken to have underlying set $X$.<|endoftext|> TITLE: Verification of a maximal antichain QUESTION [8 upvotes]: In order theory, an antichain (Sperner family/clutter) is a subset of a partially-ordered set, with the property that no two elements are comparable with each other. A maximal antichain is the antichain which is not properly contained in another antichain. Let's take the power set of $\{1,2,\ldots, n\}$ as our partially-ordered set, here the order is given by inclusion. Then my question is, for any given antichain of this partially-orded set, is there any polynomial-time algorithm (with respect to $n$) to verify that this antichain is indeed "maximal"? In other words, verifying that any subset of $\{1,2,\ldots, n\}$ is either contained in, or contains some set from the antichain. Here such algorithm should have polynomial run-time for ANY antichain. Update: To clarify, here I will treat the size of our antichain as the parameter for the verification algorithm. In other words, my question is: does there exist a verification algorithm, whose run-time is polynomial in $n$ and $m$, where $m$ is the size of the antichain. When the size of our antichain $m$ is exponential in $n$ then such algorithm is trivial (just comparing those elements one by one); but when the given antichain has O(poly(n)) size, this is my interested case. For example, when the antichain is given by $\{\{1\}, \ldots, \{n\}\}$, we certainly do not have to do the brute force comparison. REPLY [2 votes]: Remark. Originally I claimed this to be a full solution, but that was false, as shown by Emil in the comments. However, this argument proves the following weaker version. I can prove that it is co-NP-complete to decide for an input family $A$ whether there is a set $S$ that is unrelated to all sets in $A$. I'll call such families maximal. This shows that any possible polynomial time algorithm must exploit that the input family is an antichain, already for linear sized inputs. My reduction is from SAT. Given a CNF $\Psi$ on $n$ variables, we convert it into a family $A$ over $2n$ elements, such that $A$ is maximal if and only if $\Psi$ in unsatisfiable. The $2n$ elements will come in pairs, which I denote by $i$ and $i'$. The complement of every pair is contained in $A$ regardless of $\Psi$, so $\overline{11'}\in A$, $\overline{22'}\in A$, ..., $\overline{nn'}\in A$. Moreover, for every clause we add a set to $A$ such that if $x_i$ is in the clause, the set contains $i$, while if $\bar x_i$ is in the clause, the set contains $i'$. For example, the clause $(x_i\vee \bar x_j)$ adds the set $ij'$ to $A$. Suppose $\Psi$ is satisfiable. Then for a satisfying evaluation $x$, define the set $S$ such that $i\in S$ if $x_i$ is false and $i'\in S$ if $x_i$ is true. It is straight-forward to check that $S$ is not in relation with any element of $A$. Suppose that $A$ is not maximal. Take a set $S$ not in relation with any element of $A$. Define $x_i$ to be true if $i\notin S$ and false if $i'\notin S$, otherwise arbitrarily. This definition is indeed correct, as $\overline{ii'}\in A$ implies that $i,i'\in S$ is not possible. It is straight-forward to check that $x$ is a satisfying evaluation of $\Psi$.<|endoftext|> TITLE: A set theoretic question arising from trying to understand a sheaf cohomology question QUESTION [7 upvotes]: I'm trying to understand the footnote to Example 5.3 in Wiegand - Sheaf cohomology of locally compact totally disconnected spaces which is about constructing a locally compact Hausdorff and totally disconnected space whose sheaf cohomology with constant sheaf $\mathbb Z/2\mathbb Z$ coefficients doesn't vanish. An explicit covering is given by compact open sets with no three intersecting and the sheaf cohomology is shown to coincide with the Čech cohomology over any covering by compact open sets. So things boil down to computing the Čech cohomology in this sheaf with respect to this covering. In the end, since we are dealing essentially with boolean algebras, it turns into the following problem which I don't immediately see how to solve. If $Y$ is a set, let $B(Y)$ be the boolean algebra of finite and cofinite subsets of $Y$. Let $S$ be a countably infinite set and $T$ an uncountable set. Define a mapping $\Phi\colon B(T)^S\times B(S)^T\to 2^{S\times T}$ as follows. If $f\colon S\to B(T)$ and $g\colon T\to B(S)$ are maps, send $(f,g)$ to the subset of $S\times T$ consisting of all pairs $(s,t)$ with $s\in g(t)$ or $t\in f(s)$, but not both. Then I think the footnote is equivalent to the claim that $\Phi$ is not onto. Question: why is $\Phi$ not onto? REPLY [8 votes]: This is not onto for any uncountably infinite $T$, even one much smaller than the power set (if the continuum hypothesis is false). Fix $X \in 2^{ S \times T}$ such that the induced map $h \colon T \to 2^S$ where $h(t) = \{ s \mid (s,t) \in X\}$ (i.e. taking vertical fibers of $X$) has uncountable image. (For example we can take $T=2^S$ and choose $X$ to be $\{(s,t)| s\in T\}$ as LSpice suggested, so $h$ is the identity, or take $T$ to be a smaller uncountable subset of $2^S$ if one exists.) Assume that $X$ does arise from the image of some $S$ and $T$ - we wil derive a contradiction from this. Let $U$ be the set of $s \in S$ such that $f(s)$ is cofinite (rather than finite). Then for each $s$, for all but finitely many $t$, we have $t \in f(s)$ if and only if $s\in U$. Since there are only countably many $s$, for all but countably many $t$, we have $t \in f(s)$ if and only if $s \in U$. Thus, for all but countably many $t$, we have $s \in h(t)$ if and only if $s \in g(t)$ or $s\in U$, but not both. If $g(t)$ is finite then $h(t)$ is equal to $U$ up to finite error, and if $g(t)$ is cofinite then $h(t)$ is equal to the complement $U^c$ up to finite error. In either case, there are countably many possibilities for $h(t)$, which together with the countably many $t$ excluded at first, gives countably many possibilities for $h(t)$ in total, contradicting our assumption.<|endoftext|> TITLE: Two non constant meromorphic functions over a connected compact Riemann surface, could not be algebraically independent QUESTION [11 upvotes]: Let $M$ be a connected compact Riemann surface. Let $f, g$ be two nonconstant meromorphic functions. Why is there a two-variable complex polynomial $F(x,y)$ that vanishes for $(x, y)=(f, g)$, (in other words $F(f,g)=0$)? REPLY [18 votes]: Let $F$ be a polynomial of degree at most $n$. For a point $x$ where $f$ has a pole of order $a$ and $g$ has a pole of order $b$, $F(f,g)$ has a pole of order at most $n\max(a,b)$. Locally near $x$, we can write $F(f,g)$ as a Laurent series $$c_{ -N} z^{-N} + c_{1-N} z^{1-N} + \dots + c_{-1} z^{-1}+ c_0 + c_1 z + \dots $$ where $N = n \max(a,b)$ and $c_{-N}, \dots, c_{-1}$ are linear functions on $F$. There are finitely many points where $f$ or $g$ has a pole. Let $d$ be the sum of $\max(a,b)$ over these points. Then we have $nd$ linear functions of the form $c_{-k}$ at these points. If all these linear functions vanish, then $F(f,g)$ has no poles, and thus is constant. So as long as the kernel of this set of $nd$ linear functions has dimension $>1$, there will be two linearly independent polynomials $F$ with $F(f,g)$ constant, and taking a linear combination, there will be anontrivial $F$ with $F(f,g)=0$. Since the dimension of the space of polynomials of degree $\leq n$ is $\frac{n(n+1)}{2}$, the dimension of the kernel is at least $\frac{ n (n+1)}{2} - nd$, hence is $>1$ for n sufficiently large with respect to $d$.<|endoftext|> TITLE: On circles and ellipses drawn on an infinite planar square lattice QUESTION [5 upvotes]: Consider a plane with a square lattice formed by all points with both coordinates as integers. As can be easily seen, a simple parabola can be found that passes through infinitely many of the square lattice points. However, Given any positive integer n, can we always find a sufficiently large circle drawn on the plane that passes through at least n lattice points? Can such circles be found if the center is not to be a lattice point? What if we require the circle to pass through exactly n lattice points? Question 1 has a natural restatement if instead of circles, we look at ellipses (either all with a given eccentricity e or with e that can be freely chosen). The ellipses need not be axis parallel. And what can one say if the lattice of points has as unit cell not a square but a general parallelogram? Note 1: Lattice points on the boundary of an ellipse discusses a related question. Additional Question (added after Prof. Elkies's affirmative answer to questions 1 to 3): What happens in 3D and higher dimensions? REPLY [7 votes]: Such circles are known as Schinzel Circles. See also Kulikowski's Theorem for the sphere. According to zbmath Kulikowski proved his theorem in arbitrary dimension.<|endoftext|> TITLE: What functors are classified by slices of $\infty$-categories? QUESTION [11 upvotes]: Suppose I have a functor $f\colon C\to D$ between $\infty$-categories (I'll assume $C$ and $D$ are small.) Then I can form the slice categories and restriction functors $$ D_{f/}\to D\qquad \text{and} \qquad D_{/f}\to D. $$ These maps are left fibrations and right fibrations, respectively. By the theory of "straightening/unstraightening", these fibrations are classified by certain functors $$ D\to \mathcal{S} \qquad\text{and} \qquad D^{\mathrm{op}} \to \mathcal{S} $$ to the $\infty$-category $\mathcal{S}$ of $\infty$-groupoids. This raises the question: Q. What are these functors? Actually, it's pretty clear what the answer should be: A. They are inverse limits of (co)representable functors. Explicitly, we should be able to describe these functors by $$ d\mapsto \mathrm{lim}_{c\in C^{\mathrm{op}}}\operatorname{Map}_D(f(c),d) \qquad \text{and}\qquad d\mapsto \mathrm{lim}_{c\in C}\operatorname{Map}_D(d,f(c)) $$ respectively. In the 1-categorical analogue this an elementary argument. The case of $C=1$ (slices over objects, corresponding to (co)representable functors) is "well-known" (see 5.8 of Cisinski's Higher categories and homotopical algebra). So my real question is: Q'. What is a proof or reference for this fact? REPLY [5 votes]: I like Maxime's argument better, but here's another. As you say, the case when $C=\bullet$ is well-known. But we can reduce to that case! The map $D_{/f} \to D$ is pulled back from $\mathsf{Fun}(C, D)_{/f} \to \mathsf{Fun}(C,D)$ along the diagonal map $D \to \mathsf{Fun}(C,D)$. Under (un)straightening, pulling back corresponds to precomposition, and the functor on $\mathsf{Fun}(C,D)$ represented by $f$ restricts to the functor you described.<|endoftext|> TITLE: For which $n$ can $S_n$ act transitively on $n+k$ elements? QUESTION [7 upvotes]: It is known that the symmetric group $S_n$ can act transitively on $n+1$ elements if and only if $n=5$. Are there similar classifications for $S_n$ acting transitively on $n+k$ elements, where $k$ is fixed? For example, is it known whether there are only finitely many such $n$ for every $k$? REPLY [9 votes]: Fix $k > 0$. Suppose that $n > 6$ and $\frac{n(n-3)}{2} > k$. If $[S_n : H] \leq n+k$, then $H$ is one of the following: $S_n$, $A_n$, $S_{n-1}$, or $A_{n-1}$. So in particular if $[S_n : H] = n+k$, then $k = n$ and $H = A_{n-1}$. See Theorem 5.2B in "Permutation Groups" by Dixon and Mortimer, and also this Math.SE question.<|endoftext|> TITLE: A question on fractional derivatives QUESTION [6 upvotes]: I know practically nothing about fractional calculus so I apologize in advance if the following is a silly question. I already tried on math.stackexchange. I just wanted to ask if there is a notion of fractional derivative that is linear and satisfy the following property $D^u((f)^n) = \alpha D^u(f)f^{(n-1)}$ where $\alpha$ is a scalar. In the case of standard derivatives we would have $\alpha = n$. Thank you very much. REPLY [16 votes]: There are basically no interesting solutions to this equation beyond first and zeroth order operators, even if one only imposes the stated constraint for $n=2$. First, we can depolarise the hypothesis $$ D^u(f^2) = \alpha_2 D^u(f) f \quad (1)$$ by replacing $f$ with $f+g, f-g$ for arbitrary functions $f,g$ and subtracting (and then dividing by $4$) to obtain the more flexible Leibniz type identity $$ D^u(fg) = \frac{\alpha_2}{2}( D^u(f) g + f D^u(g) ). \quad (2)$$ There are now three cases, depending on the value of $\alpha_2$: $\alpha_2 \neq 1,2$. Applying (2) with $f=g=1$ we then conclude that $D^u(1)=0$, and then applying (2) again with just $g=1$ we get $D^u(f)=0$. So we have the trivial solution $D^u=0$ in this case. $\alpha_2=2$. Then $D^u$ is a derivation and by induction we have $D^u(f^n) = n D^u(f) f^{n-1}$, just as with the ordinary derivative, so we just have $\alpha_n=n$ for all $n$ with no fractional behavior. $\alpha_2=1$. Applying (2) with $g=1$ we obtain (after a little bit of algebra) $D^u(f) = mf$ where $m := D^u(1)$. Thus $D^u$ is just a multiplier operator, which obeys $D^u(f^n) = D^u(f) f^{n-1}$, thus $\alpha_n=1$ for all $n$. Thus there are no linear solutions to your equation other than the usual derivations (e.g., $D^u(f) = a(x) \frac{d}{dx} f$ for any smooth symbol $a$) and multiplier operators $D^u(f) = mf$, i.e., first order and zeroth order operators. On the other hand, fractional derivatives $D^u$ tend to obey a "fractional chain rule" $$ D^u( F(f) ) = D^u(f) F'(f) + E$$ for various smooth functions $F,f$, where the error $E$ obeys better estimates in various Sobolev spaces than the other two terms in this equation. In particular, for $F(t) = t^n$, we would have $$ D^u(f^n) = n D^u(f) f^{n-1} + E$$ for a "good" error term $E$. For instance, taking $u=n=2$ with $D$ the usual derivative, we have $$ D^2(f^2) = 2 D^2(f) f + E \quad (3)$$ with $E$ the "carré du champ" operator $$ E := 2 (Df)^2.$$ Note that the error $E$ is controlled uniformly by the $C^1$ norm of $f$ but the other two terms in (3) are not. See my previous MathOverflow answer at https://mathoverflow.net/a/94039/766 for some references and further discussion.<|endoftext|> TITLE: Point-line incidence bounds over positive characteristic fields QUESTION [6 upvotes]: I am aware of work on point-line incidence bounds over $\mathbb{R}$, $\mathbb{C}$, and finite fields, in particular various versions of the Szemeredi-Trotter bounds. I would like to know if work along these lines has been done over function fields like $\mathbb{F}_p(t)$. I looked around some and asked some people, but I didn't find anything. What I'm actually really interested in is arbitrary fields. Results on $\mathbb{C}$ often generalize to arbitrary characteristic zero fields, and results on finite fields imply results on algebraically closed fields of characteristic $p$. So that's why I ask about $\mathbb{F}_p(t)$. REPLY [5 votes]: The current best Szemeredi-Trotter bound over finite fields also holds over arbitrary fields. It is in the wonderful paper "An Improved Point-Line Incidence Bound Over Arbitrary Fields" by Sophie Stevens and Frank de Zeeuw. Unfortunately, this current best bound of $m^{11/15}n^{11/25}$ is closer to the trivial $m^{3/4}n^{3/4}$ than to the conjectured $m^{2/3}n^{2/3}$. In a somewhat different direction, model theorists generalize these bounds to things such as o-minimal frameworks and distal. See Chernikov, Galvin, and Starchenko and also Basu and Raz<|endoftext|> TITLE: Independence result where probabilistic intuition predicts the wrong answer? QUESTION [6 upvotes]: In some of his writings, Paul Cohen gave an informal, motivational discussion about the word generic (as it is used in forcing). While very suggestive, the discussion leaves the meaning of the word ambiguous, and could lead someone to guess that it is related to probability theory. Indeed, there was a recent MO question along these lines. As the comments and answers to that question make clear, generic is actually closely related to Baire category and not to measure theory. As one learns in an analysis course, it is perfectly possible for a comeager set to have measure zero and for a meager set to have positive measure. This got me wondering. Is there a simple/natural example of an independence result where, if you were to appeal to probabilistic intuition, you would guess wrongly what happens, because the generic event has probability zero? A good example should (A) be an independence result that is natural-sounding and easy for a non-set theorist to understand and (B) have an obvious and tempting—but wrong—line of reasoning based on conflating measure with category. REPLY [10 votes]: If the Borel-Cantelli lemma counts as probabilistic intuition, then here's an example. Think of the real $x$ that you adjoin to a ground model as a sequence of $0$'s and $1$'s, and let $f(n)$ be the length of the $n$-th run of consecutive $1$'s in $x$. If the bits in $x$ were chosen by independent flips of a fair coin (or even of a biased coin as long as both sides of the coin have positive probability), then the inequality $f(n)>n$ would (with probability $1$) hold for only finitely many $n$ (by Borel-Cantelli). But for a Cohen-generic $x$, that inequality holds for infinitely many $n$. In fact, for any function $g:\omega\to\omega$ in the ground model, $f(n)>g(n)$ for infinitely many $n$.<|endoftext|> TITLE: Are Spanier-Whitehead duals of general spaces expressible through some generalization of normal bundles? QUESTION [13 upvotes]: The question is inspired by an answer to The concept of Duality It is explained in that answer that the Spanier-Whitehead dual of a compact manifold is given by the Thom spectrum of normal bundles of the manifold. Spanier-Whitehead duals exist more generally for e.g. all finite CW-complexes. Is there known a generalization of the notion of normal bundle to e. g. all finite CW-complexes which would generalize the above description of the Spanier-Whitehead dual? REPLY [14 votes]: Let $X$ be a finite complex. Then the functor $$\lim_X:\operatorname{Fun}(X,\operatorname{Sp})\to \operatorname{Sp}$$ sending a local system of spectra $E$ to its limit preserves all colimits. Indeed it preserves all finite colimits by stability, and it preserves all filtered colimits by the finiteness of $X$. Therefore it is of the form $$\lim_X E \cong \operatorname*{colim}_X(E\otimes \omega_X)$$ for a certain local system of spectra $\omega_X$ (this is by the universal property of the presheaf category and the fact that $\operatorname{Fun}(X,\operatorname{Sp})$ is the stabilization of the presheaf category). The local system $\omega_X$ is called the dualizing sheaf of $X$. Now if we let $\mathbb{S}_X$ be the constant local system at the sphere spectrum $\mathbb{S}$ we have $$\mathbb{D}X_+=\operatorname{map}(\Sigma^\infty_+X,\mathbb{S})\cong\lim_X \mathbb{S}_X\cong \operatorname*{colim}_X \omega_X$$ where the right hand side is some sort of generalized Thom spectrum. When $\omega_X$ is invertible (i.e. all its stalks are spheres), it is called the Spivak normal fibration of $X$, and $X$ is said to be a Poincaré complex. Note that $\omega_X$ is rather explicit: it follows formally from the definition that $$\omega_X(x)=\lim_{z\in X^{op}} \Sigma^\infty_+ \operatorname{Map}_X(z,x)$$ where $\operatorname{Map}_X(z,x)$ is the space of paths from $z$ to $x$. Moreover one can show that for a closed topological manifold $X$ we have a natural equivalence $$\omega_X(x)\cong \mathbb{D}\left(X/X\smallsetminus\{x\}\right)$$ where $X/X\smallsetminus\{x\}$ is the cofiber of the inclusion $X\smallsetminus\{x\}\subseteq X$. This is equivalent to the construction explained in Gregory Arone's comment. A reference for this material is J. R. Klein, The dualizing spectrum of a topological group, Mathematische Annalen 319 (2001), no. 3, 421–456, DOI 10.1007/PL00004441. Another useful reference are the lecture notes for Jacob Lurie's class on Algebraic L-theory and manifold topology. In particular Lecture 26 is relevant.<|endoftext|> TITLE: Bounding degree of rational curves in the exceptional locus QUESTION [5 upvotes]: Let $f\colon \mathbb{P}^2\dashrightarrow\mathbb{P}^2$ be a rational map ($\mathrm{deg}(f)$ may be high), let $\Gamma\subset\mathbb{P}^2\times\mathbb{P}^2$ be the closure of the graph. Let $x\in\mathbb{P}^2$ be a point in the source, then the fiber $\Gamma|_x$ can be viewed as a union of rational curves in the target $\mathbb{P}^2$. Do we know if $\Gamma|_x\subset\mathbb{P}^2$ necessarily have a component of low degree (say, degree 1)? REPLY [7 votes]: The answer is no. Let us consider the case where $f$ is birational. In this case, for each point $x\in \mathbb{P}^2$ such that $f$ is not defined, the intersection of $\Gamma$ with $x\times \mathbb{P}^2$ contains all curves of $\mathbb{P}^2$ that are contracted by the inverse $f^{-1}$. The degree of these curves can be arbitrary large. As an example, you can consider some so-called "Halphen map", that is a birational map of $\mathbb{P}^2$ that preserves a pencil of curves of genus $1$ (a simple case is given by a pencil given by two general cubics). You blow-up the $9$ points $p_1,\ldots,p_9$ of intersection and consider a translation given by $E_1-E_2$, where $E_1$ and $E_2$ are the exceptional divisors of $p_1$ and $p_2$ (which are two sections of the pencil). If you consider this map $f$ on $\mathbb{P}^2$, the degree of $f^n$ grows quadratically with $n$ and there are exactly $9$ points not defined for $f^n$, namely $p_1,\ldots,p_n$. There are exactly $9$ curves contracted by $f^{-n}$ and these are irreducible of degree that is also growing. For $n$ large, the degree of the curves contracted is then unbounded. There is an explicit family given in Proposition 6.4 of https://algebra.dmi.unibas.ch/blanc/articles/degencremona.pdf where the degree of $f^n$ is $36n^2+1$, and where $f$ contracts $7$ curves of degree $12n^2$, one of degree $12n^2-6n$ and one of degree $12n^2+6n$.<|endoftext|> TITLE: How many cells needed to build the classifying space $BG$? QUESTION [16 upvotes]: Let $G$ be a finitely presented group of cohomological dimension $n$. Apart from the unresolved ambiguity pertaining to the Eilenberg--Ganea conjecture, it is known that we can find an $n$-dimensional model of the classifying space $BG = K(G,1)$. It is also not hard to see that $BG$ needs to have cells in every dimension $j \leq n$: otherwise $EG^{(j-1)}$ would be a simply-connected acyclic space, hence already equal to $EG$, which is impossible by the assumption on $\text{cd}(G)$. My question now is: what is the smallest number of $j$-cells that $BG$ can have? For $G = \mathbf{Z}^n$ the count is $\binom{n}{j}$, and in general the number of $1$- and $2$-cells are the minimal number of generators and relations of a presentation of $G$, so can be $2$ (except for $G = \mathbf Z$ where $n = 1$) and $1$, respectively. REPLY [7 votes]: A group $G$ is of type $\mathcal{F}_n$ if it has a $K(G,1)$ with finite $n$-skeleton. Let $F_2$ be the free group on $2$ generators. Consider the kernel of the map $F_2\times F_2 \times \cdots \times F_2 \to \mathbb{Z}$, sending each generator to $1$. Then the kernel is of type $\mathcal{F}_{n-1}$ but not of type $\mathcal{F}_n$. This was proved by Stallings for $n=2$ and Bieri for $n>2$, see the discussion and links here. Taking $n\geq 3$, one gets finitely presented groups of finite cohomological dimension which do not have finite $n$-skeleton. Nevertheless, one can ask your question for groups of cohomological dimension $n$ and type $\mathcal{F}_n$. For surface groups, and more generally 1-relator groups without torsion, this is known. A 1-relator group $G$ without torsion has presentation complex a $K(G,1)$, and cannot have a complex with fewer cells unless it is free. For 3-manifold groups, this is a well-studied question when the $K(G,1)$ is a manifold and the cell structure is a handle structure. In this case, the number of cells is determined by the Heegaard genus of the manifold. There are known algorithms to compute this. However, the rank and the Heegaard genus can differ. So there can be a $K(G,1)$ with fewer $1$-cells than the Heegaard genus. Most likely the rank of $3$-manifolds (so the minimal number of $1$-cells in a $K(G,1)$) is computable for $3$-manifolds, but it has only been proved in special cases. I'm not sure though whether the minimality of the number of 2-cells has been addressed. The minimal number of 3-cells is $1$. An interesting possibility is that for a CW complex with a minimal number of $2$-cells, the number of $3$-cells or 1-cells might not be minimal. In general, I'm not sure how much is known about this question, I'm just addressing the few special cases that I am familiar with to show how complex the answer can be.<|endoftext|> TITLE: Complement to a union of spheres in a sphere QUESTION [5 upvotes]: Take $S^n$ and consider the union $Z$ of $k_1$ circles, $k_2$ 2-dimensional spheres, ..., $k_{n-2}$ $(n−2)$-dimensional spheres, embedded in $S^n$ in an unknotted way, with no mutual intersection and no mutual linking. I want to understand the complement $S^n \setminus Z$. To that end, I calculated its homology via Mayer-Vietoris and I think that complement is homotopy equivalent to $(S^{n-1})^{k_1+k_2+\dots+k_{n-2}-1} \vee (S^{n-2})^{k_1} \vee \dots \vee (S^{1})^{k_{n-2}}$. Is it true? Where can I find a reference? The problem is that we cannot deduce this isomorphism from knowing the homology... Is there any way to see this explicitly (if it's true)? Originally asked as Math Stack Exchange question 3991178. REPLY [3 votes]: Your answer is correct, if "unknotted" is equivalent to: being homeomorphic to a union of standard sphere complements within disjoint convex balls. Philosophically, you could imagine giving a proof by saying that such a space is a connected sum of sphere complements, and understand enough about them to prove it inductively. Here is an argument that is a little more homotopy-theoretic, which calculates the answer inside a closed ball of large radius and then gets the answer for a sphere by gluing on a disc. You need a basic calculation to get started. For $k \leq n-2$, the complement $C_k$ of the standard k-sphere (embedded in the first $(k+1)$ coordinates) in the standard n-ball of radius 2 has an explicit determination of its homotopy type. It has a deformation retraction down to the union of the (n-1)-sphere of radius 2, together with the $(n-k-1)$-dimensional ball of radius 2 in the last $(n-k-1)$ coordinates. (This is a "draw a straight line from the nearest point on the $k$-sphere" retraction, I believe.) This space formed by attaching an $(n-k-1)$-dimensional cell along a map $S^{n-k-2} \to S^{n-1}$; this is necessarily nullhomotopic, so the space is homotopy equivalent to $S^{n-1} \vee S^{n-k-2}$, as the homology suggests. Under this, the inclusion from the outside boundary is homotopic to the inclusion of the first wedge factor. The complement $Y$ of $j$ nonintersecting convex balls in a closed ball of radius $R$ is homotopy equivalent to the complement of $j$ points in $R^n$, and this makes it homotopy equivalent to $\bigvee_{i=1}^{n} S^{n-1}$. We can use a degree calculation in homology to be more explicit: the inclusion of each boundary component of one ball is homotopic to the inclusion of a wedge factor, while the inclusion of the outside boundary is homotopic to the "pinch" map $S^{n-1} \to \bigvee S^{n-1}$. The complement $Z$ of $k_1$ 1-spheres, $k_2$ 2-spheres, etc, in a closed ball of radius $R$ is the pushout: $$ Y \leftarrow \coprod S^{n-1} \rightarrow \coprod_i \left(\coprod^{k_i} C_{k_i}\right). $$ Up to homotopy equivalence, the previous identifications say that we have the pushout of $$ \bigvee^{\sum k_i} S^{n-1} \leftarrow \coprod^{\sum k_i} S^{n-1} \rightarrow \coprod \left(\coprod^{k_i} S^{n-1} \vee S^{n-1-k_i}\right), $$ with the map on the left collapsing basepoints and the map on the right being the coproduct of wedge inclusions. The result is that we identify basepoints in all the coproduct factors, and so the resulting space is homotopy equivalent to $$ \bigvee \left(\bigvee^{k_i} S^{n-1} \vee S^{n-1-k_i}\right) \cong \bigvee^{\sum k_i} S^{n-1} \vee \bigvee \left(\bigvee^{k_i} S^{n-1-k_i}\right). $$ Moreover, the inclusion of the outside boundary sphere is the fold map from $S^{n-1}$ to the first wedge of $(n-1)$-spheres. Now finally, the complement $X$ of $k_1$ 1-spheres, etc, in the $n$-sphere, is the pushout $$ Y \leftarrow S^{n-1} \rightarrow D^n $$ where the left-hand map is the inclusion of the outside boundary sphere of radius $R$. Up to homotopy equivalence, this is the pushout of $$ \bigvee^{\sum k_i} S^{n-1} \vee \bigvee \left(\bigvee^{k_i} S^{n-1-k_i}\right) \leftarrow S^{n-1} \rightarrow D^{n-1}. $$ The left-hand map is the pinch map into the first wedge factor. Up to homotopy equivalence, gluing in this cell eliminates one sphere factor from the wedge; the resulting space is the one you were hoping for.<|endoftext|> TITLE: tangent bundle on noncommutative manifold QUESTION [8 upvotes]: Using the Serre-Swan's theorem, one can do vector bundle theory on noncommutative manifold $(A,H,D)$, by replacing vector bundle by finitely generated projectve module $M$. For the construction of tangent bundle, one can use derivation, but I am not sure how to define derivation on the module. The idea seems to be define derivation on $A$ satisfy Leibniz rule(see here). Is this the canonical way to do this? REPLY [4 votes]: Here are some thoughts which essentially complement Branimir's answer. In many naturally occurring examples, in particular the theory of quantum groups, it is actually more natural to start with a noncommutative cotangent bundle and to then construct the spectral triple from it. More explicitly, start with a pre-$C^*$-algebra $A$ whose noncommutative geometry you are trying to understand. Look for a dga (differential graded algebra) $\Omega^{\bullet}$ for which $A = \Omega^0$. Endow $\Omega^{\bullet}$ with some type of ``Riemannian metric'', which is to say an inner product $\langle \cdot,\cdot \rangle$. Confirm that the differential d of the dga is adjointable $\langle \cdot,\cdot \rangle$. Denote by d$^*$ the adjoint of d. Complete $\Omega^{\bullet}$ to a Hilbert space. Check that the closure of the symmetric operator $D:=$d$+$d$^*$ is a satisfies the axioms of a spectral triple. I should stress that this is a rough plan of attack, and will only work in situations which are, in some sense, close to the classical situation. For example, $q$-deformed $C^*$-algebras, in particular examples coming from Drinfeld-Jimbo quantum groups. The prototypical example here is the Podles sphere $\mathcal{O}_q(S^2)$, the degree zero part of a $\mathbb{Z}$-grading on $\mathcal{O}_q(SU_2)$, which $q$-deforms the algebra of algebraic functions of the $2$-sphere. This pre-$C^*$-algebra admits an essentially unique $\mathcal{O}_q(SU_2)$-covariant dga, called the Podles calculus. The calculus carries a $q$-deformation of the Fubini-Study metric, with respect to which $d$ is adjointable, and which gives a spectral triple upon completion of $\Omega^{\bullet}$ and closure of $D = $d$+$d$^*$. The dga constructed from this spectral triple (in the sense Branimir explained above) will coincide with $\Omega^{\bullet}$. This spectral triple was first constructed by Dabrowski and Sitarz (or at least the Dolbeault-Dirac version) and later reconstructed by Majid from a more geometric point of view. This construction can be extended to a much larger class of examples called quantum flag manifolds . . . but this is very much ongoing work.<|endoftext|> TITLE: Metamathematics of buts QUESTION [50 upvotes]: Something I learned (probably in middle school) that always bothered me is that the truth value of "and" and "but" are basically the same. If you were going to assign a truth-functional interpretation of "but" in first-order logic, it would be the same as "and". There's been a explosion of logical systems that are alternatives to first-order logic, such as fuzzy logic. Is there a logical system that can distinguish "and" and "but"? REPLY [14 votes]: I'm not a mathematician, but I think normally, when you say "X but Y" you mean: $$X \wedge Y \wedge P(Y|X) TITLE: Polynomials vanishing on prescribed layers QUESTION [5 upvotes]: Given a prime $p$ and an integer $n\ge p$, what is the smallest possible degree of a polynomial $Q\in\mathbb F_p[x_1,\dotsc, x_n]$ such that $Q$ vanishes on every vector $x\in\{0,1\}^n$ of weight $w(x)=p$, but $Q(0)\ne 0$? (Here the weight of a vector is the number of its nonzero coordinates.) I am also interested in variations of this question: what is the smallest possible degree if $Q$ vanishes on every zero-one vector of weight at least $p$? Of weight divisible by $p$? The range of interest is $n=p^c$ with $c>2$, particularly large values of $c$. REPLY [4 votes]: For both modified questions, the answer is $n+1-p$. Since every polynomial that vanishes on vectors of weight $\geq p$ vanishes on vectors of nonzero weight divisible by $p$, other than $0$, it suffices to prove the upper bound for polynomials vanishing on vectors of weight $\geq p$ and the lower bound for vectors of nonzero weight divisible by $p$. For the upper bound, it suffices to take $$Q =\prod_{i=1}^{n+1-p} (1-x_i)$$ since any set of at least $p$ of $x_1,\dots, x_n$ contains at least $1$ of the $x_1,\dots, x_{n+1-p}$. For the lower bound, it suffices to observe that $Q ( 1- (x_1+ \dots + x_n)^{p-1})$ vanishes on all vectors of nonzero weight but is nonzero at $\{0,\dots ,0\} $. Thus, when we remove all squares of variables, it is a nonzero scalar multiple of $$\prod_{i=1}^{n} (1-x_i)$$ and thus has degree $n$ (or use the combinatorial nullstellensatz), so has degree $\geq n$, so $Q$ has degree $\geq n +1-p$. For the original question where $Q$ vanishes on vectors of weight $p$, the answer is $p$ if $n \geq 2p-1$. For the upper bound, it suffices to take $$Q =1 - \sum_{ \substack{ S \subseteq \{1,\dots n\}\\ |S| = p }} \prod_{i\in S} x_i = 1- e_p (x_1,\dots, x_n) $$ where $e_p$ are the elementary symmetric polynomials. For the lower bound, it suffices to handle the case $n=2p-1$, as restricting a polynomial $Q$ to the first $n$ variables by setting the remaining variables to zero preserves this possibility. But for $n=2p-1$, a vector has weight $p$ if and only if it has weight a nonzero multiple of $p$, and so the lower bound in this case follows from the lower bound for the modified problem, since $n+1-p=p$.<|endoftext|> TITLE: Stable rank one and corners of $C^\ast$-algebras QUESTION [9 upvotes]: Thanks to a result of Herman and Vaserstein in [3], Rieffel's notion of stable rank [4] coincides with the Bass stable rank [1] for every $C^\ast$-algebra $A$: we denote it by $\mathrm{sr}(A)$ and we simply call it the stable rank of $A$. The following fact yields a convenient characterization of the stable rank one case. Lemma (folklore I suppose) A $C^\ast$-algebra has stable rank one if and only if the invertible elements are dense in $\widetilde{A}$, where $\widetilde{A}$ denotes $A$ if it is unital, and the unitization $A\oplus\mathbb{C}1$ of $A$ otherwise. For short, denoting the group of all invertible elements in $\widetilde{A}$ by $G\left(\widetilde{A}\right)$, we have: $$\mathrm{sr}(A)=1\quad \Leftrightarrow\quad\overline{G\left(\widetilde{A}\right)}=\widetilde{A}.$$ Now here are the facts I'm interested in. They deal with the stable rank of the corners $pAp$ of a given $C^\ast$-algebra $A$. Proposition (Corollary V.3.1.18 in [2]) Let $A$ be a unital $C^\ast$-algebra and $p$ a full projection in $A$ (i.e. the closed ideal of $A$ generated by $p$ is all of $A$). We have $\mathrm{sr}(pAp)\ge\mathrm{sr}(A)$. If $\mathrm{sr}(A)<\infty$, then $\mathrm{sr}(pAp)<\infty$ and $$\mathrm{sr}(A)=1\quad \Leftrightarrow\quad\mathrm{sr}(pAp)=1.$$ What if we remove the assumption that $p$ is full? Question A: What are the known examples of $C^\ast$-algebras $A$ and projections $p\in A\setminus\{0,1\}$ such that $$\mathrm{sr}(A)=1<\mathrm{sr}(pAp)?$$ Question B: What are the known examples of unital $C^\ast$-algebras $A$ and projections $p\in A\setminus\{0,1\}$ such that $$\mathrm{sr}(A)=1<\min\{\mathrm{sr}(pAp),\mathrm{sr}((1-p)A(1-p))\}?$$ [1] H. Bass, K-theory and stable algebra, Publications Mathématiques de l’IHÉS 22 (1964), pp. 5–60. [2] B. Blackadar, Operator Algebras, Encyclopaedia of Mathematical Sciences 122 (2006), Springer, pp. 444-452. [3] R. Herman and L. N. Vaserstein, The stable range of $C^\ast$-algebras, Inventiones Mathematicae 77 (1984) pp. 553-555. [4] M. A. Rieffel, Dimension and stable rank in the K-theory of $C^\ast$-algebras, Proceedings of the London Mathematical Society 46 (1983), pp. 301–333. REPLY [2 votes]: Here's a community wiki answer based on Leonel Robert's comment to Jamie Gabe's answer, to answer the question in general for arbitrary Banach algebras. Given the lemma established below, the answer to questions A and B is the same: there can't exist such an algebra, as spotted by Jamie Gabe in the separable case. Immitating the argument given by Bruce Blackadar to prove Proposition 4.1 in [1] (thanks to Leonel Robert for the reference), which itself mimicks Marc Rieffel's proof of Lemma 3.4 in [2], we obtain the following property in the more general setting of Banach algebras, without assuming that $p$ be full. Lemma: Let $A$ be a Banach algebra with identity element, and let $p$ be an idempotent in $A$. If the invertible elements are dense in $A$, then the invertible elements are dense in $pAp$. Proof: Let $x\in pAp$ and let $$0<\eta<\frac{1}{\|p^\perp\|^2}.$$ We are going to work with the 2x2 Peirce decomposition of $$A=pAp\oplus pAp^\perp\oplus p^\perp Ap\oplus p^\perp Ap^\perp$$ with respect to $p$ and $p^\perp:=1-p$. By assumption, we can find an invertible element $$y=\begin{bmatrix}pyp&pyp^\perp\\p^\perp yp&p^\perp yp^\perp\end{bmatrix}\in A$$ which approximates $$x+p^\perp=\begin{bmatrix}x&0\\0&p^\perp\end{bmatrix}$$ within $\eta$, i.e. $\|z\|\le\eta$ with $z:=x+p^\perp-y$, so that $\|x-pyp\|=\|pzp\|\le\eta\|p\|^2$; $\|pyp^\perp\|=\|pzp^\perp\|\le\eta\|p\|\|p^\perp\|$; $\|p^\perp yp\|=\|p^\perp zp\|\le\eta\|p\|\|p^\perp\|$; $\|p^\perp yp^\perp-p^\perp\|=\|p^\perp zp^\perp\|\le\eta\|p^\perp\|^2$. By 4, we see that $p^\perp yp^\perp$ is invertible in $p^\perp Ap^\perp$ with inverse $\delta$ such that $$\|\delta\|=\left\|\sum_{n\ge 0}\left(p^\perp-p^\perp yp^\perp\right)^n\right\|\le\frac{1}{1-\eta\|p^\perp\|^2}.$$ Using the fact that $$\delta\cdot p^\perp yp^\perp=p^\perp yp^\perp\cdot\delta=p^\perp$$ and a routine 2x2 matrix computation, we obtain the following formula: $$ \begin{bmatrix}p&-pyp^\perp\delta\\0&p^\perp\end{bmatrix}y \begin{bmatrix}p&0\\-\delta p^\perp yp&p^\perp\end{bmatrix} =\begin{bmatrix}pyp-(pyp^\perp)\delta(p^\perp yp)&0\\0&*\end{bmatrix}. $$ Since $pyp^\perp\delta$ and $\delta p^\perp yp$ are nilpotent, the three elements on the left are invertible in $A$, and it follows that $$\alpha:=pyp-(pyp^\perp)\delta(p^\perp yp)$$ is invertible in $pAp$. It only remains to check that the above estimates yield \begin{align*} \|x-\alpha\|&\le\|x-pyp\|+(pyp^\perp)\delta(p^\perp yp)\\ &\le\eta\|p\|^2+\frac{\eta^2\|p\|^2\|p^\perp\|^2}{1-\eta\|p^\perp\|^2}\\ &=\frac{\eta\|p\|^2}{1-\eta\|p^\perp\|^2}. \end{align*} Since this upper bound tends to $0$ when $\eta\rightarrow 0$, the result follows. QED [1] Bruce Blackadar, The stable rank of full corners in $C^\ast$-algebras, Proceedings of the American Mathematical Society 132 (2004), pp. 2945–2950. [2] Marc A. Rieffel, Dimension and stable rank in the K-theory of $C^\ast$-algebras, Proceedings of the London Mathematical Society 46 (1983), pp. 301–333.<|endoftext|> TITLE: Conjugation classes of pairs of involutions in the monster group QUESTION [8 upvotes]: Let $\mathbb{M}$ be the monster group, i.e. the largest finite simple sporadic group. Question: Are the conjugation classes of pairs of involutions in $\mathbb{M}$ known? What I have found so far: According to the ATLAS of finite groups, $\mathbb{M}$ has two classes of involutions 2A and 2B. The 9 classes of pairs of 2A involutions have been known for a long time, see e.g. [1]. The 12 classes of pairs of involutions of type (2A, 2B) are given in [2]. It remains to determine the classes of pairs of 2B involutions. As far as I know, this has not yet been done. My motivation for asking this question: Let $C(2B)$ be the centralizer of a 2B involution in $\mathbb{M}$. Knowing the classes of pairs of 2B involutions means that the double cosets $C(2B) \backslash \mathbb{M} / C(2B)$ are also known. Assuming that computations in $C(2B)$ are easy, we may represent elements of $\mathbb{M}$ as words in a set of generators of $C(2B)$ and an extra generator $t \in \mathbb{M} \setminus C(2B)$, see [3], [4]. Then understanding these double cosets would be helpful for estimating the maximum length of such a word, provided that there are not too many of them. This might also help to improve my implementation [5] of the monster group. A strategy for answering the question: The number of classes of pairs of 2B involutions is at least $|\mathbb{M}| / |C(2B)|^2 > 41$. So the task of finding all classes of pairs of 2B involutions is probably tedious. But it might not be hopeless, since the following Lemma gives us some kind of descent for finding classes of pairs of involutions in a group. Lemma Let $(i_1, i_2)$ be a pair of involutions in a finite group $G$, and put $a = i_1 i_2$. Then one of the following two statements holds: $a$ has odd order and all pairs of involutions $(i_3, i_4)$ with $i_3 i_4 = a$ are conjugate to the pair $(i_1, i_2)$. $G$ contains an involution $b$ such that both, $i_1$ and $i_2$, are in the centralizer of $b$. Proof If $a$ has order $2n$ then the second statement holds for $b = (i_1 i_2)^n$. So we may assume that $a$ has odd order $n$ and that there are involutions $i_3, i_4 \in G$ with $i_3 i_4 = a$ such that $i_1$ is not conjugate to $i_3$ in the centralizer $C(a)$ of $a$. We have $i_1 a i_1 = i_3 a i_3 = a^{-1}$, so $i_1 i_3$ commutes with $a$. Let $m$ be the order of $i_1 i_3$. If $m$ were odd then we would have $i_3 = (i_1 i_3)^{-(m+1)/2} i_1 (i_1 i_3)^{(m+1)/2}$, i.e. $i_1$ and $i_3$ would be conjugate in $C(a)$. Thus $m$ is even. Put $b = (i_1 i_3)^{m/2}$. So $b$ is an involution commuting with $i_1$ and $i_3$. As a power of $i_1 i_3$ it commutes with $a$. Thus $b$ also commutes with $i_2 = i_1 a$. q.e.d. References [1] J. H. Conway. A simple construction of the Fischer-Griess monster group. Inventiones Mathematicae, 1985. [2] S. P. Norton, Anatomy of the Monster I. Curtis, R., and Wilson, R. (Eds.). (1998). The Atlas of Finite Groups - Ten Years On (London Mathematical Society Lecture Note Series). Cambridge: Cambridge University Press. [3] R. A. Wilson. The Monster and black-box groups. arXiv e-prints, pages arXiv:1310.5016, October 2013. arXiv:1310.5016. [4] M. Seysen. A computer-friendly construction of the monster. arXiv e-prints, pages arXiv:2002.10921, February 2020. [5] M. Seysen, The mmgroup API reference. REPLY [4 votes]: You could use the character table of the Monster $M$ and its maximal subgroups to find more information. For the pairs from (2a, 2a) and (2a, 2b) the orbits are in bijection with the conjugacy classes containing their products. For pairs from (2b, 2b) this is not correct, but at least you can split the problem into smaller problems and get lower bounds. For example with GAP we can see that there are at least 177 orbits: t := CharacterTable("M");; ClassNames(t)[3]; # third class is 2b # m[i] contains the number of pairs (e,f) with e,f in class 2b and ef # is a fixed element in class i m := List([1..194], i-> ClassMultiplicationCoefficient(t, 3, 3, i));; # fnd is list of indices of classes occuring in (2b)*(2b) fnd := Filtered([1..194], i-> m[i] <> 0);; Length(fnd); # there are 140 such classes Let $e,f \in$ class 2b and $ef = a \in $ class $i$. Assume that there are $k$ pairs $(e',f')$ conjugate to $(e,f)$ with $e'f' = a$. Then the orbit of $(e,f)$ has length $k |M|/|C_M(a)| = |M| / |U|$ where $U$ is the stabilizer of $(e,f)$ which is a subgroup of $C_M(a)$. So $k$ is a divisor of $|C_M(a)|$ (more precisely an index of a subgroup of $C_M(a)$). So, the number m[i] is a sum of divisors of the centralizer order of an element in class i. From this we get a least number of orbits of pairs in 2b with product in class i. lbounds := [];; for i in fnd do nam := ClassNames(t)[i]; C := SizesCentralizers(t)[i]; if C mod m[i] = 0 then minnrk := 1; else div := DivisorsInt(C); minnrk := First([1..10], j-> Length(RestrictedPartitions(m[i], div, j)) > 0); fi; Add(lbounds, [nam, minnrk]); od; lbounds; # yields: #[ [ "1a", 1 ], [ "2a", 1 ], [ "2b", 3 ], [ "3a", 1 ], [ "3b", 1 ], # [ "3c", 1 ], [ "4a", 3 ], [ "4b", 2 ], [ "4c", 4 ], [ "4d", 1 ], # [ "5a", 2 ], [ "5b", 1 ], [ "6a", 2 ], [ "6b", 1 ], [ "6c", 3 ], # [ "6d", 1 ], [ "6e", 1 ], [ "6f", 2 ], [ "7a", 1 ], [ "7b", 1 ], # [ "8a", 2 ], [ "8b", 3 ], [ "8d", 1 ], [ "8e", 2 ], [ "9a", 1 ], # [ "9b", 1 ], [ "10a", 2 ], [ "10b", 3 ], [ "10c", 1 ], [ "10d", 1 ], # [ "10e", 1 ], [ "11a", 1 ], [ "12a", 2 ], [ "12b", 1 ], [ "12c", 2 ], # [ "12d", 2 ], [ "12e", 2 ], [ "12f", 1 ], [ "12g", 1 ], [ "12h", 1 ], # [ "12i", 1 ], [ "12j", 1 ], [ "13a", 2 ], [ "13b", 1 ], [ "14a", 1 ], # [ "14b", 1 ], [ "14c", 1 ], [ "15a", 1 ], [ "15b", 1 ], [ "15c", 1 ], # [ "15d", 1 ], [ "16a", 1 ], [ "16b", 1 ], [ "16c", 1 ], [ "17a", 2 ], # [ "18a", 1 ], [ "18b", 1 ], [ "18c", 1 ], [ "18d", 1 ], [ "18e", 1 ], # [ "19a", 1 ], [ "20a", 2 ], [ "20b", 2 ], [ "20c", 1 ], [ "20d", 1 ], # [ "20e", 2 ], [ "20f", 1 ], [ "21a", 1 ], [ "21b", 1 ], [ "21c", 1 ], # [ "21d", 1 ], [ "22a", 2 ], [ "22b", 1 ], [ "24a", 1 ], [ "24b", 1 ], # [ "24c", 1 ], [ "24d", 1 ], [ "24e", 1 ], [ "24i", 1 ], [ "25a", 2 ], # [ "26a", 2 ], [ "26b", 1 ], [ "27a", 1 ], [ "27b", 1 ], [ "28a", 1 ], # [ "28b", 1 ], [ "28c", 1 ], [ "28d", 1 ], [ "29a", 1 ], [ "30a", 1 ], # [ "30b", 2 ], [ "30c", 1 ], [ "30d", 1 ], [ "30e", 1 ], [ "30f", 1 ], # [ "30g", 1 ], [ "33a", 1 ], [ "33b", 1 ], [ "34a", 2 ], [ "35a", 1 ], # [ "35b", 1 ], [ "36a", 1 ], [ "36b", 1 ], [ "36c", 2 ], [ "36d", 1 ], # [ "38a", 1 ], [ "39a", 1 ], [ "39b", 1 ], [ "40b", 1 ], [ "41a", 1 ], # [ "42a", 2 ], [ "42b", 1 ], [ "42c", 1 ], [ "42d", 1 ], [ "45a", 1 ], # [ "50a", 1 ], [ "51a", 1 ], [ "52a", 1 ], [ "52b", 1 ], [ "54a", 1 ], # [ "55a", 1 ], [ "56a", 1 ], [ "57a", 1 ], [ "60a", 2 ], [ "60b", 1 ], # [ "60c", 1 ], [ "60d", 1 ], [ "60e", 1 ], [ "60f", 1 ], [ "66a", 1 ], # [ "66b", 1 ], [ "68a", 1 ], [ "70a", 1 ], [ "70b", 1 ], [ "78a", 1 ], # [ "84a", 1 ], [ "84b", 1 ], [ "84c", 1 ], [ "105a", 1 ], [ "110a", 1 ] ] Sum(lbounds, x-> x[2]); # lower bound of 177 orbits The character tables of various maximal subgroups and the fusions of conjugacy classes are also known and could help with further investigations.<|endoftext|> TITLE: Examples of non-Kähler compact complex manifolds of dimension four with some properties QUESTION [7 upvotes]: I am looking for examples of non-Kähler compact complex manifolds of dimension four with trivial canonical class and $H^i(M, {\mathcal O}_M) = H^0(M, \Omega^i_M) =0$ for $0< i < \dim M$. In dimension two, no such manifolds exist (compact complex surfaces with trivial canonical class and $H^1(M, {\mathcal O}_M)=0$ are K3 surfaces and they are all Kähler.) In dimension three, Infinitely many topological types have been constructed by R. Friedman. My question is: "Are there any examples of such manifolds in dimension four?" It is notable that Guan constructed examples of non-Kähler compact complex symplectic fourfold with trivial canonical class. REPLY [3 votes]: Examples were recently constructed by Lee and Sano by two different constructions (which both involve smoothing varieties with normal crossings). Sano's paper also constructs examples in all dimensions $\geq 4$ with arbitrarily large second Betti number.<|endoftext|> TITLE: Formula that requires a higher complexity to be proved QUESTION [5 upvotes]: Sorry if this question is naive, I am not very well versed in recursion theory. Does it exist a formula $\phi$ such that: $\phi$ is provable in Peano arithmetic $\phi \in \Sigma^0_n$ or $\phi \in \Pi^0_n$, but provably, every proof of $\phi$ from the axioms must involve a step that does not belong to $\Sigma^0_k$ or $\Pi^0_k$ for any $k \leq n$? Or even a step that needs complexity at least $\Sigma^0_{n + m}$ or $\Pi_{n + m}$ for some $m > 1$? This may even not be well defined - it could depend on the proof system - but even a result for a particular proof system could be interesting. What I am trying to capture is the following. Young mathematicians often struggle with calculus, since for the first time they have to deal with formulas involving two quantifieres ($\forall \epsilon \exists \delta \dots$). Of course, with experience, they come to manage these formulas without trouble. But juggling many more quantifiers quickly becomes overwhelming even for expert mathematicians. Could it be the case that there exists some claim of interest to mathematicians (involving few quantifiers) but such that in order to prove it we need to consider intermediate steps that are too complex for our feeble minds? REPLY [6 votes]: First, let me mention that the only way a formula may require proofs using complex formulas is that it requires complex axioms to prove: the cut elimination theorem implies that if a $\Sigma_n$ formula is provable from $\Sigma_n$ axioms, it has a proof from these axioms in which all formulas are $\Sigma_n$; and similarly for $\Pi_n$ in place of $\Sigma_n$. (Literally, this holds for the sequent calculus. When we use a different proof system, such as some form of a Hilbert calculus, we may need to raise the complexity by a small constant.) In particular, up to small constants, no such formulas exist when Peano arithmetic is replaced in the question by a finitely axiomatizable theory such as $I\Sigma_n$ (= Peano arithmetic with the induction schema restricted to $\Sigma_n$ formulas). However, such formulas do exist for Peano arithmetic, and we might even take $\phi\in\Pi_1$ irrespective of $n$: for example, let $\phi=\mathrm{Con}_{I\Sigma_n}$ (the formalized consistency statement for $I\Sigma_n$). It is known that $\mathrm{PA}\vdash\mathrm{Con}_{I\Sigma_n}$ (see also here for a more general result), but Gödel’s second incompleteness theorem shows that $I\Sigma_n\nvdash\mathrm{Con}_{I\Sigma_n}$; thus, any proof of $\mathrm{Con}_{I\Sigma_n}$ in $\mathrm{PA}$ must use an instance of the induction schema for formulas of higher complexity than $\Sigma_n$ (and higher than $\Pi_n$, as $I\Sigma_n=I\Pi_n$).<|endoftext|> TITLE: Smallest known counterexamples to Hedetniemi’s conjecture QUESTION [19 upvotes]: In 2019, Shitov has shown a counterexample (Ann. Math, 190(2) (2019) pp. 663-667) to Hedetniemi’s conjecture, $$\chi(G \times H)=\min(\chi(G),\chi(H))$$ where $\chi(G)$ is the chromatic number of the undirected finite graph $G$. Shitov counterexample is estimated to have $|V(G)|\approx4^{100}$ and $|V(H)|\approx4^{10000}$. Has there been some effort or progress to reduce the size of the counterexample? REPLY [24 votes]: Yes, Xuding Zhu did this in Relatively small counterexamples to Hedetniemi's conjecture (J. Comb. Theory B 146 (2021) pp. 141-150, doi:10.1016/j.jctb.2020.09.005, arXiv:2004.09028) where the sizes of the graphs are $3403$ and $10501$. Marcin Wrochna has a preprint, Smaller counterexamples to Hedetniemi's conjecture, arXiv:2012.13558, that brings the sizes down to $4686$ and $30$ (as well as the chromatic number down to $5$).<|endoftext|> TITLE: Is AC equivalent over ZF to 'every fibration can be equipped with a cleavage'? QUESTION [9 upvotes]: It is well known that (working over ZF) AC implies that every fibration $p:\mathcal{E}\to\mathcal{B}$ can be equipped with a cleavage by choosing, for each arrow $u:I\to p(X)$ in the base category whose codomain is in the image of the fibration, a Cartesian arrow $\overline u(X):u^*(X)\to X$ in the overcategory above $u$ whose codomain is $X$. Does the reverse implication hold? Andrej Bauer mentions in this MO answer that he suspects the converse holds; I was wondering if anyone had a proof on hand. REPLY [16 votes]: Yes, in fact Grothendieck fibration between groupoids are enough. Let $p:Y \to X$ be any surjection. We construct the following groupoid $G$. Its set of objects is $X \amalg Y$. Its morphisms corresponds to the equivalence relation such that two elements of $y$ are equivalent if they have the same image by $p$ and an element of $y$ is equivalent to its image by $p$ in $X$ (and no distinct elements of $X$ are equivalent). We then consider $I$ the walking isomorphisms, i.e. the anti-discrete groupoid on two objects, and the functor $G \to I$ that sends element of $Y$ on one object and element of $X$ on the other. I claim that: it is a fibration. (easy to check) a cleavage produces a section of $p$: for each $x \in X$ a cartesian lift of the non-identity arrow $e \to p(x)$ corresponds to the choice of a $y \in Y$ such that $p(y)=x$. Remark: a more conceptual way to understand this is that a fibration over $I$ is the same as an "anaequivalence" (in the sense of Makkai anafunctors) and a cleavage for such a fibration gives choice of a pair of inverse functors implementing this equivalence. So "anyfibration admit a cleavage" implies that every anaequivalence is implemented by a pair of inverse functor, and in particular that every fully faithful essentially surjective functor has an inverse. We then combine this with the usual proof that the existence of inverse functors imply the axiom of choice.<|endoftext|> TITLE: What is so 'coloured' on Chromatic Homotopy Theory QUESTION [8 upvotes]: As the title suggest, I would like know the motivation/ historical background why chromatic homotopy theory is called 'chromatic'. Literally, what analogy to colors it might have. Accordings to wikipedia the origin of such homotopy theories bases on Quillen's work on cohomology theories of formal groups. This allows due to wikipedia to study that studies complex oriented cohomology theories from the 'chromatic' point of view. This involves classification of these theories by so called 'chromatic levels'. Has this 'visualization' regarding the theories as 'lying on certain levels' some funny analogy to colors regarded from physical viewpoint as a part of electromagnetic spectrum? Sound probably silly but I am quite curious what was the background for the choice of the name 'chromatic' in this context. REPLY [15 votes]: This term is surely due to Doug Ravenel. In the mid 1970's, he and collaborators Steve Wilson and Haynes Miller constructed and exploited a "chromatic spectral sequence" for computing Ext groups in the Adams-Novikov spectral sequence. This is featured in the Doug's book titled Complex cobordism and stable homotopy groups of spheres published in the early 1980s. As he relates in the intro to that book, he wanted to call it The music of the spheres, but the publisher persuaded him not to. So chromatic might be more musical than luminary! At any rate, the term refers to the different sorts of periodicity one finds in stable homotopy: discovered first in the algebra, and then later in actual maps. The relationship to complex oriented theories is critical. Quillen wrote his famous paper noting the connection with formal groups. Jack Morava realized that classification theorems for these proved by number theorists were a guide to cohomology theories that should exist - now called the Morava K-theories - periodic with different periodicies - and also a guide of how to organize our thinking about stable homotopy. Miller-Ravenel-Wilson made concrete use of this stuff and Doug was inspired to make his famous conjectures, mainly proved in the mid 1980s by the next generation to come along, ...<|endoftext|> TITLE: Relationship between enriched, internal, and fibered categories QUESTION [17 upvotes]: In this question, let $(\mathcal{V}, \otimes, [-,-], e)$ be a nice enough symmetric monoidal closed bicomplete category. The usual set-based Category theory has been generalized in many directions, let me mention some of them. $\mathcal{V}$-enriched categories. Internal categories in $\mathcal{V}$. $\mathcal{V}$-fibered categories. The literature provides comparisons between these approaches. Streicher's Fibered Categories, Sec. 4 provides a strategy to construct a fibered category out of an internal category. Enriched and internal categories: an extensive relationship by Cottrell, Fujii and Power show that when $\mathcal{V}$ is cartesian, $\mathcal{V}$-Cat and the category $\text{Cat}_d(\mathcal{V})$ of internal categories in V with a discrete object of objects are equivalent. Verity's PhD thesis Enriched Categories, Internal Categories and Change of Base, Chap.2.2 discusses how to internalize an enriched category. Even common generalizations have been suggested, like enriched indexed categories, or enriched internal categories. Question. Could you help me in understanding the most cutting edge results that relate enriched, internal, and fibered categories? I am not interested in common generalizations, just in processes, functors, adjunctions, that relate $\mathcal{V}$-categories, internal $\mathcal{V}$-categories and fibered $\mathcal{V}$ categories, as in Cottrell et al's paper. REPLY [7 votes]: Too long for a comment. Thanks for compiling this bibliography, Ivan! It seems you've got the literature at your fingertips to get a sense for the available tools. One thing to say is that in contrast to the way you've presented things, usually internal category theory and fibered category theory are viewed as being tightly connected: essentially, internal categories are to small categories as fibered categories are to locally small categories. When using either framework, one gets used to passing back and forth with the other as a matter of course. For examples of this way of thinking, see e.g. this book; I'm thinking in particular of Pare and Schumacher's article therein. In particular, the construction of a fibered category from an internal one is very standard -- it should be the same whereever you read about it. When it comes to comparing internal / fibered categories to enriched categories, there's a bit more of a conceptual jump. I actually haven't delved into this into too much detail, but I think the references that you and others have mentioned are probably good starting points. One thing I do think is worth mentioning is that the enriched vs. internal comparison comes up in $\infty$-category theory, where Segal categories are a sort of "enriched" approach to $\infty$-categories whereas complete Segal spaces are a sort of "internal" approach. The fact that the equivalences in complete Segal spaces need to be "normalized" via the completeness / univalence condition echoes the basic fact that categories enriched in categories (i.e. 2-categories) are not the same thing as categories internal to categories (i.e. double categories); in order to use the latter to model the former you need to impose some condition. In a more literal sense, the enriched vs. internal distinction is also seen when comparing simplicial categories to Horel's model for $\infty$-categories, which uses categories internal to simplicial sets. (Side note: I'd actually be interested in seeing, in non-$\infty$-land, an approach to viewing enriched categories internally which goes the route of complete Segal spaces rather than Segal categories, normalizing objects to encode isomorphism information rather than being discrete -- if this is even possible!) All of that is mostly to provide context, but also to point out that if you're interested in comparison methods, it may be worth looking at the various Quillen adjunctions which are used to compare these different models of $\infty$-categories. Similarly, $(\infty,n)$-categories can be modeled as $n$-fold complete Segal spaces, which runs into the same issue of "normalizing away the double category / 2-category distinction" issue, and again the comparisons to other more "enriched-type" models like $n$-quasicategories will be confronting enriched vs. internal comparison issues. Finally, I'll add that Cruttwell and Shulman's virtual equipments -- one of these frameworks accommodating both enriched and internal settings -- have been used quite literally to do real work $\infty$-categorically by Riehl and Verity in their $\infty$-cosmos work, where they construct a virtual equipment out of an $\infty$-cosmos. They are quite concerned with addressing model comparison results systematically, so for example they study when a morphism of $\infty$-cosmoi gives rise to a morphism of the associated virtual equipments, and when notions like pointwise Kan extension are preserved. So they're going beyond just constructing Quillen adjunctions, and working to get comparisons of the category-theoretic concepts internal to their $\infty$-cosmoi. The virtual equipment framework of course owes a lot to the proarrow equipments of Wood, which also deserve to be mentioned in your general bibliography. For that matter, Koudenburg has a number of papers developing the formal category theory of virtual equipments and change-of-base questions between them -- e.g. with the goal of things like comparing monoidal Kan extensions to ordinary ones. The constant idea which I would keep in mind, which I think is stressed in most of these works and of which you're probably well aware, is that in order to adequately translate categorical concepts from one setting to another, it's often not enough to have a 2-functor between the appropriate 2-cateogories, because the notion of discrete fibration used in the internal / fibered setting does not translate directly to the notion of $V$-presheaf in the enriched setting; for that reason you get structures like (virtual) proarrow equipments and so forth which explicitly encode both the structure of functors and the structure of bimodules. And all of that is just a long way of saying that I feel irresponsible for knowing of so many places where these issues matter and yet not myself having an answer to the question of "What are the main things one needs to do to compare different category theories, and how can one do them?" EDIT: But on second thought, maybe I've seen this done enough times that I can give you a hitlist of the kinds of things one needs to think about when performing change of base from $\mathcal C$ to $\mathcal D$, where $\mathcal C$ and $\mathcal D$ are places where one can "do category theory": You probably need at least a 2-functor $\mathcal C \to \mathcal D$ to be sure that notions like adjunctions and monads transfer (though note that in $\infty$-categorical contexts, monads are an exception to the rule that all the basic concepts of $\infty$-category theory can be expressed in a metatheory that doesn't go to $\infty$ -- you can't say what a homotopy coherent monad is in an $\infty$-cosmos just using the underlying virtual equipment. But this is more an issue of the "$\infty$" part than the "category theory" part). You need some way of keeping track of distributors/presheaves/bimodules/correspondences/profunctors in order to formulate notions like representability, pointwiseness of Kan extensions, etc. Your comparison mapping needs to act on this data too. I'd say that if you can adequately translate both types of information, then you can adequately translate "all the category theory" from $\mathcal C$ to $\mathcal D$, at least according to the wisdom that seems to have accrued through the work of all the papers cited in the questions, comments, and this answer.<|endoftext|> TITLE: An approach to showing hyperbolic groups are CAT(0) QUESTION [7 upvotes]: I've been sitting on this idea for quite a while but I'm not in academia any longer so not likely to ever tackle it on my own. The approach is as follows: $G$ acts on its boundary $\partial G$ ergo, $G$ also acts on the set of distinct, ordered triples of $\partial G$, call this $\Delta G$ $\Delta G$ factors as a hyperbolic space $\mathcal{H}G$ which is quasi-isometric to $G$, and some other (possibly compact) space, $\mathcal{C}G$ The $G$ action on $\Delta G$ "plays nice" so that there's a geometric $G$-action on $\mathcal{H}G$. Tada! The motivation behind the approach is pretty simple: $\Delta G$ is the set of triangles with points on $\partial G$. Because $G$ is hyperbolic, those triangles can be readily identified with their centers which are coarsely elements of $G$ (this is why $\mathcal{H}G \sim G$) together with some other component that can be thought of as rotations of those triangles ($\mathcal{C}G$). $\Delta G$ is like the set of 2-frames on $G$, or something similar. The distinction between $\mathcal{H}G$ and $G$ is that, by reconstructing everything from the boundary, $\mathcal{H}G$ has completely lost all local structure of $G$ that might be an impediment to CAT(0) (or CAT(-k), in fact). The details of how to factor $\Delta G$ are the biggest stumbling blocks I think. In any case, I wanted to share in case this idea actually has promise to advance this question, or at least provide an avenue to identify more specific obstructions to the $CAT(0)$ condition. To get started, are there any known techniques for recognizing factors in spaces like these, perhaps in the context of tangent bundles etc? Take care, Brad REPLY [5 votes]: This approach is quite hopeless for several reasons. First of all, let me try to make sense of what you wrote. You write: $\Delta $ factors as a hyperbolic space ${\mathcal H}$ which is quasi-isometric to $$, and some other (possibly compact) space, ${\mathcal C} G$ So far, $\Delta $ is just a topological space (yes, it has some incomplete metrics coming from Gromov-type metrics on $\partial G$, but these are not even $G$-invariant), so I read "factors" as "splits as a topological direct product." You are hoping that the splitting is such that the $G$-action projects to a proper action on ${\mathcal H}$ and that the latter can be given an invariant $CAT(0)$ metric. The latter would imply contractibility and local contractibility. But in "most" cases $\Delta $ has rich topology. For instance, for generic $G$, $\partial G$ is homeomorphic to the Menger curve which has nontrivial $H^1$ even locally. Hence, by Kunneth, $\Delta $ locally has nontrivial $H^3$. Hence, again by Kunneth, it cannot even locally split as a product of a contractible space (of positive dimension) with some other factor! Thus, there is no hope even for a local splitting (in general). Edit. [Per Henry's request.] Below, all cohomology is Chech with rational coefficients (any field will work) and dimension means rational cohomological dimension: Suppose that $A$ is a compact (and Hausdorff) 1-dimensional topological space (think of the Menger curve), $U_i\subset A, i=1, 2, 3$ is are open subsets with $H^1(U_i)\ne 0$. Then by the Kunneth formula, $H^3(W)\ne 0$, where $W=U_1\times U_2\times U_3$. Furthermore, $A^3$ has dimension 3. Now, assume that all three subsets $U_i$ are such that $W$ is disjoint from the big diagonal in $A^3$, i.e. $W\subset Z:=A^{(3)}$, where the letter is the complement to the big diagonal in $A^3$. From this, using the LES of the pair $(Z,W)$ and the fact that $H^4(Z)=0$, you see that $H^3(Z)\ne 0$. I will use it to prove that $Z=A^{(3)}$ does not split off a nontrivial contractible factor. (A similar argument works also locally.) Suppose to the contrary that $Z=B\times C$, where $C$ is contractible and $dim(C)>0$. Thus, $dim(B)+dim(C)=3$. In particular, $dim(B)\le 2$ and, hence, $H^3(B)=0$. By applying the Kunneth formula again, we get $$ H^3(Z)= H^3(B)\otimes H^0(C)= H^3(B)=0. $$ A contradiction. Even if you assume that $\partial G$ is a topological $n$-sphere, your approach is just a "naive approach to Cannon's conjecture." The problem is that you know that $\Delta $ is homeomorphic to the product of a compact with $R^{n+1}$. However, this product decomposition cannot be $G$-invariant. You can only hope to get a $G$-bundle over $R^{n+1}$. Nobody succeeded (so far) in finding such bundles directly for any $n>0$. The known cases are $n=1$ and $n\ge 5$. All require very hard and nontrivial work by first-rate topologists (Gabai, Casson, Weinberger...). The proof by Casson and Jungreis is closest to the idea of finiding an equivariant fibration, but it is based on some very special 3-dimensional topology tools. Lastly, even if you are given a $G$-bundle over a contractible space $X$, how do you propose to find a $G$-invariant CAT(0)-metric on $X$? As an exercise, consider the case of 3-manifold groups. So, you got a proper and cocompact action of your hyperbolic group on a contractible 3-dimensional manifold $X$ (the base of a fibration of $\Delta G$). How do you know that there is a $G$-invariant CAT(0)-metric on $X$? Right, you will need to quote Perelman's theorem. Edit. Few more things in this direction: What are the known tools for finding a locally CAT(0)-metric on the given topological space $Y$? There are only few: a. Combinatorial, in case when $Y$ is a cell complex with a particularly nice local properties. Once $Y$ has sufficiently large dimension, all you have along these lines is the construction of (locally) CAT(0) cube complexes. This excludes hyperbolic groups with Property (T). b. Some special differential-geometric constructions for manifolds of dimensions $\le 3$ and in the case of locally symmetric spaces. c. Some variations on cut-and-paste or branched covering constructions using the pre-existing locally CAT(0) metrics. My list is missing few more sporadic constructions but, I think, the situation is quite clear: Given a general locally contractible and aspherical compact topological space $Y$ (of sufficiently high dimension) with nontrivial hyperbolic fundamental group we simply lack any tools for constructing a locally CAT(0) metric on $Y$. Cf. this question regarding CAT(1) metrics.<|endoftext|> TITLE: Solving polynomial equations in spectra? QUESTION [11 upvotes]: Let $M$ be the mod-$p$ Moore spectrum where $p \geq 3$ is a (power of) a prime. Then $M$ satisfies the "polynomial equation" $M \wedge M \cong M \oplus \Sigma M$. Is this a general phenomenon, or is it something very special about Moore spectra? More formally, let $K = K_0^\oplus(\mathbb S_{(p)})$ be the direct-sum $K$-theory of the $p$-local sphere spectrum, i.e. the free commutative ring on equivalence classes of finite $p$-local spectra, mod the equations $[0] = 0$, $[A] + [B] = [A \oplus B]$, $[\mathbb S] = 1$, and $[A][B] = [A \wedge B]$. Then $K$ is an algebra over the ring $\mathbb Z[\Sigma,\Sigma^{-1}]$, where the action of $\Sigma$ is given by suspension. The above observation says that when $M$ is a Moore spectrum [1], the element $[M] \in K$ is integral over $\mathbb Z[\Sigma,\Sigma^{-1}]$. Question 1: What are some other examples of integral elements of $K = K^\oplus_0(\mathbb S_{(p)})$, considered as a $\mathbb Z[\Sigma,\Sigma^{-1}]$-algebra? Question 2: Is every element of $K$ integral over $\mathbb Z[\Sigma,\Sigma^{-1}]$? If not, is it perhaps the case that every finite type $n$, $p$-local spectrum is integral when $p$ is large compared to $n$? Question 3: Are there other interesting "polynomial equations" involving not-necessarily-finite spectra, or not-necessarily-finite power series? [1] When $p=2$, I believe I've read that the mod-2 Moore spectrum $M$ satisfies a polynomial equation involving powers $\leq 3$, but also the coefficent $\mathbb C \mathbb P^2$. This "reduces" the question of whether $M$ is integral over $\mathbb Z[\Sigma,\Sigma^{-1}]$ to the question of whether $\mathbb C \mathbb P^2$ is integral, but I'm not sure this is an improvement. I think I'd tend to suspect that neither of these spectra are integral over $\mathbb Z[\Sigma,\Sigma^{-1}]$. Via rather indirect means, I've convinced myself that if $A$ is a finite spectrum of type $\geq 1$, then any polynomial $f(X)$ satisfied by $A$ must lie in the ideal $(X,\Sigma-1)$. Perhaps there is a direct way to see this. Note that if we were working with exact-sequence $K$-theory, then the action of $\Sigma$ would be by $-1$, thanks to exact sequences of the form $X \to 0 \to \Sigma X$. But in direct sum $K$-theory, there's no reason for this to be the case. REPLY [3 votes]: Here's a slightly fleshed-out version of Nicholas Kuhn's argument. I wasn't able to verify the exact statement he uses, but a variant thereof. For a spectrum $X$, let $f(X)$ be the maximal $k \in \mathbb N$ such that $Sq^k$ acts nontrivially on $H^\ast(X)$ (note that we're not using $Sq^{2^k}$). If $X$ is finite, then $f(X) < \infty$. Then clearly $f(X \oplus Y) = \max(f(X),f(Y))$ $f(\Sigma X) = f(X)$ Let us show that: $f(X_1\wedge X_2) = f(X_1) + f(X_2)$ By the Cartan formula, $Sq^k(x_1 \otimes x_2) = \sum_{k_1 + k_2 = k} S^{k_1}(x_1) \otimes Sq^{k_2}(x_2)$. If $k > f(X_1) + f(X_2)$, then in each summand, at least one tensor factor has $Sq^{k_i}$ acting for $k_i > f(X_i)$, so $Sq^k$ vanishes. Thus $f(X_1 \wedge X_2) \leq f(X_1)+f(X_2)$. If $k = f(X_1) + f(X_2)$, then every summand has a tensor factor for which $Sq^{k_i}$ acts with $k_i > f(X_i)$ except for $k_i = f(X_i)$, so $Sq^k = Sq^{f(X_1)} \otimes Sq^{f(X_2)}$. So if $x_i \in H^\ast(X_i)$ are such that $Sq^{f(X_i)}(x_i) \neq 0$, then $Sq^k(x_1 \otimes x_2) \neq 0$. Thus $f(X_1 \wedge X_2) \geq f(X_1) + f(X_2)$. Now suppose that $X$ is integral, i.e. $X^n \cong \oplus_{i=0}^{n-1}\oplus_j a_{ij} \Sigma^j X^i$. Then by the above three formulas, we have $n f(X) = f(X^n) = f(\oplus_{i=0}^{n-1} \oplus_j a_{ij} \Sigma^j X^i) = \max_{i=0}^{n-1} i f(X) = (n-1) f(X)$. Therefore $f(X) = 0$, i.e. the Steenrod algebra acts trivially on $H^\ast(X)$.<|endoftext|> TITLE: p-adic versions of log concavity for graphs (or matroids) QUESTION [8 upvotes]: It was recently shown using techniques inspired by algebraic geometry (by Huh and Adiprasito-Huh-Katz) that the chromatic polynomial of a graph (or matroid) has coefficients that satisfy log-concavity. Since the coefficients are integers, we can also ask if they satisfy any p-adic identities or inequalities. Are there any conjectures or results in this direction? REPLY [2 votes]: I am not aware of any conjectures or results in this direction, and I am not so optimistic because it seems to be a hard question even for binomial coefficients. So I think your suggestion of studying coefficients of polynomials of the form $\prod_i (t - n_i)$ is good, but I don't know what to expect. Some references on $p$-adic properties of binomial coefficients: Wikipedia, divisibility properties of binomial coefficients. Erdos, Graham, Ruzsa, Straus, On the prime factors of $\binom{2n}{n}$. Spiegelhofer and Wallner, Divisibility of binomial coefficients by powers of 2. We get binomial coefficients from taking the chromatic polynomial of a tree with $n$ edges, which is $$ \chi_\text{tree}(t) = t(t-1)^n = \sum_{k=0}^n (-1)^k \binom{n}{k} t^{n-k+1} . $$ The binomial coefficients are unimodal in the archimedean norm, but are far from unimodal $p$-adically. In fact, if $n$ is even, then the $2$-valuation of the sequence $\binom{n}{0}, \binom{n}{1}, \binom{n}{2},\ldots$ will alternate between getting larger and smaller, so in a sense behaves completely "opposite" to being unimodal. At another extreme, if $n = 2^m - 1$, then the binomial coefficients $\binom{n}{k}$ are all odd, so have the same size $2$-adically. The binomial coefficients have a nice archimedean limit as $n \to \infty$, up to an appropriate rescaling, but the $p$-adic limit seems more complicated. For instance, in the paper above Spiegelhofer and Wallner show that the 2-valuations of the binomial coefficients do obey a close-to normal distribution, but in the sense where we ignore the order of the sequence $\binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \ldots$ and only count how many times numbers of a given $p$-adic size appear. Another reason it seems challenging to find a proper $p$-adic analogue: Adiprasito-Huh-Katz showed that the log-concavity in the archimedean norm comes from the Hodge-Riemann relation in degrees 0 and 1 (applied to the Chow ring of a graph / matroid), which says a certain matrix with integer entries must have indefinite signature. Namely, log concavity of $a_{i-1}, a_i, a_{i+1}$ is equivalent to $$ a_{i-1}a_{i+1} \leq a_i^2 \quad\Leftrightarrow\quad \det \begin{pmatrix} a_{i-1} & a_i \\ a_i & a_{i+1} \end{pmatrix} \leq 0. $$ So what's important is not exactly the valuation $x \mapsto |x|$ that measures the size of a number, but the fact that the integers (and reals) are totally ordered so that it makes sense to distinguish positive and negative numbers. So one approach may be to ask: Is there any reason to expect a $p$-adic version of the Hodge-Riemann relation? (I have no idea..) Generally, is there a $p$-adic analogue of the signature of a bilinear form?<|endoftext|> TITLE: Elementary proof of the exactness of Čech complex associated to a hypercovering ("Illusie's Conjecture") QUESTION [5 upvotes]: Let $\mathcal{E}$ be a sheaf of abelian groups on a topological space (or a site). For an open covering $\mathfrak{U} = (U_i)_i$, it is well known that the augmented Čech complex $0 \to \mathcal{E} \to \mathcal{\check C}^\bullet(\mathfrak{U}, \mathcal{E})$ is a resolution of $\mathcal{E}$. (Depending on the sheaf cohomology of $\mathcal{E}$ on the intersections of the members of $\mathfrak{U}$, this resolution might compute $\mathbb{R}\Gamma(\mathcal{E})$, but this need not concern us here.) A simple proof runs as follows: Let $s \in \mathcal{\check C}^{n+1}(\mathfrak{U}, \mathcal{E})(V)$ such that $ds = 0$. Then $t = (s_{i_\text{fix},i_0,\ldots,i_n})_{i_0,\ldots,i_n} \in \mathcal{\check C}^{n}(\mathfrak{U}, \mathcal{E})(V \cap U_{i_\text{fix}})$ is a local preimage of $s$ under $d$, as $(dt)_{i_0, \ldots, i_{n+1}} = s_{i_0, \ldots, i_{n+1}} - (ds)_{i_\text{fix}, i_0, \ldots, i_{n+1}} = s_{i_0, \ldots, i_{n+1}}$. As $V = \bigcup_{i_\text{fix}} (V \cap U_\text{fix})$, that's all what's needed. I'm looking for a similarly elementary proof in the case that $\mathfrak{U}$ is a hypercovering. Apparently this fact is sometimes referred to as Illusie's Conjecture. A proof involving trivial Kan fibrations and an induction reducing to the case of ordinary covers is in Section 01GA of the Stacks Project, but I'd prefer either a more direct proof or some insight as to why a much simpler argument isn't likely to exist. REPLY [3 votes]: If you are working on a topological space (or more generally a site with enough points), then it's enough to show that $E \to \check C(\mathfrak U,E)$ is a quasi-isomorphism on stalks. This reduces the claim to the case that the topological space is a point. But a hypercover of a point is (after some unpacking) a simplicial set $X$ such that every map $\partial \Delta_n \to X$ extends to $\Delta_n \to X$, i.e. a contractible Kan complex. So the claim is that such an $X$ has the cohomology of a point. Not sure if this is the type of argument you want - it is more "simplicial" than "chainy" - but on the other hand, so is the definition of a hypercover. In general the "meat" of Illusie's conjecture lies in the case when the site does not have enough points.<|endoftext|> TITLE: Partitions of convex planar regions into zonogons QUESTION [6 upvotes]: A zonogon is a centrally symmetric convex polygon. Are there convex non-zonogons that can be partitioned into a finite number of (convex) zonogons? Same as 1 with the pieces allowed to be nonconvex but centrally symmetric polygons. REPLY [2 votes]: I extend on the answer of Yaakov Baruch. Choose a generic direction as being "down". We can then speak of bottom edges and top edges of the polygon and its tiles. Suppose that the polygon $P$ is partitioned into convex tiles, so that there exists a perfect matching between the tiles that are not centrally symmetric and matched tiles are related by a central point reflection (see the figure below). Assume further that the partition is edge-to-edge, that is, no vertex of one tile meets an edge of another tile (Yaakov explained why this is no restriction). Let $e_0$ be a top edge. Note that every edge is the top-edge of at most one tile. We construct a sequence of edges $e_1,...,e_m$ of edges as follows: if $e_i$ is the top edge of a centrally symmetric tile $T$, then let $e_{i+1}$ be the "opposite" bottom edge of $T$. if $e_i$ is the top edge of not-centrally symmetric tile $T$, then let $e_{i+1}$ be the corresponding edge in the matched tile to $T$. $e_{i+1}$ is then a bottom edge of the matched tile. if $e_i$ is not the top edge of any tile, then $e_i$ is a bottom edge of $P$. Note that all edges in this sequence are of the same length and have the same orientation, and so each top edge of $P$ corresponds to an equally long and equally oriented bottom edge. Since $P$ is convex, the order of these edges at the bottom must be inverse to the order at the top. This shows that $P$ must be centrally symmetric. Note that this formulation includes the case of non-convex tiles. Every centrally symmetric tile can be decomposed into a set of convex centrally symmetric tiles and a set of pairs of matched convex tiles that are not necessarily centrally symmetric:<|endoftext|> TITLE: Reference request: Brown Ozawa and strong completely positive approximation property? QUESTION [9 upvotes]: The notion of a $C^*$-algebra being nuclear has many equivalent characterisations. These are considered in the excellent, modern textbook $C^*$-Algebras and Finite-Dimensional Approximations by Brown and Ozawa. They take as definition that the identity on the $C^*$-algebra $A$ can be point-norm approximated by ccp maps which factor through matrix algebras. The original definition is that there is only one $C^*$-norm on $A\odot B$, for any $B$. In the history of the subject, Kirchberg and Choi, Effros independently showed that $A$ is nuclear if the identity on $A$ can be point-norm approximated by finite-rank ccp maps from $A$ to itself. Kirchberg's paper calls this the strong completely positive approximation property (SCPAP) but I don't think this is common terminology now. My questions: Is this result in the book by Brown and Ozawa? (Edit: By "this result" I mean precisely: that the SCPAP implies nuclearity.) I cannot seem to find it, even though it would nicely motivate the CBAP, a weaker approximation property. Is there any direct way to get between these two definitions? Both papers which I cite take quite a long way around, going through work of Lance and tensor products. My motivation is to try to give a nice, expositionary, motivation of the CBAP from nuclearity; it would be nice to point to a book for this. REPLY [10 votes]: I am not sure if it is in Brown and Ozawa, but it is in Pisier's recent book "Tensor Products of C*-algebras and Operator Spaces" as Corollary 10.16. It may also be in his earlier Operator Spaces book, but my copy isn't with me.<|endoftext|> TITLE: Prove/disprove $(\int_{0}^{2 \pi} \!\!\cos f(x) d x)^{2}+(\int_{0}^{2 \pi}\!\!\! \sqrt{(f'(x))^{2}+\sin ^{2} f(x)}dx)^{2}\ge 4\pi^{2}$ QUESTION [11 upvotes]: This problem has been posted on Math.SE but didn't receive any correct answer after a long time. Let $f(x)$ be a differentiable function on $[0,2\pi]$ s.t. $0\leq f(x)\leq 2\pi$ and $f(0)=f(2\pi)$. Prove or disprove that $$ \left(\int_{0}^{2 \pi} \cos f(x) d x\right)^{2}+\left(\int_{0}^{2 \pi} \sqrt{(f'(x))^{2}+\sin ^{2} f(x)} d x\right)^{2} \geq(2 \pi)^{2} $$ It seems that when $f$ is an arbitrary constant, the left side equals $(2\pi)^2$ and seems to be the minimum. But how can I show that there's no other $f$ that makes the left side equal (or be less than) $(2\pi)^2$? A geometric interpretation of the inequality has been found: Consider a closed curve on a sphere: $C=\{(\cos x\cdot\sin f(x),\,\sin x\cdot\sin f(x),\,\cos f(x))\mid x\in[0,2\pi)\}$, we have its perimeter $\displaystyle L=\int_0^{2\pi}\sqrt{(f'(x))^2+\sin^2 f(x)}\,dx$ and its area $\displaystyle S=2\pi-\int_0^{2\pi}\cos f(x)\,dx$. From spherical isoperimetric inequality $L^2\ge S\left( 4\pi-S \right)$, we have $\left( 2\pi-S \right)^2+L^2\ge\left( 2\pi \right)^2$, and the equality holds iff $C$ is any circle on the sphere. In this way we get the original inequality in the sense of geometry. Now the question is, how to prove the inequality with only pure analysis methods? REPLY [5 votes]: An approach that should work is to derive the differential equation that any minimizer would have to satisfy and check that its solutions are the known ones for which equality holds. To fill in the details, one would need to show that a minimizer does exist and has the necessary regularity to make the derivation of the differential equation valid, but I think that such arguments are standard in the calculus of variations and should be applicable here. So, assume that a minimizer $f$ exists and set $$ a = \frac1{2\pi}\int_0^{2\pi}\cos f(x)\,dx\quad\text{and}\quad b = \frac1{2\pi}\int_0^{2\pi}\sqrt{f'(x)^2+\sin^2 f(x)}\ dx. $$ Note that, if $b=0$, then $f(x)$ is an integer multiple of $\pi$ and we have equality. Set this case aside and assume that $b>0$. The assumption that $f$ be a minimizer implies that if $u$ is any $2\pi$-periodic function then $I'(0)=0$ where $$ \begin{align} I(t) &= \left(\int_0^{2\pi}\cos\bigl(f(x){+}tu(x)\bigr)\,dx\right)^2\\ &\qquad\qquad+\left(\int_0^{2\pi}\sqrt{(f'(x){+}tu'(x))^2+\sin^2 (f(x){+}tu(x)\bigr)}\ dx\right)^2. \end{align} $$ Calculation (using integration by parts and the fact that $u$ and $f$ can be regarded as $2\pi$-periodic) then yields that $I'(0)/(4\pi)$ equals $$ \int_0^{2\pi} \left(-b\left(\frac{f'(x)}{\sqrt{f'(x)^2{+}\sin^2 f(x)}}\right)'+\frac{b\sin f(x)\cos f(x)}{\sqrt{f'(x)^2{+}\sin^2 f(x)}}-a\sin f(x)\right)\,u(x)\ dx. $$ Consequently, setting $\lambda = a/b$, we have that $f$ must satisfy a second-order differential equation with parameter $\lambda$ $$ \left(\frac{f'(x)}{\sqrt{f'(x)^2{+}\sin^2 f(x)}}\right)'-\frac{\sin f(x)\cos f(x)}{\sqrt{f'(x)^2{+}\sin^2 f(x)}}+\lambda\,\sin f(x) = 0.\tag1 $$ Thus, there is a $3$-parameter family of solutions to this equation with parameter, and $f$ must belong to this $3$-parameter family. Now, it just so happens that we already know a $3$-parameter family of solutions to this equation, namely $$ f(x) = \arccos\left(\frac{u{+}(p\cos x {+} q \sin x)\sqrt{(p\cos x {+} q \sin x)^2+1-u^2}}{1{+}(p\cos x {+} q \sin x)^2}\right), \tag2 $$ where $|u|<1$ and $p$ and $q$ are $3$ real numbers, and all of these solutions $f$ give $a^2+b^2 = 1$, i.e., equality in the desired inequality. One gets other $3$-parameter families by adding an integer multiple of $\pi$ to the above formula, but these can be considered equivalent. It turns out that this then gives all of the solutions except the ones that have $\sin f(x) \equiv 0$. (Allowing $|u|\ge1$ gives families of solutions that can be defined only over subintervals of $[0,2\pi]$, and these must be taken into account in the analysis as well.) Again, in order to make this argument fully rigorous, one has to prove that a minimizer does exist in the first place and prove a regularity result for solutions of the above ODE at places where $\sin f(x)$ vanishes. I think that those are doable, but I haven't checked the details.<|endoftext|> TITLE: Integer matrices which are not a power QUESTION [27 upvotes]: $\DeclareMathOperator\SL{SL}\DeclareMathOperator\Sp{Sp}$In a group $G$, an element $g$ is said to be primitive if there is no $h \in G$ and integer $n >1$ such that $g = h^n$. (For clarification, I consider finite order elements to be not primitive) I was wondering, in the case $G$ is $\SL_n(\mathbb{Z})$ or $\Sp_{2n}(\mathbb{Z})$, if there exists a criterion for primitivity of matrices. I actually even struggle to find examples of primitives matrices in these groups. In $\Sp_{2}(\mathbb{Z})=\SL_{2}(\mathbb{Z})$, the matrix $\pmatrix{1 & 1 \\ 0 & 1}$ is primitive. (This can be shown by considering its action on $\mathbb{H^2}$, for example.) But this does not generalize (easily at least) to higher dimension. For example, and quite surprisingly maybe $$ \pmatrix {1 & 0 & 0& 1\\ 0 & 1 & 0 & 0\\ 0 & 1 & -1 & 1\\ 0 & 1& -1 & 0 }^3 = \pmatrix{1 & 1 & 0& 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0& 0 & 1} $$ Anyway, it seems like some things should be known, but it is very hard to find anything on google since primitive matrix usually means something else …. I would appreciate any input. REPLY [26 votes]: I actually even struggle to find examples of primitives matrices in these groups. Here is a relatively easy sufficient condition. If $M \in SL_n(\mathbb{Z})$ is the $k^{th}$ power of some other matrix $N$ then every eigenvalue $\lambda$ of $N$ has the property that $\lambda^k$ is an eigenvalue of $M$, and conversely. If we can choose $M$ and an eigenvalue $\mu$ of $M$ such that every $k^{th}$ root of $\mu$ has the property that its minimal polynomial has degree larger than $n$, then $N$ cannot exist. For this to work $\mu$ can't be a root of unity. As an explicit example we'll take $M = \left[ \begin{array}{ccc} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right]$, the Fibonacci matrix augmented by a $-1$ to live in $SL_3(\mathbb{Z})$, whose eigenvalues are $-1$ and the golden ratios $\phi, \varphi = \frac{1 \pm \sqrt{5}}{2}$. Any $\lambda$ such that $\lambda^k = \phi, \varphi$ for $k \ge 2$ generates an extension of $\mathbb{Q}$ which contains $\mathbb{Q}(\sqrt{5})$ and hence is either equal to $\mathbb{Q}(\sqrt{5})$ or else has even degree greater than $2$ and so at least $4$. Moreover $\lambda$ is an algebraic integer and a unit. But $\phi, \varphi$ each generate the unit group of $\mathcal{O}_{\mathbb{Q}(\sqrt{5})} = \mathbb{Z}[\phi]$ (up to signs). So if $\mathbb{Q}(\lambda) = \mathbb{Q}(\sqrt{5})$ then $\lambda$ must be $\pm \phi$ or $\pm \varphi$, but this is ruled out by taking the absolute value. So the minimal polynomial of $\lambda$ must have degree at least $4$, which means $\lambda$ can't be an eigenvalue of a matrix in $SL_3(\mathbb{Z})$.<|endoftext|> TITLE: Counter example of a radical extension that is not Galois/normal over $\mathbb{Q}(\omega)$? QUESTION [7 upvotes]: Most proofs of Galois theorem stating that "an equation is solvable in radicals if and only if its Galois group is solvable," show the left to right direction by induction on the height of the radical extension. Starting with a cyclotomic field, adjoining "enough" roots of the unity (e.g. adjoining the product of all the ) and continuing with successive simple radical extensions, which are all Galois and Abelian over the previous one (because we ensured that there were enough roots of unity to start with). The key element in proving that all these extensions are solvable over the base field is then to define a solvable extension as an extension which normal closure has solvable Galois group (equivalently such that there exist an extension which Galois group is solvable) (def (a)), this makes "being a solvable extension" transitive (it is distinguished) as proved in Lang Algebra. Doing this makes it possible to complete the proof by induction, while "being Galois and solvable" (def (b)) does not go through (since Galois extensions are not distinguished). Many texts also state that the (radical) extensions arising in the above mentioned induction are not Galois in general. While this is indeed not the case in general (as with the traditional example $\mathbb{Q} \subset \mathbb{Q}(\sqrt{2}) \subset \mathbb{Q}(\sqrt[4]{2})$ since it's missing $i\sqrt[4]{2}$), one might wonder if this is true or not for radical extensions where the base field contains enough roots of unity. More precisely, the question is to determine whether the following statement is true. Let $\omega_n$ be a primitive $n^\textrm{th}$ root of the unity. Let $x = (x_1, \ldots, x_h)$ be sequence of elements in the algebraic closure of $\mathbb{Q}$, such that $x_{i+1}^{m_{i+1}} \in \mathbb{Q}(x_1, \ldots, x_{i})$ for some minimal $m_i > 1$ for all $i < h$. Is there a natural number $N$, such that in the radical extension series $$ \mathbb{Q}(\omega_{N}) = K \subset K(x_1) \subset \ldots \subset K(x_1, \ldots, x_n) $$ each $K(x_1, \ldots, x_{i+1})/K$ is Galois ? Definition Let's call $N_x$ the smallest of such natural numbers if it exists, and set $N_x = \infty$ if there is no such natural number. The above question amounts to «Is $N_x$ always finite?» In a typical proof of Galois theorem above, since the primitive root of the unity that is chosen is typically a divisor of $d_x = \prod_i m_i$ (one can also pick the lcm afaik). If it were true that $N_x | d_x$ for all $x$ then the proof above could be carried out with def (b) instead of (a). (The answers seem to indicate this is not true though.) However if $N_x$ is finite for all $x$, then for each $x$ one can use def (b) by starting the radical series with $\mathbb{Q}(\omega_{N_x})$ (instead of $\mathbb{Q}(\omega_{d_x})$ as in the proof above). I conjectured the above statement is false, i.e. that there are $x$s such that $N_x > d_x$ and even such that $N_x = \infty$. Hence I am searching for a (minimal?) counter example, i.e. a radical extension $$ \mathbb{Q}(\omega_N) = K \subset K(x_1) \subset \ldots \subset K(x_1, \ldots, x_h) $$ such that each for all $N$ such that each $K(x_1, \ldots, x_{i+1})/K(x_1, \ldots, x_i)$ is Galois, $K(x_1, \ldots, x_n)/K$ is not Galois. (Another way to state it is to start with $K$ being an extension of $\mathbb{Q}$ containing all roots of unit.) I think the following is a radical extensions of total degree $6$ over $\mathbb{Q}(\omega)$, where $\omega = \frac{\sqrt{3}}{2} + \frac{i}{2}$ is a primitive 12th root of unity and where each intermediate extension is Galois over the previous one: $$K = \mathbb{Q}(\omega) \subset F = K\left(\sqrt[3]{2}\right) \subset K\left(\sqrt[3]{2}, \sqrt{1+ \sqrt[3]{2}}\right) = E.$$ However, I would like to claim $E$ is not Galois over $K$, even though all 12th roots of unity are in $K$. Indeed a conjugate of $\sqrt{1+ \sqrt[3]{2}}$ over $K$ is a square root $\alpha$ of $1 + j\sqrt[3]{2}$ -- their minimal polynomial is $X^6 - 3 X^4 + 3 X^2 - 3$ -- but $\alpha$ would not be in $E = \textrm{span}_F\left(1, \sqrt{1+ \sqrt[3]{2}}\right).$ I reduced the problem to checking that $\frac{1 + j\sqrt[3]{2}}{1+ \sqrt[3]{2}}$ is not a square in $F$ (unless I made a mistake) and some famous online computer algebra system tells that if I compare it to $(x+y\sqrt[3]{2}+z\sqrt[3]{2}^2)^2$ then $z$ needs to be expressed using a square root of an element of $F$ that does not look like it's in $F$, but I am still not 100% sure this is correct... Does someone has indications on better ways to find a counter example, check this one or prove the theorem that I believe is wrong in the first place? (i.e. that all radical extension over a field with enough roots of unity are Galois.) EDIT 25/01/2021 I made my question more precise by showing the statement for which I am searching for a counter example. Unless I am mistaken, the answers and their comments below show that both $x_1 ^ 2 = 3, \quad x_2 ^ 2= 1 + x_1$ $y_1 ^ 3 = 2, \quad y_2 ^ 2= 1 + y_1$ are indeed counter examples of $N | d$ since $d_x = 4$ and $N_x = 8$ $d_y = 6$ and $N_y > 12$. They also show that $x$ is not a counter example that $N_x < \infty$, and I think the finiteness status of $N_y$ is unknown. REPLY [2 votes]: The polynomial $f:=(x^2-1)^3-2$ has Galois group $G \simeq S_4 \times C_2$ over $\mathbb{Q}$, so the commutator subgroup $[G,G] \simeq A_4$. Let $L$ be the splitting field (over $\mathbb{Q}$) of $f$. This contains the $12$-th roots of unity, i.e. contains your $K$. Since $[K:\mathbb{Q}]=4=[G:[G,G]]$, $L$ contains no other roots of unity and $\operatorname{Gal}(L/K)\simeq A_4$. Since $A_4$ has no normal subgroups of order $2$ the extension $E/K$ is not Galois. Certifying that $G \simeq S_4 \times C_2$ could e.g. be done using (an exact version of a small part of) Algorithm 6.3.10 in Cohen's "A Course in Computational Algebraic Number Theory" or using more basic Galois resolvents, or...<|endoftext|> TITLE: What does it matter if a group has a non-elementary hyperbolic quotient? QUESTION [5 upvotes]: My adviser recently shared a problem with me that seeks to establish non-elementary* hyperbolic quotients for mapping class groups. They told me that this could be useful for establishing results on separability or omnipotence, and that these could be relevant for examining profinite rigidity of hyperbolic 3-manifolds. Unfortunately, I'm not fully read up on these topics. In this recent paper, Behrstock, Hagen, Martin and Sisto also seek to make headway on the question of hyperbolic quotients for mapping class groups. They have a discussion in their introduction of the relevance of this question, mentioning again separability and omnipotence, profinite rigidity, and placing things in the context of the virtual Haken conjecture. Again, I'm a little ignorant of these topics and their history of this point. So my question: Q: Can anyone explain with some detail (or point me to some nice references as to) why it's relevant that mapping class groups have hyperbolic quotients? Or why it's helpful that any group has such a quotient? *A Gromov-hyperbolic group is non-elementary if it is not virtually cyclic, i.e. is infinite and not virtually $\mathbb{Z}$. REPLY [7 votes]: You don't get anything just from knowing that a non-elementary hyperbolic quotient exists. However, the problem of constructing such quotients of mapping class groups appears very difficult, and can be viewed as a step on the way to constructing other interesting classes of quotients that you might hope to study, like finite quotients or virtually abelian quotients. For instance, here are two important open problems about the finite-index subgroups of mapping class groups (let's say for closed surfaces $S$ of genus $\geq 3$, to be safe). The first is easy to understand. Virtual first betti numbers: Does $\mathrm{Mod}(S)$ have a finite index subgroup that surjects $\mathbb{Z}$? The second is morally asking: "Does every finite-index subgroup of $\mathrm{Mod}(S)$ come from a finite-sheeted cover of $S$?" Congruence subgroup property: Does every finite-index subgroup of $\mathrm{Mod}(S)$ contain a principal congruence subgroup, i.e. the kernel of a natural map $\mathrm{Mod}(S)\to\mathrm{Out}(Q)$, where $Q$ is a finite characteristic quotient of $\pi_1(S)$? I'll take for granted that these are interesting questions, but could explain more if necessary. The second one is certainly related to the profinite rigidity questions mentioned in the question. (Added:) In any case, even if you don't like either of these questions specifically, the point is that we are interested in, and don't know much about, the finite and virtually abelian quotients of mapping class groups. Note that either of these questions can be answered by finding suitable quotients of $\mathrm{Mod}(S)$: the first (positively) by finding a virtually abelian quotient, the second (negatively) by finding a finite quotient that doesn't factor through some $\mathrm{Out}(Q)$. Neither can be answered directly by a non-elementary hyperbolic quotient. However, we know a lot about hyperbolic groups, and so if we can construct hyperbolic quotients, we might hope to go further and construct hyperbolic quotients with a rich supply of finite quotients (for instance if they're residually finite), and thereby answer these questions. There's some precedent for this hope. The Virtual Haken conjecture and its cousins for hyperbolic 3-manifolds asked for rich supplies of finite quotients of hyperbolic 3-manifold groups. One of the crucial tools in Agol's proof turned out to be Wise's Malnormal special quotient theorem, which guarantees that hyperbolic 3-manifold groups have non-elementary hyperbolic quotients which are themselves residually finite. (This is a slightly backwards telling of the Virtual Haken story, but the point is that these are the tools you need.) Indeed, in some ways the starting point for the proof of the Virtual Haken conjecture was a paper of Agol, Groves and Manning, which proved it (modulo the work of Kahn--Markovic) under the hypothesis that all hyperbolic groups are residually finite. I view the work of Behrstock, Hagen, Martin and Sisto as the analogue of the Agol--Groves--Manning paper in this story.<|endoftext|> TITLE: Siegel--Walfisz for number fields QUESTION [11 upvotes]: For a number field $K$, we write $\Delta_K$ for its absolute discriminant. I was hoping for a Siegel--Walfisz type theorem of the following type: Let $A > 0$. Then for every $X > 0$, every number field $K$ and every Galois extension $L/K$ with $\Delta_L \leq (\log X)^A$ and every conjugacy class $C$ of $\text{Gal}(L/K)$ we have $$ \#\{\mathfrak{p} \text{ prime of } K : \mathfrak{p} \text{ unramified in } L, \text{Frob}_{\mathfrak{p}} = C, N_{K/\mathbb{Q}}(\mathfrak{p}) \leq X\} = \frac{\# C}{\# G} \text{Li}(X) + O_A\left(\frac{X}{(\log X)^A}\right). $$ Is there such a result available in the literature? I have been going through various references that together seem to imply that the answer is yes, but I am hoping for one reference that explicitly states the above theorem. Edit: in my application the number field $K$ is fixed and $[L : K]$ is bounded. REPLY [3 votes]: There is enough in the literature to extract a result of this form, but it might not appear explicitly. I will reference recent work of Thorner and Zaman instead of Lagarias-Odlyzko, since it gives a substantial improvement: Let $L/F$ be a Galois extension with group G, and let $C\subseteq G$ be a conjugacy class. Let $H\subseteq G$ be an abelian subgroup such that $C\cap H\neq\emptyset$, let $K$ be the fixed field of $H$, and choose $g_C\in C\cap H$. Let $Q$ be the largest norm of the conductors of the Hecke characters in the dual group of $H$. Assume that $\zeta_L(s)$ has a Siegel zero $\beta_1$. Since $\zeta_L(s)$ factors as a product of Hecke characters over $K$ in $\widehat{H}$, it follows from Landau-type arguments that $\beta_1$ must be a real simple zero of the $L$-function of exactly one such Hecke character, which necessarily has order $\leq 2$. Call this Hecke character $\chi_1$. There exist absolute and effectively computable constants $c$ and $c'$ such that if $x\geq (D_K Q [K:\mathbb{Q}]^{[K:\mathbb{Q}]})^{c}$, then $\pi_C(x) = \frac{|C|}{|G|}(\mathrm{Li}(x)-\chi_1(g_C)\mathrm{Li}(x^{\beta_1}))\Big(1+O\Big(\exp\Big[-\frac{c'\log x}{\log(D_K Q [K:\mathbb{Q}]^{[K:\mathbb{Q}]})}\Big]+\exp\Big[-\frac{(c'\log x)^{1/2}}{[K:\mathbb{Q}]^{1/2}}\Big]\Big)\Big)$. A Siegel-type argument will give an ineffective lower bound on $L(1,\chi_1)$, leading to the ineffective bound $\beta_1\leq 1-c(\epsilon)(D_K Q [K:\mathbb{Q}]^{[K:\mathbb{Q}]})^{-\epsilon}$. (Maybe this is not explicitly in the literature, but I doubt that, and if I'm wrong, it is standard to prove once you've seen the usual proof of Siegel's theorem. The biquadratic extension that you'd construct would now lie over $K$, not $F$ or $\mathbb{Q}$. The convexity bound that one would use in this argument was worked out by Rademacher.) This should give you exactly what you are looking for when you solve for how large $x$ should be for the "Siegel contribution" to be absorbed into the second error term. As far as effective bounds go (per GH from MO's question in the comments), I do not know of any effective bound better than Stark's that holds in complete generality. One can do better if, for instance, the root discriminant of $K$ is not too small. Stark's bound accounts for what happens in infinite class towers, where the root discriminant is fixed and the degree blows up. If one could rule out such a possibility (which might be inherent in the fixed degree assumption in the original post), then Stark's bound should experience a dramatic improvement. ADDED: The Siegel-type bound for the exceptional zero follows from work of Fogels. When the order of the exceptional character is 1, the bound is in Lemma 16 of "On the zeros of Hecke's L-functions. I, II." (Acta Arith.). When the order is 2, the bound is in "Über die Ausnahmenullstelle der Heckeschen L-Funktionen" (Acta Arith.). Thorner, Jesse; Zaman, Asif, A unified and improved Chebotarev density theorem, Algebra Number Theory 13, No. 5, 1039-1068 (2019). ZBL1443.11239.<|endoftext|> TITLE: On superabundant-like numbers QUESTION [9 upvotes]: Define $\sigma(N)=\sum_{d|n} d$. A superabundant number is a positive integer $u$ for which $\frac{\sigma(u)}{u} > \frac{\sigma(v)}{v}$ for every positive integer $v\frac{\sigma(m^2)}{m^2}$ for every positive integer $m TITLE: Do Schlichting's and Balmer's definitions of higher Witt groups of a scheme agree when 2 is inverted? QUESTION [23 upvotes]: My question is whether the construction of higher Witt groups of a scheme in Schlichting's Hermitian K-theory of Exact Categories agrees with the definition in Balmer's chapter in the Handbook of K-theory when 2 is inverted (i.e. the schemes are over $\mathbb{Z}[\frac{1}{2}]$, although I'd also be happy to know whether this is true when we tensor the Witt groups with $\mathbb{Z}[\frac{1}{2}]$). See below for both definitions of higher Witt groups. When 2 is not inverted, they only necessarily agree on the 0-th Witt group. More background if necessary Schlichting defines higher Witt groups of an exact category with duality $(\mathcal{E}, *, \eta)$ as the homotopy groups of the geometric realization of a Hermitian Q construction, which is roughly as follows Given an exact category $\mathcal{E}$ with duality, define the category $Q^h\mathcal{E}$ as the category where objects are symmetric spaces in $\mathcal{E}$, i.e. pairs $(X,\varphi)$ of $X\in \text{ob }\mathcal{E}$ and an isomorphism $\varphi: X\mapsto X^{*}$ such that $\varphi^*\eta_X = \varphi$ morphisms from $(X, \varphi)$ to $(Y, \psi)$ are equivalence classes of diagrams $X \xleftarrow{p}U\xrightarrow{i}Y$ where $p$ is an admissible epimorphism, $i$ an admissible monomorphism, and the restrictions of the symmetric forms on $X$ and $Y$ to $U$ agree. composition is pullback (For a more precise statement, see Definition 4.1 in p.12 of Schlichting) The $0$-th Witt group $\pi_0(|Q^h\mathcal{E}|)$ turns out to be the usual $$W_0(\mathcal{E}) = \{\text{isoclasses of symmetric spaces}\}/\{\text{metabolics}\}$$ (metabolics are defined in Schlichting, p. 6, Def 2.5, the 0th Witt group is defined in Schlichting p. 7, first paragraph of 2.2 and also in Balmer p. 7 Definition 1.1.27) We can take Witt groups of a scheme $X$ by using the exact category of vector bundles over $X$ with the usual duality. On the other hand, Balmer defines higher Witt groups of a triangulated category $\mathcal{K}$ with duality by taking $$W_0(\mathcal{K}) = \{\text{isoclasses of symmetric spaces}\}/\{\text{metabolics}\}$$ as above, and taking $$W_n(\mathcal{K}) = W_0(T^n\mathcal{K})$$ where $T^n$ denotes "applying the shift functor $n$ times" in the triangulated category, which can change the duality (see Balmer Def 1.4.1 and Def 1.4.4 for details). Now we can take Balmer Witt groups of a scheme $X$ by using the category of perfect complexes over $X$. These definitions certainly agree on $W_0$, but they don't agree in the higher Witt groups (as can be seen in Remark 4.2 of Schlichting), and I want to know whether they do agree when $2$ is inverted as I said above. Thanks! REPLY [16 votes]: No, the definition in Schlichting's first paper are not the "correct" definition of higher Witt groups (in any case they are not the analogue of Balmer's Witt groups), rather they are some shifted higher Grothendieck-Witt groups. He provided later a different definition that does coincide with Balmer's (but which is defined only when 2 is invertible). For brevity let me refer to the following three papers by Marco Schlichting: [Sch1] Hermitian K-theory of Exact Categories [Sch2] The Mayer-Vietoris principle for Grothendieck-Witt groups of schemes [Sch3] Hermitian K-theory, derived equivalences and Karoubi's fundamental theorem In [Sch3] definition of the Grothendieck-Witt spectrum is provided when 2 is invertible. From this in [Prop. 7.2, Sch3] he deduces a model for Balmer's higher Witt groups as homotopy groups of the L-theory spectrum. However, by using Proposition 6 in [Sch2] we see that the space $|Q^h\mathcal{E}|$ used in [Sch1] to define the Witt groups is not the 0-th space of the underlying spectrum. Rather it is the first space of the shifted Grothendieck-Witt spectrum $\operatorname{GW}^{[-1]}(\mathcal{E})$. Therefore we have $$\pi_i |Q^h\mathcal{E}|\cong \pi_{i-1}\operatorname{GW}^{[-1]}(\mathcal{E})$$ In particular its homotopy groups are not 4-periodic, and in general are not Witt groups.<|endoftext|> TITLE: K-equivalence ⇒ isomorphism of Chow motives? QUESTION [11 upvotes]: An old conjecture of Bondal–Orlov–Kawamata predicts that K-equivalent varieties are D-equivalent, see Kawamata's paper D-equivalence and K-equivalence for definitions. In particular this applies to birational Calabi–Yau varieties. See Progress on Bondal–Orlov derived equivalence conjecture for the Bondal–Orlov formulation of the conjecture and the known cases. These conjectures are mostly open, and considered to be an important bridge between derived categories and birational geometry. Derived categories and Chow motives play the role of universal cohomology theories, in noncommutative, and commutative worlds respectively. Do we expect that K-equivalence implies isomorphism of rational, or even integral Chow motives? Example 1. Integral Chow motives of varieties related by a standard flop are isomorphic: Q. Jiang - On the Chow theory of projectivization. This was the motivation for the question. Example 2 (from Nico Berger's comment). Birational irreducible holomorphic symplectic varieties have isomorphic integral Chow motives: U. Riess - On the Chow ring of birational irreducible symplectic varieties. Remarks. In many cases, D-equivalence is known to imply isomorphisms for rational Chow motives, but not for integral ones: examples can be found among K3 surfaces. It is also known that D-equivalence does not imply K-equivalence, even for rational surfaces: Uehara - An example of Fourier–Mukai partners of minimal elliptic surfaces. REPLY [4 votes]: K-equivalence $\implies$ isomorphism of (rational) Chow motives is a conjecture going back to this 2002 paper of Wang (Conjecture 2.2); see this overview paper of his for what else K-equivalence should imply on the cohomological level. Some important special cases, such as ordinary flops and Mukai flops are proven in this 2010 paper by Lee, Lin and Wang. In 2016 the case of semismall K-equivalences has been proved by Liu: "Motivic equivalence under semismall flops".<|endoftext|> TITLE: A question on a real sequence QUESTION [14 upvotes]: Let $\{a_n\}_{n\ge1}$ be a real sequence that decays faster than any algebraic speed, that is, $\lim_{n\to \infty} n^pa_n = 0$ for every positive integer $p$. Assume that $$\sum_{n\ge 1}(n+1)^kn^ka_n = 0$$ for every integer $k \ge 0$. Question: Can we conclude that $a_n \equiv 0$? REPLY [14 votes]: Counterexample. Consider the analytic function in the unit disk $$f(z)=\exp\left(-\sqrt{\frac{1}{1-z}}\right)=a_0+a_1z+\ldots,\quad |z|<1,$$ where the principal branch of the $\sqrt{\;}$ is used. This is the definition of our sequence $a_n$. Function $f$ extends to a $C^\infty$ function on the unit circle, which evident at every point except $z=1$, and it tends to $0=f(1)$ exponentially as $z\to 1$. So we have that the sequence $(a_n)$ has your property: $|a_n|$ tends to $0$ faster than any negative power of $n$ (as Fourier coefficients of a $C^\infty$ function). Function $f$ and all derivatives of $f$ vanish at the point $1$. This implies (by Tauber's theorem) $$\sum_{n=0}^\infty n(n-1)(n-2)\ldots (n-k)a_n=0.$$ for every $k\geq 0$. This easily implies that $$\sum_{n=0}^\infty n^ka_n=0$$ for every $k\geq 0$, and then $$\sum_{n=0}^\infty n^k(n+1)^ka_n=0$$ for every $k\geq 0$.<|endoftext|> TITLE: An Indepth Look at Isoperimetry in the Cayley Graph Generated by All Transpositions QUESTION [5 upvotes]: Let $\Omega_n$ denote the symmetric/permutation group on $n$ objects. Let $T_n \subseteq \Omega_n$ denote the set of transpositions. Drop the $n$-subscripts. Define the Cayley graph $G = (\Omega, E)$ by saying that $\sigma, \sigma' \in \Omega$ are adjacent if $\sigma^{-1} \sigma \in T$, ie they differ by a transposition. One can use the Aldous spectral gap conjecture to show that the spectral gap of (the simple random walk on) this graph is $1/n$. A standard result by multiple authors (Jerrum and Sinclair, to name two) then says that $1/n^2 \lesssim \Phi_* \lesssim 1/n$ where $\Phi_*$ is the isoperimetric constant: $$ \Phi(S) := \frac{|\partial S|}{|S|} \quad\text{and}\quad \Phi_* := \min_{|S| \le |\Omega|/2} \Phi(S),$$ where $\partial S$ is the edge boundary of the set $S$. This is a worst-case bound. I am interested in a better understanding of the isoperimetric profile $\Phi(S)$, not just for the worst-case $S$. References and the like would be appreciated. Consider this example. Define $S$ by including every permutation independently with probability $\tfrac12$. (I just mean site $\tfrac12$-percolation by this.) Fix any transposition $\tau \in T$. Define $$ \partial_\tau S := \{ \sigma \tau \mid \sigma \in S \text{ and } \sigma \tau \notin S \}. $$ One can pair up all the permutations: $(\sigma, \sigma \tau)$ where $\sigma$ ranges over a set of size $\tfrac12 |\Omega|$. This outlook shows that $$ |\partial_\tau S| \sim \textrm{Bin}(|\Omega|, \tfrac14). $$ Indeed, for every point in $S$ there is a $\tfrac12$ chance that its pair is not in $S$. Also, $|S| \sim \textrm{Bin}(|\Omega|, \tfrac12)$. Thus $ \Phi(S) \approx \tfrac12. $ So if $S$ is drawn in this sense then typically it has isoperimetric expansion roughly $\tfrac12$. This is much better than the worst-case, which is at least as bad as $1/n$. Ideally I would like to determine some characterisation of the set of 'expanding' sets, ie ones with $\Phi(S) \asymp 1$. More formally, define $\mathcal P_c := \{ S \subseteq \Omega \mid \Phi(S) \ge c \}$ for $c \ge 0$. Then $\mathcal P_0$ is just the power set of $\Omega$. I am after some characterisation of $\mathcal P_c$, or some large subset of it, for $c > 0$ fixed and $n \to \infty$. REPLY [4 votes]: This reply is a bit too long for a comment, but it responds mainly to the comment from the OP who wrote that "it surely must be the case that $\Phi(S_t)$ is typically of order 1". The mixing time bound using evolving sets need not be sharp, and there are results (see [2]) that support Benoît Kloeckner's intuition that the evolving sets are typically "round", i.e., provide sparse cuts. The full spectrum (not just the spectral gap) of random transpositions has been known long before the Aldous conjecture, see [3]. [1] Morris, Ben, and Yuval Peres. "Evolving sets, mixing and heat kernel bounds." Probability Theory and Related Fields 133, no. 2 (2005): 245-266. [2] Andersen, Reid, and Yuval Peres. "Finding sparse cuts locally using evolving sets." In Proceedings of the forty-first annual ACM symposium on Theory of computing, pp. 235-244. 2009. [3] Diaconis, Persi, and Mehrdad Shahshahani. "Generating a random permutation with random transpositions." Zeitschrift für Wahrscheinlichkeitstheorie und verwandte Gebiete 57, no. 2 (1981): 159-179.<|endoftext|> TITLE: How to rewrite mathematics constructively? QUESTION [43 upvotes]: Many mathematical subfields often use the axiom of choice and proofs by contradiction. I heard from people supporting constructive mathematics that often one can rewrite the definitions and theorems so that both the axiom of choice and proofs by contradiction aren't needed anymore. An example is the theory of locales. This is a reformulation of topology that doesn't need the axiom of choice to prove analogues of results that classically need some form of choice, such as Tychonoff's theorem. I wonder: Are there some tricks behind this "reformulation"? When I have a theorem with a classical proof, how can I reformulate it so that it is provable constructively? In case this question is too broad, I would be interested in the following example: The usual proof of Gödel's completeness theorem by Henkin isn't constructive. Is there some easy way to reformulate the theorem and the proof constructively? Or is it necessary to define, with much non-trivial work, categorical semantics and state the theorem in this new context? REPLY [5 votes]: Are you aware that the (semantic) completeness theorem for countable first-order languages is equivalent over RCA0 (a weak subsystem of second-order arithmetic) to WKL0? Essentially the non-constructive part of this lies in the fact that you want to obtain an infinite path from an infinite binary tree. This principle is truly non-constructive, since there is a computable infinite tree with no computable infinite path. So even in the countable realm, which arguably is what truly constructive mathematics should be about, the completeness theorem is non-constructive. But it is not that bad; we can obtain a model of any consistent computable FOL theory that is computable from the first Turing jump. If you are interested in the use of AC in the uncountable case, the issue lies entirely with whether the theory can be well-ordered. The completeness theorem holds for any theory over a well-orderable language. In general you can find that almost every theorem that invokes AC or Zorn's lemma can be rephrased as theorems about well-ordered stuff whose proofs need no choice. For example, every well-orderable field has an algebraic closure, and every vector space with a well-orderable spanning set has a basis. So even in the usual setting of Zermelo set theory you can easily extract the 'more constructive' version just by adding well-orderability conditions.<|endoftext|> TITLE: Mean value principle reversed QUESTION [5 upvotes]: Suppose that $\Omega \subset \mathbb R^3$ is a domain with smooth boundary $\partial \Omega$ and suppose that $0\in \Omega$. Given any $f \in C^{\infty}(\partial \Omega)$ let $u^f$ denote the unique harmonic function on $\Omega$ with Dirichlet data $f$. Finally, suppose that given any smooth $f$ there holds: $$ u^f(0)= \frac{1}{|\partial \Omega|}\int_{\partial \Omega} f(y) \,d\sigma_y.$$ Does it follow that $\Omega$ is a sphere? REPLY [6 votes]: Edit: I misread the question. New answer: The question asks whether the Poisson kernel $P_\Omega(0, \cdot)$ is constant only when the domain $\Omega$ is a ball centred at $0$. This is indeed true: let $r$ be the radius of the largest ball $B(0, r)$ contained in $\Omega$, and $R$ the radius of the smallest ball $B(0, R)$ containing $\Omega$. If $y \in \partial \Omega$ and $|y| = r$, then $P_\Omega(0, y) \geqslant P_{B(0,r)}(0, y)$. Similarly, if $z \in \partial \Omega$ and $|z| = R$, then $P_\Omega(0, z) \leqslant P_{B(0,R)}(0, z)$. If $P_\Omega(0, \cdot)$ is constant, it follows that $P_{B(0,r)}(0, y) = P_{B(0,R)}(0, z)$, and therefore $r = R$. Old answer, to a different question, whether $u_f(0) = \frac{1}{|\Omega|} \int_\Omega u_f(x) dx$ implies that $\Omega$ necessarily a ball? Yes, it does. This is sometimes called Kuran's theorem, proved by Ülkü Kuran in 1972, see [1], after preliminary works by a number of other researchers. Reference: [1] Ülkü Kuran, On the mean-value property of harmonic functions. Bull. London Math. Soc. 4 (1972): 311–312, DOI:10.1112/blms/4.3.311.<|endoftext|> TITLE: When is the category of finitely presented modules abelian? QUESTION [7 upvotes]: Let $R$ be an associative ring with identity and $\mathrm{mod}R$ be the category of finitely presented $R$-modules. I would like to know when the category $\mathrm{mod}R$ is abelian. I know that if $R$ is noetherian, then $\mathrm{mod}R$ is an abelian category. Maybe, it could be the case that if $\mathrm{mod}R$ is abelian, then $R$ must be noetherian. Is it true? REPLY [17 votes]: Wojowu's idea is right: Lemma. Let $R$ be a ring, let $\mathbf{Mod}_R$ be the category of (left) $R$-modules, and let $\mathbf{Mod}_R^{\text{fp}}$ be the subcategory of finitely presented modules. Then the following are equivalent: $R$ is left coherent, i.e. every finitely generated left ideal is finitely presented; $\mathbf{Mod}_R^{\text{fp}}$ is a weak Serre subcategory of $\mathbf{Mod}_R$; $\mathbf{Mod}_R^{\text{fp}}$ is abelian and the inclusion $\mathbf{Mod}_R^{\text{fp}} \hookrightarrow \mathbf{Mod}_R$ is exact; $\mathbf{Mod}_R^{\text{fp}}$ is abelian. Proof. Implication 3 $\Rightarrow 4$ is obvious; implication 2 $\Rightarrow$ 3 is Tag 0754; and 1 $\Rightarrow$ 2 is well known (in the commutative case, see Tags 05CW and 05CX; the proofs generalise without difficulty to the general case). For 4 $\Rightarrow$ 1, we note that the inclusion $\mathbf{Mod}_R^{\text{fp}} \hookrightarrow \mathbf{Mod}_R$ is left exact, as it is isomorphic to $\operatorname{Hom}_R(R,-)$. Now if $I \subseteq R$ is a finitely generated left ideal, then $M = R/I$ is finitely presented, i.e. $M \in \mathbf{Mod}_R^{\text{fp}}$. The morphism $R \to M$ is an epimorphism in $\mathbf{Mod}_R$, so in particular in $\mathbf{Mod}_R^{\text{fp}}$, so we get some short exact sequence $$0 \to K \to R \to M \to 0$$ in $\mathbf{Mod}_R^{\text{fp}}$. By left exactness, we conclude that $K = I$ is the ideal we started with, which therefore has to be finitely presented. Thus $R$ is left coherent as $I \subseteq R$ was arbitrary. $\square$ (I feel there might be some general reason why $\mathbf{Mod}_R^{\text{fp}} \to \mathbf{Mod}_R$ satisfies nice properties, which could shorten the above proof.) Example. Easy examples of coherent rings $R$ that are not Noetherian are valuation rings. Every finitely generated ideal of $R$ is principal, but $R$ is Noetherian if and only if the valuation is discrete.<|endoftext|> TITLE: How does assuming GRH help us calculate class group? QUESTION [19 upvotes]: It seems that, almost all computer programs assume GRH to calculate $\mathbb{Q}(\zeta_p)$ for $p > 23$. I'm very curious how assuming the GRH, helps us to calculate class groups in practice. Can anyone give an explicit example of a number field (not necessarily $\mathbb{Q}(\zeta_p)$'s), and explicit calculation of the class group, based on assuming the GRH? REPLY [36 votes]: In general, to compute the class group of a number field $K$ of degree $n$ and discriminant $\Delta$, we need to find some bound $N$ such that the class group of $K$ is generated by primes of norm at most $N$. Unconditionally, we have the Minkowski bound $$M_K = \sqrt{\lvert \Delta \rvert} \left( \frac{4}{\pi} \right)^{s} \frac{n!}{n^n},$$ where $s$ is the number of conjugate pairs of complex places of $K$. Since $M_K$ is proportional to the square root of the absolute discriminant, it is often infeasible to check all primes up to that bound. To my knowledge, no sizable asymptotic improvement on this bound has been proven unconditionally. However, assuming GRH, we have a vastly stronger bound due to Bach (1990): under GRH, the class group is generated by primes of norm at most $$B_K = 12 \log^2 \lvert \Delta \rvert.$$ A similar bound, asymptotically weaker in theory but stronger in practice for many fields of interest, is due to Belabas, Diaz y Diaz, and Friedman (2008). Let's take $\mathbb{Q}(\zeta_{29})$ as an example: the Minkowski bound is $14956520729 \approx 1.4 \times 10^{10}$, so we would need to check about 64 million primes to provably compute the class group—not totally infeasible with supercomputers but certainly a very computationally intensive task. In contrast, the Bach bound is $99190$, while the version implemented in Magma using results of Belabas, Diaz y Diaz, and Friedman gives a bound of $826$. With this latter bound, Magma can compute the class group in $1.8$ seconds on my laptop, assuming there are no elements of the class group whose representatives all have norm greater than $826$ (the existence of which would contradict GRH). References: Bach, E. “Explicit bounds for primality testing and related problems”. Mathematics of Computation 55 (1990), no. 191, pp. 355–380. [MR1023756] Belabas, K., Diaz y Diaz, F., and Friedman, E. “Small generators of the ideal class group”. Mathematics of Computation 77 (2008), no. 262, pp. 1185–1197. [MR2373197]<|endoftext|> TITLE: Invariant ring of $\textrm{Sym}^2(\wedge^2\mathbb{R}^4)$ under $\textrm{SO}(4)$ QUESTION [8 upvotes]: Consider the representation of $\textrm{SO}(4)$ on $\textrm{Sym}^2(\wedge^2\mathbb{R}^4)$ induced by the standard representation of $\textrm{SO}(4)$ on $\mathbb{R}^4$. I am interested in the ring of invariants of this representation, i.e. the ring of all polynomial functions on $\textrm{Sym}^2(\wedge^2\mathbb{R}^4)$ that are invariant. Here is a way of constructing some of these invariants. The space of $\textrm{SO}(4)$-linear maps $\wedge^2\mathbb{R}^4\to\wedge^2\mathbb{R}^4$ is $2$-dimensional: We have the identity $I$ and the Hodge star operator $\star$. Thus the coefficients of the polynomial $$\det(x\cdot I+y\cdot \star+A)\in\mathbb{R}[x,y]$$are polynomials in the entries of $A\in\textrm{Sym}^2(\wedge^2\mathbb{R}^4)$ which are $\textrm{SO}(4)$-invariant. My question is: Do these polynomials already generate the ring of invariants? REPLY [7 votes]: The answer is 'no', though I don't know an easy way to see this without doing an explicit calculation. Here is where to look though, if you want to do the calculation yourself: Things work out a bit better if one uses indeterminates $z = x+y$ and $w = x-y$. Then one has an expansion $$ \det\bigl(x{\cdot}I + y{\cdot}\star + A\bigr) = \sum_{0\le i,j\le 3} P_{ij}(A)\,z^{3-i}w^{3-j} $$ where $P_{ij}(A)$ is a polynomial in $A$ of degree $i{+}j$. One finds that $P_{00}=1$. Meanwhile, the polynomials $P_{10}$, $P_{01}$, $P_{20}$, $P_{11}$, $P_{02}$, $P_{30}$, $P_{21}$, $P_{12}$, and $P_{03}$ are independent and generate all of the polynomial invariants of degree $3$ or less. Moreover, the three quartic polynomials $P_{3,1}$, $P_{2,2}$, and $P_{1,3}$ are algebraically independent over the ring generated by the lower degree $P_{ij}$. However, the $P_{ij}$ with $i+j\le 4$ do not generate all of the invariant polynomials of degree $4$. There is one further quartic polynomial invariant, say $Q$, that is not a polynomial in the $P_{ij}$ with $i+j\le 4$. Since the remaining three coefficients $P_{3,2}$, $P_{2,3}$ and $P_{3,3}$ are all of degree 5 or higher, $Q$ does not lie in the ring generated by the $P_{ij}$. Added detail: $\mathrm{SO}(4)$ preserves the splitting $\Lambda^2(\mathbb{R}^4)= \Lambda^2_+(\mathbb{R}^4)\oplus\Lambda^2_-(\mathbb{R}^4)$ into self-dual and anti-selfdual forms, i.e., the split into the eigenspaces of $\star$, which has trace-zero and square $\star^2 = I$. It follows that we can write block decompositions $$ \star = \begin{pmatrix} I_3&0_3\\0_3&-I_3\end{pmatrix} \quad\text{and}\quad A = \begin{pmatrix} a&b\\b^T&c\end{pmatrix}, $$ where $a=a^T$, $b$ and $c=c^T$ are $3$-by-$3$ matrices. The action of $\mathrm{SO}(4)$ on $A$ can be described as $$ g\cdot A = \begin{pmatrix} g_+ag_+^T&g_+bg_-^T\\g_-b^Tg_+^T&g_-cg_-^T\end{pmatrix} $$ where $g_\pm:\mathrm{SO}(4)\to\mathrm{SO}(3)$ are homomorphisms such that $(g_+,g_-):\mathrm{SO}(4)\to\mathrm{SO}(3)\times\mathrm{SO}(3)$ is surjective, with kernel $\{\pm I_4\}$. We can write $$ \det(x{\cdot}I_6 + y{\cdot}\star + A) = \det\begin{pmatrix} zI_3+a & b\\ b^T& wI_3 + c\end{pmatrix} = \sum_{0\le i,j\le 3} P_{ij}(A)\,z^{3-i}w^{3-j}, $$ and we can refine our notation by writing $P_{ij}(A) = P_{ij}(a,b,c)$. Now, because of the way $a$, $b$, and $c$ transform under the action of $\mathrm{SO}(4)$, we see, for example, that $\mathrm{tr}(a^k)$ and $\mathrm{tr}(c^l)$ are invariant polynomials, as is $\mathrm{tr}(bb^T)$ (though $\mathrm{tr}(b)$ is not), and, indeed, we see, by inspection, that, for example, $$ P_{10} = \mathrm{tr}(a)\quad\text{and}\quad P_{01} = \mathrm{tr}(c), $$ and a little reflection shows that $$ P_{20} = \tfrac12\bigl(\mathrm{tr}(a)^2-\mathrm{tr}(a^2)\bigr) \quad\text{and}\quad P_{02} = \tfrac12\bigl(\mathrm{tr}(c)^2-\mathrm{tr}(c^2)\bigr) $$ while it's not hard to see that $$ P_{11} = \mathrm{tr}(a)\mathrm{tr}(c) - \mathrm{tr}(bb^T). $$ To go further, imagine that we have an alphabet of 4 letters, $a$, $b$, $b^T$, and $c$, and we declare a valid `word' to be string of these letters subject to the following rules: $a$ can be followed only by $a$ or $b$, $b$ can only be followed by $c$ or $b^T$, $c$ can only be followed by $c$ or $b^T$, $b^T$ can only be followed by $a$ or $b$, and finally, that the number of $b$s in the word is the same as the number of $b^T$s. Thus, for example, $a^4$, $abb^T$, and $abcb^T$ are valid words, but $abc$ is not valid. What one sees that is a vaild word $w$, when interpreted as a product of matrices, transforms under $\mathrm{SO}(4)$ into $g_\pm w g_\pm^T$, and hence $\mathrm{tr}(w)$ is an invariant polynomial for any valid word. Now we can go on: For example $P_{30} = \det(a)$ which is a weighted homogeneous polynomial in $\mathrm{tr}(a)$, $\mathrm{tr}(a^2)$, and $\mathrm{tr}(a^3)$, and similarly with $P_{03}= \det(c)$. Moreover, one finds that $$ P_{21} = P_{20}P_{01}-P_{10}(P_{10}P_{01}-P_{11}) +\mathrm{tr}(abb^T) $$ with a similar formula for $P_{12}$. Thus, all of the polynomials $P_{ij}$ for $0 TITLE: Heart of a bounded $t$-structure on the derived category of coherent sheaves QUESTION [7 upvotes]: Let $X$ be an elliptic curve and $D(X)$ the bounded derived category of $Coh(X)$, coherent sheaves on $X$. If $(D^{\leq 0}, D^{>0})$ is a bounded $t$-structure, then can we already say that the heart $D^{\leq 0}\cap D^{>0}[1]$ is equivalent to $Coh(X)$ (by a shift) ? REPLY [7 votes]: One can construct t-structures on the bounded derived category of coherent sheaves on a smooth projective curve (or higher-dimensional variety) by tilting, see Bayer's notes, Prop. 3.6.1, and the corresponding hearts are not shifts of each other, because the tilt shifts some objects but not all of them. Another way to obtain t-structures is to apply autoequivalences of the derived category to the standard heart, see Exercise 11 in Bayer's notes for the elliptic curve case. Generally t-structures on bounded derived categories of coherent sheaves are studied in the context of the Bridegland stability conditions, and stability conditions for elliptic curves have been classified in Section 9 of Bridgeland's original paper.<|endoftext|> TITLE: Need advice or assistance for son who is in prison. His interest is scattering theory QUESTION [221 upvotes]: The letter below is written by my son. I have been sending him text books and looking for answers on the internet to keep his interest up. He has progressed so far on his own and now he needs direction and assistance from a professional in mathematics. Any advice or assistance you can provide is greatly appreciated. My name is ---, I'm 25, I've been in prison for the past 6 years, and I'm self-taught in mathematics. I began with a list of courses required in a standard undergraduate curriculum and studied the required texts from each course. I covered the basics in this way before branching off into my own interests, beginning with partial differential equations and eventually landing in scattering theory. I began studying mathematics because it was fun and interesting (and passed the time), but it has since become so much more. The progress that I've made, combined with the observation that I am capable of at least understanding research in my fields of interest, has compelled me to take the next step into conducting research of my own, and my current goal is to make advances of publishable value. I am just beginning in this process, yet already I have made progress studying scattering resonances. At the moment, I'm working on a number of problems related to resonance counting. In particular, my primary focus is on "inverse resonance counting": By assuming an asymptotic formula for the resonance counting function (as well as some other results concerning distribution), my goal is to determine properties of the potential. Similarly, in the case of a surface with hyperbolic ends, the goal is to determine properties of the surface from knowledge of an exact asymptotic formula for the counting function. My primary resources at present are Mathematical Theory of Scattering Resonances by Dyatlov and Zworski, and Spectral Theory of Infinite-Area Hyperbolic Surfaces by Borthwick. I'm not sure what I'm asking for here, I just know that I am ready for the next step and seek some guidance as I enter the world of research mathematics. I encounter many problems when it comes to research, such as staying up to date on current topics, finding open problems which suit my skills and interests, and finding papers on topics I need to study more deeply. For example, right now I am in need of results on how resonances change under smooth, small changes in the potential. One of my texts mentioned the paper of P. D. Stefanov, Stability of Resonances Under Smooth Perturbations of the Boundary (1994), but I need more, and that paper makes no citation to papers of the same content. How do I find papers which are similar, or even cite this one? In short, without direct access to the internet or fellow researchers, I hit many roadblocks which are not math-related, and that can be frustrating. I'm looking for ways to make my unconventional research process go a little more smoothly. If anyone has any suggestions, please let me know here. And thanks in advance. REPLY [15 votes]: In a follow up to my previous question my son has asked me to send him this, S. V. Petras, On the Continuous Dependence of the Poles of the Scattering Matrix on the Coefficients of an Elliptic Operator, Proc. Steklov Inst. Math. 159 (1983) pgs. 135-139. I have searched the web including Google Scholar which only has a version in Russian. I found the English version of the article in our library, here is a link to a photocopy. (apologies for the poor quality of the photocopy, I was not able to make it myself because of the lockdown restrictions; I trust it is still usable.) Follow-up: I'm getting close and these two short papers sound like they might be what I need: • E. Korotyaev, Stability for Inverse Resonance Problem, Int. Math. Res. Not. 73 (2004), pgs. 3927 - 3936 • E. Korotyaev, Inverse Resonance Scattering on the Real Line, Inverse Problems 21 (2005), no. 1, pgs. 325 - 341. Dan: I have access to both these papers. Could you tell me your email address (contact info in my profile), so that I can email the pdf's to you? (I'd avoid posting them here, because of copyright restrictions.)<|endoftext|> TITLE: Year of birth of Craige Schensted QUESTION [15 upvotes]: For a paper I am writing related to the history of combinatorics, I am looking for the year of birth of Craige Eugene Schensted, the eponym for the Schensted correspondence. According to this site, a Craige Eugene Schensted was born in 1924 and is now deceased. However, I haven't seen any other indication that the Craige Schensted in whom I am interested is deceased. Moreover, I once found a website with a photo of an 18-year old Craige Schensted in 1946, which would mean that he was born in 1927 or 1928. I am no longer able to find this website. REPLY [31 votes]: This may be more than you wanted to know about Craige Schensted aka Ea Ea. Obituary Information for Ea Ea Ea, (the adopted name of Craige Val Eugene Shensted) was born in Mayville, ND, April 12, 1927, son of Roy and Helen Chance Shensted. He completed high school early at the age of 17, enlisted in the US Army and achieved the highest intelligence score ever recorded to that time. Because of his youth, he attended college until he turned 18 and then was enrolled in active military service. Discharged shortly after the conclusion of WW 2, he attended the University of Minnesota, receiving a BS in Applied Mathematics. He and Irene _____ were married after graduation. During their employment at the University of Michigan at Ann Arbor, Craige worked as a mathematical researcher in the Radio Astronomy Laboratory and invented forms of mathematics currently in use. He collaborated with Charles Titus, another mathematics faculty member, in creating strategy games “Y” “Mudcrack Y” and "Star" which were highly praised in Games magazine and are still available here: http://gamepuzzles.com/abstract.htm With a desire to leave a small footprint, in Ann Arbor the Schensteds lived simply and eventually car-free. Shortly after Craige’s retirement at age forty-four, they bought a home on Peaks Island, off Portland, ME where they lived also car-free for many years. Craige’s interests broadened to include kayaking on Casco Bay, improvisational drumming, chanting, dancing and in-depth reading and computer research especially in areas including nutrition, physics, architecture, and psychology. The Schensteds eventually divorced and during his subsequent fifteen-year relationship and collaboration with with mind-body practitioner Mariah Williams from Lunenburg, Massachusetts and later Liberty, Maine, Craige explored the commonalities between contemporary physics and the energy work Mariah practiced. He theorized that what many practitioners call human energy fields are instead fields of probability in which perceived realities may be accessed by intuition and potentially influenced by intention. Also, during that time, after reading the Descent to the Goddess, about the myth of Inanna, Mariah described the similarities between the Babylonian god Enki, later the Sumerian god Ea, and Craige chose to change his name to Ea. In 2008, after an injury on Peaks Island, Mariah moved Ea to an apartment overlooking Passagassawaukeag Bay in Belfast, Maine and he and Mariah maintained a close friendship until his death in hospice on January 22. You may find the Game/Puzzles.com bio of interest: http://www.gamepuzzles.com/ea-star.htm I am Mariah Williams and may be reached via LibertyHealingArts.com<|endoftext|> TITLE: Can locally constant real functions on a space be made into continuous functions (on a different space)? QUESTION [8 upvotes]: [This question came up while idly thinking about this other one, but it is not directly related.] Definitions: If $X$ is a topological space, let $C(X)$ stand for the $\mathbb{R}$-algebra of continuous real-valued functions $X\to\mathbb{R}$ where $\mathbb{R}$ has its usual (Euclidean) topology, and let $D(X)$ stand for the $\mathbb{R}$-algebra of locally constant functions $X\to\mathbb{R}$, or equivalently, continuous functions $X\to\mathbb{R}_{\mathrm{disc}}$ where $\mathbb{R}_{\mathrm{disc}}$ stands for $\mathbb{R}$ endowed with the discrete topology. Both $C$ and $D$ can be seen as contravariant functors from the category of topological spaces to the category of $\mathbb{R}$-algebras (by taking a continuous map $f\colon X\to Y$ to the ring homomorphism defined by right-composition by $f$). Motivations for the following questions: Quite a lot is known about $C(X)$ (and its subalgebra $C^*(X)$ of bounded functions), as witnessed, e.g., by the classic book by Gillman & Jerison, Rings of Continuous Functions (and also its “sequel” of sorts, by Fine, Gillman & Lambek, Rings of Quotients of Rings of Functions). For example, $C(X)$ uniquely determines $X$ up to isomorphism when $X$ is [completely regular and] realcompact, and when this is the case, $X$ can be recovered as the closure of the image of $X$ under the evaluation map $X \to \mathbb{R}^{C(X)}$, or as the set of maximal ideals of $C(X)$ whose quotient ring is $\mathbb{R}$ endowed with the Zariski topology. However, that I know of, there is no satisfactory purely algebraic characterization of $\mathbb{R}$-algebras of the form $C(X)$. Now I realized to my horror that I had no idea about the analogous questions for $D(X)$ (and its subalgebra $D^*(X) := D(X) \cap C^*(X)$). I could ask a million questions, the gist of which would be “where can I learn about the same things concerning $D(X)$ that Gillman & Jerison teaches us about $C(X)$?”, so if somebody knows the answer to that, please do tell; but since MathOverflow requires that I ask something more specific than “every question answered in Gillman & Jerison”, let me ask something slightly different, namely whether we can reduce the study of $D(X)$ to that of the well understood $C(X)$. Namely: Questions: Given a topological space $X$, does there always exist a topological space $\nabla X$ such that $C(\nabla X) = D(X)$? Better: can we find a functor $\nabla$ (from topological spaces to topological spaces) such that $C\circ\nabla = D$ and a natural transformation $\eta \colon 1_{\mathbf{Top}} \to \nabla$ such that $C(\eta_X)\colon C(\nabla X) = D(X) \to C(X)$ is the inclusion, and perhaps, say, that $\eta$ is an isomorphism on discrete spaces? Even better: is $\nabla$ left-adjoint to the inclusion functor of the full subcategory of topological spaces $X$ for which $D(X)=C(X)$ (the existence of this left adjoint would answer all questions positively)? [edit: see below] PS: I just remembered that spaces such that $D(X)=C(X)$ are known as “P-spaces”, but this doesn't really help. It does, however, allow me to restate the strongest version of my question as: “the the full subcategory of P-spaces reflective?”. Also, maybe I should be formulating my questions in the language of locales rather than topological spaces? Edit: Simon Pepin Lehalleur has just pointed out to me that an arbitrary product of P-spaces can fail to be a P-space (Gillman & Jerison, exercise 4K(6)), so we can't hope for them to form a reflective subcategory. This doesn't invalidate weaker forms of the question, however. Remark: Even in the case of $X=\mathbb{Q}$ (with the usual (=order) topology), I don't know whether there exists a space $\nabla\mathbb{Q}$ such that $C(\nabla\mathbb{Q}) = D(\mathbb{Q})$. REPLY [6 votes]: $D(\mathbb{Q})$ is not isomorphic to $C(X)$ (as an $\mathbb{R}$-algebra) for any topological space $X$. For both $D(X)$ and $C(X)$, you can recover the (extended) uniform norm from the $\mathbb{R}$-algebraic structure with $$ \left\| x \right\| = \sup\{|\lambda| : x - \lambda 1\text{ is not invertible}\},$$ where $1$ is the multiplicative identity of the algebra and $\lambda$ ranges over $\mathbb{R}$. $D(\mathbb{Q})$ is not metrically complete under the (extended) metric induced by the uniform norm, but $C(X)$ always is, so $D(\mathbb{Q})$ can't be isomorphic to any $C(X)$ as an $\mathbb{R}$-algebra.<|endoftext|> TITLE: A scaled fractional ''Sobolev inequality'' QUESTION [5 upvotes]: Does a fractional interpolation inequality similar to $$ \int_{B_R(0)} |u| dx \le C R^{2} \sqrt{\log(2R)} \Big( \int_{\mathbb R^2}\int_{\mathbb R^2} \frac{|u(x)-u(y)|^2}{|x-y|^{2+2s}} dxdy + \int_{B_1(0)} |u|^2 dx \Big)^{1/2} $$ (where $B_R(0), B_1(0) \subset \mathbb R^2$) hold? Maybe changing the powers of $R$ or of the logarithm if needed? Is it known in the literature? REPLY [2 votes]: Then, unless there are more typos, what you request is very cheap and no logarithmic factor is needed. WLOG, $u\ge 0$ (just consider $|u|$ instead, if not). Denote by $A_1$ the average of $u$ over the unit disk centered at $0$ and by $A_r$ ($r=2,4,8,16,\dots$) the average of $u$ over the annulus $\frac r2\le|x|\le r$. Then, assuming that the expression in parentheses on the $RHS$ is $1$, we get $A_0\le C$ and $|A_r-A_{\frac r2}|^2 \frac {r^4}{r^{2+2s}}\le C$ (just restrict the double integration to $x\in A_{\frac r2}, y\in A_r$ and use the trivial upper bound on the denominator and Cauchy-Schwarz). Hence $|A_r-A_{\frac r2}|\le Cr^{s-1}$ and, summing a geometric progression, we get that $A_r\le C'$ for all $r$, i.e., the integral of $u$ over any disk centered at the origin is at most constant times its area. Edit: the details for the difference estimate Let $Q_r$ be the region (annulus or disk) over which $A_r$ is taken. Then $$ |A_r-A_{r/2}|=\frac{1}{Area(Q_r)Area(Q_{r/2})}\left|\iint_{x\in Q_r, y\in Q_{r/2}}(u(x)-u(y))\,dxdy\right|\le \\ (2r)^{1+s}\frac{1}{Area(Q_r)Area(Q_{r/2})}\iint_{x\in Q_r, y\in Q_{r/2}}\frac{|u(x)-u(y)|}{|x-y|^{1+s}}\,dxdy \le \\ (2r)^{1+s}\left[\frac{1}{Area(Q_r)Area(Q_{r/2})}\iint_{x\in Q_r, y\in Q_{r/2}}\left(\frac{|u(x)-u(y)|}{|x-y|^{1+s}}\right)^2\,dxdy\right]^{1/2} \\ \le\frac{(2r)^{1+s}}{\sqrt{Area(Q_r)Area(Q_{r/2})}}\left[\iint_{x\in\mathbb R^2, y\in \mathbb R^2}\frac{|u(x)-u(y)|^2}{|x-y|^{2+2s}}\,dxdy\right]^{1/2} $$ and it remains to note that $\sqrt{Area(Q_r)Area(Q_{r/2})}\ge cr^2$.<|endoftext|> TITLE: Is $\mathbb{Z}$ universally definable in any number fields other than $\mathbb{Q}$? QUESTION [10 upvotes]: In 2009, Jochen Koenigsmann showed that $\mathbb{Z}$ is universally definable in the field $\mathbb{Q}$. My question is, are there any other number fields in which $\mathbb{Z}$ is universally definable? Or failing that, what is the lowest complexity definition of $\mathbb{Z}$ known for a number field other than $\mathbb{Q}$? REPLY [12 votes]: Koenigsmann's result was generlized by Jennifer Park to number fields, giving a universal definition of the ring of integers $\mathcal{O}_K$ in $K$. Then there is a series of results proving that $\mathbb{Z}$ is existentially definable in $\mathcal{O}_K$ for certain $K$, starting I think with results by Jan Denef (totally real number fields, or quadratic extension of a totally real field). Combining the two, one gets an $\exists\forall$-definition of $\mathbb{Z}$ in $K$ at least for these number fields. It is possible though that one also has an $\forall$-definition of $\mathbb{Z}$ in $K$ for some number fields $K\neq\mathbb{Q}$, but I don't remember right now.<|endoftext|> TITLE: Local limit theorem for random walks on $\mathbb Z^d$ QUESTION [7 upvotes]: I'm looking for a reference for the following claim. Let $W(n)$ be a centered random walk on $\mathbb Z^d$ with $W(0)=0$. Suppose that $W(n)$ has a finite second moment. Let $n\ge 1 $ and $k \in \mathbb Z ^d$ such that $||k||_2 \le \sqrt{n}$ and $\mathbb P ( W(n)=k)>0$. I want to show that in this case $\mathbb P (W(n)=k)\ge c n^{-\frac{d}{2}}$ for some positive $c$ that is independent of $n$ and $k$. I found references that give the asymptotic behavior of $\mathbb P (W(n)=k)$ under the assumption that $W(n)$ is irreducible or aperiodic but I didn't find a reference for the general case above. REPLY [5 votes]: $\newcommand\R{\mathbb R}\newcommand\Z{\mathbb Z}\newcommand\De\Delta$This follows almost immediately from the multidimensional local limit theorem (MLLT) proved by Meĭzler, D. G.; Parasyuk, O. S.; Rvačeva, E. L. in the paper "On a many dimensional local limit theorem of the theory of probability." (Russian) Ukrain. Mat. Žurnal 1, (1949). no. 1, 9--20. The sufficiency part of their result can be stated as follows. Let $Y_1,Y_2,\dots$ be iid random vectors with values in $\Z^d$. Let $S$ denote the support set of (the distribution of) $Y_1$. Suppose that the lattice generated by the set $S-S$ is the entire set $\Z^d$. Let $a$ and $C$ denote, respectively, the mean and the covariance matrix of (the distribution of) $Y_1$. (Note that the condition on $S$ implies that the matrix $C$ is invertible.) Let $Z_n:=Y_1+\dots+Y_n$. Then \begin{equation} P(Z_n=z)=\frac1{(2\pi n)^{d/2}\sqrt\De}\,\Big[\exp\Big\{-\frac12\,Q\Big(\frac{z-na}{\sqrt n}\Big)\Big\}+o(1)\Big] \tag{1} \end{equation} uniformly over all $z\in\Z^d$, where $\De$ is the determinant of $C$ and $Q$ is the quadratic form with the matrix $C^{-1}$. The paper by Meĭzler et al is in Russian; however, I think it is not hard to read it using Google Translate, say. Unfortunately, there is a mistake in the statement of the result in that paper, where $C$ is defined, not as the covariance matrix $EY_1Y_1^T-EY_1\,EY_1^T$ of $Y_1$, but as $EY_1Y_1^T$ (as usual, we identify $\Z^p$ with the set $\Z^{p\times1}$ of all $p\times 1$ column matrices with integral entries). However, the proof actually seems to be a correct proof of the correct version of the result -- see e.g. formula (10) in that paper. Let now $X_1,X_2,\dots$ be the iid jumps of your random walk, so that $W_n:=W(n)=X_1+\dots+X_n$. Let $L$ be the minimal lattice in $\Z^d$ such that the support of the distribution of $X_1$ is contained in $c+L$ for some $c\in\Z^d$. For this minimal lattice $L$, take indeed any $c\in\Z^d$ such that the support of the distribution of $X_1$ is contained in $c+L$. Let $(b_1,\dots,b_r)$ be any basis of the lattice $L$ (where necessarily $1\le r\le d$), so that $L=\Z b_1+\dots+\Z b_r$. For each natural $j$, let $Y_j$ the $r\times 1$ column matrix of the coordinates of the random vector $X_1-c$ with respect to the basis $(b_1,\dots,b_r)$ of $L$, so that $$Y_j=M(X_j-c)$$ for some matrix $M$. Moreover, all values of $Y_j$ are in $\Z^r$. Furthermore, the lattice generated by the set $S-S$ is the entire set $\Z^r$, where $S$ still denotes the support set of (the distribution of) $Y_1$. Note also that $a=EY_j=EM(X_j-c)=-Mc$, since $EX_j=0$. It follows by (1) that uniformly in $k=nc+l\in nc+L$ with $\|k\|_2\le\sqrt n$ \begin{align*} P(W_n=k)&=P(W_n=Ml) \\ &=\frac1{(2\pi n)^{r/2}\sqrt\De}\,\big[\exp\big\{-\frac1{2n}\,Q(Ml+nMc)\big\}+o(1)\big] \\ &=\frac1{(2\pi n)^{r/2}\sqrt\De}\,\big[\exp\big\{-\frac1{2n}\,Q(Mk)\big\}+o(1)\big] \\ &\ge\frac K{n^{r/2}} \end{align*} for some positive real constant $K$ independent of $n,k$. On the other hand, $P(W_n=k)>0$ only if $k\in nc+L$. Thus, \begin{align*} P(W_n=k)\ge\frac K{n^{r/2}}\ge\frac K{n^{d/2}} \end{align*} for all $k\in\Z^d$ such that $P(W_n=k)>0$, as desired.<|endoftext|> TITLE: Sequence of epimorphisms of residually finite groups stabilizes QUESTION [10 upvotes]: Let $G_1 \to G_2 \to \cdots$ be a sequence of epimorphisms of finitely generated residually finite groups. Does it eventually stabilize? That is, are all but finitely many epimorphisms actually isomorphisms? Note that finitely generated residually finite groups are Hopfian, so this excludes the simple counterexample of each $G_i$ being a fixed group and each epimorphisms being a fixed one onto itself. The analogous result holds when the groups are residually free: this is Proposition 6.8 in Charpentier Guirardel "Limit groups as limits of free groups". The proof only uses the fact that residually free groups are residually $SL_2(\mathbb{C})$, and it seems that it can be adapted to the case where each $G_i$ is residually $GL_n(\mathbb{C})$ for a fixed $n$. It seems unlikely that this holds for general residaully finite groups: the Jordan-Schur Theorem implies that for a general finite group the minimal degree $n$ such that it embeds into $GL_n(\mathbb{C})$ can be arbitrarily large. Is there another way to adapt the proof? Is there a counterexample? REPLY [7 votes]: In the same vein as dodd's answer, a counterexample can also be deduced from the second Houghton group $H_2$, which is defined as the group of bijections $L^{(0)} \to L^{(0)}$ that preserves adjacency and non-adjacency for all but finitely pairs of vertices in the bi-infinite line $L$. A presentation of $H_2$ is $$\left\langle \sigma_i (i \in \mathbb{Z}), t \left| \array{ \sigma_i^2=1, \ i \in \mathbb{Z} \\ [\sigma_i,\sigma_j]=1, \ |i-j| \geq 2}, \ \array{\sigma_i\sigma_{i+1}\sigma_i= \sigma_{i+1}\sigma_i \sigma_{i+1} = 1, \ i \in \mathbb{Z} \\ t\sigma_it^{-1}= \sigma_{i+1}, \ i \in \mathbb{Z}} \right. \right\rangle$$ where $t$ corresponds to a unit translation and $\sigma_i$ to the permutation $(i,i+1)$. Now, truncate the presentation and define $G_n$ via $$\left\langle \sigma_i (i \in \mathbb{Z}), t \left| \array{ \sigma_i^2=1, \ i \in \mathbb{Z} \\ [\sigma_i,\sigma_j]=1, \ n \geq |i-j| \geq 2}, \ \array{\sigma_i\sigma_{i+1}\sigma_i= \sigma_{i+1}\sigma_i \sigma_{i+1} = 1, \ i \in \mathbb{Z} \\ t\sigma_it^{-1}= \sigma_{i+1}, \ i \in \mathbb{Z}} \right. \right\rangle.$$ By using the relations $t\sigma_it^{-1}=\sigma_{i+1}$ in order to remove the generators $\sigma_0,\sigma_{-1},\ldots$ and $\sigma_{n+2},\sigma_{n+3},\ldots$, we find the following presentation of $G_n$: $$\left\langle \sigma_1, \ldots, \sigma_{n+1}, t \left| \array{ \sigma_i^2=1, \ 1 \leq i \leq n+1 \\ [\sigma_i,\sigma_j]=1, \ |i-j| \geq 2}, \ \array{\sigma_i\sigma_{i+1}\sigma_i= \sigma_{i+1}\sigma_i \sigma_{i+1} = 1, \ 1 \leq i \leq n \\ t\sigma_it^{-1}= \sigma_{i+1}, \ 1 \leq i \leq n} \right. \right\rangle.$$ Observe from this presentation that $G_n$ decomposes as an HNN extension of $$\left\langle \sigma_1,\ldots, \sigma_{n+1} \left| \array{ \sigma_i^2=1, \ 1 \leq i \leq n+1 \\ [\sigma_i,\sigma_j]=1, \ |i-j| \geq 2}, \ \sigma_i\sigma_{i+1}\sigma_i= \sigma_{i+1}\sigma_i \sigma_{i+1} = 1, \ 1 \leq i \leq n \right. \right\rangle,$$ which turns out to be isomorphic to the symmetric group $\mathfrak{S}_{n+2}$, where the stable letter conjugates $\langle \sigma_1,\ldots, \sigma_n \rangle$ to $\langle \sigma_2, \ldots, \sigma_{n+1} \rangle$. Thus, as the HNN extension of a finite group, $G_n$ must be virtually free. The conclusion is that the canonical quotient maps $G_1 \twoheadrightarrow G_2 \twoheadrightarrow \cdots$ defines a sequence of epimorphisms between virtually free groups that does not stabilise. Remark: By reproducing the above argument almost word for word with the lamplighter group $\mathbb{Z}_2 \wr \mathbb{Z}$ instead of the Houghton group $H_2$ provides the same conclusion. The reason is that these groups have a similar structure: they are of the form $C \rtimes \mathbb{Z}$ for some locally finite Coxeter group $C$ where $\mathbb{Z}$ acts on $C$ via an isometry of the graph defining $C$. (Loosely speaking, all the other groups of this form can be recovered from $\mathbb{Z}_2 \wr \mathbb{Z}$ and $H_2$, so there is no other interesting examples in this direction.)<|endoftext|> TITLE: Finite non-empty coproduct in the absolute prismatic site QUESTION [6 upvotes]: Let $(R/A)_\Delta$ be the prismatic site over $R$ relative to a prism $(A, I)$, then it is known that $(R/A)_\Delta$ admits finite non-empty coproduct, for instance, by Cor. 5.2 in Bhatt's lecture notes V on prismatic cohomology. My question is, if we consider the absolute prismatic site $R_\Delta$, does $R_\Delta$ still always have finite non-empty coproduct? If we restrict to the existence of finite non-empty self coproducts, it seems to me that the answer is yes, as it has been suggested (implicitly) in work: Arthur-César Le Bras and Johannes Anschütz: Prismatic Dieudonné theory. Or an upcoming work of Bhatt and Scholze, see the notes of Scholze on this topic in the RAMpAGe Seminar. A more concrete question is the following: Let $R=O_K$ be a complete discrete valuation ring of mixed characteristic with perfect residue field $k$, and let $(\mathfrak{S},(E))$ be a Breuil-Kisin prism in $R_\Delta$, i.e., $\mathfrak S=W(k)[\![u]\!]$ and $E \in W(k)[u]$ is the Eisenstein polynomial of a uniformizer of $O_K$. In the above two works, they claim the self product of $(\mathfrak{S},(E))$ in $R_\Delta$ exists and is equal to a prismatic envelop of $\mathfrak{S}\otimes_{W(k)}\mathfrak{S}$. Why is the tensor product over $W(k)$? I know for $(R/A)_\Delta$, the similar construction takes tensors over $A$, but it is clear that $(\mathfrak{S},(E))$ is not a prism over $(W(k),(p))$. REPLY [6 votes]: Yes, it does admit nonempty finite coproducts. If you have two prisms $(A_1,I_1)$ and $(A_2,I_2)$ with maps $R\to A_i/I_i$, you need to find the initial prism $(A,I)$ with maps from both $(A_i,I_i)$ such that the two induced maps $R\to A_i/I_i\to A/I$ agree. For this, start with $A_0=A_1\hat{\otimes}_{\mathbb Z_p} A_2$ (where the tensor product is $(p,I_1,I_2)$-adically completed, say) and then take a suitable prismatic envelope to ensure the required conditions. You raise an interesting point, that if $R=\mathcal O_K$, then one can take the tensor product over $W(k)$ instead of $\mathbb Z_p$, where $k$ is the residue field of $\mathcal O_K$. The reason is that for any prism $(A,I)$ with a map $\mathcal O_K\to A/I$, you get in particular a map $W(k)\to A/I$, and this lifts uniquely to a map $W(k)\to A$: This is using that the $p$-completed cotangent complex of $W(k)/\mathbb Z_p$ vanishes, or some more concrete assertion involving Teichmüller lifts. The map $W(k)\to A$ is also necessarily a map of $\delta$-rings. So all those $A$'s live canonically over $W(k)$, so one can as well directly take the tensor product there.<|endoftext|> TITLE: What is the Galois group of one ultrapower over another ultrapower? QUESTION [6 upvotes]: Let $F$ be a field, let $E$ be a field extension of $F$, and let $U$ be an ultrafilter. Then my question is, what is the relationship between the Galois groups $Gal(\Pi_U E/\Pi_U F)$ and $Gal(E/F)$? Or if that's too general, is it at least possible to say something in the case when $E$ is a number field and $F=\mathbb{Q}$? REPLY [9 votes]: $\newcommand{\Gal}{\operatorname{Gal}}$If $E/F$ is a finite Galois extension, then $\Gal(\prod_UE/\prod_UF)$ is canonically isomorphic to $\Gal(E/F)$. Indeed, by the primitive element theorem, $E=F(\alpha)$ for some $\alpha\in E$. This means every element of $E$ can be written as a polynomial in $\alpha$ with coefficients in $F$ of degree smaller than $d=\deg\alpha$. Let $\alpha^*$ be the image of $\alpha$ in the ultrapower. Then we have $\prod_UE=(\prod_UF)(\alpha^*)$, as is straightforward from the description using polynomials above. Therefore any element automorpmism of $\prod_UE/\prod_UF$ is determined by the image of $\alpha^*$, which on $U$-most indices must coincide with some conjugate of $\alpha$, and hence is induced by an automorphism of $E/F$. Conversely, any automorphism of $E/F$ induces an automorphism of the ultrapowers. If $E/F$ is Galois but not finite, then $\prod_UE/\prod_UF$ need not a Galois extension, in fact it can fail to even be algebraic extension. This happens whenever $U$ is a nonprincipal ultrafilter on $\mathbb N$ (or more generally for any $U$ which is not countably complete). Indeed, in that case for any $n\in\mathbb N$ we can pick $\alpha_n\in E$ which has degree at least $n$ over $F$. Then $[(\alpha_n)]\in\prod_UE$ will not satisfy a polynomial equation of degree $\leq n$ over $\prod_UF$ for any $n$ by Łoś, so will be transcendental. You can still ask for the automorphism group of this extension, but I'm not sure it will have a nice explicit description. A natural guess would be that the automorphism group is $\prod_U\Gal(E/F)$, but I have a suspicion it will be larger than that. If $E/F$ is infinite and $U$ is countably complete, then at the very least you can get that $\prod_UE/\prod_UF$ is algebraic and Galois. At least in the case of $E,F$ countable you should be able to get an isomorphism like in the first part of my answer, but I have no idea about general case. Edit: as Andreas Blass points out in a comment below, if $U$ is countably complete, then $\prod_U E\cong E$ if $E$ is countable. More generally, if $U$ is $\kappa$-complete, then $\prod_UE\cong E$ whenever $|E|<\kappa$. This in particular will hold if $U$ is countably complete and $|E|$ is below the first measurable.<|endoftext|> TITLE: Which direction of the adjoint functor theorem is most useful? QUESTION [6 upvotes]: In the daily life of a working mathematician which direction of the adjoint functor theorem is more useful? Unpacking, does one find it more useful to: a) prove that a functor admits an adjoint and conclude that it preserves limits/colimits, OR b) prove that a functor preserves limits/colimits and conclude that it admits an adjoint? I guess I should also include a third option: c) neither a) nor b) is true in general, it really depends on what part of math you work in. REPLY [10 votes]: In 1-category theory, the easy direction (a) is invoked all the time. The hard direction (b) doesn't have to be formally invoked very often, because most adjoints can be constructed by hand (and even if you do initially construct an adjoint via the Adjoint Functor Theorem, usually you'll want to get a more explicit understanding of it as you move forward anyway). However, (b) is still used all the time as a heuristic consideration: if you want to know whether a functor admits an adjoint and it's not immediately obvious how to construct one, usually the thing to do is to check limit/colimit preservation. In $\infty$-category theory, the situation is different (EDIT: At least superficially? Perhaps more deeply? See Mike Shulman's important objections in the comments below). There, (a) is still just as important, but (b) (in various incarnations) is invoked quite frequently. The reason is that in $\infty$-category theory, it is often difficult to construct functors explicitly! This is because it doesn't suffice to say what the functor does on objects and morphisms and check a funcotriality condition -- rather, you've got to specify higher coherence data all the way up. Adjoint functor-type theorems are used as ready-made packages where all of this coherence data can be supplied automatically. This is a central insight of Lurie and really one of the major factors making the whole theory useful.<|endoftext|> TITLE: Is the surreal number $\omega(\sqrt{2}+1)+1$ a prime? QUESTION [15 upvotes]: In the 1986 book An Introduction to the Theory of Surreal Numbers, Gonshor, on page 117, notes that it is an open problem whether $\omega(\sqrt{2}+1)+1$ is a prime, using the standard definition of integral number he introduces earlier in the chapter. I'll give a little background, as requested, using Gonshor as a reference. The surreal numbers are a superclass of the infinitesimals, though I'm not familiar with their exact sizes. A surreal number is a function from an initial segment of the ordinals into $\{+,-\}$. (An ordinal sequence which terminates) Then, given two surreals, $a$, $b$, let $a < b$ if $a(\alpha) < b(\alpha)$ where $\alpha$ is the first place they differ, with the convention $- < 0 < +$, where $0$ is an abuse of notation that simply means that the function is undefined at that point. So, $(++) > (+) > (+-)$. Some investigation gives that $0\sim (),\ 1 \sim (+),\ 2 \sim (++), \ldots$ Most of the book is spent on embedding the reals and infinitesimals in the surreals. Then you can define the generalized integers in the surreals: $a$ is an integer if the exponents in normal form of $a$ are non-negative, and if a $0$ exponent occurs, then the real coefficient is an integer. Normal form is a bit complicated to define, but basically you write numbers in terms of $\omega$ and $\epsilon$ where $\omega$ is the first infinite ordinal, so $(+++...)$ times and $\epsilon = 1/\omega$, which also exists in the surreals. So, $1/3\omega^2+3$ is a generalized integer, and so is $\sqrt{\omega}+2$, but not $0.5+\omega^{-1}$, as both terms violate the definition. For the definition of prime, $1=(+)$ is still a unit, so primality amounts to proving that a surreal has more than two factors (itself and $1$). $\omega$ is factorizable since $n \in \mathbb{N}$ and $\omega/n$ are both generalized integers. That trick doesn't work with $\omega(\sqrt{2}+1)+1$. Is this still an open problem? The book is a couple decades old (picked it up at the library), and searching for surreal numbers doesn't turn up many results. Thanks REPLY [12 votes]: The answer is no. The factorization is $(\sqrt{\sqrt{2}+1}\omega^{1/2} - \sqrt{2\sqrt{\sqrt{2}+1}}\omega^{1/4}+1)(\sqrt{\sqrt{2}+1}\omega^{1/2} + \sqrt{2\sqrt{\sqrt{2}+1}}\omega^{1/4}+1)$<|endoftext|> TITLE: What are some of the earliest examples of analytic continuation? QUESTION [15 upvotes]: I'm wondering how Riemann knew that $\zeta(z)$ could be extended to a larger domain. In particular, who was the first person to explicitly extend the domain of a complex valued function and what was the function? REPLY [10 votes]: (Expanded 1/26/21 First let me point out for non-native English speakers that the use of the article 'a' in the phrase 'a complex-valued function' means that the question is not solely in reference to the Riemann or any other zeta function. It includes any function whose domain is some set of the reals, so I interpret the question as "Who is the first to have published an extension of the domain of a significant function from some set of the reals to some continuous domain of the complex, and what was that function?" To me, the exact meaning of the term analytic continuation and whether it is unique or not is a different question. The first sentence and several of the comments focus on the Riemann zeta function. Riemann did not stand alone and his interests were much broader than the sometimes almost obsessive focus today on the RH might imply. His interests encompassed pretty much all of complex analysis, so it was natural for him to consider extensions of real functions to complex functions. Hard to believe (smacks of some type of regional bias) that no mathematician before Euler, woke up one morning and thought, "What if I modify my real formulas to include that crazy square root of -1?" Roger Cotes was primed to meaningfully do so with his interest in astronomy and celestial mechanics; familiarity with the work of his colleague Newton on the series reps of the trig functions, their inverses, the calculus, and Newtonian mechanics; use of the logarithmic tables introduced at the beginning of the 1600s by Napier to deal with computations with large numbers encountered in surveying the Earth and the skies; and work on interpolation (Cotes' and Newton's). Let me stress again that Cotes was familiar with Newton's compositional inversion of power series (one formula includes the associahedron version of the Lagrange inversion formula for formal series, see Ferraro below), including that for the exponential function, and, as noted by Griffiths' comment to the post "The making of the logarithm" by Freiberger: Without these tables of logarithms there would be no theory from Nicholas Mercator of the area under a symmetrical hyperbola equaling the log of the distance along the x axis, nor of Isaac Newton's reversion of the hyperbola formula to achieve the infinite series for the antilogarithm $e^x$. (Mercator maps, beginning to see the dots?) In fact, Ferraro discusses on pages 74 and 75 of "The Rise and Development of the Theory of Series up to the Early 1820s" how Newton inverted the power series for the logarithm $-\ln(1-x)$ to obtain the power series of the antilogarithm $1- e^{-x}$. (Newton with his superb mastery of geometry and analysis would surely have noted the simple inverse function theorem relation here between the derivatives of the two series as well.) Consequently, it seems natural that at the birth of calculus and its association with power series and compositional inverses, Cotes wrote down in 1714, when Euler was seven years old, $$ ix = \ln[ \;\cos(x) + i \sin(x) \;]$$ a nascent version of Euler's 1748 fabulous formula (cf. Wikipedia) $$ e^{i\theta} = \cos(\theta) + i \sin(\theta).$$ An obvious check with the derivative (or fluxions) verifies the formula without explicit use of the exponential $$ \frac{d}{dx} (ix +constant) = i = \frac{d}{dx} \; \ln[ \;\cos(x) + i \sin(x) \;]= \frac{-\sin(x) + i \cos(x)}{\cos(x) + i \sin(x)},$$ which I'm sure was SOP for Newton and Cotes--application of the chain rule, a.k.a. inverse function theorem in this case, $dx = df(f^{-1}(x)) = f'(f^{-1}(x)) \; (f^{-1})'(x) \; dx$, which indeed makes the formula obvious. In "The history of the exponential and logarithmic concepts," Cajori explains how John Bernoulli considered the solutions of a differential equation transformed from the reals to the imaginary in 1702 and gives Cotes' derivation of his formula, which Cotes published in 1714 and 1722. (Edit 4/28/21: Nahin in An Imaginary Tale gives Cotes' derivation also and some more info on Cotes.) Cajori also claims that subsequently Euler did not shy from using imaginary numbers. Euler's formula as written today had to wait for the development by Euler and colleagues of the symbolic rep of the exponential function $\exp(z) = e^z$ with $e$ being Euler's constant, sometimes referred to as Napier's constant since it occurred in Napier's log tables. This was after much calculus underlying the log had been explicated by Huygens and others. The exponential function was sometimes even referred to as the 'antilogarithm', reflecting the log's priority, as noted in the log post. Cote's logarithmic formula is an extension from the positive reals to the realm of complex numbers of the argument of the logarithm in a rather more difficult way than simply replacing $n$ in the series rep of $\zeta(n)$ by real numbers on the real line and then to other numbers in the complex plane. According to the Wikipedia article on Cotes, he published an important theorem on the roots of unity (and gave the value of one radian for the first time) in 1722 in "Theoremata tum logometrica tum triogonometrica datarum fluxionum fluentes exhibentia, per methodum mensurarum ulterius extensam" (Theorems, some logorithmic, some trigonometric, which yield the fluents of given fluxions by the method of measures further developed). He understood trig rather well, and from this perspective, both Cotes and Euler's formulas can be regarded as the continuation of the solutions of $|x| = 1$ into the complex plane. The solutions define the very simple function with domain 1 and -1 and range 1, which is then analytically continued as a circle of radius 1 in the complex domain--a type of interpolation (hover over the interpolation link in the Wiki on Roger Cotes) satisfying a simple functional equation $|f(x)|=1$. (Other examples of types of interpolation/analytic continuation from functions with discrete integer domains to those with continuous complex domains (related to Newton and sinc/cardinal series interpolations) are given in this MO-Q and this MSE-Q.) From a broader perspective Cotes' log formula is a clear example of analytic continuation of the log as a mapping from the real numbers to the real to a mapping of the complex to the complex. Cotes was, of course, aware that (indeed utilized, and would have taken for granted that anyone familiar with the log knew also), for $u,v > 0$, $$\ln(u)+\ln(v) = \ln(uv),$$ so he wrote down the most difficult part of the analytic continuation of the log from the positive reals to the complex (albeit not explicitly accounting for multiplicity) $$\ln(r) + ix = \ln[\; r\; (\;\cos(x) + i \; \sin(x)\;) \;].$$ Refs in Wikipedia: John Napier,The History of Logarithms, Logarithm, Roger Cotes, Euler's identity, Euler's Formula. In addition to Euler summation with complex arguments, Euler was the first to extend the factorial to the gamma function for complex arguments to develop a fractional calculus with his hybrid Mellin-Laplace integral rep for the gamma function (see "The Euler legacy to modern physics" by Dattoli and Del Franco and the MSE-Q noted above). Euler's integral for the beta function allows the same for the generalized binomial coefficients, which Newton (again, colleague of Cotes) had done for the extension to the reals of the integer binomial coefficients. Unfortunately, Euler didn't fully understand the extension to complex numbers (Argand and Wessel come later) otherwise he would have scooped Cauchy, Liouville, and Riemann on the calculus of complex analysis. For a prehistory of the Riemann zeta function, see "Aspects of Zeta-Function Theory in the Mathematical Works of Adolf Hurwitz" by Oswald and Steuding. The authors don't say whether 's` is real or complex in their discussion of the prehistory of zeta. It would have been natural for Euler and others before Riemann to consider $s$ complex. Euler had the association to powers of pi for even integer arguments of zeta that would have suggested a connection to the complex via both his fabulous formula and his reflection formula for the gamma function, but then he had nothing much to glean from this perspective without Riemann's Mellin transform rep. through which Riemann was the first to really tease out new properties of zeta, to apply Euler's reflection formula to give the Hankel contour continuation of zeta from the right half-plane to the full complex plane, and to develop a clever algorithm to determine the non-trivial zeros, among other developments. A red herring seems to be some short-sighted effort to force an artificial dichotomy between interpolation and analytic continuation. I use Cotes'(and Newton's) interest and skill at interpolation in the real realm (certainly related to approximating celestial orbits) to indicate he was predisposed to make analytic continuations. In addition, there is no dichotomy. In several MO and MSE questions, I show how interpolation is related to analytic continuation of the factorial to the gamma function, the Bernoulli numbers to the Riemann zeta, the Bernoulli polynomials to the Hurwitz zeta, and the classic calculus of integer powers of the derivative op to complex non-integer values, among other interpolations/ACs (e.g., start at this MO-Q or this MO-Q). These can be related to sinc function/cardinal series interpolations, binomial expansion interpolation, and/or Newton interpolation and probably others (e.g., this MO-Q). Some more sophisticated associations are related to Mahler's theorem and the ref in the answer to this MO-Q. One aspect of Riemann's gifts was his insight on how this is related to the Mellin transform. (For accessibility bias, see Khaneman and Tversky.)<|endoftext|> TITLE: 3-manifold with fundamental group $\mathbb Z$ QUESTION [5 upvotes]: Let $M$ be a compact $3$-manifold with nonempty boundary. If $\pi_1(M)=\mathbb Z$, can we prove that $M$ is homeomorphic to $S^1 \times D^2$? REPLY [19 votes]: No. For example, take a copy of $S^1 \times S^2$ and remove the interior of a closed, nicely embedded, three-ball. You will need to add the hypothesis of irreducibility (to rule out "punctures" as in my example immediately above) and the hypothesis of orientability (to rule out the solid Klein bottle). These hypotheses, plus the disk theorem, gives the desired result. See Proposition 3.4 of Hatcher's three-manifold notes, Exercise 5.3 in Hempel's book, or Exercise I.32 in Jaco's book.<|endoftext|> TITLE: Homotopy coherent colimits in chain complexes QUESTION [5 upvotes]: In remark 1.2.6.2 (HTT), Lurie states that Another possible approach to the problem of homotopy coherence is to restrict our attention to simplicial (or topological) categories C in which every homotopy coherent diagram is equivalent to a strictly commutative diagram. For example, this is always true when C arises from a simplicial model category (Proposition 4.2.4.4). I am working with homotopy coherent diagrams from $\mathbb{Z}_{\ge 0}$ (thought as a poset) to $\text{Ch}_{\ge 0}(\mathbb{Z})$. Here $\text{Ch}_{\ge 0}(\mathbb{Z})$ stands for nonnegatively graded chain complexes of abelian groups, thought as an $\infty$ category. Since I want to compute the colimit of such a diagram, I am wondering the following: Is $\text{Ch}_{\ge 0}(\mathbb{Z}) $ a simplicial model category? In particular, in light of Corollary 4.2.4.7, it seems like for any diagram $N(\mathbb{Z}_{\ge 0}) \to N(\text{Ch}_{\ge 0}(\mathbb{Z})) $ I can find a simplicial functor $\mathbb{Z}_{\ge 0} \to \text{Ch}_{\ge 0}(\mathbb{Z})$ such that the $\infty$ -colimit of the former is equivalent to the homotopy colimit of the latter. Also, it seems like $\mathbb{Z}_{\ge 0}$ being a discrete category, such hocolimit should be computable from a discrete information. In other words, let $sk_0 : \text{sSet-Cat} \to \text{Cat}$ be the functor that takes the $0$ skeleton hom-wise and $LS: \text{Cat} \to \text{sSet-Cat}$ the functor that takes the trivial simplicial set home wise. Note that $LS$ is left adjoint to $sk_0$. Let $J$ be an ordinary category and $C$ a simplicial model category. Let $F: LS(J) \to C$ be a simplicial functor corresponding to $F': J \to sk_0(C)$. Is it true that the homotopy colimit of $F$ and of $F'$ are equivalent? I hope I have not said too much stupid things :) REPLY [11 votes]: The result is not only true for simplicial model categories, but for plain combinatorial model categories too - this is Higher Algebra 1.3.4.25.. In fact, for this you can reduce to the case of simplicial model categories, by noting that a combinatorial model category is always Quillen equivalent to a simplicial one. This is for instance corollary 6.4 in Dugger's Universal homotopy theories It follows that for any small category $I$ and any combinatorial model category $C$, any homotopy coherent diagram $I\to C[W^{-1}]$ can be lifted to a strict diagram $I\to C$; where $C[W^{-1}]$ is the $\infty$-categorical localization. Also note that there is a simpler approach in your particular situation : $Ch_{\geq 0}(\mathbb Z)$ is Quillen equivalent (in fact, equivalent ! but the equivalence is compatible with the model structures) to $sAb$, simplicial abelian groups, via the usual Dold-Kan correspondance, and the latter is in fact simplicially enriched. As for colimits, this will depend on $I$, in fact it's also a theorem in HTT that these can be computed as homotopy colimits in $C$ (HTT 4.2.4.1 in the simplicial case, HA 1.3.4.24. in the non simplicial case), in the usual model category theoretic sense. Now $\mathbb Z_{\geq 0}$ is filtered, so since filtered colimits are exact in abelian groups, it follows that if you have a filtered diagram $I\to Ch_{\geq 0}(\mathbb Z)$, then its colimit and its homotopy colimit agree, and in particular the localization functor $Ch_{\geq 0}(\mathbb Z)\to D_{\geq 0}(\mathbb Z)$ preserves these filtered colimits, where I let $D_{\geq 0}(\mathbb Z)$ denote the localization at quasi-isomorphisms (i.e. the underlying $\infty$-category) Note that the exactness of filtered colimits is true in abelian groups and more generally $R$-modules, but it is not true in general abelian categories. This is to some extent related to how $\mathbb Z_{\geq 0}$ is "discrete", because it follows from the previous paragraph that the higher homology groups of the homotopy colimit of a functor $I\to Ab\to Ch_{\geq 0}(\mathbb Z)$ are computed as the left derived functors of the colimit functor; and those are related to the nerve of $I$ as a space. The key-word for this is "higher colimit", or perhaps you'll be luckier with "higher limit" As for your last paragraph, $LS$ is not left adjoint to $sk_0$, rather to evaluation at $0$ (also known as "limit along $\Delta^{op}$", which has an initial object). But quite generally nonetheless, you can in fact compare a simplicial model category and its underlying plain model category. This is also somewhere in HTT I believe, where it is stated that the homotopy coherent nerve of the full subcategory of cofibrant-fibrant objects of a (sufficiently nice) simplicial model category is equivalent to the localization of its underlying plain model category at the weak equivalences. So you should be able to compare homotopy colimits in one and homotopy colimits in the other. (I could not find it in HTT but it is 1.3.4.20. in HA) Hopefully that answers your questions<|endoftext|> TITLE: Two-point Helly QUESTION [7 upvotes]: Suppose that for a finite collection of planar convex sets $\mathcal F$ the following holds. For any six members of $\mathcal F$ there are two points such that every set contains (at least) one of the points. Does it follow that all members of $\mathcal F$ can be stabbed by two points? I am sure that this is known, and probably even the more general problem of determining the optimal value instead of six for $k$ stabbing points in $\mathbb R^d$. Related question with many links to related problems: A curious generalization of Helly's theorem. REPLY [9 votes]: This is true for special families of convex sets, for example axis parallel rectangles, but it is false for general convex sets, even if $6$ is replaced by any other finite number. This was shown by M. Katchalski and D. Nashtir in the paper On a conjecture of Danzer and Grünbaum (Proc. Amer. Math. Soc. 124 (1996), 3213-3218, doi: 10.1090/S0002-9939-96-03806-3).<|endoftext|> TITLE: Two $\infty$-categories of chain complexes QUESTION [7 upvotes]: In the literature, I've mostly seen two quasicategories coming from $\text{Ch}_R$: By considering $\text{Ch}_R$ with weak equivalences $\mathcal W = \text{quasi-isomorphisms}$, we can consider its Dwyer-Kan localization $L^H(\text{Ch}_R)$, a simplicial category. Then, from the Quillen equivalence $$|-|:\text{sSet}_{\text{Joyal}}\leftrightarrows\text{sSet-Cat}_{\text{Bergner}}:N_\Delta,$$ we can define, taking a fibrant replacement if necessary, $$ N_\Delta(L^H(\text{Ch}_R))\in\text{quasicategories} $$ In the Dold-Kan equivalence $N:\text{sAb}\leftrightarrows\text{Ch}_R:\Gamma.$, $\Gamma$ is a monoidal functor (between cartesian categories), thus it induces a functor $$ \Gamma_*:\text{Ch}_R\text{-Cat}\to \text{sAb-Cat}. $$ The underlying set of a simplicial abelian group is a Kan complex thus $$ N_\Delta(\Gamma_*(\text{Ch}_R))\in\text{quasicategories} $$ Are the quasicategories $N_\Delta(L^H(\text{Ch}_R)$ and $N_\Delta(\Gamma_*(\text{Ch}_R)$ equivalent? Is this trivial? Related: how to make the category of chain complexes into an $\infty$-category. REPLY [11 votes]: The two categories you describe are not equivalent in the fashion that you hope. No matter what kind of simplicial category $C$ is, the quasicategory $N_\Delta(C)$ has an explicit description of its homotopy category: namely, it has the same objects as $C$, and $$ Hom_{hN_\Delta C}(X,Y) = \pi_0 Hom_C(X,Y). $$ (This is true even if we need to take a fibrant replacement of $C$ first.) Consider first case (1). Whenever $C$ is a category with a subcategory $W$ of weak equivalences, the Dwyer-Kan localization $L^H C$ has morphism spaces that are explicitly described: the 0-simplices are composites of morphisms in $C$ with formal inverses of morphisms in $C$, and the 1-simplices are natural transformations ("hammocks") built out of commutative diagrams involving these morphisms in $C$ or $W$. Taking $\pi_0$, we find that the homotopy category is formed by taking the morphisms in $C$ and formally inverting the morphisms in $W$: it is the localization. $$ hN_\Delta L^H C \cong C[W^{-1}] $$ Now consider case 2. If $C$ is a dg category, then we can also be explicit about the simplices in $Hom_{\Gamma_* C}(X,Y)$. Namely, a 0-simplex is a 0-cycle $f$ in the Hom-complex $\underline{Hom}_C(X,Y)$, and an edge from $f$ to $g$ is a 1-chain $h$ in $\underline{Hom}_C(X,Y)$ with $\partial h = g-f$. Taking $\pi_0$, we find that the homotopy category is formed by taking the morphisms in $C$ and modding out by the relation of chain homotopy equivalence $$ Hom_{hN_\Delta \Gamma_* C}(X,Y) \cong H_0 \underline{Hom}_C(X,Y). $$ This makes it the "chain homotopy category". Applied to $Ch(R)$, the first construction is the derived category $D(R)$, and the second construction is the chain homotopy category $K(R)$. There is a natural functor $K(R) \to D(R)$, but it is not an equivalence: for example, there is a map of complexes from $(\dots \to 0 \to \Bbb Z \to \Bbb Z \to 0)$ to $(\dots \to 0 \to 0 \to \Bbb Z/2 \to 0)$ that is a quasi-isomorphism; there are no nonzero maps of chain complexes in the opposite direction in $Ch(\Bbb Z)$ and hence no nonzero maps in $K(\Bbb Z)$ either. I believe that if you restrict to the subcategory of "cofibrant" complexes (e.g. bounded-below complexes of projectives) then you get an equivalence. Most of the versions of this result that I know, however, are for simplicial model categories (which $Ch_R$ is not), and so I don't have a reference handy.<|endoftext|> TITLE: What's motivic about $\mathbb{A}^1$-homotopy theory? What's motivic about correspondences? QUESTION [20 upvotes]: I come today to mathoverflow to showcase some genuine confusion about the motivic world. I want to ask some questions before actually starting to study the subject, to build some sense of direction. I take Grothedieck's orginal idea of motives to be that of an abelian category through which every good cohomology should factorize. Skimming through the Morel-Voevodskij paper, $\mathbb{A}^1$-homotopy theory is what you get when you: start with the category of smooth $k$-schemes of finite type; take Nisnevich topology on it; take simplicial sheaves for this topology; put a model structure where weak equivalences are stalkwise; finally localize to make the projections $\mathbb{A}^1 \times{X} \to X$ weak equivalences. I will call this category $\text{MV}_k,$ and its homotopy category $\text{H}(k).$ I also know from here that the model category defined above is Quillen equivalent to some localization (imposing sheaf conditions and $\mathbb{A}^1 \times{X} \xrightarrow{\sim} X$) of the category of all simplicial presheaves on smooth schemes of finite type with the projective model structure; this latter category enjoys a universal property being a localization of the universal model category on smooth schemes of finite type over $k.$ What's the relationship of this category with motives? Skimming through the motivic cohomology book, I see that the triangulated category of motives is defined in the following way: first one takes the additive category of correspondences $\text{Corr}_k,$ then takes $\text{Ab}$-enriched preasheaves on $\text{Corr}_k^{\text{op}},$ then takes sheaves for the Nisnevich topology, takes the derived category of this, and then localizes at $\mathbb{A}^1$-weak equivalences. This category I guess would be denoted $\text{DM}^{\text{eff}}_{\text{Nis}}(k,\mathbb{Z}).$ For some general commutative ring $R,$ one takes instead $\text{Ab}$-enriched presheaves and sheaves of $R$-modules to define $\text{DM}_{\text{Nis}}^{\text{eff}}(k,R).$ I hope I got this right. What is the relationship between this latter construction and the first one? I also would like to understand better the intuition behind the use of the category $\text{Corr}_k$. An elementary correspondence between $X$ and $Y$ is an irreducible closed subset of $X \times{Y}.$ In $\text{Corr}_k$ objects are smooth separated schemes of finite type and the set of morphisms between $X$ and $Y$ is the free abelian group on elementary correspondences. So we are choosing certain set of spans between two objects and then taking the free abelian group on it. I also understand that considering elementary correspondences is a way to enlarge the category $Sm_k$ in a such a way that every morphism $f:X \to Y$ is sent to its graph. I see that correspondences are used in the original formulation of Grothedieck's pure motives. So they were there from the very first idea of motives. How should I think of correspondences? What is the connection between the idea of correspondence and Grothendieck's original idea of motives? Why, if the goal is to build a(n abelian) category through which good cohomology theories factorize, we begin the construction starting with correspondences? REPLY [13 votes]: There is no need a priori to define these categories of motives starting from correspondences. The stable homotopy theory of schemes $SH$ may be characterized by a universal property saying that it is the universal setting in which one may define the six operations; see Drew and Gallauer. Usually, if you have a system of coefficients $D$ in which the six operations are defined, we have in particular: for each morphism of scheme $f:X\to Y$, a pull-back functor $f^*:D(Y)\to D(X)$ with a right adjoint $f_*$ as well as a push-forward with compact support functor $f_!:D(X)\to D(Y)$ with right adjoint $f^!$ satisfying a bunch of properties, among which we ask for homotopy invariance (i.e. full faithfulness of $f^*$ in the case where $f$ is the structural map of a vector bundle). The universal property of $SH$ means that $SH$ has the six operations and is initial for this, as far as we ask that $D$ takes values in presentable stable $\infty$-categories. More precisely, the assignment $(f:Y\to X)\mapsto f_!f^!(1_X)$ defines a functor from smooth $X$-schemes to $D(X)$ for each $X$, and this extends uniquely into a compatible family of functors $SH(X)\to D(X)$. For example, we may take $D(X)=D(Sh(X_{\acute{e}t},\Lambda))$ with $\Lambda$ any ring of positive characteistic invertible $\mathcal{O}_X$. OR we may consider its $\ell$-adic counterpart $D(X,\mathbb{Z}_{\ell}):=D(Sh(X_{\acute{e}t},\mathbb{Z}_{(\ell)}))/D(Sh(X_{\acute{e}t},\mathbb{Q}))$ with $\frac{1}{\ell}\in\mathcal{O}_X$. If we focus on $\mathbb{Q}$ coefficients, we may consider the $\mathbb{Q}$-linear version of $SH$, and consider one of the following properties: we want representable cohomology theories to be graded in the usual sense (because, due to the interpretation of quadratic forms in $SH$, there is an element $\varepsilon$ which is a non-trivial square root of $1$ but which is not always equal to $-1$). we want the projective bundle formula to hold (equivalently, we want Chern classes to exist in our cohomology theories). we want étale descent (or hyperdescent). we want étale descent and proper descent (or hyperdescent). we want (finite) correspondences to act on cohomology theories. It happens that one of these properties hold in a $\mathbb{Q}$-linear $D$ if and only if all of them do. Or, equivalently, if we force one of these properties to hold in the $\mathbb{Q}$-linear version of $SH$, all the other ones hold as well. Each of these properties drives us to some theory of of motives. If you insist on finite correspondences, you will go in the direction of $DM(X,\mathbb{Q})$. If you insist on the graded thing, you will consider Morel's $SH(X)_{\mathbb{Q},+}$. If you prefer étale descent, you get what Ayoub likes to call $DA(X,\mathbb{Q})$. If you insist on proper descent, you get Voevodsky's $DM_h(X,\mathbb{Q})$, constructed from $h$-sheaves. One way to make Chern classes appear naturally is to use representability algebraic $K$-theory in $SH$, and this leads to we are called Beilinson motives here. But, as shown in loc. cit., all these constructions are equivalent (except for $DM(X,\mathbb{Q})$ where we also need $X$ to be geometrically unibranch because finite correspondences are tricky to control over non-normal schemes), and many of them do not mention correspondences in their definition. And these theories are nice because Chow groups naturally are representable in there, so that we get the lovely picture that classical intersection theory has something to do with the universal cohomology theory (this is a super fancy and efficient way to construct cycle classes). The reason why you may see people focusing on correspondences is because they appear naturally via Poincaré duality (Gysin maps), but also mainly because they want to be able to compute: the construction of motivic categories eventually demand that we invert the Tate object, and that leads to categories which are far from usual derived categories, thus making concrete computations easily out of reach. In versions with correspondences, and if we restrict ourselves to motives over a field, the constructions that involve finite correspondences usually imply that the inversion of the Tate object is less dramatic, so that we can still control what is happening. This kind of results are called cancellation theorems and usually are technical (I mean that the proof requires non-formal arguments which only hold over fields and that such cancellation results are known to be false over higher dimensional base schemes). The cancellation theorem for $DM$ is proved by Voevodsky, which implies the full faithfulness of the functor $DM^{eff}_{Nis}(k)\to DM(k)$. Another reason to go through correspondences is that, with integral coefficients, having a nice theory of Chern classes is not a mere property, but a genuine structure, and making this precise in a coherent way is challenging (especially if we do not work in a context with étale descent). For a conceptual approach to correspondences in $SH$ (for $R$-modules in $SH$ with $R$ a motivic ring spectrum), I would recommend to have a look at the work of Elden Elmanto, Marc Hoyois, Adeel A. Khan, Vladimir Sosnilo, and Maria Yakerson. The fact that one can prove a cancellation theorem with an appropriate notion of finite correspondences for $SH$ is also due to Voevodsky and has been developed in full by the aforementioned authors as well as by Alexey Ananyevskiy, Grigory Garkusha and Ivan Panin.<|endoftext|> TITLE: What can $I\Delta_0$ prove? QUESTION [9 upvotes]: What combinatorial and number-theoretic propositions can $I\Delta_0$ prove? Obviously there are an infinitude of them, but what are some well known theorems that can be proved in $I\Delta_0$, if any? REPLY [9 votes]: Since no one else is biting, I'll answer, and thanks to comments I now this is accurate: $I\Delta_0$ can prove several basic theorems: Every square equals 0 or 1 mod 4 No prime has a rational square root The only solutions to $x^3+y^3=z^3$ or $x^4+y^4=z^4$ are trivial Every $x$ is divisible by a prime $p$ with $p \le x$ (The standard proofs can be reproduced in $I\Delta_0$, since they do not require any lists or sequences or products thereof. The last claim is proved in $I\Delta_0$ in a paper by D'Aquino.) $I\Delta_0$ seems not to be able to prove that: there are arbitrarily large primes every prime of the form $4m+1$ can be written as $a^2+b^2$ (The first is a well-known open problem due to Wilkie) $I\Delta_0$ cannot prove that: the functions $x^{\log x}$, $x!$, or $x^y$ are total there are solutions to the Pell equation $x^2-Ny^2=1$ (The $x^{\log x}$ is due to Parikh; the Pell equation result is due to D'Aquino.) But $I\Delta_0(exp)$, i.e. the theory $I\Delta_0$ in the language with exponentiation, seems to prove every theorem published in the Annals of Mathematics whose statement involves only finitary mathematical objects (This is a well-known conjecture due to Friedman.)<|endoftext|> TITLE: General Sylvester's linear matrix equation QUESTION [8 upvotes]: For what conditions on $A$, $B$ and $C$ (square matrices of size $n$) would there be a unique solution to $$ ABX + AXC + XBC = D, $$ for any $D$? Can one expect a characterization similar to the Sylvester Theorem, which states that there always exists a unique solution to $AX + XB = C$, for any $C$, if and only if $A$ and $-B$ do not share an eigenvalue? And then, can this be extended to equations of the form $ABCX + ABXD + AXCD + XBCD = E$, and beyond? REPLY [5 votes]: Can one expect a characterization similar to the Sylvester Theorem As far as I know, no, apart from very special cases where the coefficients can be triangularized simultaneously. There is a big gap in difficulty between the 2-term case (where there is a canonical form for pairs of matrices, an $O(n^3)$ algorithm, etc.) and the 3-term case, when there is basically nothing apart from turning it into a $n^2 \times n^2$ linear system. Even in your case where your coefficients have a special form no particular simplifications spring to mind. As far as I know, a nice characterization is an unsolved (and very likely unsolvable) problem.<|endoftext|> TITLE: Stone duality for the algebra of Boolean functions such that $f(\top,\dots,\top) = \top$, or: What does the presheaf topos on $FinSet_\ast$ classify? QUESTION [12 upvotes]: $\newcommand\FinSet{\mathit{FinSet}}\newcommand\FinBool{\mathit{FinBool}}\newcommand\FreeFinBool{\mathit{FreeFinBool}}\newcommand\Set{\mathit{Set}}\newcommand\Psh{\mathit{Psh}}$It's well-known that the topos of presheaves on the category $\FinSet$ of finite sets is the classifying topos for boolean algebra objects. The argument goes like this: By Stone duality, the functor $n \mapsto 2^n$ is an equivalence of categories $\FinSet \simeq \FinBool^{\text{op}}$, where $\FinBool$ is the category of finite Boolean algebras, which are necessarily projective (so we may use $\FinBool$ and $\FreeFinBool$, which is dual to the category $\FinSet_{2^\bullet}$ of sets with cardinality a finite power of $2$, pretty much interchangeably). Since the theory of Boolean algebras is a finitary algebraic theory, finite-product-preserving functors $\FinBool^{\text{op}} \to \mathcal E$ are identified with Boolean algebra objects in $\mathcal E$, for any category $\mathcal E$ with finite products and split idempotents. If $\mathcal E = \Set$, then any finite-product-preserving functor $\FinBool \to \Set$ is automatically flat; the proof uses the fact that every finitely-presentable Boolean algebra is projective. So a geometric morphism $\Psh(\FinSet_{\neq \emptyset}) \to \Set$ is just a Boolean algebra. Then because $\Psh(\FinSet_{\neq \emptyset})$ is a presheaf topos and the theory of Boolean algebras is algebraic, this classifying topos identification extends to all Grothendieck toposes $\mathcal E$. But what happens when we perturb the input a bit? For example, what happens when we consider presheaves on the category $\FinSet_\ast$ of finite pointed sets? The category $\FinSet_\ast^{op}$ is the idempotent completion of a Lawvere theory $\mathcal T = \FinSet_{\ast,2^\bullet}$ with half as many operations of each arity as the Lawvere theory for Boolean algebras ($2^{2^n - 1}$ instead of $2^{2^n}$). Indeed, by identifying the basepoint of a finite pointed set with $\top$, we can regard $\mathcal T$ as the Lawvere theory whose $n$-ary operations are all Boolean algebra homomorphisms $f: 2^n \to 2^1$ such that $f(\top,\dotsc,\top) = \top$ — i.e. if all inputs are true, the output must also be true. I believe that $\mathcal T$ is generated by the operations $\top, \wedge, \vee, \to$, but I'm not sure what a complete set of relations would be. As a Lawvere theory, $\mathcal T$ may be rather messy. But we're considering something more refined: a finite-product-preserving functor $A: \mathcal T \to \Set$ is not automatically flat. Flatness is equivalent to requiring that $A$ is a filtered colimit of finitely-generated free algebras. In particular, every finitely-generated subalgebra of $A$ is a subalgebra of a finitely-generated free algebra. It follows that $A$ will satisfy any universal statement satisfied by all finitely-generated free $\mathcal T$-algebras — not just the algebraic ones. I think (but I'm not certain) that the converse is also true — so that $A$ is flat if and only if it satisfies the universal theory of free $\mathcal T$-algebras. Questions: What is an axiomatization of the algebraic theory of the logical connectives $\top, \wedge, \vee, \to$ (i.e. the algebraic theory of connectives $f$ such that $f(\top, \dotsc, \top) = \top$, i.e. the equational theory of the structures $\{\top,\bot\}^n$, in the language $(\top,\wedge,\vee,\to)$)? What is an axiomatization of the universal theory of these logical connectives (i.e.— I think — the theory classified by the presheaf topos $\Psh(\FinSet_\ast)$, i.e. the universal theory of the structures $\{\top,\bot\}^n$, in the language $(\top,\wedge,\vee,\to)$)? How do these theories compare to the theory of Boolean algebras — what are some examples of algebras or algebra homomomorphisms which don't come from Boolean algebras or homomorphisms thereof? Is there a form of Stone duality for these algebras, and how do the corresponding spaces compare to Stone spaces? More fundamental than the presheaf toposes $\Psh(\FinSet)$ and $\Psh(\FinSet_\ast)$ are the presheaf toposes $\Psh(\FinSet^{\text{op}})$ and $\Psh(\FinSet_\ast^{\text{op}})$ which classify objects and pointed objects respectively. Unfortunately, these facts don't seem to shed much light on the less fundamental toposes I'm interested in here. For instance, it doesn't seem to be the case that the tensor product of the Lawvere theories $\FinSet^{\text{op}}_\ast$ and $\FinSet$ is $\FinSet_\ast$ — I believe that tensor product is just the terminal Lawvere theory (note that the "op"'s for the Lawvere theories are reversed from the "op"'s for the presheaf categories). The operations $(\top,\wedge,\vee,\to)$ being considered here are almost the same as the operations of a Heyting algebra — the difference being that a Heyting algebra has a constant $\bot$ as well. So, based on the axioms for a Heyting algebra, here are some guesses: Guess for (1): The following algebraic axioms, which are satisfied (thanks to Andreas Blass for catching a mistake in the comments below!), might be a complete set of algebraic axioms: $(\top,\wedge,\vee)$ form an upper-bounded distributive lattice; $x \to x = \top$; $x \wedge (x \to y) = x \wedge y$; $y \wedge (x \to y) = y$; $x \to (y \wedge z) = (x \to y) \wedge (x \to z)$; $(x \vee y) \to z = (x \to z) \wedge (y \to z)$. As pointed out by მამუკა ჯიბლაძე in the comments below, we also need the following axiom which goes beyond the Heyting axioms, and is equivalent to the interval $[y,\top]$ being a Boolean algebra with $(-) \to y$ for negation: $x \vee (x \to y) = \top$ REPLY [8 votes]: I might be missing something, but I think you are overcomplicating things. I clain that your topos classifies the theory $T$ of pairs $(B,\phi)$ where $B$ is a boolean algebra and $\phi : B \to \{0,1\}$ is a boolean algebra morphism. Indeed, this theory is given by a finite limit sketch, so it is classified by the the category of presheaf on the opposite of the category of finitely presented $T$-models. A $T$-model $(B,\phi)$ is finitely presented exactly when the boolean algebra $B$ is finite, hence the category of finitely presented $T$-model is equivalent to $FinSet_*$ by the duality between finite boolean algebra and finite sets. The theory you describe in your "Guess" is an attempt at axiomatizing the properties fiber $\phi^{-1}(1)$ as an Heyting algebra. This is definitely possible as the negation induce a bijection between the two fiber $\phi^{-1}(0)$ and $\phi^{-1}(1)$ so you can reconstruct $B$ out of one of the fibers and you can express the fact that $B$ is a boolean algbera over $\{0,1\}$ in terms of these fibers.<|endoftext|> TITLE: Proof that a Cartesian category is monoidal QUESTION [15 upvotes]: If $\mathcal C$ is a category with products and a terminal object, then $\mathcal C$ is monoidal. This seems obvious, but wherever I look for a proof or a reference it simply states that the proof is "pretty involved". I found this 8 years old question on MathOverflow: Monoidal structure on a category with products and with terminal object The answer given there uses the fact that Set is monoidal, along with the Yoneda lemma. Is there any other reference that proves it more "generally" (i.e without resorting to Set), or a standard reference? REPLY [3 votes]: One can find a completely explicit, constructive proof of this in the agda-categories library too, with explicit equational proofs for all pieces. There's an html version for easy web readability too (go to the almost bottom of that file). The symmetric monoidal category is worked out too. The advantage of doing it this way is that, if you ever wonder why some specific detail is true, it's all here to be seen. No details omitted.<|endoftext|> TITLE: Non-degenerate points on a Jacobian surface QUESTION [8 upvotes]: Let $C$ be a (hyperelliptic) genus $2$ curve over a number field $K$ with a $K$-rational Weierstrass point $\infty$. We embed $C$ in its Jacobian $J$ via $\infty$. Question: Is there a quadratic extension $L/K$ and a point $x\in C(L)$ which is non-degenerate in $J$, i.e. such that $\mathbb{Z}x$ is dense in $J$? If $J$ has only finitely many abelian subvarieties, the answer is obviously yes. What about the case where $J$ is isogeneous to the square of an elliptic curve $E$? There are infinitely many elliptic curves on $J$ given by isogenies of $E$ and a priori, it seems possible to me though quite implausible that they could cover all quadratic points on $C$. Possible general statement: I would expect that this is part of a much more general "unlikely intersection" statement for any hyperbolic curve $C$ embedded in an abelian variety and demanding $[L:K] TITLE: Moduli of rational equivalent classes of 0-cycles QUESTION [5 upvotes]: Let $X$ be a smooth variety over a field $k$ and I'd like to think about $CH_0(X)$ the 0-Chow group i.e. the group of rational equivalent classes of 0-cycles. I'm wondering if there is any reasonable formulation to make sense of "family"/"moduli" of rational equivalent classes of 0-cycles, forming an fppf sheaf over $\text{Spec} k$? If there is such a sheaf, is there any chance the sheaf is actually representable? A quick literature search shows some work on certain formulations of "Chow schemes". However, they seem to parametrize the 0-cycles, instead of the rational equivalent classes. But this seems to suggest that now we only need to construct a subsheaf of principal 0-cycles and then we can consider the quotient sheaf. A known example: if X is in addition one dimensional and proper, we can speak of the Picard scheme which represents the relative Picard functor. I would really like to know what happens when the dimension goes higher. REPLY [3 votes]: When $X$ is a smooth projective variety over an algebraically close field $k$, then $CH_0(X)_{tors} = Alb(X)_{tors}$, a result due to Rojtman in https://www.jstor.org/stable/1971109?seq=1#metadata_info_tab_contents in characteristic zero, and with variants by Bloch (http://www.numdam.org/item/CM_1979__39_1_107_0/) and Milne (http://www.numdam.org/item/CM_1982__47_3_271_0/) for torsion away and at the characteristic respectively. In general and not restricted to torsion, even for the simplest case of surfaces over $\mathbb{C}$, this is not really possible. Mumford proved that as soon as $p_g > 0$ the albanese map is not, in general, injective https://projecteuclid.org/euclid.kjm/1250523940 but it is expected that the map should be injective when $p_g = 0$ and this is known as Bloch's conjecture which has been verified for many surfaces but it still pen in general.<|endoftext|> TITLE: Is there a pseudofinite group with a quantifier-free instance of the order property? QUESTION [7 upvotes]: Recall that a group $G$ is pseudofinite if every first-order sentence $\varphi$ (in the language of groups) satisfied in $G$ is also satisfied in some finite group. Also recall that an instance of the order property in $G$ is a pair of sequences of finite tuples of elements of $G$, $\{\bar{a}_i\}_{i<\omega}$ and $\{\bar{b}_{j}\}_{j < \omega}$ (with any two $\bar{a}_i$'s the same length and any two $\bar{b}_{j}$'s the same length), together with a formula $\varphi(\bar{x},\bar{y})$ such that for any $i, j < \omega$, $G$ satisfies $\varphi(\bar{a}_i,\bar{b}_j)$ if and only if $i2$, an infinite extra-special $p$ group is pseudofinite, and the quantifier-free formula $xy=yx$ witnesses the independence property (and so witnesses the order property too). Details: I am basically just quoting from the Appendix in Definable envelopes in groups having a simple theory by Milliet. Let $p>2$ be prime. An extra-special $p$ group is a group $G$ such that $g^p=1$ for all $g\in G$, $Z(G)$ is cyclic of order $p$, and $Z(G)=[G,G]$. It is well-known that there is a unique countably infinite extra-special $p$-group. In fact, Felgner showed that the theory of infinite extra-special $p$-groups is a well-defined complete $\aleph_0$-categorical first-order theory in the language of groups. The following is a concrete construction of the countable model. Let $V$ be a vector space over $\mathbb{F}_p$ of dimension $\aleph_0$, and let $\langle\cdot,\cdot\rangle$ be a non-degenerate skew-symmetric bilinear form on $V$. Let $G=V\times\mathbb{F}_p$ and define the group operation $$ (u,a)\ast (v,b)=(u+v,a+b+\langle u,v\rangle). $$ Then $(G,\ast)$ is the countably infinite extra-special $p$-group. Now, since the bilinear form is skew-symmetric, and $p>2$, it follows that $(u,a)\ast(v,b)=(v,b)\ast (u,a)$ if and only if $\langle u,v\rangle=0$. So we can witness the independence property for the formula $\varphi(x,y)$ given by $x\ast y=y\ast x$. Specifically, we fix $n$ and find $g_1,\ldots,g_n\in G$ such that for any $X\subseteq\{1,\ldots,n\}$ there is some $h_X\in G$ so that $g_i\ast h_X=h_X\ast g_i$ if and only if $i\in X$. Let $u_1,\ldots,u_n$ be linearly independent vectors in $V$ and set $g_i=(u_i,0)$. Since the bi-linear form is non-degenerate, the maps $v\mapsto \langle u_i,v\rangle$ are linearly independent. So we can find $v_X\in V$ such that $\langle u_i,v_X\rangle =0$ if $i\in X$ and $\langle u_i,v_X\rangle =1$ if $i\not\in X$. Now let $h_X=(v_X,0)$ and we have the desired elements of $G$. Note that we are really just using the fact that $G$ interprets the structure $V$ (with the bilinear form), which has the independence property witnessed by the formula $\langle x,y\rangle =0$. (Actually these theories are bi-interpretable.) Finally, to see that this theory is pseudofinite, one only needs to know that there are arbitrarily large finite extra-special $p$-groups, which is also well-known. Then any non-principal ultraproduct is a model of the theory. Actually this is a $1$-dimensional asymptotic class in the sense of Macpherson and Steinhorn (see Proposition 3.11 of One-dimensional asymptotic classes of finite structures). Edit. The theory of infinite extra-special $p$ groups is simple, so there is no formula with the strict order property. I don't know an example for SOP. There is no $\aleph_0$-categorical example, since no $\aleph_0$-categorical pseudofinite theory has the strict order property (see Proposition 1.3 this paper by Alex Kruckman).<|endoftext|> TITLE: Sum of order polynomials of a set of posets QUESTION [6 upvotes]: Let $n\in \mathbb{Z}_{>0}$. For every subset $S\subseteq \left[ n-1\right]$ we define a poset $P_S=\left([n],\le_{P_S}\right)$ given by the covering relation $\lessdot$ which is defined as \begin{align*} &\forall i\in S & i+1&\lessdot i \\ &\forall i\in\left[n-1\right]\setminus S & i&\lessdot i+1\\ \end{align*} For instance if $n=3$ we have the posets Useful Definitions A function $f\colon P_S\to [m]$ is called order-preserving for $P_S$, if for any $x,y\in P_S$ $$x<_{P_S}y\implies f(x)\le f(y),$$ where $m\in\mathbb{Z_{>0}}$. We denote with $\Omega(P_S,m)$ the order polynomial of $P_S$ for $m$, i.e. $$\Omega(P_S,m)=\#\left\{ f\colon P_S\to [m]\,\mid f\;\; \text{is order-preserving}\right\}$$ We call an order-preserving bijection $\omega\colon P_S\to[n]$ natural labeling of $P_S$. For any permutation $w\in\mathcal{S}_n$ we denote with $\mathsf{Des}(w)$ the descent set of $w$, i.e. $$\mathsf{Des}(w)=\left\{ i\in\left[n\right]\mid w\left(i\right)>w\left(i+1\right)\right\} $$ We denote with $A_S$ the set of permutations that corresponds to the natural labelings of $P_S$, i.e. $$A_{S}=\left\{ w\in\mathcal{S}_{n}\mid w=\left(\omega\left(1\right),\omega\left(2\right),\dots,\omega\left(n\right)\right)\text{ for some natural labeling }\omega\text{ of }P_{S}\right\} $$ The question We want to calculate, for any given $m\in\mathbb{Z}_{>0}$ the sum of the order polynomials on the subsets $S\subseteq[n-1]$, i.e. $$\sum_{S\subseteq[n-1]}\Omega\left(P_{S},m\right).$$ My progress It seems that the requested sum is equal to $$m(1+m)^{n-1},$$ but I did not manage to show it. It is not hard to show that the set of all the natural labelings for a given subset $S$ has the same cardinality as the set of all permutations of $[n]$ with descent set equal to $S$, i.e. $$\#A_S=\#\left\{ \omega\colon P_{S}\to[n]\mid\omega\;\text{natural labeling}\right\} =\#\left\{ w\in\mathcal{S}_{n}\mid\mathsf{Des}\left(w\right)=S\right\} .$$ It is well known that $$\Omega\left(P_{S},m\right)=\sum_{w\in A_{S}}\binom{m+n-\mathsf{des}\left(w\right)-1}{n}.$$ I tried to use these facts in order to calculate the requested sum but I failed... (P.S. I posted this question on Math Stack Exchange some days ago, without getting an answer. If this question does not fit here, I will agreeable delete it.) REPLY [8 votes]: The key result is equation (1) of the paper here, which expresses the chromatic polynomial $\chi(G,m)$ of a graph $G$ as a sum of strict order polynomials $\overline{\Omega}(\overline{\mathcal{O}},m)$ of the transitive (and reflexive) closures $\overline{\mathcal{O}}$ of the acyclic orientations $\mathcal{O}$ of $G$. If we take $G$ to be an $n$-vertex path, then we get $$ \chi(G,m) = \sum_{S\subseteq[n-1]}\overline{\Omega}(P_S,m). $$ Now $\chi(G,m)=m(m-1)^{n-1}$. Moreover, by the reciprocity theorem for order polynomials, $$ \overline{\Omega}(P_S,m) = (-1)^n\Omega(P_S,-m), $$ and the result follows.<|endoftext|> TITLE: Reflection principle vs universes QUESTION [81 upvotes]: In category-theoretic discussions, there is often the temptation to look at the category of all abelian groups, or of all categories, etc., which quickly leads to the usual set-theoretic problems. These are often avoided by using Grothendieck universes. In set-theoretic language, one fixes some strongly inaccessible cardinal $\kappa$ -- this means that $\kappa$ is some uncountable cardinal such that for all $\lambda<\kappa$, also $2^\lambda<\kappa$, and for any set of $<\kappa$ many sets $S_i$ of size $<\kappa$, also their union is of size $<\kappa$. This implies that the stage $V_\kappa\subset V$ of "sets of size $<\kappa$" is itself a model of ZFC -- by applying any of the operations on sets, like taking powersets or unions, you can never leave $V_\kappa$. These sets are then termed "small", and then the category of small abelian groups is definitely well-defined. Historically, this approach was first used by Grothendieck; a more recent foundational text is Lurie's work on $\infty$-categories. However, their use has always created somewhat of a backlash, with some people unwilling to let axioms beyond ZFC slip into established literature. For example, I think at some point there was a long discussion whether Fermat's Last Theorem has been proved in ZFC, now settled by McLarty. More recently, I've seen similar arguments come up for theorems whose proofs refer to Lurie's work. (Personally, I do not have strong feelings about this and understand the arguments either way.) On the other hand, it has also always been the case that a closer inspection revealed that any use of universes was in fact unnecessary. For example, the Stacks Project does not use universes. Instead, (see Tag 000H say) it effectively weakens the hypothesis that $\kappa$ is strongly inaccessible, to something like a strong limit cardinal of uncountable cofinality, i.e.: for all $\lambda<\kappa$, one has $2^\lambda<\kappa$, and whenever you have a countable collection of sets $S_i$ of size $<\kappa$, also the union of the $S_i$ has size $<\kappa$. ZFC easily proves the existence of such $\kappa$, and almost every argument one might envision doing in the category of abelian groups actually also works in the category of $\kappa$-small abelian groups for such $\kappa$. If one does more complicated arguments, one can accordingly strengthen the initial hypothesis on $\kappa$. I've had occasion to play this game myself, see Section 4 of www.math.uni-bonn.de/people/scholze/EtCohDiamonds.pdf for the result. From this experience, I am pretty sure that one can similarly rewrite Lurie's "Higher Topos Theory", or any other similar category-theoretic work, in a way to remove all strongly inaccessible cardinals, replacing them by carefully chosen $\kappa$ with properties such as the ones above. In fact, there seems to be a theorem of ZFC, the reflection principle (discussed briefly in Tag 000F of the Stacks project, for example), that seems to guarantee that this is always possible. Namely, for any given finite set of formulas of set theory, there is some sufficiently large $\kappa$ such that, roughly speaking, these formulas hold in $V_\kappa$ if and only if they hold in $V$. This seems to say that for any given finite set of formulas, one can find some $\kappa$ such that $V_\kappa$ behaves like a universe with respect to these formulas, but please correct me in my very naive understanding of the reflection principle! (A related fact is that ZFC proves the consistency of any given finite fragment of the axioms of ZFC.) On the other hand, any given mathematical text only contains finitely many formulas (unless it states a "theorem schema", which does not usually happen I believe). The question is thus, phrased slightly provocatively: Does the reflection principle imply that it must be possible to rewrite Higher Topos Theory in a way that avoids the use of universes? Edit (28.01.2021): Thanks a lot for all the very helpful answers! I think I have a much clearer picture of the situation now, but I am still not exactly sure what the answer to the question is. From what I understand, (roughly) the best meta-theorem in this direction is the following (specialized to HTT). Recall that HTT fixes two strongly inaccessible cardinals $\kappa_0$ and $\kappa_1$, thus making room for small (in $V_{\kappa_0}$), large (in $V_{\kappa_1}$), and very large (in $V$) objects. One can then try to read HTT in the following axiom system (this is essentially the one of Feferman's article "Set-theoretic foundations of category theory", and has also been proposed in the answer of Rodrigo Freire below). (i) The usual ZFC axioms (ii) Two other symbols $\kappa_0$ and $\kappa_1$, with the axioms that they are cardinals, that the cofinality of $\kappa_0$ is uncountable, and that the cofinality of $\kappa_1$ is larger than $\kappa_0$. (iii) An axiom schema, saying that for every formula $\phi$ of set theory, $\phi\leftrightarrow \phi^{V_{\kappa_0}}$ and $\phi\leftrightarrow \phi^{V_{\kappa_1}}$. Then the reflection principle can be used to show (see Rodrigo Freire's answer below for a sketch of the proof): Theorem. This axiom system is conservative over ZFC. In other words, any theorem in this formal system that does not refer to $\kappa_0$ and $\kappa_1$ is also a theorem of ZFC. This is precisely the conclusion I'd like to have. Note that $V_{\kappa_0}$ and $V_{\kappa_1}$ are models of ZFC, but (critically!) this cannot be proved inside the formal system, as ZFC is not finitely axiomatizable, and only each individual axiom of ZFC is posited by (iii). One nice thing about this axiom system is that it explicitly allows the occasional arguments of the form "we proved this theorem for small categories, but then we can also apply it to large categories". A more precise question is then: Do the arguments of HTT work in this formal system? Mike Shulman in Section 11 of https://arxiv.org/abs/0810.1279 gives a very clear exposition of what the potential trouble here is. Namely, if you have a set $I\in V_{\kappa_0}$ and sets $S_i\in V_{\kappa_0}$ for $i\in I$, you are not allowed to conclude that the union of the $S_i$ is in $V_{\kappa_0}$. This conclusion is only guaranteed if the function $i\mapsto S_i$ is also defined in $V_{\kappa_0}$ (or if $I$ is countable, by the extra assumption of uncountable cofinality). In practice, this means that when one wants to assert that something is "small" (i.e. in $V_{\kappa_0}$), this judgment pertains not only to objects, but also to morphisms etc. It is not clear to me now how much of a problem this actually is, I would have to think more about it; I might actually imagine that it is quite easy to read HTT to meet this formal system. Shulman does say that, with this caveat, the adjoint functor theorem can be proved, and as Lurie says in his answers, the arguments in HTT are of similar set-theoretic complexity. However, I'd still be interested in a judgment whether the answer to the question is "Yes, as written" or rather "Probably yes, but you have to put some effort in" or in fact "Not really". (I sincerely hope that the experts will be able to agree on roughly where the answer falls on this spectrum.) A final remark: One may find the "uncountability" assumption above a bit arbitrary; why not allow some slightly larger unions? One way to take care of this is to add a symbol $\kappa_{-1}$ with the same properties, and ask instead that the cofinality of $\kappa_0$ is larger than $\kappa_{-1}$. Similarly, one might want to replace the bound $\mathrm{cf} \kappa_1>\kappa_0$ by a slightly stronger bound like $\mathrm{cf} \kappa_1>2^{\kappa_0}$ say. Again, if it simplifies things, one could then just squeeze another $\kappa_{1/2}$ in between, so that $\mathrm{cf} \kappa_{1/2}>\kappa_0$ and $\mathrm{cf} \kappa_1>\kappa_{1/2}$. This way one does not have to worry whether any of the "standard" objects that appear in some proofs stay of countable size, or whether one can still take colimits in $V_{\kappa_1}$ when index sets are not exactly of size bounded by $\kappa_0$ but have been manipulated a little. PS: I'm only now finding all the relevant previous MO questions and answers. Some very relevant ones are Joel Hamkins' answers here and here. REPLY [14 votes]: Answering this question depends strongly on exactly what you want from Higher Topos Theory, because expressing high logical strength is a different goal from expressing an aptly unified logical framework for algebraic geometry and number theory.  Unified strong foundations for general categorical mathematics are one fine goal, and seem to be the goal of many contributors here.  For that goal everything said in comments and answers to this question is relevant. But apt work in geometry and number theory does not call for vast logical strength. While HTT is more intertwined with universes than SGA, neither HTT nor SGA makes real use of the (very strong) axiom scheme of replacement. Thus they can use "universes" radically weaker than Grothendieck's. As a typical and germane example, Grothendieck he made just one appeal to the axiom scheme of replacement.  That is in his quite crucial proof that every AB5 category with a generating set has enough injectives.   And this use of replacement turns out to be eliminable.  It worked, but Grothendieck actually did not need it to get his result. To expand on Grothendieck's use of replacement: Reinhold Baer in the 1940s used transfinite induction (which requires the axiom scheme of replacement) to prove modules (over any given ring) have enough injectives.  He was consciously exploring new proof techniques and got a good result.  Grothendieck's Tohoku cast that proof in a form showing every AB5 category with a small set of generators has enough injectives--and a few years later Grothendieck found this was exactly the theorem he needed for topos cohomology.   Baer and Grothendieck both had practical goals, not tied to foundation concerns, but both wanted to get foundations right too.  And they did.  But it turns out they could have gotten those same theorems, correctly, without replacement, by nearly the same proofs, by specifying large enough function sets to begin with (using power set, but not replacement).  There are results that genuinely require the replacement axiom scheme.  But those results rarely occur outside of foundational research. A lot of people coming from very different angles (some logicians, some disliking logic) since the 1960s have remarked that in the context of algebraic geometry and number theory, the high logical strength  of Grothendieck's universe axiom, is an actually unused by-product of Grothendieck's desire for a unified framework for cohomology.  That can now be made quite precise:  The entire Grothendieck apparatus including not just derived functor cohomology of toposes but the 2-category of toposes, and derived categories, can be formalized in almost exactly the same way as it was formalized by Grothendieck, but at logical strength far below Zermelo-Fraenkel or even Zermelo set theory.    The same is true for HTT.  You can get it without inaccessible universes or reflection as long as you do not need the vast (and rarely used) strength of replacement.  The proof has not actually been given for HTT.  It has been for Grothendieck's uses of universes. It seems clear the same will work for HTT. The logical strength needed has been expressed indifferent ways:  Simple Type Theory (with arithmetic), Finite Order Arithmetic, the Elementary Theory of the Category of Sets, Bounded Quantifier Zermelo set theory.  Roughly put, you posit a set of natural numbers, and you posit that every set has a power set, but you do not posit unbounded iteration of power sets.  A fairly naive theory of universes can be given conservative over any one of these (the way Godel-Bernays set theory is conservative over ZFC) and adequate to all the large structure apparatus of the Grothendieck school.<|endoftext|> TITLE: 3 questions about basics of Martin-Löf type theory QUESTION [11 upvotes]: I started to read the HoTT book. I'm now on chapter 1 and I have several questions concerning not even homotopical, but "regular" type theory. On page 24, where the universes are introduced, there is a sequence: $$\mathcal U_0:\mathcal U_1:\mathcal U_2:\cdots$$ Everything here makes sense, but I don't understand what is $\mathcal{U}_0$. Maybe I've glossed over the definition? Later in the same section it is stated that there is no type containing all $\mathcal{U}_i$. Since I'm not yet free from the set-theoretic perspective, I can't help but wonder: can we define something like $\mathcal{U}_{\omega_0}$ and so on? Finally, I have a more vague question. I've got some experience with C++ and so I explained to myself Martin-Löf types using data types from C++. However, I have doubts that this approach is even remotely correct. Can somebody help me to understand if this analogy is correct or not? REPLY [15 votes]: Universe levels usually trip up newcomers to type theory since there is no straightforward intuition for them. What I found helpful is to think of them as a merely technical device to prevent impredicativity, and only dive deeper into the technicalities when necessary. The first recognition is that we need a universe: A basic judgment of MLTT is that something is a type: $A \ type$. When making a statement about all types, we need to refer to a collection of types, i.e., a universe $\mathcal{U}$. Consider the type $B :\equiv \Pi_{A:\mathcal{U}} \mathsf{Id}_\mathcal{U}(A, A \times \top)$. If we assumed that types like $B$ also lived in $\mathcal{U}$, we could devise devious Russell-style paradoxes, so hence we let $B$ live in a different universe $\mathcal{U}'$. It is no problem to assume that $A$ also lives in $\mathcal{U}'$. Renaming both universes to $\mathcal{U}_0$ and $\mathcal{U}_1$ and repeating these considerations leads to a cumulative hierarchy of universes $\mathcal{U}_0, \mathcal{U}_1, \mathcal{U}_2, ...$. So to answer your first question directly: $\mathcal{U}_0$ is the basic universe, and universes later in the hierarchy are introduced at will to prevent impredicativity. Note that cumulativity means that there is a difference between the term-type and the type-universe relationship: Any term lives in exactly one type, e.g., $0 : \mathbb{N}$. A type $A : \mathcal{U}_i$ lives in infinitely many universes, namely all $\mathcal{U}_j$ for $j \geq i$. There are different approaches to formally introduce universes and manage the relation to the typing judgment, most common are "Tarski-" and "Russell-style" approaches. You probably don't have to worry about that, as the exact implementation is mostly irrelevant when using universe levels in a formalization project. I'm not aware of any rigorous approaches to introduce something like $\mathcal{U}_{\omega_0}$. Since few type theorists are enthusiastic set theorists, I don't think anyone has had the motivation to do such a thing. Paradoxes abound very quickly, and one has to be very careful to not introduce Girard- and Hurkens-style paradoxes. (Dan has pointed out that people around Michael Rathjen and Anton Setzer have worked on the proof-theoretic strength of various type theories, and consider more interesting universe hierarchies in this context.) If you are an avid C++ programmer, it's completely fine to test out your intuitions in that language. You can't expect any formal rigour, but Bartosz Milewski has written a whole book expressing some more abstract programming languages principles in C++, and that seems to work surprisingly well. This book also introduces all examples in Haskell, which might provide a nice bridge to languages that have a richer type theoretic foundation.<|endoftext|> TITLE: Lagrangian of Reshetikhin-Turaev TFT's QUESTION [13 upvotes]: One of the results from the Reshetikhin-Turaev package is that given a modular tensor category $\mathscr{C}$ one can construct a TFT $Z$. In the case where $\mathscr{C}$ is the category of positive energy representations of the loop group, it is accepted that $Z$ coincides with the Chern-Simons theory. Now, the latter has a well-known Lagrangian. Is there a way to recover this Lagrangian from the RT construction? More generally, can one always have a Lagrangian for any of these RT theories? How are these constructed? REPLY [8 votes]: It's an open conjecture by Moore and Seiberg (originally in the context of conformal field theory) that every MTC can be obtained from Chern-Simons theory of simple Lie groups with known constructions. The constructions in the CFT context include orbifolding, coset and chiral algebra extension, which correspond to gauging symmetry and condensation for MTCs. Note that their original statement is that every MTC is the Chern-Simons theory of compact Lie groups, which I believe is equivalent to the above. If the conjecture is true, there will be a Lagrangian description for each Reshetikhin-Turaev TQFT, although not unique. There have been some math/physics works on potential "exotic" MTCs, basically counter-examples of the Moore-Seiberg conjecture. Two examples were considered in https://arxiv.org/abs/0710.5761, but one quickly shot down, and the other, quantum double of the even sectors of the Haagerup subfactor, still remains.<|endoftext|> TITLE: Are generic filters that produce the same forcing extension related by a ground-model automorphism? QUESTION [7 upvotes]: Suppose $M$ is a countable transitive model of some fragment of $\mathbf{ZFC}$, $\mathbb{P}\in M$ is a forcing notion and $G, H$ are $\mathbb{P}$-generic such that $M[G]=M[H]$. Does it then follow that there is some automorphism $\pi:\mathbb{P}\longrightarrow\mathbb{P}$ such that $\pi\in M$ and $\pi[G]=H$? If the answer is no, are there natural restrictions one could impose on $\mathbb{P}$ (maybe apart from $\mathbb{P}$ being finite) such that the above sentence holds? REPLY [12 votes]: This works much better in terms of complete Boolean algebras. If $\mathbb B$ and $\mathbb B'$ are complete Boolean algebras and a $\mathbb B$-generic filter $G$ and a $\mathbb B'$-generic filter $G'$ generate the same forcing extension, then there are $b\in G$ and $b'\in G'$ such that the part of $\mathbb B$ below $b$ is isomorphic to the part of $\mathbb B'$ below $b'$ by an isomorphism in the ground model, that sends the restriction of $G$ to the restriction of $G'$. I believe this fact is in Serge Grigorieff's paper "Intermediate submodels and generic extensions in set theory" [Ann. Math. Second Series, Vol. 101, No. 3 (May, 1975), pp. 447-490]. It's certainly possible to rewrite this in terms of posets instead of complete Boolean algebras, but the result looks messy to me.<|endoftext|> TITLE: Diffeomorphism groups of h-cobordant manifolds QUESTION [19 upvotes]: Do we have specific examples of h-cobordant smooth manifolds $M$ and $M'$ such that $\operatorname{BDiff}(M) \not \simeq \operatorname{BDiff}(M')$? Perhaps something can be said in terms of K-theory by analyzing bundles of h-cobordisms? In fact, if it helps to stabilize, I would be happy to know that $\operatorname{BDiff}_{\partial}(M \times I) \not \simeq \operatorname{BDiff}_{\partial}(M’ \times I)$. REPLY [6 votes]: This is regarding your second question. In dimensions $\geq 5$, where the $s$-cobordism theorem applies, $h$-cobordisms are invertible in the following sense: if $W : M \leadsto M'$ is an $h$-cobordism, then there exists an $h$-cobordism $W' : M' \leadsto M$ such that $W' \circ W \cong M \times [0,1]$, and $W \circ W' \cong M' \times [0,1]$. In other words, $W$ embeds into $M \times [0,1]$ relative to $M \times \{0\}$ and so on. This can be used to obtain maps $$B\mathrm{Diff}_\partial(M \times [0,1]) \overset{W \circ -}\to B\mathrm{Diff}_\partial(W) \overset{W' \circ -}\to B\mathrm{Diff}_\partial(W' \circ W) \cong B\mathrm{Diff}_\partial(M \times [0,1])$$ which are homotopy inverses to each other. But similarly $$B\mathrm{Diff}_\partial(M' \times [0,1]) \overset{- \circ W}\to B\mathrm{Diff}_\partial(W) \overset{- \circ W'}\to B\mathrm{Diff}_\partial(W \circ W') \cong B\mathrm{Diff}_\partial(M' \times [0,1])$$ are homotopy inverses, and so $$B\mathrm{Diff}_\partial(M \times [0,1]) \simeq B\mathrm{Diff}_\partial(W) \simeq B\mathrm{Diff}_\partial(M' \times [0,1]).$$ In the comments @archipelago suggests that one should think about the case of block diffeomorphisms. Consider the (semi-simplicial) group $\widetilde{\mathrm{Diff}}(W)$ of block diffeomorphisms of $W$ which do not fix the boundaries pointwise but preserve each of the two boundary components setwise. There are maps $$B\widetilde{\mathrm{Diff}}(M) \leftarrow B\widetilde{\mathrm{Diff}}(W) \to B\widetilde{\mathrm{Diff}}(M') \tag{1}$$ given by restriction, whose fibres are $B\widetilde{\mathrm{Diff}}_M(W)$ and $B\widetilde{\mathrm{Diff}}_{M'}(W)$. Now by the same kind of reasoning as above (namely gluing on $W'$) we have $$B\widetilde{\mathrm{Diff}}_M(W) \simeq B\widetilde{\mathrm{Diff}}_{M'}(M' \times [0,1])$$ but this is just the (classifying space of the) space of block concordances of $M'$, and spaces of block concordances are contractible by a kind of Alexander trick. Thus the two maps in (1) are equivalences, so indeed $$B\widetilde{\mathrm{Diff}}_\partial(M) \simeq B\widetilde{\mathrm{Diff}}_\partial(M').$$ (I don't know the answer to your original question about $B{\mathrm{Diff}}_\partial(M) \overset{?}\simeq B{\mathrm{Diff}}_\partial(M')$, but it seems quite interesting and the above suggests also considering the related question about concordances of $M$ versus those of $M'$.) EDIT: My argument in the case of block diffeomorphisms is slightly fallacious. The fibres of the maps in (1) need not be connected, as diffeomorphisms of $M$ (or $M'$) need not extend over $W$. However, it is true that each path-component of these fibres is contractible, as I said, because it can be identified with a space of block-concordances. The conclusion of the argument is thus that $B\widetilde{\mathrm{Diff}}(M)$ and $B\widetilde{\mathrm{Diff}}(M')$ have a common covering space, so e.g. have equivalent universal covers. Remarkably, their fundamental groups can be different: this has been proved by Samuel Muñoz Echániz, and will appear in his forthcoming PhD thesis.<|endoftext|> TITLE: How to write a good MathSciNet review? QUESTION [73 upvotes]: When reviewing for MathSciNet, I routinely find myself just paraphrasing and abbreviating the introduction provided by the author, and occasionally adding a few words about the quality of the research or the cleverness of the argument (which the authors themselves would not be able to write for obvious reasons). There seems to be very little added value in doing this, since the paper already has the introduction (which has the added benefit of being written by someone who has intimate knowledge of the paper) as well as abstract (which will in most cases be sufficient to decide if the paper is worth reading). Of course, I can imagine edge cases when someone can't quite decide if the paper is worth delving into based on the abstract alone, while the introduction is for some reason difficult to read or the paper is difficult to access. But I can't help feeling that there should be more to it. I would love to hear opinions about what makes a MathSciNet review useful, and how to achieve it. Edit to add: As YCor correctly points out, the same question applies with ZBMath or any other place that hosts public reviews in place of MathSciNet. To avoid creating a question which is a moving target, I will refrain from making edits above. REPLY [23 votes]: I'm the Executive Editor at Mathematical Reviews. I'm impressed by some of the answers, particularly those from Timothy Chow and Denis Serre, which give as good an answer as I could hope for. As Timothy Chow and Joe Silverman point out, having links to other items in MathSciNet related to the current paper is especially helpful to users. Kimball quotes the salient bits of our Guide for Reviewers. Thank you for doing that. People tend to forget that the Guide exists. When I talk with people in person at a conference, I compress the goal of a good review down to: describe something about the context of the result (where did it come from?); describe the main result (what is it?); describe the main technique(s) or method(s) (how did they do it?). Doing all this concisely can be difficult, of course. The AMS Bulletin still publishes reprints of reviews. They are generally tied to one of the articles in the issue. While they are generally above average, they are not necessarily meant to be the best of the best. For a time, I was posting particularly good reviews that I came across. You can find them via this link: exceptional reviews. I should probably get back to posting examples of good reviews. I second Yemon Choi's recommendation for Kimball's collection of (his version of) exceptional reviews.<|endoftext|> TITLE: Eigenvarieties and functoriality QUESTION [11 upvotes]: In Langlands' review of Hida's book "$p$-adic automorphic forms on Shimura varieties", he discusses a nexus of 4 areas of modern number theory: automorphic representations, motives, spaces of $p$-adic Galois representations, and a fourth less well-defined area which he describes in some cases as spaces of representations of $p$-adic Hecke rings. These two final objects should correspond to each other in some exact way through the $p$-adic Langlands conjectures 'in families'. My understanding is that the fourth area is now underway via the theory of eigenvarieties. Is it indeed the case that eigenvarieties (are conjectured to) play this role in the Langlands set-up? Langlands states many conjectural relationships between this fourth area and the other 3, and I would like to find out to what extent these relationships have been satisfied by the theory of eigenvarieties. The first two questions concern the relationship between automorphic representations and eigenvarieties, and the third concerns that between $p$-adic Galois representations and eigenvarieties. Are there known examples/a general theory for functoriality of eigenvarieties not simply arising from known cases of functoriality for automorphic representations, and would functoriality at the level of eigenvarieties get us any closer to the global conjectures? An automorphic representation $\pi$ on an adelic group $G(\mathbb{A}_F)$ has attached to it a pair $(\{A(\pi_{\mathfrak{p}})\}, {}^{\lambda}H_{\pi})$, consisting of the set of Frobenius-Hecke conjugacy classes $\{A(\pi_{\mathfrak{p}})\}$ contained in some subgroup $^{\lambda}H_{\pi}\subset {}^L G$ of the L-group, which is a reductive algebraic group over $\mathbb{C}$. This pair should be equal to the Frobenius-Hecke conjugacy classes of the corresponding motive, $M$, along with the reductive algebraic group $ ^{\mu}H_M$, attached by the Tannakian theory. In the world of Galois $p$-adic representations we get Frobenius conjugacy classes in $p$-adic groups. For the points on the eigenvariety over a prime $\mathfrak{q}$, do we attach conjugacy classes in some reductive algebraic group over $F_{\mathfrak{q}}$, in comparison with those attached to the corresponding $\mathfrak{q}$-adic Galois representation? If this is the case, are we conjecturing some rationality results on $ ^{\lambda} H_{\pi}$, or just on the classes $\{A(\pi_{\mathfrak{p}})\}$, so that they can be considered in the $p$-adic world? I suppose this is a generalisation on the results of Shimura on the finite dimension of the Hecke field... Where do we stand on the relation between eigenvarieties and deformation spaces for $p$-adic Galois representations? I suppose these are in the area of the $ \mathcal{R}=\mathbb{T}$ theorems. Now that we have constructed some (all?) eigenvarieties, do we know that their dimensions of these varieties match with the Galois side? Can we construct the $p$-adic Langlands correspondence in families even for $GL_1$? Apologies for the broad question, I am not very familiar with the area. Thank you in advance! REPLY [4 votes]: You have asked a lot of questions at once, and it is impossible to give more than a hint at a small subset of these questions. I think the general theme here is: the existence of eigenvarieties doesn't "create information from nowhere" about the core questions of global Langlands (functoriality and reciprocity); but it allows you to "move information from place to place" -- allowing you to propagate instances of functoriality/reciprocity from some restricted class of automorphic forms to a bigger class. (1) All of the cases of functoriality of eigenvarieties I'm aware of take some kind of "classical" functoriality as an input (proved either via classical automorphic methods, e.g. comparison of trace formulas, converse theorems, etc, or by fancier methods coming from modularity lifting etc). You can sometimes get new, purely p-adic functoriality results out of the combination of classical functoriality + eigenvariety deformation; e.g. there are some (rather fragmentary) results by Kilford giving a p-adic Jacquet--Langlands correspondence for weight 1 modular forms, which is a purely p-adic thing not seen on the classical side. But it's maybe not terribly satisfying because you don't actually get much information about the resulting maps. (2) What you're describing -- the collection of Frobenius conj classes -- is a sort of "skeleton" on which you can build $\ell$-adic Galois reps for all $\ell$. However, you wouldn't expect the $\ell$-adic guys to deform over eigenvarieties for all $\ell$. As you correctly guessed, it is only the p-adic Galois reps which deform over the p-adic eigenvariety -- the $\ell$-adic reps for $\ell \ne p$ don't deform well. The cases where we can construct eigenvarieties are indeed a subset (probably strict!) of the cases where we know some kind of finite generation of Hecke eigenvalue fields a la Shimura, so it makes sense to consider these objects p-adically. (3) Eigenvarieties actually play a rather crucial role in lots of work on the "reciprocity" side of global Langlands (going back + forth between Galois and automorphic objects). The general theme here is that one constructs Galois reps attached to automorphic forms of "sufficiently regular weight" using etale coh. of Shimura varieties, and then uses deformation over an eigenvariety to fill in the "bad" weights where the etale coh argument breaks down (e.g. non-regular limits of discrete series, like weight 1 modular forms for GL2). See e.g. Chenevier's article in the Paris Book Project. However, it's important here to be aware that the eigenvariety can't see all automorphic forms, it only detects those $\pi$ for which $\pi_{\mathfrak{p}}$ for $\mathfrak{p} \mid p$ are "finite-slope", i.e. subquotients of principal series (and you need to make a choice of one among the Weyl-group orbit of characters from which $\pi_{\mathfrak{p}}$ is induced, which we call a "p-refinement" or "p-stabilisation"). This needs to be reflected on the Galois side; so an eigenvariety should match up with a slightly modified deformation space for p-adic Galois reps -- parametrising Galois reps with some additional "triangulation" data. However, we're rather a long way from proving this for general reductive groups; for $GL_2 / \mathbf{Q}$ there is an almost complete (but not 100% complete) picture but things get very hard very quickly once you step away from that. (For GL1, all representations are vacuously finite-slope, so that case is a little misleadingly easy.) (Footnote: I think we're a long way from constructing "all eigenvarieties" just yet! We can pick up those $\pi$ which have non-vanishing $(\mathfrak{g}, K)$-cohomology, or those which have non-vanishing $(\mathfrak{p}, K)$-cohomology if $G$ admits a Shimura variety; but that leaves out lots of interesting automorphic reps which we have no way of seeing at present.)<|endoftext|> TITLE: Characterization of the family of simple groups PSL(2,q) by tensor multiplicity QUESTION [9 upvotes]: Let $G$ be a finite group and $(\chi_i)$ its irreducible characters. Then $\forall i,j,k, \exists!n_{i,j}^k \in \mathbb{N}_{\ge 0}$ such that $$\chi_i \chi_j = \sum_k n_{i,j}^k \chi_k.$$ Let the tensor multiplicity of $G$ be $m(G):= \max_{i,j,k} (n_{i,j}^k)$; for ex. $m(A_5) = 2$ and $m(A_6) = 3$. Theorem: Let $q > 5$ be a prime-power, then $$m(\mathrm{PSL}(2,q)) = \left\{ \begin{array}{ll} 2 & \text{ if } q \text{ even,} \\ 3 & \text{ if } q \text{ odd.} \end{array} \right.$$ proof: see (for example) the generic computation of $(n_{i,j}^k)$ for $\mathrm{PSL}(2,q)$ in my talk at 30:00. $\square$ This post is about the converse of above theorem. Let $G$ be a non-abelian finite simple group with $m(G) \le 3$. Question: Is it true that $G \simeq \mathrm{PSL}(2,q)$ for some prime-power $q$? It was checked by GAP for $|G|<10^7$ (see Appendix). Remark: The classification below suggests that the family $(\mathrm{PSL}(2,q))$ is the only one, among the usual infinite families of (non-abelian) finite simple groups, where the tensor multiplicity is bounded above. This would provide an other interesting characterization of this family. Appendix We improved our way to check as suggested by Mikko Korhonen in comment. Let $G$ be a finite group, $d_+(G)$ be the maximum among the degrees of complex irreducible characters of $G$, and $\Sigma(G)$ their sum. They are exactly $27$ non-abelian finite simple groups $|G|<10^7$ not isomorphic to some $\mathrm{PSL(2,q)}$. The classification below lists them ordered according to $d_+^2(G)/\Sigma(G)$, which is less than or equal to $m(G)$. Computation Each element of the list reads as $d_+^2(G)/\Sigma(G)$, its numerical approximation, $G$ and $|G|$. gap> classification(10000000); [ [ 256/63, 4.06349, PSU(3,3), 6048 ], [ 1125/208, 5.40865, PSU(3,4), 62400 ], [ 13/2, 6.5, PSL(3,3), 5616 ], [ 18432/2107, 8.74798, PSU(3,7), 5663616 ], [ 175/18, 9.72222, A7, 2520 ], [ 2048/203, 10.0887, PSL(3,4), 20160 ], [ 175/16, 10.9375, A8, 20160 ], [ 279/25, 11.16, PSL(3,5), 372000 ], [ 729/64, 11.3906, PSp(4,3), 25920 ], [ 3025/218, 13.8761, M11, 7920 ], [ 8281/484, 17.1095, Sz(8), 29120 ], [ 10368/553, 18.7486, PSU(3,5), 126000 ], [ 7225/271, 26.6605, PSp(4,4), 979200 ], [ 321489/11210, 28.6788, PSU(3,8), 5515776 ], [ 43681/1464, 29.8367, J_1, 175560 ], [ 704/21, 33.5238, M12, 95040 ], [ 23328/683, 34.1552, A9, 181440 ], [ 5472/155, 35.3032, PSL(3,7), 1876896 ], [ 28224/709, 39.8082, J_2, 604800 ], [ 16384/319, 51.3605, PSp(6,2), 1451520 ], [ 321489/5356, 60.0241, A10, 1814400 ], [ 152100/2519, 60.3811, PSp(4,5), 4680000 ], [ 21175/258, 82.0736, M22, 443520 ], [ 173056/1875, 92.2965, G(2, 3), 4245696 ], [ 20800/207, 100.483, PSL(4,3), 6065280 ], [ 775/7, 110.714, PSL(5,2), 9999360 ], [ 28672/227, 126.308, PSU(4,3), 3265920 ] ] Below is the alternative classification suggested by Goeff Robinson in comment, considering $d_+(G)/c(G)$, with $c(G)$ the class number of $G$: gap> classification2(10000000); [ [ 16/7, 2.28571, PSU(3,3), 6048 ], [ 13/4, 3.25, PSL(3,3), 5616 ], [ 75/22, 3.40909, PSU(3,4), 62400 ], [ 35/9, 3.88889, A7, 2520 ], [ 81/20, 4.05, PSp(4,3), 25920 ], [ 5, 5., A8, 20160 ], [ 11/2, 5.5, M11, 7920 ], [ 31/5, 6.2, PSL(3,5), 372000 ], [ 32/5, 6.4, PSL(3,4), 20160 ], [ 192/29, 6.62069, PSU(3,7), 5663616 ], [ 91/11, 8.27273, Sz(8), 29120 ], [ 72/7, 10.2857, PSU(3,5), 126000 ], [ 176/15, 11.7333, M12, 95040 ], [ 12, 12., A9, 181440 ], [ 340/27, 12.5926, PSp(4,4), 979200 ], [ 209/15, 13.9333, J_1, 175560 ], [ 16, 16., J_2, 604800 ], [ 256/15, 17.0667, PSp(6,2), 1451520 ], [ 81/4, 20.25, PSU(3,8), 5515776 ], [ 228/11, 20.7273, PSL(3,7), 1876896 ], [ 390/17, 22.9412, PSp(4,5), 4680000 ], [ 189/8, 23.625, A10, 1814400 ], [ 385/12, 32.0833, M22, 443520 ], [ 1040/29, 35.8621, PSL(4,3), 6065280 ], [ 832/23, 36.1739, G(2, 3), 4245696 ], [ 224/5, 44.8, PSU(4,3), 3265920 ], [ 1240/27, 45.9259, PSL(5,2), 9999360 ] ] Code classification:=function(n) #for Geoff way: classification2:=function(n) local it,LL,g,A,L,l,dmax,S,c,cc; it:=SimpleGroupsIterator(2520,n);; #2520=|A_7| LL:=[];; for g in it do A:=StructureDescription(g);; if Length(A)<6 or List([1..6],i->A[i])<>"PSL(2," then #PSL(2,q) excluded L:=CharacterDegrees(g);; l:=Length(L);; dmax:=L[l][1];; S:=Sum(L,i->i[1]*i[2]); #Goeff way: Sum(L,i->i[2]); cc:=dmax^2/S;; Add(LL,[cc,Float(cc),g,Order(g)]);; fi; od; Sort(LL); return LL; end;; Recall that $A_7$ is the smallest non-abelian finite simple group not isomorphic to some $\mathrm{PSL}(2,q)$. REPLY [5 votes]: The check of $\textrm{max}\{\textrm{deg}(\chi)\mid \chi \in \textrm{Irr}(G)\} > 3 k$ could be extended to more finite simple groups (including all sporadics) by using the character tables which are available in GAP: simpnames:= AllCharacterTableNames( IsSimple, true, IsAbelian, false); LL := [];; for nam in simpnames do Print(nam," "); t := CharacterTable(nam); d := Maximum(List(Irr(t),Degree)); k := NrConjugacyClasses(t); Print([d,k,Float(d/k)],"\n"); Add(LL, [d/k,Float(d/k),nam, Size(t),d,k]); od; Sort(LL, {a,b} -> a[4] <= b[4]); Filtered(LL, a-> a[1] <= 3); From the result I would guess the following statement: G is non-abelian simple with max. character degree $\leq 3k$ iff G = PSU(3,3) or PSL(2,q). Sketch of proof: G sporadic: checked above (GAP, resp. ATLAS) G alternating: (note that $A_5\cong PSL(2,4)$ and $A_6 \cong PSL(2,9)$ are among the exceptions) Symmetric group $S_n$ has representation for partition (2,2,2, ...., 2,2 or 1) and the hook formula and Stirling approximation of $n!$ yield that its degree grows with $n$ like $e^n$, while the number of conjugacy classes of $S_n$ (= number of partitions of $n$) grows like $e^{const \sqrt{n}}$. (The numbers for the alternating group differ at most by a factor $2$). G of Lie type of rank l: they always have the Steinberg representation of degree $q^{N}$ where $N$ is the number of postive roots of $G$, which is roughly $N \sim l^2$ their number of conjugacy classes is bounded by $< c q^l$ for some constant c (so only rank $l=1$, $PSL_2(q)$, or very small $l$ and $q$ will give exceptions) So, asymptotically the statement looks ok, one needs to be a bit more precise to get the list of small exceptions.<|endoftext|> TITLE: Isn't there an algebraic topology criterion for embeddability of an *abstract* simplicial complex in a Euclidean space of *specified* dimension? QUESTION [6 upvotes]: Some textbooks on algebraic topology define a simplicial complex as a special kind of subset of a Euclidean space (e.g., A. Wallace, p.3). Others define an abstract simplicial complex as a special kind of combinatorial structure (e.g., E. H. Spanier, p. 108). There exists a functor from the category of abstract simplicial complexes and simplicial maps to the category of topological spaces and continuous functions. My question is, what is an algebraic criterion for deciding whether or not there exists an embedding of the realization of an abstract simplicial complex into a Euclidean space of specified dimension. For example (J. R. Munkres, p. 18) exhibits an abstract simplicial complex whose realization is a Klein bottle, which cannot be embedded in $R^3$. If a concrete 2-dimensional finite simplicial complex consists of point, line segment, and triangular subsets of 3-dimensional Euclidean space, then obviously there is more information about these simplices and their relationships than is contained in the corresponding abstract simplicial complex. In particular, there are the lengths of the line segments, the angles between connected line segments (one shared point), and the dihedral angles between connected triangles (one shared edge). This is a finite collection of data. My actual motivation for the titular question is the following question. Given an abstract finite simplicial complex together with "lengths" of 1-dimensional abstract simplices, "angles" between connected 1-dimensional abstract simplices, and "dihedral angles" between connected 2-dimensional abstract simplices, is there an algorithm to decide whether there is an embedding of its simplicial realization in 3-dimensional Euclidean space? REPLY [9 votes]: It is not exactly clear to me what you mean by an algebraic criterion. But Uli Wagner and Martin Tancer have, with various coauthors and in several papers, tried to give an answer to the question of deciding embeddability, see for instance Filakovský, Wagner, and Zhechev - Embeddability of Simplicial Complexes is Undecidable or de Mesmay, Rieck, Sedgwick, and Tancer - Embeddability in $\mathbb R^3$ is NP-hard. Of course, they are more interested in the complexity aspect, but an intermediate step is always finding a criterion for embeddability, and then checking it.<|endoftext|> TITLE: Is the complex structure of $\mathbb CP^n$ unique? QUESTION [22 upvotes]: Let $\mathbb CP^n$ denotes the complex projective space of dimension $n$, we have a standard complex structure of $\mathbb CP^n$, and my question is: is this complex structure unique? Or equivalently, let $X$ be a complex manifold diffeomorphic to $\mathbb CP^n$, is $X$ biholomorphic to $\mathbb CP^n$? What I know is from p45 of Morrow&Kodaira's book 《complex manifolds》: $\mathbb CP^n$ is rigid. But this fact only ensures that small deformations don't change the complex structure of $\mathbb CP^n$, we did not even know whether the large deformations change the complex structure of $\mathbb CP^n$, or more generally, whether the same diffeomorphic type of $\mathbb CP^n$ admits different complex structures? For dimension 1, I have learnt from some book that the answer is yes. For dimension 2, cited form Yau's 1977 paper 《Calabi's conjecture and some new results in algebraic geometry》, as a corollary of Yau's solution of Calabi's conjecture, the complex structure of $\mathbb CP^2$ is unique. But for higher dimensions, is this problem solved? or any progress has been made? REPLY [10 votes]: I would also like to mention an interesting related result of T. Fujita (which is not cited in the referenced survey article). "On topological characterizations of complex projective spaces and affine linear spaces", Proc. Japan Acad. Ser. A Math. Sci. 56 (1980), no. 5, 231–234. Theorem: Let X be a smooth Fano n-fold with cohomology ring isomorphic to $H^{*}(\mathbb{CP}^n,\mathbb{Z})$ and $n \leq 5$. Then $X \cong \mathbb{CP}^n$.<|endoftext|> TITLE: Milnor lattice and Du Val singularity QUESTION [8 upvotes]: I am reading this paper: https://arxiv.org/abs/0810.2687 by A. J. de Jong and Robert Friedman. In the proof of Theorem 4.10, a singularity of the following type shows up $$y^2=x^3+z^{6d-1}.$$ When $d=1$, this is exactly the type of $E_8$ in Du Val singularities. And the Milnor lattice and Dynkin diagram can be obtained by blowing up. In general, looks like the corresponding Milnor lattice is isomorphic to $(2d-2)U\oplus d(-E_8)$. I am wondering how to verify this in this case. In particular, can this be deduced from the case when $d=1$? Thanks! REPLY [3 votes]: For singularities of the form $\{z^d = f(x,y)\}$ (which, I believe, are called suspension singularities), the Milnor fibre $M$ has a nice topological description as the $d$-fold cover of $B^4$ branched over the Milnor fibre $F \subset B^4$ of the curve singularity $\{f(x,y) = 0\}$ at $(0,0)$. Now there are several ways of computing the signature from here: Lefschetz fibrations: there's a Lefschetz fibration on $B^4$ with page $F$ and a monodromy $\mu = \delta_1 \circ \dots \circ \delta_m$ factorisation as a product of (right-handed) Dehn twists. Now $M$ has a Lefschetz fibration with fibre (diffeomorphic to) $F$ and monodromy (conjugated to) $\mu^d$, factored as $(\delta_1\dots\delta_m)^d$. This factorisation gives (with a bit of work) an explicit basis of the second homology and its intersection form. handle diagrams: $F$ is obtained by pushing in a surface in $S^3$ whose boundary is a link. Akbulut and Kirby give you a recipe to build a handle decomposition of the $d$-fold cover of $B^4$ cyclically branched over such objects, which in turn gives a basis for $H_2$ and the intersection form in that basis. Levine–Tristram signatures: as explained in Kauffman's book On knots, the Levine–Tristram signatures of a knot $K \subset S^3$ compute the signature of cyclic cover of $B^4$ branched over a surface whose boundary is $K$. (More precisely, you need to look at signatures at the $d$-th roots of unity to know about the $d$-fold cover. In the case at hand, we also get to compute the rank of $H_2$ (which is $(d-1)b_1(F)$). If we know that the link of the singularity is a homology sphere (as in the case at hand), usually the classification of unimodular indefinite forms is enough to conclude. To stay in the algebro-geometric context, Durfee's paper The signature of smoothings of surface singularities seems relevant. In the case of your singularities, the first and third approach are quite easy to implement. For the monodromy approach, it is convenient to look at the singularity as a $6k-1$-fold suspension (i.e. $d = 6k-1$, $f(x,y) = x^2+y^3$); here $F$ has genus 1 and the factorisation is the product of two Dehn twist along two curves generating $H_1(F)$. For the signature approach, either of the three choices ($d = 2, d = 3, d = 6k-1$) is feasible; I'd say that the easiest is looking at $d=6k-1$, and then you have to compute signatures of the torus knots $T(2,3)$, which is quite easy. (For instance, this might be done explicitly in Lickorish's An introduction to knot theory.)<|endoftext|> TITLE: Case study: what does it take to formulate and prove Quillen's small object argument in ZFC? QUESTION [9 upvotes]: I'm getting a bit lost over at Peter Scholze's interesting question about removing the dependence on universes from theorems in category theory. In particular, I'm being forced to admit that I don't really know when replacement is being invoked, never mind when it is invoked "in an essential way". So I'd like to work through a reasonably concrete example of the phenomenon. I understand that replacement should "really" be thought of as the axiom which allows for transfinite recursion. My sense is that category theory tends not to use recursion in a heavy-duty way (although, more so than other branches of mathematics, it does have plenty of definitions which at least prima facie have nontrivial Levy complexity. For instance, I think the formula $\phi(x,y,z,p,q)$ saying that the set $z$ and functions $p: z \to x$ and $q: z \to y$ are a categorical product of the sets $x,y$ is syntactically $\Pi_1$, and the statement that binary products exist in the category of sets is syntactically $\Pi_3$ (ignoring bounded quantifiers of course)). The following theorem is, I think, one of the notable exceptions to the category-theoretic-non-use-of-recursion: Theorem [Quillen] "The small object argument": Let $\mathcal C$ be a locally presentable category, and let $I \subseteq Mor \mathcal C$ be a small set of morphisms. Let $\mathcal L \subseteq Mor \mathcal C$ be the class of retracts of transfinite composites of cobase-changes of coproducts of morphisms in $I$, and let $\mathcal R \subseteq Mor \mathcal C$ comprise those morphisms weakly right orthogonal to the morphsims of $I$. Then $(\mathcal L, \mathcal R)$ is a weak factorization system on $\mathcal C$. For the proof, see the nlab. Basically, factorizations are constructed by transfinite recursion. The recursion seems "essential" to me because new data is introduced at each stage of the construction. Formalization: I think this theorem and its proof are straightforwardly formalizable in MK, where the category-theoretic "small/large" distinction is interpreted as MK's "set/class" distinction. I don't feel qualified to comment on whether the proof works in NBG, but the statement at least makes sense straightforwardly. When it comes to formalizing in ZFC, we have choices to make regarding the small/large distinction: One option is to introduce a "universe" $V_\kappa$ (which, if we're actually trying to work in ZFC, will be a weaker sort of universe than usual). We'll interpret "small" to mean "in $V_\kappa$". We won't consider "truly large objects" -- everything we talk about will be a set -- in particular, every category we talk about will be set-sized, even if not "small" per se. We'll interpret "locally presentable category" to mean "$\kappa$-cocomplete, locally $\kappa$-small category with a strong $\kappa$-small, $\lambda$-presentable generator for some regular $\lambda < \kappa$" (I don't know if it makes a difference to say that $V_\kappa$ thinks $\lambda$ is a regular cardinal). Another option is to not introduce any universe, and just interpret "small" to mean "set-sized". In this case, any "large" object we talk about must be definable from small parameters. So we define a category to comprise a parameter-definable class of objects, a parameter-definable class of morphisms, etc. This might seem restrictive, but it will work fine in the locally presentable case, since we can define a locally presentable category $\mathcal C$ to be defined, relative to parameters $(\lambda, \mathcal C_\lambda)$ (where $\lambda$ is a regular cardinal and $\mathcal C_\lambda$ is a small $\lambda$-cocomplete category), as the category of $\lambda$-Ind objects in $\mathcal C_\lambda$. Now, for the theorem at hand, approach (2) seems cleaner because the necessary "tranlsation" is straightforward, and once it is done, the original proof should work without modification. I think the main drawbacks of (2) come elsewhere. For instance it will probably be a delicate matter to formulate theorems about the category of locally presentable categories. In general, there will be various theorems about categories which have clean, conceptual formulations and proofs when the categories involved are small, but which require annoying technical modifications when the categories involved are large. It's for such reasons that approaches more like (1) tend to be favored for large-scale category-theoretic projects. So let's assume we're following approach (1). The question then becomes: Question 1: Exactly what kind of universe do we need to formulate and prove the above theorem following approach (1)? Question 2: How many such universes are guaranteed to exist by ZFC? Presumably, the answer to Question 2 will be that there are a lot of such universes -- enough so that we can do things like, given a category, pass to a universe large enough to make that category small and invoke the theorem for that universe. Question 3: How far into the weeds must we go to answer Questions 1 and 2? Do we have to analyze the proof of the Theorem in a deep way? Is there an easy rubric of criteria which allow us to glance at the proof and, for 99% of theorems like this, easily say that it "passes" without delving into things too much? Or is there even some formal metatheorem we can appeal to such that even a computer could check that things are fine? REPLY [2 votes]: Jacob Lurie's comment gives an answer to Question 1. Namely, assuming that the estimates I gave in my comment are correct, in order to formulate and prove the theorem it will suffice to suppose that $\kappa$ is regular and that for every $\mu < \kappa$, there exists $\rho < \kappa$ such that $\mu \ll \rho$ (meaning that $\mu' < \mu, \rho' < \rho \Rightarrow (\rho')^{\mu'} < \rho$). Perhaps this property of $\kappa$ might be viewed as a "form" of replacement. But really, what we have is two conditions on $\kappa$ which are purely set-theoretic rather than metamathematical, so that the answer to Question 1 is something much cleaner than I had supposed. This allows us to address Question 2. Presumably, the result is that ZFC proves that there are lots and lots of $\kappa$ satisfying the two conditions above. When it comes to Question 3, it would seem that in this approach we do in fact need to delve pretty deeply into the proof. In fact, it seems that in order to carry out this approach, we must add some genuine mathematical content to the proof, and actually prove a stronger statement. The further questions then become Will it generally be possible to "constructivize" "most" category-theoretic theorems in this way, or will other issues show up in the course of the "ZFC-ify category theory" project? If the answer to (1) is "yes" (or if it's generally "no" and we restrict our attention to the cases where it's "yes"), then "how much extra work" would such a project really be? My guess is that the answer to (1) is that when it comes to the use of transfinite recursion in category theory, it will indeed typically be the case that the use of replacement can be eliminated in a similar way to this, but that more importantly I've missed the point: as Jacob Lurie argues in response to Peter Scholze's question, the thornier issues with ZFC-ifying category theory are not to do with transfinite recursion but rather with being able to freely go back and forth between "big categories" and "little categories" in various ways. My guess is that the answer to (2) is that for "most" category-theoretic uses of transfinite recursion, it should actually be pretty straightforward to "constructivize" them so that they fit in a "baby universe" with the properties above or something similar, and that with just a little bit of practice, one could develop the ability to verify almost at a glance that it's possible, though still on a theorem-by-theorem basis. But I'd love to be proven wrong and shown a theorem in category theory where this sort of approach fails! Finally, this leaves as an open question whether there's a "more automatic" way of doing all of this -- perhaps with a weaker conclusion than "our universe need not satisfy any form of replacement at all".<|endoftext|> TITLE: Higher homotopy groups of irreducible 3-manifolds QUESTION [6 upvotes]: A 3-manifold $M$ is irreducible if every embedded 2-sphere bounds a 3-ball. Thanks to Papakyriakopoulos's sphere theorem, irreducibility is the same as having $\pi_2(M)=0$. Does irreduciblity imply that the manifold is in fact aspherical, i.e. that $\pi_k(M)=0$ for all $k \geq 2$? (Or maybe I should say that the universal cover $\tilde M$ is aspherical, but the question is the same in terms of homotopy groups.) As pointed out by @Matt Zaremsky in the comments, there is an obvious counterexample in $S^3$. But perhaps this is the only counterexample, or the counterexamples are easy to classify? Given all of the tools we have about geometric classification of 3-manifolds, I expect someone would have a quick answer. I'm just not enough of an expert to make those arguments myself. REPLY [12 votes]: An irreducible 3-manifold $M$ is aspherical if and only if it's not a finite quotient of $S^3$, which in turn is equivalent to having infinite fundamental group. Essentially you've already outlined the proof: the universal cover $\tilde M$ is a simply-connected 3-manifold with trivial $\pi_2$, and so also $H_2(\tilde M) = 0$; if $\tilde M$ is not compact, then $H_3(\tilde M) = 0$ (because of non-compactness) and $H_k(\tilde M) = 0$ for higher $k$ (because it's a 3-manifold), so by the Hurewicz theorem $\tilde M$ is aspherical. On the other hand, $\tilde M$ is compact if and only if $\pi_1(M)$ is finite. In this case, Perelman proved that $\tilde M$ is $S^3$. 3-manifolds covered by $S^3$ are classified, and they correspond to finite subgroups of $SO(4)$. I think that Scott's The geometries of 3-manifolds has a precise statement. REPLY [7 votes]: It's wrong for finite fundamental group, as then the universal cover is closed and has nonvanishing $\pi_3$ by Hurewicz. It's true for infinite fundamental group, again by Hurewicz applied to the universal cover.<|endoftext|> TITLE: Can you cover a genus a billion hyperbolic surface with 15 balls? QUESTION [10 upvotes]: Here's a question I was wondering about this week. Not sure how interesting it is, but I thought it was kind of curious. Question: Given $k$, is there a number $N=N(k)$ such that if a closed orientable hyperbolic surface X is the union of at most $k$ embedded metric balls, then the genus of $X$ is at most $N$? (Here, an embedded metric ball is an pathwise isometric embedding of a ball in $\mathbb H^2$.) Some half baked comments: It's well known that any $X$, of any genus, can be covered by $3$ topological balls. You can convince yourself of this pretty easily: draw a couple embedded arcs that cut $X$ into disks, thicken them to get the first two balls, and then make the third ball by taking all the complementary disks and connecting them together in the pattern of some sort of dual spanning tree. So, the metric content above is essential. If you make a graph G by taking a vertex for every ball and an edge for every connected component of intersection of two balls, then $\pi_1 G $ will surject on $\pi_1 X$. So, if you could bound the number of connected components of the intersection of each pair of balls, you can bound the genus. However, I think two balls in a hyperbolic surface of genus g can have something on the order of g connected components, which isn't helpful. For an example, take a right angled regular polygon and double it along every other side. This gives a surface with boundary, and then you can glue on whatever to make it closed. If you take the largest embedded balls around the two copies of the center of the polygon, they'll have a component of intersection for every edge of the polygon. Another reason you're never going to get a universal bound on the number of connected components of intersections is that this would give you that $N(k)$ is linear in $k$, which I think is impossible. Namely, there are some examples out there of genus g surfaces where the injectivity radius is everywhere at least $a \log g$, for some fixed constant a>0. (See e.g. https://arxiv.org/pdf/math/0505007.pdf, where they give examples with $a=2/3$, although their examples may have cusps, I haven't checked. There are other closed examples too though.) You can then construct a maximal set $S $ of points in $X$ such that the distance between any two points is at least $a \log g$. The balls of radius $a/2 \log g$ around the points of S are disjoint and have volume at least $g^{a/2}/100$ or something, so there are at most $1000 g^{1-a/2}$ points in $S$, by Gauss Bonnet. By maximality of S the radius $a \log g$ balls around the points of S cover X. They are embedded, but there are only around $g^{1-a/2}$ of them. That is, you can cover a genus $g$ surface with a number of balls that's sublinear in $g$. REPLY [4 votes]: Consider a regular hyperbolic $4n+2$-gon with vertex angle of genus $3$. I claim we can glue $3$ copies of this polygon together to form an oriented hyperbolic surface of genus $n$, without gluing any polygon to itself. This will form a decomomposition of the surface with $4n+2$ vertices, $6n+3$ edges, and $3$ faces, for an Euler characteristic of $4n+2 + 3 - (6n + 3 ) = 2-2n$. I will describe it by first describing the vertices, then the edges, then the cyclic ordering of the edges at each vertex, and finally the faces. The vertices are indexed by the numbers $0$ to $4n+1$, mod $4n+2$. There are edges from vertex $i$ to vertices $i+1, i-1, i+2n+1$. Thus each vertex has three edges. The three edges leaving $i$ towards vertices $i+1, i-1, i+2n+1$ are in clockwise order. One face has boundary vertices $0,1,2,3,\dots, 4n+1$ in order (clockwise around the face. Another has boundary the vertices $$ 2n+1 , 2n , 4n+1, 4n, 2n-1, 2n-2, 4n-1, 4n-2, 2n-3,2n-4, \dots, 1,0$$ in order (clockwise around the face). The last one has boundary the vertices $$ 4n+1, 2n, 2n-1, 4n, 4n-1 , 2n-2, 2n-3, \dots, 2n+1,0$$ in order (clockwise around the face). Each edge meets two of the three faces and each vertex meets all three faces. So indeed the polygons glue to a smooth hyperbolic surface. Now take a disc whose center is the center of each polygon, whose radius is at least the distance from the center to a vertex but less than the distance from a center to a vertex plus half the length of an edge). The disc will contain the polygon, hence three will cover the surface, but each disc will only extend a short distance from each polygon into the next and thus will not intersect itself.<|endoftext|> TITLE: Complex-doubly periodic function in two variables? QUESTION [6 upvotes]: I am looking for a function $f:\mathbb C^2 \rightarrow \mathbb C^2$ that satisfies the two equations $$\partial_{z_2}f_1(z_1,z_2) + \partial_{z_1} f_2(z_1,z_2)=0 \text{ and }$$ $$\partial_{\bar z_1}f_1(z_1,z_2) - \partial_{\bar z_2} f_2(z_1,z_2)=0$$ and in addition, is doubly-periodic in both its complex variables $z_1,z_2$. Does such a function exist and if not, why? I would not even know how to start building such a function. In particular, I would like to have $$f_1(z_1+1,z_2)=f_1(z_1,z_2+1)=f_1(z_1,z_2)$$ and $$f_1(z_1+i,z_2) = e^{2\pi i k_1}f_1(z_1,z_2)$$ and $$f_1(z_1,z_2+i) = e^{2\pi i k_2}f_1(z_1,z_2)$$ for some fixed $k_1,k_2 \in \mathbb R.$ Please let me know if you do have any questions. I had some typos in there, but hopefully everything is coherent now. REPLY [4 votes]: The answer is that the only solutions have the form $$ f = (f_1,f_2) = \bigl(c, h(\,\overline{z}_1, z_2)\bigr) $$ where $h:\mathbb{C}^2\to\mathbb{C}$ is holomorphic and $c$ is a constant, which must equal zero unless $k_1$ and $k_2$ are integers. The argument is as follows: The first equation implies that there exists a function $g:\mathbb{C}^2\to\mathbb{C}$ such that $$ f_1 = \frac{\partial g}{\partial z_1} \quad\text{and}\quad f_2 = -\frac{\partial g}{\partial z_2}. $$ Substituting this into the second equation implies that $g$ must satisfy $$ \frac{\partial^2 g}{\partial z_1\partial\overline{z}_1} + \frac{\partial^2 g}{\partial z_2\partial\overline{z}_2} = 0. $$ In other words $g$ is a harmonic function on $\mathbb{C}^2$. Since $g$ is harmonic, so is its derivative with respect to $z_1$, i.e., $f_1$. The periodicity conditions imposed on $f_1$ imply that $f_1$ is bounded, and a bounded harmonic function on $\mathbb{C}^2$ is constant. Thus, $f_1 = c$ for some constant $c\in\mathbb{C}$. Obviously, $c$ must be zero unless $k_1$ and $k_2$ are integers. Since $f_1$ is constant, the given equations on $f_2$ reduce to $$ \frac{\partial f_2}{\partial z_1} = \frac{\partial f_2}{\partial\overline{z}_2} = 0. $$ Hence $f_2 = h(\overline{z}_1,z_2)$ for some homorphic function $h:\mathbb{C}^2\to\mathbb{C}$. Remark: It wasn't clear from the OP's question whether the OP wanted $f$ to be 'doubly-periodic' or just $f_1$, nor was it clear exactly what the OP meant by 'doubly-periodic' because, normally, the 'doubly-periodic' condition wouldn't have the exponential factors in its definition.<|endoftext|> TITLE: Conjugacy classes of monoids II: Abelianising a monoid, wrongly QUESTION [5 upvotes]: $\newcommand{\unsim}{\mathord{\sim}}$Let $G$ be a group. What is $$ G/\left(ab\sim ba\ \middle|\ a,b\in G\right)? $$ Answer: not $G^{\mathrm{ab}}$, but the set of conjugacy classes of $G$. When passing to monoids, the situation gets more complicated: the equivalence relations generated by the following declarations are all equivalent for $M$ a group, but not for $M$ a monoid: (1) Conjugacy. $a\sim_{1}b$ iff there exists an invertible $g\in M$ such that $gag^{-1}=b$; (2) Conjugacy, II. $a\sim_{2}b$ iff there exists a non-necessarily invertible $m\in M$ such that $ma=bm$; (3) "Commutativity". $ab\sim_{3}ba$ for each $a,b\in M$; In an MO answer, Tom Leinster explained how to describe $M/\unsim_{1}$ and $M/\unsim_{2}$: First, let $[\mathbb{N},M]$ be the category where $\mathrm{Obj}([\mathbb{N},M])=M$; A morphism $m\longrightarrow m'$ is an element $g$ of $M$ such that $gm=m'g$. Then we have bijections $$ \begin{align*} M/\unsim_{1} &\cong [\mathbb{N},M]/\{\text{isos}\},\\ M/\unsim_{2} &\cong \pi_{0}([\mathbb{N},M]). \end{align*} $$ Question: Given a monoid $M$, what is the set $M/\unsim_{3}=M/\left(ab\sim ba\ \middle|\ a,b\in M\right)$? REPLY [7 votes]: Defining conjugacy for monoids is a dicey subject because many different notions that are equivalent for groups are different for monoids and it is not clear which of these is interesting. The one you call 3 is probably the most commonly studied one, although it varies depending on the context how useful its. I am not sure of a category theoretic description of the third relation because I am semigroup theorist and not a category theorist but I will say that if you move to representations or algebras, then it is one of the more natural notions. This is because any trace on the monoid algebra factors through the quotient of $KM$ by this relation. If $K$ is a field, then the equivalence classes of $\sim_3$ form a basis for the $0$-Hochschild cohomology $HH_0(KM)$ and that seems to me a good reason already to think about it. For finite von Neumann regular monoids, one has that two elements are equivalent under $\sim_3$ if and only if all complex characters of the monoid agree on them. This is not true for finite monoids in general, which have an extra relation that you need to add.<|endoftext|> TITLE: Homology of the free loop space of generalized flag varieties QUESTION [5 upvotes]: Is it known whether for a generalized complex flag variety $X$ (that is, $G/P$ for a complex semisimple Lie group $G$ and a parabolic $P$), the homology of the free loop space $H_*(\Lambda X, \mathbb{Q})$ is degree-wise finite-dimensional? Is it known at least for type A (i.e. classical) flag varieties? This is known to be true for some concrete examples, such as complex projective spaces (Ziller et al.), and complete flag varieties of rank 2 (Burfitt-Grbić), but I was unable to find any statements for the general case. REPLY [9 votes]: Serre proved that for any simply-connected $X$, if $X$ has finitely generated homology groups in each degree, then the loop space of $X$ has finitely generated homology groups in each degree. (Proposition 9 of chapter IV of Homologie singulière des espaces fibrés. Applications. Ann. of Math., 54, 1951, p. 425-505.) So the only fact you need about generalized flag varieties is that they are simply-connected compact manifolds.<|endoftext|> TITLE: "Non-categorical" examples of $(\infty, \infty)$-categories QUESTION [25 upvotes]: This title probably seems strange, so let me explain. Out of the several different ways of modeling $(\infty, n)$-categories, complicial sets and comical sets allow $n = \infty$, providing mathematical definitions of $(\infty, \infty)$-categories. I've asked people a few times for interesting examples of $(\infty, \infty)$-categories that could fit into these definitions, and I've always gotten the answer: the $(\infty, \infty)$-category of (small) $(\infty, \infty)$-categories. This is not a bad example, and I think it's cool, but I would like to know what kinds of examples are out there other than just categories of categories. For example, for $(\infty, n)$-categories with $n$ finite, "non-categorical" examples include $(\infty, n)$-categories of bordisms as well as the Morita $(\infty, n)$-category of $E_{n-1}$-algebras in an $(\infty, 1)$-category: people care about bordisms and $E_{n-1}$-algebras before learning that they have this higher-categorical structure. I'm interested in hearing about examples like these for $(\infty, \infty)$-categories. It doesn't matter a lot to me whether something's been rigorously shown to be an example of one of these models or not; and maybe your favorite example is a different kind of $(\infty, \infty)$-category, such as the ones discussed in Theo's question from several years ago; that's also welcome. What would be really neat is an example of a new phenomenon at the $n = \infty$ level, so an example of an $(\infty, \infty)$-category that's not similar to an $(\infty, n)$-category example for any $n$, but that seems like a lot to ask for. In addition to Theo's question that I linked above, this question by Alec Rhea and this question by Giorgio Mossa are also relevant, asking similar questions for $n$ finite. REPLY [10 votes]: As mentioned by Simon Henry: The $(\infty,\infty)$-category of cobordisms. (Not constructed, but if you did it you could presumably have any of the usual bells and whistles you might want.) To clarify Simon Henry's comment: The statement is that that $(\infty,\infty)$-category of cobordisms in the coinductive setting is an $\infty$-groupoid by Cheng's theorem (so it's whatever Thom spectrum you expect by GMTW). In the inductive setting, Cheng's theorem doesn't hold. So non-invertible $(\infty,\infty)$-TFT's should be a thing. I think nobody's formally written down this $(\infty,\infty)$-category -- I assume because $(\infty,n)$-TFTs are hard enough so there's not much demand for it. Please challenge that assumption! One nice thing about complicial sets (and I guess also comical sets) is that they (ought to) naturally put you in the (more general) inductive setting, and you might hope they'd be a good place to construct these (∞,∞)-categories. Anyway, this ticks a few boxes: The inductive / coinductive distinction is arguably a "new phenomenon" (though maybe it's just a "new complication"), and this example already illustrates how it works. It's a super-canonical example, and should be super-interesting for all the reasons its lower brethren are. It's not a category of categories.<|endoftext|> TITLE: Relation between flatness and integrability of an algebraic connection QUESTION [13 upvotes]: Long time listener, first time caller! Suppose that I have a locally free sheaf $\mathcal{E}$ on an smooth algebraic variety $X/k$. Let $\Delta^{(1)}\subset X\times X$ denote the first-order neighbourhood of the diagonal, with projection maps $p_1,p_2:\Delta^{(1)}\to X$. Then there are a few different ways one can describe a connection on $\mathcal{E}$: As a $k$-linear map $D:\mathcal{E}\to\mathcal{E}\otimes\Omega^1_X$ that satisfies the Leibniz rule. As an $\mathcal{O}_X$-linear splitting $s$ of the first jet bundle exact sequence $0 \to \mathcal{E}\otimes\Omega^1_X \to J^1(\mathcal{E})\to \mathcal{E}\to 0$. (Recall that $J^1(\mathcal{E}) = p_{1\ast}p_2^\ast\mathcal{E}$.) As an isomorphism $\phi:p_1^\ast\mathcal{E}\simeq p_2^\ast\mathcal{E}$ which restricts to the identity map on the diagonal $\Delta\subset X\times X$. Passing between these different descriptions of a connection is fairly straightforward: given $D$ one obtains a splitting $s$ by taking $s = D+ j^1$ where $j^1$ is the $k$-linear "jet prolongation" splitting of the jet bundle exact sequence; given a splitting $s$ one obtains an isomorphism $\phi$ by pulling back along $p_1$ and applying the counit of the $(p_1^\ast,p_{1\ast})$ adjunction. So far, so good. Now, there are various types of "integrability" conditions one might be interested in for a connection: Flatness: The connection is flat if the curvature $F(D)\in\Omega_X^2(End(\mathcal{E}))$ vanishes. Integrability: Let $\Delta^{(1)}_3\subset X\times X\times X$ be the first order neighbourhood of the small diagonal with projections $q_1,q_2,q_3$, and let $\phi_{ij}:q_i^\ast\mathcal{E}\simeq q_j^\ast\mathcal{E}$ be the isomorphisms induced by $\phi$. Then the connection is integrable if it satisfies the cocycle condition $\phi_{23}\circ\phi_{12} = \phi_{13}$. Formal lifting: The connection can be lifted to as isomorphism $p_1^\ast\mathcal{E}\simeq p_2^\ast\mathcal{E}$ on every finite order neighbourhood of the diagonal $\Delta^{(n)}$. My question is as follows: What is the relation between these three integrability conditions? I believe I was once told that integrability $\Rightarrow$ formal lifting in characteristic zero, although I've never seen a proof (and have thus far failed to cook one up myself). I naively thought that there would be a straightforward relation between flatness and integrability, but if my calculations are correct the connection $D=d_{dR} + t^1dt^2$ on $\mathbb{A}_k^2$ provides a simple example of an integrable nonflat connection. (Of course, my calculations could always be wrong!) The curvature of a connection lies in $\Omega_X^2(End(\mathcal{E}))$, while the obstruction to lifting from the first- to second-order neighbourhood of the diagonal lives in $H^1(X;\text{Sym}^2(\Omega_X^1)\otimes\mathcal{E}nd(\mathcal{E}))$ (to get the obstruction to lifting from $\Delta^{(n)}$ to $\Delta^{(n+1)}$ replace $\text{Sym}^2$ with $\text{Sym}^{n+1}$). Should I take this as a hint that flatness and formal lifting are unrelated, or is there some sort of relation between the curvature and this lifting obstruction? Edited to add: Since it isn't explicit in the answer below or the comments that follow: the correct relation here is that for $X$ smooth and in characteristic zero, flatness is equivalent to formal lifting together with the cocycle condition. REPLY [6 votes]: (In characteristic zero) Flatness implies the other two definitions; integrability and formal lifting are very weak conditions (in fact if I haven't made a mistake I think this notion of integrability is automatic. As such it seems like a weird definition to me; I usually use the words "integrable" and "flat" as synonyms.) First of all, let me rephrase the notion of flatness a few times. Take $X$ affine for simplicity (to do the general case turn everything into a sheaf, with a little bit of care). Let $\mathcal{D}_X$ be the ring of differential operators on $X$. It can be presented as the universal envelopping algebroid (over $\mathcal{O}_X$) of the Lie algebroid $T_X$ of vector fields. Concretely, this means giving a $\mathcal{D}_X$-module $M$ is equivalent to giving a $k$-linear map $T_X\otimes M\rightarrow M,$ so that for two vector fields $v_1,v_2$ and $m\in M$, we have $v_1\cdot(v_2\cdot m)-v_2\cdot(v_1\cdot m)=[v_1,v_2]\cdot m,$ where $\cdot$ denotes the action of $T_X$ on $M$. I claim that, if $M=\mathcal{E}$, this is exactly the data of a flat connection on $\mathcal{E}$. The map $T_X\otimes M\rightarrow M$ is equivalent to a map $M\rightarrow M\otimes\Omega^1_X$, aka the map defining our connection. The expression $$v_1\cdot(v_2\cdot m)-v_2\cdot(v_1\cdot m)-[v_1,v_2]\cdot m$$ defines a map $T_X\otimes T_X\otimes M\rightarrow M$. It can be checked that this map is $\mathcal{O}_X$-multilinear and antisymmetric on the first two , so it actually lies in $$\operatorname{Hom}_{\mathcal{O}_X}(\Lambda^2T_X\underset{{\mathcal{O}_X}}{\otimes}M,M)\cong\Omega^2_X(\operatorname{End}(M))$$ and you can check that this is the curvature of your connection. So the connection is flat exactly when this gives a $\mathcal{D}_X$-module structure. Let $\hat{\Delta}$ and $\hat{\Delta}_3$ be the formal neighborhoods of $\Delta\subseteq X\times X$ and $\Delta_3\subseteq X\times X\times X$. Then a map $\mathcal{D}_X\underset{\mathcal{O}_X}{\otimes}M\rightarrow M$ is equivalent to the data of an isomorphism $\hat{\phi}:\hat{p}_1^*M\cong\hat{p}_2^*M$ over $\hat{\Delta}$. The action of $\mathcal{D}_X$ on $M$ is compatible with the algebra structure on $\mathcal{D}_X$ if and only if $\hat{\phi}$ satisfies a cocycle condition. In particular we see that flatness implies your other conditions. To prove these statements, the key is the following. A map $\hat{p}_1^*M\rightarrow\hat{p}_2^*M$ is equivalent to a map $M\rightarrow\hat{p}_{1*}\hat{p}_2^*M\cong\mathcal{O}_{\hat{\Delta}}\otimes_{\mathcal{O}_X}M.$ (Note here that we have a left and a right $\mathcal{O}_X$ action on $\mathcal{O}_\hat{\Delta}$. The right one is absorbed into the tensor product, while the left one defines the $\mathcal{O}_X$-module structure on $\mathcal{O}_{\hat{\Delta}}\otimes_{\mathcal{O}_X}M$.) Furthermore, convolution defines a coalgebra structure on $\mathcal{O}_{\hat{\Delta}}.$ If one dualizes this coalgebra, one gets exactly $\mathcal{D}_X$, and you can use this to prove the above interpretation of $\mathcal{D}$-modules. (This is a little painful and has a number of details that need to be checked, so I am avoiding doing it...) Now let me explain why the other conditions are very weak. For formal lifting, as you note, the obstruction lives in $H^1$, so it automatically vanishes if $X$ is affine. I believe it can be a nontrivial condition when $X$ is not affine. On the other hand, here is an argument that all connections are integrable in your sense. We would like to check that the map $\phi_{31}\circ\phi_{23}\circ\phi_{12}$ is the identity over $\Delta_3^{(1)}$. This is true on the diagonal. It is also true on the image of any of the three natural embeddings of $\Delta^{(1)}\rightarrow\Delta_3^{(1)}$. I claim this is sufficient to see that it is the identity over all of $\Delta_3^{(1)}$. Let $N_{123}$ be the normal bundle of $\Delta_3$ inside $X\times X\times X$. We let $N_{12}$ be the subbundle given by the normal vectors inside $\Delta\times X$. Define $N_{13}$ and $N_{23}$ similarly. An endomorphism of the pullback of $\mathcal{E}$ over $\Delta_3^{(1)}$ which restricts to the identity on $\Delta_3$ can be identified with a map $N_{123}\otimes\mathcal{E}\rightarrow\mathcal{E}$, and the original endomorphism is the identity if and only if this resulting map is zero. So we see that the maps $N_{ij}\otimes\mathcal{E}\rightarrow\mathcal{E}$ must be trivial. As the map $N_{12}\oplus N_{23}\oplus N_{31}\rightarrow N_{123}$ is surjective, this implies that $N_{123}\otimes\mathcal{E}\rightarrow\mathcal{E}$ is trivial, as desired.<|endoftext|> TITLE: Ordered Bell numbers QUESTION [12 upvotes]: The ordered Bell numbers (also known as Fubini numbers, sequence A000670 in OEIS) count the number of ordered partitions of an n-element set. Experimentally I have found the following expression for the n-th ordered Bell number $a_n$: $$a_n = \sum_{\sigma \in S_n}\prod_{i=1}^n \binom{i}{\sigma(i)-1}$$ where the sum ranges over all permutations of $\{1,2,\ldots,n\}$. Even though there are $n!$ terms in the sum, only $2^{n-1}$ are non-zero. More generally, letting $S_n^m$ denote the set of permutations of $\{1,2,\ldots,n\}$ with exactly $m$ fixed points, I believe the following is also true: the number of ordered partitions of an n-element set having exactly $m$ blocks of cardinality one is given by $$\sum_{\sigma \in S_n^m}\prod_{i=1}^n \binom{i}{\sigma(i)-1}$$. For example, for $m=0$ the formula appears to yield OEIS sequence A032032. Is this known? Any ideas how to prove it or references to an existing proof? REPLY [4 votes]: I would accept Sam's and lambda's comments as the answer. For the record, I'll just flesh it out a bit for the first formula. In terms of compositions of $n$, the following is all but self-evident $$a_n = \sum_{n_1+n_2+\ldots+n_k=n} \binom{n}{n_k}\binom{n-n_k}{n_{k-1}}\binom{n-n_k-n_{k-1}}{n_{k-2}}\ldots \binom{n-n_k-n_{k-1}-n_{k-2}-\ldots-n_2}{n_1}$$ where the sum is over all compositions of $n$. Now define a mapping from the compositions of $n$ to the permutations $\sigma$ with $\sigma(i) \le i+1$ for $1 \le i \le n$ as follows: Map composition $n_1+n_2+\ldots+n_k = n$ to \begin{align} \sigma(n_1) &= 1\\ \sigma(n_1+n_2) &= n_1+1\\ \sigma(n_1+n_2+n_3) &= n_1+n_2+1\\ &\ldots\\ \sigma(n_1+n_2+\ldots+n_k) &= n_1+n_2+\ldots+n_{k-1}+1\\ \sigma(i) &= i+1\ \text{for}\ i \not\in \{n_1,n_1+n_2,\ldots,n_1+n_2+\ldots+n_k\} \end{align} After checking this mapping is indeed a 1-1 correspondence between the compositions of $n$ and the permutations $\sigma$ with $\sigma(i) \le i+1$, we can now rewrite \begin{align} &\binom{n}{n_k}\binom{n-n_k}{n_{k-1}}\binom{n-n_k-n_{k-1}}{n_{k-2}}\ldots \binom{n-n_k-n_{k-1}-n_{k-2}-\ldots-n_2}{n_1} = \\ &\binom{n_1+n_2+\ldots+n_k}{n_k}\binom{n_1+n_2+\ldots+n_{k-1}}{n_{k-1}} \binom{n_1+n_2+\ldots+n_{k-2}}{n_{k-2}}\ldots \binom{n_1}{n_1} =\\ &\binom{n_1+n_2+\ldots+n_k}{n_1+n_2+\ldots+n_{k-1}} \binom{n_1+n_2+\ldots+n_{k-1}}{n_1+n_2+\ldots+n_{k-2}} \binom{n_1+n_2+\ldots+n_{k-2}}{n_1+n_2+\ldots+n_{k-3}}\\ &\ldots\binom{n_1}{0}=\\ &\binom{n_1+n_2+\ldots+n_k}{\sigma(n_1+n_2+\ldots+n_k)-1} \binom{n_1+n_2+\ldots+n_{k-1}}{\sigma(n_1+n_2+\ldots+n_{k-1})-1}\\ &\binom{n_1+n_2+\ldots+n_{k-2}}{\sigma(n_1+n_2+\ldots+n_{k-2})-1} \ldots \binom{n_1}{\sigma(n_1)-1} = \prod_{i=1}^n \binom{i}{\sigma(i)-1} \end{align} The more general formulas follow by noticing that the mapping from compositions to permutations described above is a bijection between compositions with exactly m ones and permutations with exactly m fixed points.<|endoftext|> TITLE: Explicit different proofs of the same identity type in MLTT QUESTION [7 upvotes]: This question is similar to (but more specific than) this one: When are two proofs of the same theorem really different proofs I do not know very much about homotopy type theory, but I am trying to understand how essentially different proofs of the same identity claim may naturally appear in Martin-Löf's intentional intuitionistic type theory. So, I am looking for an (explicit, as simple as possible) example of a type $A$ with terms $a,b: A$ and identity proofs $p,q: \mathrm{Id}_A(a,b)$ such that the type $\mathrm{Id}_{\mathrm{Id}_A(a,b)}(p,q)$ is empty. (I am not sure I have by this correctly formulated my original intention. Homotopically, I want two non-equivalent paths, but I want it in type theory form.) REPLY [11 votes]: Martin-Löf type theory contains no such type because it is consistent with uniqueness of identity proofs which states precisely that what you are looking for is not there. Martin-Löf type theory is also consistent with the univalence axiom. In the presence of this axiom we may convert the two equivalences $\mathrm{id} : \mathrm{bool} \to \mathrm{bool}$ and $\mathrm{not} : \mathrm{bool} \to \mathrm{bool}$ to terms $\mathrm{ua}(\mathrm{id}), \mathrm{ua}(\mathrm{not}) : \mathrm{Id}_{\mathcal{U}}(\mathrm{bool}, \mathrm{bool})$. Then the identity type $$\mathrm{Id}_{\mathrm{Id}_{\mathcal{U}}(\mathrm{bool}, \mathrm{bool})}(\mathrm{ua}(\mathrm{id}), \mathrm{ua}(\mathrm{not}) )$$ is empty: given any $p$ of this type, we may obtain from it to a proof of equality of $\mathrm{id}$ and $\mathrm{not}$ and from that we would get an element of $\mathrm{Id}_{\mathrm{bool}}(\mathrm{false}, \mathrm{true})$. Another possibility is to use the circle $S^1$ in which case there is no path between $\ell$ and $\ell \circ \ell$, where $\ell : \mathrm{Id}_{S^1}(\mathrm{base}, \mathrm{base})$ is the generating loop.<|endoftext|> TITLE: Can we show that a functor is a fibration without choosing a cleavage? QUESTION [6 upvotes]: Is there a standard method for showing that a functor $F:\mathcal{C}\to\mathcal{D}$ is a fibration, aside from constructing a cleavage? In the proof of the Grothendieck construction, the fibration we obtain from an indexed category $\Psi:\mathcal{B}^{op}\to\mathfrak{Cat}$ is automatically cloven since we're constructing a specific Cartesian arrow $(u,1_{\Psi(u)(Y)})$ for each arrow $u:I\to J\in\mathcal{B}$ and object $(J,Y)\in\int\Psi$ above $J$. Every time I want to show that a functor is a fibration, I end up constructing Cartesian arrows parametrized as above and thusly showing that it's a cloven fibration -- is this by necessity? Any method of showing that a functor is a fibration without choosing a cleavage is welcome, but in particular something similar to the adjoint functor theorem for fibrations would be cool. That is, a statement along the lines of If $F:\mathcal{C}\to\mathcal{D}$ is a functor and $\mathcal{C}$ is blah and $\mathcal{D}$ is bloop and $F$ preserves/reflects blorps then $F$ is a fibration. REPLY [9 votes]: Just as an example, given a category $\mathcal{C}$ with finite limits, showing $\mathrm{cod}\colon \mathcal{C}^\mathbf{2}\to \mathcal{C}$ is a fibration does not involve choosing a cleaving. All that you need is that a pullback square exists for each piece of relevant data. A cleaving would be a specified choice of pullback square for each cospan. More generally, any time that one uses a universal property to show the existence of a cartesian lift, then you aren't exactly constructing a cleaving, since you don't have to specify precisely which item with the universal property you are using. I'm thinking also of the result at the nLab page on Grothendieck fibrations: A functor $p \colon E \to B$ is a cloven fibration if and only if the canonical functor $i \colon E \to B\downarrow p$ has a right adjoint $r$ in $\mathbf{Cat} / B$. where instead of asking that that the adjoint is given, one just has that an adjoint functor theorem is applicable. "Constructing" the adjoint is (probably) equivalent to choosing a cleaving.<|endoftext|> TITLE: Rational homology of $\Omega^2 ( \mathbb CP^n \vee S^d)$ QUESTION [11 upvotes]: What is the rational homology of $\Omega^2 ( \mathbb CP^n \vee S^d)$? Here $\Omega$ denotes based loop space. REPLY [19 votes]: Let us first consider the homotopy fiber of the map $f\colon\mathbb CP^n\vee S^d \to \mathbb CP^\infty$ which is the inclusion on $\mathbb CP^n$ and is trivial on $S^d$. The homotopy fiber of the map $\mathbb CP^n\to \mathbb CP^\infty$ is $S^{2n+1}$. The homotopy fiber of the map $*\to \mathbb CP^\infty$ is $S^1$, and the homotopy fiber of the trivial map $S^d\to \mathbb CP^\infty$ is $S^1\times S^d$. It is well-known that taking homotopy pushout commutes with homotopy fibers, therefore the homotopy fiber of $f$ is equivalent to the homotopy pushout of the following diagram. $$ S^{2n+1}\leftarrow S^1 \to S^1\times S^d. $$ I will assume that $n>0$. In this case the map $S^1\to S^{2n+1}$ can only be null homotopic. It follows that the homotopy fiber is equivalent to $$S^{2n+1}\vee S^1_+ \wedge S^d\simeq S^{2n+1}\vee S^d \vee S^{d+1}$$ (for the second equivalence, assume $d\ge 1$). Applying $\Omega^2$ we get a homotopy fibration sequence $$\Omega^2(S^{2n+1}\vee S^d\vee S^{d+1})\to \Omega^2 (\mathbb CP^n \vee S^d)\to \Omega^2 \mathbb CP^\infty\simeq \mathbb Z.$$ This is a fibration sequence over a homotopically discrete space $\mathbb Z$. Moreover, it is a fibration of (double) loop spaces, so all the homotopy fibers are homotopy equivalent to each other. It follows that there is a homotopy equivalence of spaces $$ \Omega^2 (\mathbb CP^n \vee S^d)\simeq \mathbb Z \times \Omega^2(S^{2n+1}\vee S^d\vee S^{d+1}). $$ So in order to calculate the homology of $\Omega^2 (\mathbb CP^n \vee S^d)$ (with any coefficients), it is enough to calculate the homology of $\Omega^2(S^{2n+1}\vee S^d\vee S^{d+1})$ (with same coefficients). Let me assume for simplicity that $n\ge 1$ and $d\ge 2$. In this case we can write $$\Omega^2(S^{2n+1}\vee S^d\vee S^{d+1})\simeq \Omega^2\Sigma^2(S^{2n-1}\vee S^{d-2}\vee S^{d-1}).$$ The rational homology of spaces of the form $\Omega^2\Sigma^2 X$ is well-understood. If $X$ is path connected, then it is isomorphic to a certain type of free algebra on $X$. I think some people will call it the free Gerstenhaber algebra and some will say it is the free Poisson algebra. I am not sure what is the most accepted terminology. To get a description of $H_*(\Omega^2 \Sigma^2 X;\mathbb Q)$ as a vector space, first take the free Lie algebra on $H_*(X;\mathbb Q)$, where the Lie bracket has degree $+1$, and then take the free graded commutative algebra on that. I believe the result about $H_*(\Omega^2\Sigma^2 X)$ is due to Fred Cohen ("The homology of $C_{n+1}$-Spaces", LNM 533). You can find a perhaps somewhat more modern explanation and proof in this paper of Alexander Berglund (Corollary 8 in particular).<|endoftext|> TITLE: Strange behavior of $x_{n+1}=x_n +\lambda \sin x_n$ QUESTION [11 upvotes]: Consider a sequence $(x_n)$ satisfying $x_{n+1}=x_n +\lambda \sin x_n$. You would expect the sequence $x_n$ to depend on $x_0$ and to exhibit a chaotic, Brownian-type behavior, and indeed it does pretty much all the time. However, if $\lambda=8$ (also true if $\lambda$ is very close to $8$), we have $x_n \sim \pm 2\pi n$. The sign depends on the initial value $x_0$. Assuming $x_0=2$ and $\lambda=8$, we have $x_{2n}-x_{2n-1}\sim \alpha=7.939712...$ and $x_{2n-1}-x_{2n-2}\sim \beta=-1.65653...$ with $\alpha + \beta = 2\pi$. Also, $\alpha$ is solution of $$2\pi=\alpha +\alpha\cos\alpha -\sqrt{\lambda^2-\alpha^2}\sin\alpha.$$ I am wondering if this non-chaotic behavior also happens with other values of the parameter $\lambda$, and when the sign alternates (depending on $x_0$) in the asymptotic formula $x_n \sim \pm 2\pi n$. The sign is very sensitive to $x_0$. Are there other unexpected (non-chaotic) behavior for this sequence, depending on $\lambda$ and $x_0$? For instance, if $x_0$ is large (say $x_0=67$) and $1<\lambda<3$, then $x_n$ converges very rapidly so the sequence looks flat. If $x_0=67, \lambda=7.99$, we have the expected chaotic behavior. If $x_0=67, \lambda=8$ we have the behavior described earlier. And with $\lambda>8.02$ we are back to chaotic behavior. Now if $x_0=67, \lambda=4$, then $x_n$ stays in a flat, narrow band, constantly oscillating. Generalizations I added a lot of material in this article. It mostly deals with the basins of attractions in the 2-dimensional case. The picture below (taken from that article) features some of these basins. References See Denis Serre's answer below. My discussion of the case $\lambda=8$, as well as the exact formula for $\alpha,\beta$, might be new. Other references include Chaotic Synchronization and Antisynchronization in Coupled Sine Maps. Maistrenko V. at al. International Journal of Bifurcation and Chaos, Vol. 15, No. 07, pp. 2161-2177 (2005). See here. Basins and Critical Curves Generated by A Family of Two-Dimensional Sine Maps. Nasr-Eddine Hamri, Yamina Soula. Electronic J. of Theoretical Physics 7, No. 24 (2010) 139–150. See here. REPLY [23 votes]: This is exactly the dynamics studied by V. I. Arnold, which exhibits what is known as Arnold's tongues. See this link.<|endoftext|> TITLE: Is physical stamina important for doing mathematics? QUESTION [6 upvotes]: Recently, I read an interesting story about A. Weil and J.P. Serre. The general gist of the story is as follows: During the Autumn of 1955, an international symposium on algebraic number theory was held at Nikko [a small city around 150 km north of Tokyo]. The Japanese mathematicians invited Weil and Serre on a trip to Lake Chuzenji. Upon arriving, Weil and Serre stripped down, and started swimming in the cold lake. The Japanese mathematicians followed suit, but they quickly gave up because of the cold. After a while, the two mathematicians came back out and started running. The Japanese mathematicians ran after the duo, but once again, they gave up. Eventually, Weil came back, smiled at the Japanese mathematicians, and said, "Math is all about physical stamina." (My translation.) I think the last Weil quote sums up the main message of the story. My Question. Based on people's experience, is this a true statement? If possible, I would like to hear what physical activities people pursue, and in what ways that has helped in doing mathematics. I hope this is not too off-topic. I've never heard something like this from mathematicians I know personally, so I was curious to hear what the community thinks. Edit. As per helpful comments, I would like to restrict the question to research mathematics (rather than mathematics in general). REPLY [5 votes]: Apparently this is one of those opinion-based questions that are destined to be closed soon. Still, I'll share my perspective. IMHO it doesn't matter at all (for doing mathematics, at least) if you can outrun a cheetah or to come up as a winner in a wrestling match with an elephant. What is really required is an ability to concentrate and to think for an extended period of time without developing a headache, falling asleep, or going in circles. I doubt it burns any noticeable amount of calories because the best position I know for doing it is lying on the sofa. If that is the "stamina" we are talking about, I agree, but in all other respects, any physical shape that doesn't give you an obvious trouble and provides enough oxygenation to the brain would, probably, do. Another possible meaningful interpretation is that perseverance is at least as important as brilliance. I'm not quite sure I believe in Edison's statement that "genius is one percent inspiration and 99 percent perspiration", but I certainly agree that trying to ride the flashes of insight alone often won't bring you very far. As to the rest, it is in the human nature to show off a bit now and then, and, for a mathematician it is a little difficult to impress a non-mathematician (or even a mathematician in a different area) with his or her professional achievements, so I suspect many of us are in the habit of developing some marketable "side skills" and going into sports is certainly an option (though by no means the only one: somebody can choose to go into, say, arts instead and, if he ever becomes as famous as Weil, leave us pondering on MO if mathematics is all about the powers of imagination and the ability to reflect and transform all that we see and feel into sophisticated creations in the parallel world). Just my two cents.<|endoftext|> TITLE: On the Euler number of an orbit space QUESTION [6 upvotes]: Let $X$ be a differentiable manifold and $G$ a finite group acting differentiably on $X$. The following formula for the Euler number $\text{e}(X/G)$ of the orbit space $X/G$ appears to be well-known: \begin{equation*}\tag{1} \text{e}(X/G) = \dfrac{1}{\lvert G \rvert} \sum_{g \in G} \text{e}(X^g) \end{equation*} where, for each $g \in G$, $X^g$ is the subspace of elements fixed by $g$. Instances of this formula can be found in [3] and [5], but without proof or link to such. A quite elaborate search on the web did not produce one, either. Following the allusion to Lefschetz numbers in [3], a proof might be given, I suppose, along the following lines. Consider the canonical projection $\pi : X \rightarrow X/G$ and its induced map $\pi^* : \text{H}^*(X/G;\Bbb{Q}) \rightarrow \text{H}^*(X;\Bbb{Q})$ on rational cohomology. According to a celebrated result of Grothendieck in [4], this map is injective and maps $\text{H}^*(X/G;\Bbb{Q})$ isomorphically onto the $G$-invariants $\text{H}^*(X;\Bbb{Q})^G$ (this follows from the Leray SS for the map $\pi$, which collapses). Now for any representation of $G$ on a finite dimensional vector space $V$ over a field of characteristic 0, one has that \begin{equation*} P := \dfrac{1}{\lvert G \rvert} \sum_{g \in G} g : V \rightarrow V \end{equation*} is the projector onto the $G$-invariants $V^G$ and so, by taking traces, \begin{equation*} \dim V^G = \text{tr}\, P = \dfrac{1}{\lvert G \rvert} \sum_{g \in G} \text{tr}\, g. \end{equation*} Therefore, for each $i$, \begin{equation*} \dim \text{H}^i(X/G;\Bbb{Q}) = \dim \text{H}^i(X;\Bbb{Q})^G = \dfrac{1}{\lvert G \rvert} \sum_{g \in G} \text{tr}\, g^i, \end{equation*} where $g^i$ is the endomorphism induced by $g$ on $\text{H}^i(X;\Bbb{Q})$. Taking the alternating sum and interchanging the summation order leaves us with \begin{equation*} \begin{aligned} \text{e}(X/G) &= \sum_i (-1)^i \dim \text{H}^i(X/G;\Bbb{Q}) \\ &= \dfrac{1}{\lvert G \rvert} \sum_{g \in G} \sum_i (-1)^i \text{tr}\, g^i \end{aligned} \end{equation*} and so \begin{equation*}\tag{2} \text{e}(X/G) = \dfrac{1}{\lvert G \rvert} \sum_{g \in G} L(g), \end{equation*} where \begin{equation*} L(g) := \sum_i (-1)^i \text{tr}\, g^i \end{equation*} is the Lefschetz number of the endomorphism $g : X \rightarrow X$. But now we are in business; (1) boils down to the sixty-four dollar question \begin{equation*}\tag{3} \forall g \in G: L(g) = \text{e}(X^g). \end{equation*} A high-brow reference for this is is the general Lefschetz fixed point theorem Theorem (2.12) of [2] (which is a synthesis of the main theorem Theorem (6.7) of [1] and the Localization Theorem (1.1) of [2]), which, after plodding through the quite intricate intricacies of the $K$-theory apparatus surrounding the Index Theorem, should grind out the desired answer (3), which is, in a sense a localized version of the Gauss-Bonnet Theorem. What makes me unhappy with this is that the derivation of (1) up to stage (3) is maybe on the level of a first year graduate course in Algebraic Topology, wherea the application of the full force of the Index Theorem in (3) appears like the proverbial cracking of a peanut with a sledgehammer. Now it is known that the Gauss-Bonnet Theorem is much more elementary than the Index Theorem (which just illustrates it but is not needed for its proof), and also localization results come in various flavours, so my question is whether there exists a much more elementary proof of (1) somewhere in the world. [1] Atiyah, M.F. & Singer, I.M., The Index of Elliptic Operators: I, Ann.of Math, Sec.Ser., Vol. 87, No. 3 (May, 1968), pp. 484-530. [2] Atiyah, M.F. & Segal, G.B., The Index of Elliptic Operators: II, Ann.of Math, Sec.Ser., Vol. 87, No. 3 (May, 1968), pp. 531-545. [3] Bryan, J. & Fulman, J., Orbifold Euler characteristics and the number of commuting n-tuples in symmetric groups, [math/9712248] on $\mathtt{arxiv.org}$ [4] Grothendieck, A., Sur quelques points d'algèbre homologique, II, Tohoku Math. J. (2) Volume 9, Number 3 (1957), 119-221. [5] Hirzebruch, F. & Höfer, T., On the Euler number of an orbifold, Math. Ann. 286, 255-260 (1990) REPLY [5 votes]: Here is a variant of Greg Arone's fine answer. Following tom Dieck [p.227 of his book Transformation Groups], call a function $\chi$ from finite $G$-CW complexes to $\mathbb Q$ (or any abelian group) additive, if $\chi(X) = \chi(Y)$ if $X$ and $Y$ are $G$-homotopy equivalent, $\chi(\emptyset) = 0$, and $\chi(X \cup Y) = \chi(X) + \chi(Y) - \chi(X \cap Y)$. Because of how one builds $G$-CW complexes, an immediate lemma is that two additive functions agree on all $X$ if they agree on all the single orbit $G$-sets $G/H$. Both the left and right side of your putative equation are easily seen to be additive functions. Thus they are equal if they are equal on finite $G$-sets, and this is Burnside's formula.<|endoftext|> TITLE: How to show simply that $e^{\frac{x}{2}}\int^\infty_0 e^{-t}t^{n-\frac{1}{2}}\cos(2\sqrt{xt})dt=O\big(\frac{n!}{\sqrt{n}}\big)$? QUESTION [6 upvotes]: Can we prove, without using Laguerre polynomials, that $f_n(x)=O(\frac{n!}{\sqrt{n}})$ i.e. that $$ \exists C>0, \exists N\in\mathbb N, \forall x\geq0, \forall n\geq N :\ \big| f_n(x) \big|\leq C \frac{n!}{\sqrt{n}}, $$ where $$ f_n(x)=e^{\frac{x}{2}}\int^\infty_0 e^{-t}t^{n-\frac{1}{2}}\cos(2\sqrt{xt})dt,\quad x \geq 0\;? $$ Proof (by using Laguerre polynomials): it's easy to show that $$ f_n(x)=\sqrt \pi e^{-x/2} n! L^{(-1/2)}_n(x) $$ and we know that $$ L^{(-1/2)}_n(x)=O\Big(e^{x/2}\frac{1}{\sqrt{n}}\Big). $$ My reference is page 9 formula 18 $$L^{\alpha }_n(x)=O\Big(e^{\frac x2}x^{\frac{-\alpha}2 -\frac 14}n^{\frac{\alpha}2 -\frac 14} \Big) .$$ or see The polynomials' asymptotic behaviour for large n However, I'd like to not use this simple argument. REPLY [4 votes]: Let, $u^2=x$ and $v^2=t$, and let, $f_n(x)e^{-\frac{x}{2}}=\phi_{n}(u)=2\int_{0}^{\infty} e^{-v^2}v^{2n}\cos(2uv) dv$ Hence, $\phi''_{n}(u)=-2.4\int_{0}^{\infty} e^{-v^2}v^{2(n+1)} \cos(2uv)dv =(-4)\phi_{n+1}{u}$ Using this we get, $\phi_{n}(u)=2.\frac{(-1)^n}{2^{2n}}\frac{d^{2n}}{du^{2n}}\phi_0(n)$ Now, using contour integral we can find $\phi_{0}(u)=\int_{0}^{\infty} e^{-v^2} \cos(2uv) dv=\frac{\sqrt{\pi}}{2}e^{-u^2}$. From Rodrigue's formula of Hermite polynomial and the previous expression we get $\phi_n(u)=\sqrt{\pi}\frac{e^{-u^2}}{2^{2n}}H_{2n}(u)$ $\Rightarrow f_{n}(x=u^2)=\sqrt{\pi}e^{\frac{-u^2}{2}}2^{-2n}H_{2n}(u)$. Now, we have an asymptotic of Hermite polynomial like this, $e^{\frac{-x^2}{2}}H_n(x) ≈\frac{2^n}{\sqrt{\pi}}\Gamma(\frac{n+1}{2})\cos(x\sqrt{2n}-\frac{n\pi}{2})(1-\frac{x^2}{2n+1})^{-1/4}$ https://en.m.wikipedia.org/wiki/Hermite_polynomials We get, $f_n(x)≈\frac{n!}{\sqrt{n}}\cos(\sqrt{4xn}-\frac{2n\pi}{2})(1-\frac{x}{4n+1})^{-1/4}$.<|endoftext|> TITLE: how to see the Gysin map explicitly in an easy situation QUESTION [10 upvotes]: Let $C$ be a smooth projective curve and let $U \subset C$ be an open affine subset, with closed complement $S$ consisting of a finite number of points. I am trying to see explicitly the Gysin map in algebraic de Rham cohomology $$ H^0_{dR}(S) \longrightarrow H^2_{dR}(C). $$ By explicitly, I mean covering $C$ by $U$ and another affine open subset, say $V$, containing $S$ and computing $H^2_{dR}(C)$ a la Cech. In this way, every element can be represented by a global section of $\Omega^1$ on $U \cap V$. On the other hand, elements of $H^0_{dR}(S)$ are functions on $S$. So I am trying to find a way to produce a differential on $U \cap V$ out of a function on $S$. I have the intuition that logarithmic derivative should play a role here but I am stuck. Can anyone help? Thanks! REPLY [5 votes]: Let me first give an answer over $\mathbb C$ in the analytic topology. Let $V$ be a small neighborhood of the set $S$, and compute cohomology using the cover $C = U \cup V$. Then $U \cap V$ is a union of $\vert S \vert$ punctured disks. The function on $S$ prescribes a set of residues of a holomorphic 1-form on $U \cap V$, and any two holomorphic 1-forms with the same set of residues are cohomologous. So this is your representative of the de Rham cohomology class. In general you may choose a point $\infty \notin S$ and set $V = C \setminus \{\infty\}$. Part of the Riemann-Roch theorem is that there exists a meromorphic 1-form on $C$ with simple poles only along $S \cup \{\infty\}$, with residues along $S$ prescribed by the given function on $S$; the residue at $\infty$ is necessarily minus the sum of the other residues. This gives a holomorphic 1-form on $U \cap V$ which represents the de Rham cohomology class.<|endoftext|> TITLE: About definition of homotopy colimit of Kan and Bousfield QUESTION [8 upvotes]: In Bousfield and Kan's book"Homotopy Limits, Completions and Localizations",they define homotopy direct limit for system of pointed simplicial sets(Ch XII S2 2.1 P327), while they define homotopy inverse limit for system of simplicial sets without pointed condition(P 295), why do they insist this condition for direct limit? REPLY [18 votes]: You can define homotopy limits and colimits in pointed as well as unpointed spaces. It so happens that the two notions of homotopy limit coincide, basically because the forgetful functor from pointed to unpointed spaces is a right Quillen adjoint. That is, the homotopy limit of a diagram of pointed spaces is the same whether taken in the pointed or unpointed category. On the other hand, pointed and unpointed homotopy colimits are not equivalent, and it is important to be clear which one you are using. To see why this is plausible, think of products and coproducts. The categorical product is the same in the pointed and unpointed categories. But the pointed coproduct is the wedge sum while the unpointed coproduct is disjoint union. More generally, the pointed homotopy colimit is the quotient of the unpointed homotopy colimit by the classifying space of the indexing category.<|endoftext|> TITLE: de Rham-invariants of a Riemannian metric QUESTION [7 upvotes]: $\DeclareMathOperator{\Sym}{Sym}$For $N>0$, consider the $O_N$-representations $V = \mathbb R^N$ and $M_n = \ker (\Sym^n{V}\otimes\Sym^2 V\to \Sym^{n+1} V\otimes V)$ (the irreducible $GL_n$-representation corresponding to the partition $(n+2) = n + 2$). There is an equivariant map $\rho_n:M_{n}\to M_{n-1}\otimes V$, induced from the "desymmetrization" $\Sym^{n} V\to \Sym^{n-1} V\otimes V$. From this, we can produce a derivation on the (graded) symmetric algebra $\Sym^* (\bigoplus_{n\ge 2} M_n)\otimes\Lambda^* V$ which vanishes on the degree $1$ generators coming from $V$, and on the degree $0$ generators coming from $M_n$ is given by $\rho_n$. Since $$ M_{n+1}\xrightarrow{\rho_{n+1}} M_n\otimes V\xrightarrow{\rho_n\otimes\operatorname{id}_V}M_{n-1} V\otimes V^{\otimes 2} $$ factors through $M_{n-1} V\otimes \Sym^2 V$, this derivation squares to $0$, producing a cochain complex $C_N$. Moreover, there are obvious $O(N)$-equivariant inclusions $C_N\to C_{N+1}$. Question What is $$ \varinjlim_{N\to\infty}H^*(C_N)^{O(N)}, $$ i.e. the $O(N)$-invariant part of the cohomology of this complex in the infinite-dimensional limit $N\to \infty$? Background/Motivation The representations $M_n$ arise as the possible $n$-th partial derivatives of the metric tensor in normal coordinates. More invariantly, the symmetrization of the iterated covariant derivative $\nabla^n R$ of the Riemann curvature tensor takes values in the bundle associated to $M_{n-2}$. The individual terms of the complex are then diffeomorphism-invariant formulae for building a $k$-form out of the metric tensor (the existence of normal coordinates shows that any such formula must be given by contracting indices of products of covariant derivatives of the curvature tensor), and the differential on the complex is given by the de Rham differential. In particular, one can identify the Chern-Weyl representatives of the Pontryagin classes explicitly: Using the identification $M_2\cong \Sym^2(\Lambda^2 V)$, there is an equivariant map $\Sym^k M_2\to \Sym^k(\Lambda^2 V)\otimes\Sym^k(\Lambda^2 V)\to \Lambda^{2k} V\otimes \Sym^k(\Lambda^2 V)$, and for $k$ even the subspace $\Sym^k(\Lambda^2 V)^{O_n}$ contains the $k/2$-th power of the invariant pairing on $\Lambda^2 V$; taking adjoints, one obtains an invariant element $p_{k/2}\in \Lambda^{2k} V\otimes \Sym^k M_2$, which (up to combinatorial factors) gives the trace of the $k$-th power of the curvature tensor. Using the degree grading of the complex $C_N$ (where $V$ has degree $1$ and $M_n$ has degree $(n+2)$) it's easy to see that this class defines a nontrivial cocycle in $C_N$. Thus a more precise question is Question* Is the inclusion $$ \mathbb R[p_1,p_2,\dots]\to \varinjlim_{n\to\infty}H^*(C_n)^{O(n)}, $$ an isomorphism? A class which is not in the image would give an interesting (i.e. not coming from a characteristic class of the tangent bundle) de Rham invariant of a Riemannian metric. Note that such invariants exist for each finite dimension $N$, at least when restricting to $SO(N)$-invariants (e.g. the product of the scalar curvature and the volume form), but this seems unlikely in the infinite-dimensional limit. REPLY [2 votes]: $\DeclareMathOperator{\Sym}{Sym}\DeclareMathOperator{\Map}{Map}$After Robert Bryant's helpful comment, I was able to find a positive answer to Question* as Theorem 1.2 in the article Gilkey, Peter B. Local invariants of an embedded Riemannian manifold. (English) Ann. Math. (2) 102, 187-203 (1975). Let me summarize the argument here: Using the injection $M_n\hookrightarrow V^{\otimes n+2}$ and H. Weyl's determination of the $O(n)$-invariants of $V^{\otimes k}$, the equivariant maps $\Sym^k \bigoplus_{n\ge 2} M_n\to \Lambda^p V$ are spanned by the ones given by antisymmetrizing $p$ of the indices and pairing the remaining ones using the invariant bilinear form. There is a "weight grading" on the space of invariants, where the subspace $\Map\left(\bigotimes_{k} M_{n_k},\Lambda^p V\right)^{O(n)}$ has weight $r -\sum_k n_k$ (in terms of Riemannian metrics in normal coordinate form, this describes the behaviour under rescaling of the metric) which is preserved by the differential. Observe that the antisymmetrization of any three indices vanishes on $M_n$, so that we must have $r\le 2k$; it follows that there are no elements of positive weight and that for weight $0$ we must have $n_i = 2$, which is readily checked to correspond to the construction of (polynomials in) the Pontryagin classes I gave above. It remains to see that the subcomplex of negative weight has vanishing cohomology. For this, Gilkey extends to the complex $\Map(\bigoplus_{k_1,k_2}\Sym^{k_1} V\otimes\Sym^{k_2} \bigoplus_{n\ge 2} M_n,\Lambda^* V)^{O(n)}$ and cites an argument from another paper that the cohomology of this complex vanishes in a suitable range depending on the dimension; as far as I can tell, this uses the Poincaré lemma on the $\Sym^* V$-component together with an inductive/filtration argument. Interpreting the original complex as diffeomorphism-invariant differential operators $g\mapsto P(g)$ from Riemannian metrics to differential forms, he then considers their behaviour under infinitesimal conformal transformations $g\mapsto g+ \epsilon hg$; if $P$ is a cocycle, i.e. takes values in closed forms, this defines a cocycle $Q:h,g\mapsto \partial_\epsilon P(e^{\epsilon h}g)|_{\epsilon = 0}$ of the above complex, so that the above argument gives another operator $h,g\mapsto R(h,g)$ such that $Q =\mathrm dR$. We obtain \begin{align*} P(e^{\epsilon h}g) &= P(g) + \epsilon Q(h,g)\\ &= P(g) + \epsilon \mathrm d R(h,g) \end{align*} Taking $h = 1$, the left-hand side is $(1 + \epsilon\operatorname{wt}(P))P(g)$; if this weight is nonzero, comparing coefficients shows that $P = \mathrm d\frac{R(1,g)}{\operatorname{wt}(P)}$.<|endoftext|> TITLE: Congruences for "colored partitions" a la Ramanujan QUESTION [13 upvotes]: Let $t\in\Bbb{N}$ and consider the sequences $p_t(n)$ defined by $$\sum_{n\geq0}p_t(n)x^n=\prod_{i\geq1}\frac1{(1-x^i)^t}=(x;x)_{\infty}^{-t}.$$ The numbers $p_t(n)$ can be regarded as enumerating partitions of $n$ into parts of $t$ colors. Furthermore, $p_t(n)=\sum_{\lambda\vdash n}\prod_{j\geq1}\binom{k_j+t-1}{t-1}$ where $\lambda=1^{k_1}2^{k_2}\cdots$ and each $k_j\geq0$. Note also that $p_1(n)=p(n)$ is the usual unrestricted partition of $n$. Ramanujan's famous congruences state $$\begin{cases} p(5n+4)\equiv0\mod 5, \\ p(7n+5)\equiv0\mod 7, \\ p(11n+6)\equiv0\mod 11. \end{cases}$$ Following this tradition, I ask: QUESTION. Do these congruences hold true? $$\begin{cases} p_4(5n+4)\equiv0\mod 5, \\ p_4(7n+5)\equiv0\mod 7, \\ p_3(11n+7)\equiv0\mod 11. \end{cases}$$ POSTSCRIPT. I thank Gjergji Zaimi for his response shown below. I now exhibit all the congruences that I have found. Let's see how many are already covered in the literature. $$\begin{cases} p_t(5n+1)\equiv0\mod 5 \qquad {} \text{if $t\equiv0\mod5$}, \\ p_t(5n+2)\equiv0\mod 5 \qquad {} \text{if $t\equiv0,2\mod5$}, \\ p_t(5n+3)\equiv0\mod 5 \qquad {} \text{if $t\equiv0,2,4\mod5$}, \\ p_t(5n+4)\equiv0\mod 5 \qquad {} \text{if $t\equiv0,1,2,4\mod5$}. \end{cases}$$ $$\begin{cases} p_t(7n+1)\equiv0\mod 7 \qquad {} \text{if $t\equiv0,\mod7$}, \\ p_t(7n+2)\equiv0\mod 7 \qquad {} \text{if $t\equiv0,4\mod7$}, \\ p_t(7n+3)\equiv0\mod 7 \qquad {} \text{if $t\equiv0,6\mod7$}, \\ p_t(7n+4)\equiv0\mod 7 \qquad {} \text{if $t\equiv0,4,6\mod7$}, \\ p_t(7n+5)\equiv0\mod 7 \qquad {} \text{if $t\equiv0,1,4\mod7$}, \\ p_t(7n+6)\equiv0\mod 7 \qquad {} \text{if $t\equiv0,4,6\mod7$}. \end{cases}$$ $$\begin{cases} p_t(11n+1)\equiv0\mod 11 \qquad\text{if $t\equiv0\mod11$}, \\ p_t(11n+2)\equiv0\mod 11 \qquad\text{if $t\equiv0,8\mod11$}, \\ p_t(11n+3)\equiv0\mod 11 \qquad\text{if $t\equiv0,10\mod11$}, \\ p_t(11n+4)\equiv0\mod 11 \qquad\text{if $t\equiv0,8\mod11$}, \\ p_t(11n+5)\equiv0\mod 11 \qquad\text{if $t\equiv0,8\mod11$}, \\ p_t(11n+6)\equiv0\mod 11 \qquad\text{if $t\equiv0,1,10\mod11$}, \\ p_t(11n+7)\equiv0\mod 11 \qquad\text{if $t\equiv0,3,8\mod11$}, \\ p_t(11n+8)\equiv0\mod 11 \qquad\text{if $t\equiv0,5,8,10\mod11$}, \\ p_t(11n+9)\equiv0\mod 11 \qquad\text{if $t\equiv0,7,8,10\mod11$}, \\ p_t(11n+10)\equiv0\mod 11 \qquad\text{if $t\equiv0,10\mod11$}. \end{cases}$$ REPLY [15 votes]: Yes, these are all true and they are all in the literature. The first two congruences are part of infinite families of the form $$\begin{cases} p_{\ell-1}(\ell n+a)\equiv 0\mod \ell, \\ p_{\ell-3}(\ell n+b)\equiv0\mod \ell, \end{cases}$$ where $\ell \geq 5$ is a prime and $a,b$ are such that: $24a+1$ is a quadratic nonresidue, $8b+1$ is either a quadratic nonresidue or divisible by $\ell$. Both congruences are immediate consequences of the Euler pentagonal theorem and Jacobi triple product respectively. This is explained in section 2 of the paper: I. Kiming and J. Olsson, Congruences like Ramanujan's for powers of the partition function (Arch. Math., 59(4):348-360, 1992) doi: 10.1007/BF01197051, where they are called the non-exceptional case. The third identity is a bit more recent. It was proven in Lin, B.L.S. Ramanujan-style proof of $p_{−3}(11n+7)≡0 \pmod{11}$ (Ramanujan J 42, 223–231 (2017)) doi:10.1007/s11139-015-9733-5. It was generalized to the following infinite family $$p_{\ell-8}\left(\ell n+\frac{\ell^2-1}{3}-\ell \left\lfloor\frac{\ell^2-1}{3\ell}\right\rfloor\right)\equiv 0 \mod \ell.$$ For this last congruence and many other exceptional infinite families see the paper Locus, M., Wagner, I. Congruences for Powers of the Partition Function (Ann. Comb. 21, 83–93 (2017)) doi:10.1007/s00026-017-0342-4 arXiv:1604.07495.<|endoftext|> TITLE: What's the matrix of logarithm of derivative operator ($\ln D$)? What is the role of this operator in various math fields? QUESTION [11 upvotes]: Babusci and Dattoli, On the logarithm of the derivative operator, arXiv:1105.5978, gives some great results: \begin{align*} (\ln D) 1 & {}= -\ln x -\gamma \\ (\ln D) x^n & {}= x^n (\psi (n+1)-\ln x) \\ (\ln D) \ln x & {}= -\zeta(2) -(\gamma+\ln x)\ln x. \end{align*} I wonder, what is its matrix, or otherwise, is there a method of applying it to a function? What is its intuitive role in various fields of math? REPLY [5 votes]: The interpretation of a $\ln(D)$ depends on the interpolation that one chooses of the usual derivative operator and its positive integer powers to a fractional integro-derivative operator (FID), i.e., an interpretation of $D$ exponentiated by any real (or complex number via analytic continuation), which in turn, depends on the functions the FID is to act upon. The extension described below produces B & Ds three identities and is consistent with the properties that Pincherle imposed on any legitimate family of FIDs (see this MO-Q on a 1/2 derivative and this MO-Q on fractional calculus). It can be defined by the action on a 'basis set' of entire functions in the complex variable $\omega$ as $$D_x^{\alpha} \; H(x) \; \frac{x^{\omega}}{\omega!} = H(x) \frac{x^{\omega-\alpha}}{(\omega-\alpha)!} ,$$ where $H(x)$ is the Heaviside step function, and $\alpha$ and $\omega$ may be any complex numbers with the usual identification in the theory of generalized functions and distributions of $$(-1)^n \delta^{(n)}(x) = H(x) \frac{x^{-n-1}}{(-n-1)!},$$ with $n=0,1,2,3,...$. Note this has little to do with a Fourier transform over the real line or any pseudo-diff op/symbol associated with such. In particular, $D^{\alpha}$ here is NOT associated with multiplication by $(i 2 \pi f)^{\alpha}$ in frequency space. Elsewhere I show various equivalent convolutional reps of this FID as 1) a FT over a circle via a transformation of a regularized Cauchy complex contour integral, 2) the analytic continuation of the integral rep of the Euler beta function either through a blow-up into the complex plane of the integral along the real line segment or regularization via the Hadamard finite part or via the Pochhammer contour, 3) the Mellin interpolation of the standard derivative operator via the action of the generating function $e^{tD_x}$, an operator application of Ramanujan's master formula, or 4) a sinc function/cardinal series interpolation of the generalized binomial coefficients. Let's see how viable the above definition of the FID is; its connection to an infinitesimal generator (infinigen) of the FID and the three B & D identities; a connection to the formalism of Appell Sheffer polynomial sequences and, therefore, symmetric polynomial/function theory; and matrix reps of the infinigen and FID. If we assume that an infinitesimal generator $IG$ exists such that $$ e^{\alpha \; IG} \; H(x) \; \frac{x^{\omega}}{\omega!} = D_x^{\alpha} \; H(x) \; \frac{x^{\omega}}{\omega!} = H(x) \frac{x^{\omega-\alpha}}{(\omega-\alpha)!} = e^{-\alpha D_{\omega}} \; H(x) \; \frac{x^{\omega}}{\omega!},$$ then formally $$D_{\alpha} \; e^{\alpha IG} \; H(x) \; \frac{x^{\omega}}{\omega!} |_{\alpha =0} = IG \; H(x) \; \frac{x^{\omega}}{\omega!} = \ln(D_x) \; H(x) \; \frac{x^{\omega}}{\omega!}$$ $$ = D_{\alpha} \; H(x) \; \frac{x^{\omega-\alpha}}{(\omega-\alpha)!} |_{\alpha =0} = -D_{\omega} \;\frac{x^{\omega}}{\omega!}$$ $$ = [\; -\ln(x) + \psi(1+\omega) \;] H(x) \; \frac{x^{\omega}}{\omega!} $$ $$ = [ \; -\ln(x) + \psi(1+xD_x) \;] \; H(x) \; \frac{x^{\omega}}{\omega!}, $$ and the infinigen is $$ \ln(D_x) := IG = -\ln(x) + \psi(1+xD_x),$$ where $\psi(x)$ is the digamma function, which can be defined over the complex plane as a meromorphic function and is intimately related to the values of the Riemann zeta function at $s = 2,3,4,...$. Some reps (that give the same identities as in B & D) are $$IG \; f(x)=\frac{1}{2\pi i}\oint_{|z-x|=|x|}\frac{-\ln(z-x)+\lambda}{z-x}f(z) \; dz$$ $$=(-\ln(x)+\lambda) \; f(x)+ \int_{0}^{x}\frac{f\left ( x\right )-f(u)}{x-u}du$$ $$ = [\; -\ln(x)+ \frac{\mathrm{d} }{\mathrm{d} \beta}\ln[\beta!]\mid _{\beta =xD} \; ] \; f(x)=[ \; -\ln(x)+\Psi(1+xD) \;] \; f(x)$$ $$ = [ \; -\ln(x)+\lambda - \sum_{n=1}^{\infty } (-1)^n\zeta (n+1) \; (xD)^n \;] \; f(x)$$ where $\lambda$ is related to the Euler-Mascheroni constant via $\lambda=D_{\beta} \; \beta! \;|_{\beta=0}$. Other reps and other ways of arriving at the reps above are given in the refs below. Let's look at a way via the formalism of Appell Sheffer polynomial sequences, which settles any issues of convergence upon exponentiation of the explicit diff op formula for the infinigen and allows connections to the theory of symmetric polynomials/functions. The relevant Appell sequence of polynomials $p_n(z) = (p.(z))^n$ has the exponential generating function, entire in the complex variable $t$, i.e., with its Taylor series globally convergent, $$\frac{1}{t!} \; e^{zt} = e^{a.t} \; e^{zt} = e^{(a.+z)t} = e^{p.(z)t} = \sum_{n\geq 0} p_n(z) \frac{t^n}{n!}$$ with the reciprocal polynomial sequence defined in four consistent ways $\hat{p}(z)$ 1) $t! \;e^{zt} = e^{\hat{a}.t} \; e^{zt} = e^{(\hat{a}.+z)t} = e^{\hat{p}.(z)t} $, an e.g.f., 2) $M_p \cdot M_{\hat{p}} = I $, in terms of the lower triangular coefficient matrices of the two sequences in the monomial power basis $z^n$ with unit diagonal, 3) $p_n(\hat{p}.(z)) = \hat{p}_n(p.(z)) = (a. + \hat{a.}+z)^n = 1$, an umbral convolutional inversion, 4) $D_z! \; z^n = e^{\hat{a.}D_z} \; z^n = (\hat{a.}+z)^n = \hat{p}_n(z)$, an operational generator. It follows that the raising op of the Appell polynomials $p_n(z)$ defined by $$R_z \; p_n(z) = p_{n+1}(z)$$ is given by $$ R_z \; p_n(z) = \frac{1}{D_z!} \; z \; D_z! \; p_n(z) = \frac{1}{D_z!} \; z \; p_n(\hat{p}.(z))$$ $$ = \frac{1}{D_z!} \; z \; z^n = \frac{1}{D_z!} \; z^{n+1} = p_{n+1}(z),$$ an operator conjugation, or 'gauge transformation', of the raising operator $z$ for the power monomials. In addition, with the operator commutator $[A,B] = AB - BA$, $$R_z = \frac{1}{D_z!} \; z \; D_z! = z + [\frac{1}{D_z!},z] \; D_z! .$$ Now re-enter Pincherle and the eponymous operator derivative, which Rota touted for the finite operator calculus. The Graves-Pincherle derivative derives its power from the Graves-Lie-Heisenberg-Weyl commutator $[D_z,z] = 1$ from which, by normal re-ordering, implies for any function expressed as a power series in $D_z$ $$[f(D_z),z] = f'(D_z) = D_t \; f(t) \; |_{t = D_z}.$$ This is an avatar of the Pincherle derivative (PD) that follows from the action $$[D^n,z] \; \frac{z^{\omega}}{\omega!} = [\;\frac{\omega+1}{(\omega+1-n)!} - \frac{1}{(\omega-n)!}\;] \; z^{\omega+1-n} = n \; D_z^{n-1} \; \frac{z^{\omega}}{\omega!},$$ but the PD is valid for more general lowering and raising (ladder) ops that satisfy $[L,R]= 1$. Then $$R_z = \frac{1}{D_z!} \; z \; D_z! = z + [\frac{1}{D_z!},z] \; D_z! = z + D_{t = D_z}\; \ln[\frac{1}{t!}] $$ $$ = z - \psi(1+D_z).$$ With the substitution $ z = \ln(x)$ $$R_z = R_x = \ln(x) - \psi(1+ x D_x) = -IG = -\ln(D_x).$$ The raising op is defined such that $$ e^{t \; R_z} \; 1 = \sum_{n \geq 0} \frac{t^n}{n!} R_z^n \; 1 = e^{tp.(z)} = \frac{1}{t!} \; e^{zt},$$ an entire function for $t$ complex; therefore, $$e^{-t \; IG} \;1 = e^{t \;R_x} \; 1 = e^{t \; p.(\ln(x))} = \frac{x^t}{t!},$$ so $$e^{-(\alpha+\beta) \; IG} \;1 = e^{(\alpha+\beta) \; R_x} \; 1 = e^{(\alpha+\beta) \; p.(\ln(x))} = \frac{x^{\alpha+\beta}}{(\alpha+\beta)!}, $$ $$ = e^{-\alpha \; IG} e^{-\beta \; IG} \;1 = e^{-\alpha \; IG} \; \frac{x^\beta}{\beta!} , $$ and we can identify that indeed $$e^{-\alpha \; IG} = D_x^{-\alpha}$$ and $$IG = \ln(D_x).$$ Now apply the PD to $\ln(D)$, as a check of the formalism and an avenue to a matrix rep, giving formally $$ [\ln(D),x] = [\ln(1-(1-D)),x] = \frac{1}{1-(1-D)} = \frac{1}{D} = D^{-1}.$$ This is given an explicit meaning by evaluating the commutator for a general function $g(x)$ analytic at the origin (which generalizes to our 'basis' set) using the integral rep for $R_x = -\ln(D_x)$, giving $$[\ln(D_x),x] \; g(x) = [-R_x,x] \; g(x) = (-\ln(x)+\lambda) \; [x,g(x)]$$ $$ + \int_{0}^{x}\frac{xg(x)-ug(u)}{x-u} \; du - x \int_{0}^{x}\frac{g(x)-g(u)}{x-u} \; du$$ $$ = \int_{0}^{x} \; g(u) \; du = D_x^{-1} g(x).$$ So, we have $$[\ln(D_x),x] = [-R_x,x] = D_x^{-1} = [-\ln([-R_x,x]),x]$$ and $$-R_x = \ln(D_x) = -\ln(D_x^{-1}) = -\ln([-R_x,x]),$$ implying $$e^{R_x} =\exp[\ln([-R_x,x])] = [-R_x,x] = D_x^{-1}.$$ In addition, with $$\bigtriangledown^{s}_{n} \; c_n=\sum_{n=0}^{\infty}(-1)^n \binom{s}{n}c_n,$$ then $$R_x = -\ln(D_x) = \ln(D_x^{-1}) = \ln[1-(1-D_x^{-1})]$$ $$ = - \sum_{n \geq 1} \frac{1}{n} \; \bigtriangledown^{n}_{k} D_x^{-k}, $$ where $$D_x^{-1} \frac{x^{\omega}}{\omega!} = \frac{x^{\omega+1}}{(\omega+1)!}.$$ The finite difference op series is embedded in the derivative $D_{\alpha =0}$ of the Newton interpolator $$ \frac{x^{\alpha+\omega}}{(\alpha+\omega)!} = \bigtriangledown^{\alpha}_{n}\bigtriangledown^{n}_{k}\frac{x^{\omega+k}}{(\omega+k)!}$$ $$ = \bigtriangledown^{\alpha}_{n}\bigtriangledown^{n}_{k} D_x^{-k} \;\frac{x^{\omega}}{\omega!}$$ $$ = [1-(1-D_x^{-1})]^{\alpha} \; \;\frac{x^{\omega}}{\omega!} = D_x^{-\alpha}\;\frac{x^{\omega}}{\omega!}. $$ For $\alpha = -m$ with $m = 1,2,...$ and $\omega = 0$, this Newton interpolator gives $$D^m_x \; H(x) = \delta^{(m-1)}(x) = H(x) \; \frac{x^{-m}}{(-m)!} = \bigtriangledown^{-m}_{n}\bigtriangledown^{n}_{k} D_x^{-k} \; H(x)$$ $$ = \sum_{n \geq 0} (-1)^n \binom{-m}{n} \bigtriangledown^{n}_{k} \; H(x) \frac{x^k}{k!} = H(x) \; \sum_{n \geq 0} (-1)^n \binom{-m}{n} \; L_n(x)$$ $$ = H(x) \; \sum_{n \geq 0} \binom{m-1+n}{n} \; L_n(x), $$ which agrees in a distributional sense with the Laguerre polynomial resolutions of $f(x) = \delta^{(m-1)}(x)$ in the formulas of this MO-Q since, with $c_n = f_n$ in the notation there, $$ f(x) = \sum_{n \geq 0} c_n \; L_n(x)$$ with $$\sum_{n \geq 0} t^n \; c_n = \frac{1}{1-c.t} = \int_0^{\infty} e^{-x} \sum_{n \geq 0} t^n \; L_n(x) f(x) \; dx$$ $$ = \int_0^{\infty} e^{-x} \frac{e^{-\frac{t}{1-t}x}}{1-t} f(x) \; dx = \int_0^{\infty} \frac{e^{-\frac{1}{1-t}x}}{1-t} f(x) \; dx,$$ so, for the $m$-th derivative of the Heaviside function, $$\frac{1}{1-c_{m,.}t}= \int_0^{\infty} e^{-x} \frac{e^{-\frac{t}{1-t}x}}{1-t} f(x) \; dx = \int_0^{\infty} \frac{e^{-\frac{1}{1-t}x}}{1-t} \delta^{(m-1)}(x) \; dx = \frac{1}{(1-t)^{m}},$$ and, therefore, the coefficients of the Laguerre series resolution of the $m$-th derivative of the Heaviside function are $$c_{m,n} =(-1)^n \binom{-m}{n} = \binom{m-1+n}{n},$$ in agreement with the Newton interpolator. Applying $D_x^{-1}$ iteratively to both sides of this identity establishes convergent interpolations for $\omega = 1,2,3,...$, and acting on the power basis within the binomial expansion of $\frac{x^{\omega}}{\omega!} = \frac{(1-(1-x))^{\omega}}{\omega!}$ should give convergent expressions as well. Similarly for $\omega=0$, we have the Laplace transform (or more accurately, the modified Mellin transform central to Ramanujan's master formula via which the FIDs may be cast as Mellin interpolations of the standard derivatives), $$\frac{1}{1-c.t} = \int_0^{\infty} \frac{e^{-\frac{1}{1-t}x}}{1-t} \frac{x^{\alpha}}{\alpha!} \; dx = (1-t)^{\alpha},$$ for $Re(\alpha) > -1$, giving $$c_n = (-1)^n \binom{\alpha}{n}.$$ This Laplace transform and, therefore, the Newton interpolator can be analytically continued in several standard ways (e.g., blow-up from the real line to the complex plane via a Hankel contour, Hadamard finite part) to the full complex plane for $\alpha$. For the negative integer exponents, the Hankel contour contracts to the usual Cauchy contour rep for differentiation. The Hadamard-finite-part approach allows the Newton interpolator to be appropriately modified strip by strip to give the intended results. Returning to the finite difference rep for $\ln(D_x)$, action of the infinigen on 1 then gives, for $x > 0$, $$\ln(D_x) 1 = \sum_{n \geq 1} \frac{1}{n} \; \bigtriangledown^{n}_{k} D_x^{-k} 1$$ $$ = \sum_{n \geq 1} \frac{1}{n} \; \bigtriangledown^{n}_{k} \frac{x^k}{k!}$$ $$ = \sum_{n \geq 1} \frac{1}{n} \; L_n(x) = -\ln(x)-.57721... , $$ where $L_n(x)$ are the Laguerre polynomials, in agreement with the first equation of B & D in the question. Plots of the results of evaluation of the operator series truncated at $n=80$, or so, acting on $x^2$ and $x^3$ match the analytic results as well. The matrix rep $M$ of the action of this integration op $D_x^{-1}$ on $x^n$ is simple enough in the power basis--a matrix with all zeros except for the first subdiagonal, or superdiagonal, depending on left or right matrix multiplication, with elements $(1,1/2,1/3,...)$. The matrix rep for $R_x$ is then $$ R_M = \ln[I-(I-M)] = - \sum_{n \geq 1} \frac{1}{n} \; \bigtriangledown^{n}_{k} M^k. $$ Exponentiating, $$D_x^{-\beta} = \exp(-\beta R_x)= (1-(1-D_x^{-1} ) )^{\beta} = \bigtriangledown^{\beta}_{n} \bigtriangledown^{n}_{k} (D_x^{-1})^k.$$ The associated matrix rep is $$ \exp(-\beta R_M)= \bigtriangledown^{\beta}_{n} \bigtriangledown^{n}_{k} M^k.$$ (I haven't checked these matrix computations numerically as I normally would since my MathCad disc is in storage in another state.) To act on non-integer powers of $x$, you must represent them as superpositions of the integer power basis as in the binomial expansion $$x^{\alpha} = [1 - (1-x)]^{\alpha} = \bigtriangledown^{\alpha}_{n} \bigtriangledown^{n}_{k} x^k .$$ Alternatively, return to the $z$ rep and write down the matrix rep of the raising op $R_z$. This is a simple transformation of the infinite lower triangular Pascal matrix augmented with a first superdiagonal of all ones. OEIS A039683 has an example of the matrix equivalent of a raising op in the monomial power basis, also known as a production matrix in another approach (Riordan?) to polynomial sequences. Better in this case to switch to the divided power basis $z^n/n!$. Then the augmented Pascal matrix becomes the simple summation matrix of all ones. Multiply along the n-th diagonal by $c_n$ where $(c_0,c_1,..) = (1-\lambda,-\zeta(2),...,(-1)^k \; \zeta(k+1),...)$ to generate the matrix rep for the raising op, but since, e.g., $x^2=e^{2z}$, this quickly becomes a messy algorithm to apply compared to the finite difference rep. Further references (not exhaustive): Riemann zeta and fractional calculus, an MO-Q Digamma / Psi function, Wiki OEIS A238363 on log of the derivative operator OEIS A036039 on the cycle index polynomials and symmetric functions Zeta functions and the cycle index polynomials, an MO-Q On the raising op for FIDs, an MSE-Q OEIS A132440 on a matrix infinigen OEIS A263634 on partition polynomial reps for Appell raising ops Ref for another interp of a log of a derivative, a pdf Interpolation/analytic continuation of the factorials to the gamma fct, MSE-Q Raising ops for Appell sequences, a blog post Example of Mellin interpolation of $e^{tD}$, MO-Q More on interpolation/analytic continuation of differential ops, a blog post Two analytic continuations of the coefficients of a generating function, MO-Q FIDs and confluent hypergeometric functions, an MO-Q Note on the Pincherle derivative, a blog post FIDs and interpolation of binomial coefficients, a blog post FIDs, interpolation, and travelling waves, a blog post<|endoftext|> TITLE: Proving that the primitives of the Laguerre functions are uniformly bounded QUESTION [5 upvotes]: Let $(L_k)_{k\geq 0}$ be the Laguerre polynomials. These polynmials are orthogonal with respect to the inner product: $$\langle f,g\rangle = \int_0^\infty f(x)g(x)\mathrm e^{-x}\,\mathrm dx.$$ Hence, the functions $\psi_k(x) = \sqrt{2} L_k(2x) \mathrm{e}^{-x}$ form a basis of $\mathrm L^2(\mathbb R_+)$ called the Laguerre basis. I am interested in the functions $\Psi_k(x) = \int_0^x \psi_k(t)\,\mathrm dt$. I can see numerically that these functions are uniformly bounded by $\sqrt 2$ (actually, it seems that they are bounded by $\sqrt 2(1-\mathrm e^{-x})=\Psi_0(x)$, but I just need the constant), but I fail to prove it. Does anybody know if these functions have already been studied by someone? And if not, does anybody have an idea on how to prove it? Since the closed form of the Laguerre polynomials is: $$ L_k(x) = \sum_{j=0}^k \binom{k}{j}\frac{(-x)^j}{j!}, $$ I can show that $\Psi_k$ tends to $(-1)^k\sqrt{2}$ when $x$ goes to $+\infty$, so I need to prove that it's the maximum/minimum of $\Psi_k$. I tried to use the recurrence relations satisfied by the Laguerre polynomials. For instance, $(\Psi_k)$ satisfy the relation: $$ \Psi_{k+1}(x) = \frac{2x\,\psi_k(x) - \Psi_k(x) + k \Psi_{k-1}(x)}{k+1}. $$ They also satisfy the relation $\Psi_{k+1}+\Psi_k = \psi_k - \psi_{k+1}$. I tried to use both relations, and nothing came out, but maybe it is the right way. REPLY [3 votes]: OK, here is the argument. We want to show that the partial integrals of $u(x)=L_n(x)e^{-x/2}$ are not (much) larger in the absolute value than the full integral. We'll just use the differential equation $$ xL_n''+(1-x)L_n+nL_n=0\,. $$ Plugging $L_n=e^{x/2}u$, we get $$ x(u''+u'+\tfrac 14 u)+(1-x)(u'+\tfrac 12u)+nu=xu''+u'+(n+\tfrac 12-\tfrac 14x)u=0 $$ Now we want to kill the first derivative, so we define $v(x)=u(x^{3/2})x^{1/2}$ (so that partial integrals of $v$ are the same as of $u$ up to the constant $2/3$ factor). We have $$ v'(x)=u'(x^{3/2})\tfrac 32x +u(x^{3/2})\tfrac 12 x^{-1/2} $$ and $$ v''=u''(x^{3/2})\tfrac 94 x^{3/2}+\tfrac 94 u'(x^{3/2})-u(x^{3/2})\tfrac 14 x^{-3/2} $$ so, by the differential equation for $u$, $$ v''=-[\tfrac 94(n+\tfrac 12-\tfrac 14x^{3/2})+\tfrac 14 x^{-3/2}]u(x^{3/2}) $$ and, writing $u(x^{3/2})=\frac{v(x)}{x^{1/2}}$, we finally get $$ v''+\Phi v=0 $$ where $$ \Phi(x)=\tfrac 94\left(\tfrac{n+\frac 12}{x^{1/2}}-\tfrac 14x\right)+\tfrac 14 x^{-2} $$ We don't really care much what this $\Phi(x)$ is. All that will matter is that it is a decreasing convex function that starts positive at $0$ and ends up negative near $+\infty$. Since $\Phi$ is decreasing, each next hump of $v$ between the zeroes is larger in area than the previous one (the rightmost hump stretches to $+\infty$), so the maximal absolute value of the partial integrals is attained either when we take the full integral (which is fine for us), or if we stop after the penultimate hump. Thus we just need to show that the second option is dominated by the first one. If the humps have the areas $A_00$. It suffices to show that $A_{n-1}\le cA_{n}$ with some absolute $c<1$ because then $|I|\ge (1-c)A_n$ and $|I\mp A_n|=|A_n-|I||\le \frac c{1-c}|I|$. Let now $z$ be the last zero of $v$. Assume for definiteness that the last hump is positive. Let $\Psi(x)=\Phi(z)-\lambda(x-z)$ be the unique linear function such that the solution $w$ of $w''+\Psi w=0$ shot up from $x=z,w(x)=0$ stays positive on $[z,+\infty)$ and tends to $0$ at $+\infty$, so $w$ is just the Airy function, properly shifted and stretched. Let us make $w'(z)$ just a tiny bit less than $v'(z)$. Since the Wronskian $W(v,w)=\det\begin{bmatrix} v & w\\v' & w'\end{bmatrix}$ vanishes at $z$ and $+\infty$ and $W'=(\Phi-\Psi)vw$, we must have $\int_z^{\infty}(\Phi-\Psi)vw=0$, so $\Phi-\Psi$ changes sign. Since $\Phi$ is convex and $\Psi$ is linear, it is possible only if first $\Phi<\Psi$ and later $\Phi>\Psi$, so $W(v,w)<0$ all the way between $0$ and $+\infty$. By our choice of derivative at $z$, $w$ stays below $v$ near $z$. If it ever breaks up, we should have $v=w>0, v'\le w'$ at the first breaking point, so $W(v,w)=vw'-wv'\ge 0$, which has been ruled out. Thus the area of the hump of the rescaled Airy function $w$ with $w'(z)=v'(z)$ is smaller than $A_n$. Since $\Phi>\Psi$ to the left of $z$, we conclude that $A_{n-1}$ is less than the adjacent hump of $w$. Thus $A_{n-1}/A_n\le c$ where $c$ is the ratio of the areas of the penultimate and the ultimate humps of the Airy function. It is a fixed constant less than $1$, so we are done. In reality, $c\approx 0.365\dots<\frac 12$, so $\frac c{1-c}<1$ and the maximal integral is the full one but, as I said, I currently have no proof of it. Edit: Details requested by the OP. I'm using the following comparison principle. Suppose that $\Phi>\Psi$ and we shoot two solutions $v,w$ of the equations $v''+\Phi v=0$ and $w''+\Psi w=0$ from the same point $a$ with the initial data $v(a)=w(a)=0, v'(a)=w'(a)>0$. Then we have $v\le w$ as long as both $v$ and $w$ stay positive. In particular, if both $v$ and $w$ make positive humps, the hump of $v$ is contained in that of $w$, so its area is not larger. The proof is the same Wronskian story as I wrote for the comparison of the infinite humps, only simpler. Assume first that $v'(a)$ is just a tiny bit below $w'(a)$. Then $v$ starts lower and can break through $w$ at some point $b$ with $v(b)=w(b)>0$, $v|_{(a,b]},w|_{(a,b]}>0$ only if $v'(b)\ge w'(b)$. But then we have $W(v,w)(a)=0$, $W(v,w)\le 0$ while $W'(v,w)=(\Phi-\Psi)vw>0$ on $(a,b)$, which is a clear contradiction. This immediately answers your question about the humps of $v$ and $w$ to the left of the last zero of $v$ (only there we shoot from the right endpoint). As to two adjacent humps, just reflect the left one about the common zero and use the decreasing property of $\Phi$ to compare (if $a$ is the common zero of two adjacent humps, then you just need the inequality $\Phi(a-t)\ge \Phi(a+t)$ for $t>0$).<|endoftext|> TITLE: Representing finite sums of rational powers of 2 QUESTION [10 upvotes]: Let $X \subseteq \mathbb{R}$. Let $A$ and $B$ be finite subsets of $X$. The statement $$\sum_{a \in A} 2^a = \sum_{b \in B}2^b \iff A = B $$ is true if $X = \mathbb{N}$ or $X = \mathbb{Z}$; this follows from the uniqueness of finite binary represntation (for naturals and dyadic rationals). However, the statement is false if $X = \mathbb{R}$ as, for example, $2^0+2^2 = 5 = 2^1 + 2^{\log_2{3}}$. My question: what about the rationals? Is this statement true for $X = \mathbb{Q}$? In other words, is there a unique representation of the form $\sum_{a \in A}2^a$ for all $s \in S$, where $$S = \left\{\sum_{a \in A}2^a : A \subset \mathbb{Q} \text{ is finite}\right\}.$$ Bonus question: is there better, more concise notation for $S$? (Edit) Bonus answer: we can write $S = \mathbb{N}[2^{-1/2},2^{-1/3},...]$. This is similar to the notation for dyadic rationals $\mathbb{N}[2^{-1}] = \{n2^{-m}: n,m \in \mathbb{N} \}$. REPLY [15 votes]: If $A$ and $B$ are distinct finite subsets of $\mathbb Q$ which are not both subsets of $\mathbb Z$, let $d$ be the least common denominator and $m$ the minimum of $A \cup B$. Thus $\sum_{a \in A} 2^a - \sum_{b \in B} 2^b = 2^m P(2^{1/d})$ where $P$ is a polynomial with coefficients in $\{-1,0,1\}$. Now for this to be $0$, $P(z)$ must be divisible (as a member of $\mathbb Z[z]$) by $z^d-2$, which is the minimal polynomial of $2^{1/d}$. But the lowest nonzero coefficient of any nonzero multiple of $z^d-2$ is divisible by $2$, so this is impossible. REPLY [4 votes]: It is true. Without loss of generality, we may assume that $A \cap B$ is empty because any element that lies in both can be removed from both, preserving the property $\sum_{ a\in A} 2^a = \sum_{ b\in B} 2^b$. Let $c$ be the minimal element of $A \cup B$. Let $d$ be the lcm of the denominators appearing in $A \cup B$. Then for every $a \in (A\cup B)$, $ 2^a \in 2^c \mathbb Z [ 2^{1/d} ]$. If we have $$ \sum_{ a\in A} 2^a = \sum_{ b\in B} 2^b \in 2^c \mathbb Z [ 2^{1/d} ] $$ then $$ \sum_{ a\in A} 2^a \equiv \sum_{ b\in B} 2^b \mod 2^{c+1/d} \mathbb Z [ 2^{1/d} ] $$ We have $2^a=0$ mod $2^{c+1/d}$ if $a>c$, but $2^c \neq 0 \mod 2^{ c+ 1/d}$ so $\sum_{ a \in A} 2^a \equiv 2^c$ if $c \in A$ and $0$ otherwise. Since $c$ is either in $A$ or $B$ but not the other, the two sides are not congruent, hence not equal. We can express the set as $$\bigoplus_{ \substack{ a,b\in \mathbb N \\0 \leq a < b \\ \gcd(a,b)=1}} 2^{ \frac{a}{b} } \mathbb Z[1/2]^+ $$ where $ \mathbb Z[1/2]^+$ represents the positive dyadic rationals. Because of this positivity condition, there is no simpler description, e.g. involving the rings $\mathbb Z[ 2^{1/b}, 1/2]$.<|endoftext|> TITLE: Reference for coefficients of equivariant Eilenberg-MacLane spectra QUESTION [5 upvotes]: I would like to have proper references in a paper that I'm writing down. This concerns computations of the coefficients of equivariant Eilenberg-MacLane spectra over the cyclic group of order 2 (denoted here by $Q$), so I would like to know all of the examples appearing already in the literature. The ones I am aware of are (I'll list the underlying Mackey functors): the constant Mackey functor $\underline{\mathbb{F}}_2$ - these computations are built on the unpublished work of Stong and appear (among others) in Lewis's article on $RO(Q)$-graded cohomology of complex projective spaces ; the constant Mackey functor $\underline{\mathbb{Z}}$ - its coefficients are computed in Dugger's paper on Atiyah-Hirzebruch Spectral Sequence for $KR$-theory; the Burnside Mackey functor $\underline{A}$ - the description of $RO(Q)$-graded abelian structure is also built on the work of Stong and may also be found in Lewis's paper. Are there any other examples of such calculations? REPLY [7 votes]: It's done for an arbitrary Mackey functor, for the group $C_p$ for any prime, as Theorem 8.1 here: https://hopf.math.purdue.edu/Ferland-Lewis/FerlandLewis.pdf<|endoftext|> TITLE: Is analytification of regular holonomic D modules a fully faithful functor? QUESTION [5 upvotes]: It seems to me that by the algebraic Riemann Hilbert functor(which factors through the analytification map) and also the analytic Riemann Hilbert functor that the (derived) category of (algebraic) regular holonomic D-modules embeds into the (derived) category of analytic regular holonomic D-modules. Is this correct? I want to make sure I'm not missing anything. Is it also correct to say that this is a nontrivial fact and needs much of the theory of Riemann Hilbert correspondence to prove? REPLY [5 votes]: Yes, for a smooth algebraic variety $X$, the analytification functor $D^b(\mathcal D_X)_{rh} \to D^b(\mathcal D_{X^{an}})_{rh}$ is fully faithful. As you note, it is a consequence of the usual algebraic and analytic versions of the Riemann-Hilbert correspondence for $D$-modules: analytic $D$-modules correspond to complexes of sheaves which are constructible with respect to analytic stratifications and the algebraic category corresponds to the subcategory for which the stratification is algebraic. This is stated, for example, as Proposition 7.8 in this paper of Brylinski. As to how much of the theory of Riemann-Hilbert one needs to prove this fact, here are my thoughts. One way to prove the algebraic version of Riemann-Hilbert is by first proving the analytic version then proving the fully faithfulness of analytification statement in your question. If I understand correctly this is the approach taken in Brylinski's paper (there is a comment to that effect on page 174 of the book by Hotta-Tanisaki-Takeuchi). So in some sense, the statement in your question seems kind of orthogonal to the analytic version of the Riemann-Hilbert correspondence at least. To prove the fully faithfullness of analytification, first consider the case where $X$ is projective, in which case it follows from a suitable application of GAGA theorems (see Theorem 7.1 in that paper of Brylinski). Now, for a general smooth algebraic variety $X$, consider a nice compactification $j:X \hookrightarrow \overline{X}$. Then for a r.h. algebraic D-module $M$ on $X$ we can look at the $D$-modules $j_\ast M$ which will be r.h. on $\overline{X}$ thus reducing to the projective case (note that in the algebraic setting, unlike the analytic, the definition of regularity includes being regular "at $\infty$"). This proof requires some theory of regular singularities, but is independent of certain other fundamental facts involved in RH. For example, there is the basic fact that an analytic flat connection on $X$ extends uniquely to a meromorphic connection on $\overline{X}$ with regular singularities (I think this is due to Deligne?). There is also the constructibility of the solution/de Rham complex. I would say these essential aspects of Riemann-Hilbert are kind of orthogonal to (or at least independent of) the comparison of algebraic and analytic theories in your question.<|endoftext|> TITLE: Is there any use for $\sin(\sin x)$? QUESTION [26 upvotes]: The convention that $\sin^2 x = (\sin x)^2$, while in general $f^2(x) = f(f(x))$, is often called illogical, but it does not lead to conflicts because nobody uses $\sin(\sin x)$. But is this really true? Or is there a real-world application in which $\sin(\sin x)$ occurs? Or maybe something a bit more general, like $\sin(C \sin x)$ for some constant $C \neq 0$? REPLY [14 votes]: The characteristic function for the Poisson distribution is given by $$ e^{\lambda\left(e^{i\theta}-1\right)}=e^{\lambda\left(\cos\theta-1\right)}\left(\cos(\lambda\sin\theta)+i\sin(\lambda\sin\theta)\right) $$ REPLY [6 votes]: Valerii Salov, in Notation for Iteration of Functions, Iteral, has written a 23 page (!) paper on how to denote the iteration of a function, specifically for the issue raised by the OP in Inevitable Dottie Number. Iterals of cosine and sine. The notation proposed by Salov uses the first Cyrillic letter И of the Russian word Итерация for iteration, as a variation on the product symbol $\prod$, so that И$_v^n f$ is the n-fold iteration of the function $f(x)$ for initial value $x=v$. Some examples of the use of this compact notation: This can be typeset in MathJax as a LARGE version of Unicode character 1048: $$\large{ {\LARGE\unicode{1048}}_0^n(x+1)=n,\ {\LARGE\unicode{1048}}_{x=1}^n ax=a^n,\ {\LARGE\unicode{1048}}_{1}^\infty \frac{1}{x+1}=\lim_{n\to\infty} {\LARGE\unicode{1048}}_{1}^n \frac{1}{x+1}=\frac{2}{1+\sqrt{5}} }$$<|endoftext|> TITLE: Does there exist a constant $C$, such that, $\Pr[\max_k|\sum_{i\neq k}X_i|\ge t]\le C\Pr[|\sum_i X_i|\ge t]$ for all independent symmetric variables? QUESTION [5 upvotes]: Let $X_1, \ldots, X_n$ be independent symmetric variables. Now I would like to know whether there exists a constant $C$, such that $$ \Pr[\max_{k \in [n]} |\sum_{i \in [n] \setminus \{k\}} X_i| \ge t] \le C\Pr[|\sum_{i \in [n]} X_i| \ge t] $$ Now similar inequalities are Levy's inequalities \begin{align} \label{eq:levy-sum} \Pr[\max_{1 \le l \le n} |\sum_{i = 1}^l X_i| \ge t] &\le 2 \Pr[|\sum_{i = 1}^n X_i| \ge t] \\ \label{eq:levy-individual} \Pr[\max_{1 \le l \le n} |X_l| \ge t] &\le 2 \Pr[|\sum_{i = 1}^n X_i| \ge t] \end{align} Unfortunately, I am unable to use the techniques used to prove Levy's inequality in my case. The best result I have been able to prove is the following \begin{align*} \Pr[\max_{k \in [n]} |\sum_{i \in [n] \setminus \{k\}} X_i| \ge t] &\le \Pr[\max_{k \in [n]} |X_k| + |\sum_{i \in [n]} X_i| \ge t] \\&\le \Pr[\max_{k \in [n]} |X_k| \ge t/2] + \Pr[|\sum_{i \in [n]} X_i| \ge t/2] \\&\le 3\Pr[|\sum_{i \in [n]} X_i| \ge t/2] \end{align*} The last inequality follows from using the second of Levy's inequalities. But here we lose a factor of 2 on $t$, which I would like to avoid. REPLY [6 votes]: Alas, no such inequality can hold. Suppose that the symmetric $X_i$ take values $\pm 1$ and $t=n-1$. Then $$ \Pr[\max_{k \in [n]} |\sum_{i \in [n] \setminus \{k\}} X_i| \ge t] =(n+1)2^{1-n} $$ but $$\Pr[|\sum_{i \in [n]} X_i| \ge t]=2^{1-n} \,. $$<|endoftext|> TITLE: Random domino tilings: Is this distribution well-defined, and how can it be sampled from? QUESTION [5 upvotes]: I'd like to ask questions about a "random domino tiling of the plane". However, it's not quite obvious how to go about precisely specifying what this means. My first instinct was to do something like "the center of a random tiling of a large square". More formally, consider the following property for a random distribution on grid-aligned domino tilings in the plane: For each $r>0$, the distribution of domino positions within distance $r$ of the origin is the limit of the distributions on that region among randomly-selected tilings of $2N\times 2N$ origin-centered squares as $N$ goes to infinity. Clearly, if a distribution satisfying this property exists, it is unique, and seems like a reasonable definition to use. However, while intuitive, it is not clear to me how to show that the centers of random tilings of large squares converge in the necessary sense. The most pressing question about this distribution is whether it actually exists, which I strongly suspect is the case. Conditional on that being true, I have several followup questions: Is the same distribution obtained if we replace "square" with "torus" or "aztec diamond"? Random tilings of the latter are substantially easier to describe and generate (see the Arctic Circle theorem). Is the distribution translation-invariant? Again, I strongly suspect this is the case, but I don't know how I'd go about proving it. If so, how far into a large square do we need to go to see this distribution? E.g., is it the case that a patch of a random $2N\times 2N$ tiling at distance $\log(N)$ from the border asymptotically looks like a patch at the center? What exact probabilities of different configurations does it have? For instance, what are the odds that the origin is not a vertex of any domino? Concretely, at the center of a $2k\times 2k$ grid we have probabilities of $1,\frac19,\frac{361}{841},\frac{139129}{811801},\ldots$, which is approximately $1,0.1111,0.4293,0.1714,\ldots$. Is there a reasonably efficient algorithm to sample from finite portions of this distribution? Ideally in a constructive manner, i.e. an algorithm which when initialized with a random seed spits out progressively more and more dominoes around the origin that extend to a tiling of the plane. Possibly relevant is the fact that a random tiling of a $2N\times 2N$ square is given in the limit by starting with any tiling and randomly flipping any two dominoes joined in a $2\times 2$ square (see e.g. Laslier and Toninelli 2012). This was previously posted on Math StackExchange here, without much progress. In the comments, a paper of P. W. Kasteleyn was linked which may be relevant. REPLY [7 votes]: This distribution is the maximal entropy Gibbs measure for domino tilings of the plane. Burton and Pemantle (https://projecteuclid.org/euclid.aop/1176989121) proved many important facts about this distribution, including some remarkable formulas for specific probabilities. (But beware of typos: if I remember right, a few formulas have typos in them.) Rick Kenyon (http://www.numdam.org/article/AIHPB_1997__33_5_591_0.pdf) showed how to compute all probabilities of local configurations explicitly, via determinantal formulas. This is enough to answer your questions about tilings of squares or tori: everything converges to maximal entropy statistics and can be computed explicitly. For more general regions, it depends on the tilt of the height function, but you should get maximal entropy statistics whenever the height function is flat. To resolve the local statistics problem in full generality, you need two things: a classification of Gibbs measures, and the knowledge that the local statistics are translation-invariant in the continuum limit. Scott Sheffield (http://www.numdam.org/issue/AST_2005__304__R1_0.pdf) completed the classification of all ergodic, translation-invariant Gibbs measures, including the ones with smaller entropy. Amol Aggarwal (https://arxiv.org/abs/1907.09991) has a recent paper that proves translation invariance in the limit for lozenge tilings. That’s very close to the domino case, but extending it to that case would take a little more (see page 7 of Aggarwal’s article). For the special case of the Aztec diamond, Chhita, Johansson, and Young (https://projecteuclid.org/euclid.aoap/1427124128) obtained this local convergence result. I’m not sure about sampling algorithms. My initial thought is to relate the problem to spanning forests (see Burton and Pemantle’s article) and use Wilson’s cycle popping algorithm (https://dl.acm.org/doi/10.1145/237814.237880), but I haven’t thought this through carefully and I’m not sure what’s known about sampling from Gibbs measures.<|endoftext|> TITLE: Why does Riesz's Representation Theorem apply in quantum mechanics? QUESTION [8 upvotes]: $\DeclareMathOperator\tr{tr}$One begins with a quantum mechanical system, i.e. a unital $C^*$-algebra $A$. It is common to begin the discussion with embedding $A$ into the algebra of bounded operators $\mathcal{B}$ on some Hilbert space $H$. A state is defined as a positive linear functional $\varphi: A\rightarrow \mathbb{C}$ taking $1$ to $1$. Since neither $A$ nor $\mathcal{B}$ is a Hilbert space, we can't use Riesz's Representation Theorem directly. In Physics texts that I have encountered, however, a state is often cast as a trace $1$ operator $\alpha$ in $\mathcal{B}$, and its action on an observable $\rho$ is by $\tr(\rho\alpha)$. Recall that observables are self dual, and so $\tr(\rho\alpha)=\tr(\rho^* \alpha)$, which is highly suggestive of the subalgebra of $\mathcal{B}$ given by the Hilbert–Schmidt operators, which is a Hilbert space with the inner product $\alpha\cdot\beta=\tr(\alpha^*\beta)$. Indeed if you restrict $\varphi$ to the algebra of Hilbert–Schmidt operators, then $\varphi$ can be associated with a trace $1$ operator via Riesz's Representation Theorem. Questions What is going on here? Are we saying that states (defined as positive linear functionals taking $1$ to $1$) are completely determined by their restriction to the subalgebra of Hilbert–Schmidt operators in $A$? Or are there two competing definitions of "states" here? And if so, what is the merit of having these two different definitions? In https://math.stackexchange.com/questions/77820/a-question-about-pure-state they appear to suggest that not all pure states are represented by projections to one dimensional subspaces of $H$. This confuses me, because I thought those are exactly the pure states. Is this somehow an issue with diverging definitions, related perhaps to my first question? REPLY [12 votes]: $\DeclareMathOperator\Ann{Ann}\DeclareMathOperator\Tr{Tr}$My answer is somewhat complementary to Nik Weaver's, and admitedly more focused on Question 2 since I have nothing more to add to the latter regarding Question 1. When you deal with a not necessarily commutative C${}^*\!$-algebra $\mathfrak{A}$, the Riesz representation theorem is no longer the tool one uses to represent states. What is usually done is the so-called Gel'fand–Naimark–Segal (GNS) construction of a ${}^*\!$-representation $\pi_\omega:\mathfrak{A}\rightarrow\mathfrak{B}(\mathscr{H}_\omega)$ of $\mathfrak{A}$ in a Hilbert space $\mathscr{H}_\omega$ starting from a state $\omega$ on $\mathfrak{A}$. Let us recall the GNS construction for convenience. We assume for now that $\mathfrak{A}$ has a unit $\mathbb{1}$. Recall that a state on $\mathfrak{A}$ is a linear map $\omega:\mathfrak{A}\rightarrow\mathbb{C}$ such that $\omega(a^*a)\geq 0$ for all $a\in\mathfrak{A}$ and $\omega(\mathbb{1})=1$. This implies that $$\mathfrak{A}\times\mathfrak{A}\ni(a,b)\mapsto\omega(a^*b)$$ is a positive semidefinite Hermitian sesquilinear form on $\mathfrak{A}$, and therefore satisfies the Cauchy–Schwarz inequality $$\lvert\omega(a^*b)\rvert^2\leq\omega(a^*a)\omega(b^*b)\leq\|a\|^2\omega(b^*b)\ ,\quad a,b\in\mathfrak{A}\ $$ (the latter inequality comes from the fact that $b=\|a\|^2\mathbb{1}-a^*a$ is a positive element of $\mathfrak{A}$, i.e. it has the form $b=c^*c$ for some $c\in\mathfrak{A}$) This means that the so-called annihilator $\Ann \omega$ of $\omega$ $$\Ann \omega=\{a\in\mathfrak{A} \mathrel\vert \omega(a^*a)=0\}$$ is a left ideal (hence a vector subspace) of $\mathfrak{A}$, consisting of the zero-seminorm elements of $\mathfrak{A}$ with respect to the seminorm $\|\cdot\|_\omega$ on $\mathfrak{A}$ induced by this sesquilinear form: $$\|a\|_\omega=\sqrt{\omega(a^* a)}\ ,\quad a\in\mathfrak{A}\ .$$ Hence, the latter induces a complex scalar product on $\mathfrak{A}/\Ann \omega$ - if $[a],[b]$ are the respective equivalence classes of $a,b\in\mathfrak{A}$ modulo $\Ann \omega$, we write $$\langle[a],[b]\rangle=\omega(a^*b)\ .$$ Moreover, since $\Ann \omega$ is a left ideal, $\mathfrak{A}$ has a natural left action on $\mathfrak{A}/\Ann \omega$ as $$\pi_\omega(a)[b]=[ab]\ .$$ This defines a ${}^*\!$-representation $\pi_\omega$ on $\mathfrak{A}/\Ann \omega$ which satisfies $$\|\pi_\omega(a)[b]\|\leq\|a\|\,\|[b]\|=\|a\|\,\|b\|_\omega\ ,\quad a,b\in\mathfrak{A}\ ,$$ thanks to the Cauchy–Schwarz inequality. This means that $\pi_\omega$ extends uniquely to a ${}^*\!$-representation of $\mathfrak{A}$ in the Hilbert space $\mathscr{H}_\omega=\overline{\mathfrak{A}/\Ann \omega}$ by bounded linear operators therein. The state $\omega$ is then represented in $\mathscr{H}_\omega$ by the unit-norm element $\Omega_\omega=[\mathbb{1}]$, for $\omega(a)=\langle\Omega_\omega,\pi_\omega(a)\Omega_\omega\rangle$ for all $a\in\mathfrak{A}$. As you can see, the GNS construction "kind of" identifies $\mathfrak{A}$ with (a dense subspace of) the Hilbert space $\mathscr{H}_\omega$ - that is, modulo $\Ann \omega$. If the state is faithful, i.e. $\Ann \omega=\{0\}$, then $\mathfrak{A}$ is indeed identified with (a dense subspace of) $\mathscr{H}_\omega$ - this is equivalent to $\pi_\omega$ being injective, i.e. faithful, and to $\pi_\omega$ being isometric. This happens regardless of $\mathfrak{A}$ having Hilbert-Schmidt elements or not. However, even though any nonzero C${}^*\!$-algebra possesses a good deal of faithful states (this is a consequence of the Hahn-Banach theorem and leads to the important Gel'fand-Naimark theorem identifying abstract C${}^*\!$-algebras with closed ${}^*\!$-subalgebras of bounded linear operators in a Hilbert space), not all states of $\mathfrak{A}$ are faithful. Depending on which kind of C${}^*\!$-algebra $\mathfrak{A}$ and reference state $\omega$ you have, another given state $\eta$ on $\mathfrak{A}$ may or may not be representable as trace-class operators $\rho_\eta$ in $\mathscr{H}_\omega$ with unit trace, that is, given a linear map $\eta:\mathfrak{A}\rightarrow\mathbb{C}$ such that $\eta(a^*a)\geq 0$ for all $a\in\mathfrak{A}$ and $\eta(\mathbb{1})=1$ there may or may not be a $0\leq\rho_\eta\in\mathfrak{B}(\mathscr{H}_\omega)$ with $\Tr(\rho_\eta)=1$ such that $\eta(a)=\Tr(\rho_\eta\pi_\omega(a))$ for all $a\in\mathfrak{A}$. A situation where this is true regardless of which $\omega$ you choose is when $\mathfrak{A}$ is finite dimensional (i.e. a ${}^*\!$-algebra of matrices). This remains true for $\mathfrak{A}=\mathfrak{B}(\mathscr{H})$ with a separable Hilbert space $\mathscr{H}$ - i.e. a type-I factor (as mentioned at the end of Nik Weaver's answer), which is the case of the algebra of observables for physical systems with finitely many degrees of freedom. On the other hand, for the kind of C${}^*\!$-algebras that appear as algebras of observables for physical systems with infinitely many degrees of freedom (e.g. thermodynamic limits of quantum statistical systems and quantum field theory), this is usually false — as examples, one may cite thermal equilibrium states at different temperatures (in the thermodynamic limit), distinct pure thermodynamic phases of a non-pure thermal equilibrium state, different superselection sectors in quantum field theory, etc. In physical terms, infinitely many degrees of freedom usually is a manifestation of locality. More precisely, one usually considers the self-adjoint elements of $\mathfrak{A}$ as local observables measured within certain space(-time) regions (or limits of Cauchy sequences thereof). This means that states $\eta$ of the form $\eta(a)=\Tr(\rho_\eta\pi_\omega(a))$ may be seen as states "accessible" from $\omega$ through "physically allowed local operations", at least to an arbitrary degree of accuracy. Any other state is seen as "disjoint" from $\omega$. In infinitely extended space(-time) regions, there are usually plenty of mutually disjoint states on $\mathfrak{A}$.<|endoftext|> TITLE: Existence of complex function? QUESTION [7 upvotes]: Motivated by a similar question Complex-doubly periodic function in two variables?, I would like to ask if there exists a non-zero function $(z_1,z_2) \mapsto f(z_1,z_2)$, where $z_1,z_2 \in \mathbb C$ are two complex variables, that satisfies $$ (\partial_{z_2} + \partial_{z_1})f =0 \text{ and } (\partial_{\bar z_1} - \partial_{\bar z_2})f=0.$$ Notice that such a function $f$ is necessarily harmonic. $$ (\partial_{\bar z_2} \partial_{z_2} + \partial_{\bar z_1} \partial_{z_1} )f=0.$$ In addition, I require for $k_1,k_2 \in \mathbb R$ the periodicity conditions \begin{gather*} f (z_1+1,z_2 ) = f(z_1,z_2 ), \quad f (z_1+i,z_2 ) = f(z_1,z_2 ), \\ \text{and}\quad f (z_1,z_2+1 ) = e^{ik_1} f(z_1,z_2), \quad f (z_1,z_2+i ) = e^{ik_2} f(z_1,z_2). \end{gather*} Such a function $f$ must necessarily have poles, unless $f$ is constant, so what I am asking here is if there exists a function $f$ that satisfies the above differential equations up to a set of lower dimension where the functions exhibits poles and in addition all the periodicity conditions. REPLY [4 votes]: The answer is 'yes' there do exist such functions that are non-constant with singularities only along surfaces $\Sigma\subset\mathbb{C}^2$, and here is how one can understand them: First, it helps to change coordinates, though, perhaps, a little more subtly than Fedor Petrov suggested: Let $$ y_1 = \tfrac i2({\overline z}_1+{\overline z}_2 - z_1 + z_2) \quad\text{and}\quad y_2 = \tfrac12({\overline z}_1+{\overline z}_2 + z_1 - z_2) $$ Then $(y_1,y_2):\mathbb{C}^2\to\mathbb{C}^2$ is a diffeomorphism. Using the complex structure on $\mathbb{C}^2$ for which $y_1$ and $y_2$ are holomorphic coordinates, we find that $$ \frac{\partial f}{\partial{\overline y}_1} = \frac{\partial f}{\partial{\overline y}_2} = 0, $$ so that $f$ is a holomorphic function of $y_1$ and $y_2\,$. Moreover, we have the periodicity relations $$ f(y_1{+}1,y_2)= f(y_1,y_2{+}1) = f(y_1,y_2) $$ while $$ f(y_1{+}i,y_2)= \mathrm{e}^{ik_1}f(y_1,y_2) \quad\text{and}\quad f(y_1,y_2{+}i) = \mathrm{e}^{-ik_2}f(y_1,y_2) $$ Now, one can construct the general meromorphic $f$ that satisfies these condition as follows: First, let $p_1$ and $p_2$ be (not-identically-vanishing) meromorphic functions on $\mathbb{C}$ that satisfy the period relations $$ p_1(y+1) = p_1(y)\quad\text{and}\quad p_1(y+i) = \mathrm{e}^{ik_1}\,p_1(y) $$ and $$ p_2(y+1) = p_2(y)\quad\text{and}\quad p_2(y+i) = \mathrm{e}^{-ik_2}\,p_2(y). $$ Techniques for constructing meromorphic functions on $\mathbb{C}$ satisfying such double-periodicity relations are well-known, using $\vartheta$-series or the theory of elliptic curves. The point is that, if $C = \mathbb{C}/\mathbb{Z}[i]$ is the square torus (which is an elliptic curve), then the above relations on meromorphic functions on $\mathbb{C}$ essentially describe the meromorphic sections of two flat complex line bundles $E_1$ and $E_2$ over $C$, i.e., elements of $\mathrm{Pic}_0(C)$. Given $p_1$ and $p_2$, then the general meromorphic $f$ that satisfies the above conditions can be written as a product $$ f(y_1,y_2) = p_0(y_1,y_2)\,p_1(y_1)\,p_2(y_2), $$ where $p_0$ comes from any meromorphic function on $C\times C$, i.e., $p_0$ is a meromorphic function on $\mathbb{C}^2$ that satisfies $$ p_0(y_1{+}m,y_2)= p_0(y_1,y_2{+}m) = p_0(y_1,y_2) $$ for any Gaussian integer $m\in\mathbb{Z}[i]$. Note that the 'singularities' of $f$ occur along surfaces in $\mathbb{C}^2$ that are complex curves in the $y$-coordinates whose images in $C\times C$ are algebraic curves.<|endoftext|> TITLE: The "stubborn" solutions to sums of three cubes QUESTION [25 upvotes]: It is conjectured (see [1]) that for any integer $k\not\equiv \pm 4\pmod 9$ there are infinitely many integer solutions to $$ a^3+b^3+c^3=k. $$ Numerical investigations of this conjecture show that for some $k$ solutions are easily found (for example, $510=101^3-100^3-31^3$), while other $k$ kind of "resist" the numerical search. Notable example is $$ (−80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3=42 $$ found by A. Booker and A. Sutherland in 2019. But is this phenomenon known to exist for infinitely many $k$? More precisely, let $$ s(k)=\min_{a^3+b^3+c^3=k}\max\{|a|,|b|,|c|\}. $$ and $S(x)=\max\limits_{\substack{k\leq x\\ k\not\equiv \pm 4\pmod 9}} s(k)$. Obviously, we don't even know if $S(114)$ is finite, so there is no upper bound for $S$ known. But what can be said about lower bounds for $S(x)$ for large $x$? Obviously, $S(x)\gg x^{1/3}$. Is it at least true that $$ \limsup_{x\to +\infty} \frac{S(x)}{x^{1/3}}=+\infty? $$ I think that a solid case for existence of "stubborn" integers would be a bound of the form $S(x)\gg x^{f(x)}$ with $f(x)\to +\infty$ as $x\to +\infty$ or at least $S(x)\gg x^A$ for some large $A$ (like $A=100$, for example), but I don't know if anything like this is within the reach of current techniques or even likely to be true. REPLY [30 votes]: This lim sup indeed goes to $\infty$. We can prove this using exactly the strategy Lucia suggested. We will count the number of $x,y,z$ in a box with $x^3+y^3+z^3$ not a cubic residue modulo $p$ for a large finite list of primes $p$, all congruent to $1$ mod $3$. We can similarly count the number of $n$ not a cubic residue modulo $p$ for the same finite list of primes. From the pigeonhole principle, we can deduce that some $n$ is not a cubic residue for any $x,y,z$ in the box. For $p$ congruent to $1$ mod $3$, the number of solutions of $x^3 + y^3 + z^3 + w^3=0 $ in $\mathbb F_p$ is $$ p^3 + 6 p^2-6p $$ and the number of solutions to $x^3 + y^3 + z^3 = 0$ is $$1 + (p-1 ) (p - a_p + 1) = p^2 - a_p (p-1) $$ where $|a_p| < 2 \sqrt{p}$ so the number of $x,y,z\in \mathbb F_p$ such that $x^3+y^3+z^3$ is a perfect cube is $$ \frac{ p^3 + 6 p^2 -6p + 2 ( p^2 - a_p (p-1) )}{3} $$ and thus the number of $x,y,z$ such that $x^3+y^3+z^3$ is not a perfect cube is $$ p^3 - \frac{ p^3 + 6 p^2 -6p + 2 ( p^2 - a_p (p-1) )}{3} = \frac{ 2 p^3 - 8 p^2 - 2 a_p (p-1)}{3} .$$ Now let $c>0$ be an integer, and assume that $n>0$ is divisible by $\prod_{i=1}^k p_i^3$. Then the number of $x,y,z$ with $|x|, |y|, |z| < c n^{1/3} $ such that $x^3+y^3 + z^3$ is not a perfect cube mod any of $p_1,\dots p_k$ is $$8 c^3 n \prod_{i=1}^{k} \frac{ 2 p_i^3 - 8 p_i^2 - 2 a_{p_i} (p_i-1) + 6p_i }{3p_i^3},$$ while the number of numbers less then $n$ that are not perfect cubes mod any of the $p_1,\dotsc, p_k$ is $$ 2 n \prod_{i=1}^k \frac{ 2 (p_i-1)}{ 3 p_i }, $$ Therefore, by the pigeonhole principle, there must be some number $0$ sufficiently large, which lets the lim sup go to $\infty$, we need only ensure that the product goes to $\infty$ as the list $p_1,\dotsc, p_k$ grows to include all primes congruent to $1$ mod $3$. This is not hard to do as each factor is approximately $\frac{1}{ (1-p_i^{-1})^3}$ and the divergence follows from the fact that the primes congruent to $1$ mod $3$ have Dirichlet density $1/2 > 0$. If we take $p_1,\dots, p_k$ the $k$ smallest primes congruent to $1$ mod $3$, and $n^{1/3}$ to be $\prod_{i=1}^k p_i$, I believe this method gives a lower bound like $\limsup_{ x\to\infty} \frac{S(x)}{ x^{1/3} \sqrt{ \log \log x }} >0$.<|endoftext|> TITLE: Orbits space of real-analytic planar foliations QUESTION [5 upvotes]: Consider a foliation of $\mathbb{R}^2$, say coming from the trajectories of a vector field $X$. Its orbit space (the quotient of $\mathbb{R}^2$ by the relation "lying on the same trajectory") is seldom Hausdorff. Such foliated structures have been intensively studied, and a complete $C^r$-classification is due to Haefliger and Reeb in the case where $X$ is regular on a simply connected region (thanks to the same topological niceness of the plane used in the proof of Bendixon-Poincaré's theorem). Almost any reasonable one-dimensional, simply-connected non-Hausdorff manifold can be realized as the orbit space of a foliation. The two main sources of non-separability of orbits are: Saddle singularities: the stable and unstable (half-)manifolds cannot be separated. Limit cycles: the limit cycle cannot be separated from the accumulating trajectories. I believe that orbits space coming from real-analytic foliations should have a "nicer" structure. I expect also that the work of Kaplan, Haefliger, Reeb dating back from the 40--50's should have been generalized to the analytic setting. Is that so? Is there any special structure / characterization on the (non-Hausdorff) analytic orbits space of a real-analytic planar foliation that I should be aware of (and where can I find it)? A special case of particular interest is where the the vector field $X$ is the realification of a holomorphic vector field on $\mathbb C\simeq \mathbb R^2$. Now there are no limit cycles. The topology of the phase-portrait looks simpler and the orbit space also. Is there any known characterization of the analytic one-dimensional (non-Hausdorff) manifolds that can arise in this very special case? REPLY [2 votes]: You wrote "I believe that orbits space coming from real-analytic foliations should have a "nicer" structure". I think that this nicer structure arises when we consider a more technical "Leaf space" so called "Groupoid foliation" not mereley the orbit space. The orbit space has a very poor topology. For example it can not distingishe two topological different foliation of the torus with different slopes $"\sqrt 2"$ and $"\pi"$ see "The" kronecker foliation or "a" kronecker foliation? Note that the holomorphicity can not do any thing special if we resist on the plain orbit space and do not consider the groupoid foliation. Because the simplest holomorphic map in the world, the constant map $z'=c$ produce a complicated kronecker foliation. But the k theory of $C^*$ algebra of the corresponding groupoid foliation, is the true useful tool for studing the invisible features of the foliation. So with consideration of groupoid foliation $G(M,F)$, it is always a Haussdorf space if the foliation is real analytic. In general $G(M,F)$ is $k+n$ (not necessarily Hausdorf) manifold where $F$ is a $k$ -dimensional foliation of an n manifold Your question reminds me of an MO question of mine as follows. I have neither deleted that question nor I can find it in my question list. Is there a non analytic foliation whose groupoid foliation is diffeomorphic to an $S^3 $ with two north pole, a compact 3 dimensional analogy of a line with two origin? https://ncatlab.org/nlab/show/line+with+two+origins We identify two disjoint $S^3$ at all its points exept at north pole: In the disjoint union $S^3\times\{0\} \coprod S^3\times \{1\}$ we identify $(x,0)$ with $(x,1)$ for all $x\in S^3\setminus \{N\}$ so we obtain a 3-sphere with two north pole. Any way if you give a precise reference to papers of Haefliger et al one can search for thses papers to realize which kind of orbit space they are working with. P.S: for the holomorphic foliation $Z'=e^Z$ we have infinite number of points(in the leaf space according to your terminology) with the following property: There are infinite paires such that each pair consite of two points which can not seperate from each other. But is there an entire holomorphic function(non vanishing) for which the corresponding foliation does not admit infinit number of paires with the above property, but any way the leaf space is non Haussdorf? On the other extrem is there a non vanishing entire function whose corresponding leaf space(according to your terminology) which contains infinite number of points which mutually are non seperable? Note: You wrote: "Limit cycles : the limit cycle cannot be separated from the accumulating trajectories" But surprisingly in the groupoid foliation a limit cycle can seperates from other trajectories!. The only case it can not seperate is that "it is center from exterior and it is limit cycle from the interior or it is exteror limit cycle and interior center. Of course this can not happen in real analytic case.<|endoftext|> TITLE: Characterization of pretty compact spaces QUESTION [7 upvotes]: This is a cross post from MSE. I believe that the following problem have already been considered by some sophisticated topologist. Definition 1. A non-compact Hausdorff topological space $X$ is called almost compact if its Stone–Čech compactification coincides with its one point compactification. An example of almost compact space is $[0,\omega_1)$ for the first uncountable ordinal $\omega_1$. All almost compact spaces are locally compact and pseudocompact. Definition 2. A compact Hausdorff space $X$ is called pretty compact if $X\setminus\{p\}$ is almost compact for all non-isolated points $p\in X$. I know that Stonean spaces are pretty compact. By a result of van Douwen, Kunen and van Mill (There can be $C^*$-embedded dense proper subspaces in $\beta\omega - \omega$) $\beta\mathbb{N}\setminus \mathbb{N}$ is consistently pretty compact. What are other examples of pretty compact spaces? Does there exist any characterization of pretty compact spaces or at least a strong necessary condition? REPLY [9 votes]: A partial answer: other examples of pretty compact spaces are uncountable powers of $\{0,1\}$ and $[0,1]$, and in general products of uncountably many non-trivial compact Hausdorff spaces. See Problem 3.12.24(c) in Engelking's General Topology, or Glicksberg, Stone-Čech compactifications of products. If $a$ is in the product take $b$ in the product that differs everywhere from $a$. Then $\Sigma(b)$ is a subset of $X\setminus\{a\}$. As the product is $\beta\Sigma(b)$ it is also $\beta(X\setminus\{a\})$ (general result: if $X\subseteq Y\subseteq\beta X$ then $\beta X=\beta Y$). As noted in the question extremally disconnected compact spaces have the same property; as these are quite different from product spaces finding a characterization that is not a direct translation seems difficult.<|endoftext|> TITLE: Is there any limited access to MathSciNet for retired mathematics faculty? QUESTION [19 upvotes]: I am a retired mathematics professor and AMS member continuing to do research and publish papers. Unfortunately, my former university (39 years) allows library access only to Emeritus Professors so I have no access to JSTOR or MathSciNet, putting me at somewhat of a disadvantage. Needless to say, living on a retirement pension puts individual subscriptions beyond my means. Have other academic retirees found workarounds to similar situations? REPLY [25 votes]: Zbmath is now completely open, and hence it is a free alternative to Mathscinet.<|endoftext|> TITLE: Two-term recurrence relation QUESTION [9 upvotes]: We consider the following system of recurrence relations for $n \in \mathbb Z$ and $\vert \lambda \vert=1$ with $\lambda \in \mathbb{C}$ $$a_{n+1} = \lambda a_{n-1}+ \lambda^* a_n + \lambda^* n b_n $$ $$b_{n+1} = \lambda^* b_{n-1}+ \lambda b_n - \lambda n a_n. $$ Here, $\lambda^*$ is the complex conjugate of $\lambda.$ I am interested in non-zero initial conditions under which $a_n,b_n$ tend to zero for $n \rightarrow \pm \infty.$ Observation: If there is a limit $a_n \rightarrow a$ and $b_n \rightarrow b$, then indeed by the last term in each row, it has to be zero and $a_n =b_n=o(n)$ REPLY [2 votes]: I assume that the initial conditions $a_0,a_1,b_0,b_1$ and that $n\to +\infty$. Let $A(x):=\sum_{n\geq 0} a_n x^n$ and $B(x):=\sum_{n\geq 0} a_n x^n$. Then the recurrence relations become: $$\begin{cases} A(x) - a_1x - a_0 = \lambda x^2 A(x) + \lambda^* x (A(x)-a_0) + \lambda^* x^2 B'(x), \\ B(x) - b_1x - b_0 = \lambda^* x^2 B(x) + \lambda x (B(x)-b_0) - \lambda x^2 A'(x). \end{cases}$$ That is, we have a system of 2 linear first-order ODE: $$ \begin{bmatrix} A'(x)\\ B'(x)\end{bmatrix} = \begin{bmatrix} 0 & -\lambda^* x^{-2}+x^{-1}+\lambda^{*2}\\ \lambda x^{-2} - x^{-1} - \lambda^2 & 0\end{bmatrix} \cdot \begin{bmatrix} A(x)\\ B(x)\end{bmatrix} + \begin{bmatrix} \lambda^*(b_0x^{-2} + (b_1-\lambda b_0)x^{-1})\\ -\lambda(a_0x^{-2} + (a_1-\lambda^*a_0)x^{-1})\end{bmatrix}, $$ which may be analyzed with the standard methods.<|endoftext|> TITLE: Derivation of formal power series QUESTION [8 upvotes]: The basic idea of this question is to see if there is any other derivations than 'formal derivations'. Let $\mathbb{K}$ be a field. Given a commutative $\mathbb{K}$-algebra $A$, a derivation of $A$ is a $\mathbb{K}$-linear map $D:A\rightarrow A$ satisfying $D(ab)=D(a)b+ aD(b)$. Consider the case when $\mathbb{K}=\mathbb{R}$ and the algebra $A$ is the formal power series $\mathbb{R}[[x]]$. One derivation of $\mathbb{R}[[x]]$ is $p(x)\frac{\partial}{\partial x}$, defined formally as $$ p(x) \frac{\partial}{\partial x} \left(\sum_{n=0}^\infty c_n x^n \right)=p(x)\cdot \left(\sum_{n=0}^\infty n\cdot c_n x^{n-1}\right)$$ for any $p(x)\in \mathbb{R}[[x]]$. My question is that, are there any derivations that is not of the form $p(x)\frac{\partial}{\partial x}$ as above? If yes, is there any reference/proof/example for existence of such derivations? This question arose when I was thinking of vector fields of a $\mathbb{Z}$-graded manifold as a derivation of smooth functions. With my notion of graded manifolds, smooth functions are formal power series of virtual coordinates. When I was thinking about it, since $\mathbb{R}[[x]]$ is not generated by $\{x\}$ as an algebra, there should be some other derivations but I could not make it clear. Any help to figure it out would be greatly appreciated. REPLY [6 votes]: Every $\mathbb{K}$-derivation $D$ of $\mathbb{K}[[x_1,\dots,x_n]]$ has the standard form $$D(f)=\sum_{i=1}^n p_i \frac{\partial f}{\partial x_i},$$ where of course $p_i=D(x_i)$. Indeed, let $\mathfrak{m}$ be the maximal ideal of $\mathbb{K}[[x_1,\dots,x_n]]$: then clearly $D(\mathfrak{m}^{N+1})\subset \mathfrak{m}^N$ for all $N\geq0$, so $D$ is continuous for the $\mathfrak{m}$-adic topology. The claim follows because the formula holds when $f$ is a polynomial, $\mathbb{K}[x_1,\dots,x_n]$ is dense in $\mathbb{K}[[x_1,\dots,x_n]]$, and $\mathbb{K}[[x_1,\dots,x_n]]$ is Hausdorff.<|endoftext|> TITLE: Is an $H_0^1$ function continuous to the boundary if it is continuous in the interior? QUESTION [5 upvotes]: Suppose $\Omega$ is a bounded domain in $\mathbb R^3$ with Lipchitz boundary $\partial\Omega$, and $u\in H_0^1(\Omega)\cap C(\Omega)$. Is $u$ continuous to the boundary i.e. do we have $u \in C( \overline{\Omega})$? In other words, is is true that $H_0^1 (\Omega)\cap C(\Omega)\subset C(\overline \Omega)$? Depending on the answer(s), I may have some follow-up questions (for what it's worth). Thank you any and all in advance. Edit: It seems the answer is no so I am adding follow-up questions: Can I get that $u$ is bounded and/or attains its maximum on $\overline\Omega$? REPLY [5 votes]: The answer to the follow-up question is negative too. For consider the half-ball $\Omega=\{x\,;\,x_3>0,\,|x|<1\}$. Choose a number $\alpha\in(1,\frac32)$, and a function $\phi\in C^\infty({\mathbb R}^3)$ such that $\phi(x)\equiv1$ for $|x|<\frac13$, while $\phi(x)\equiv0$ for $|x|>\frac23$. Then the function $u(x)=r^{-\alpha}x_3\phi(x)$ belongs to $H^1(\Omega)\cap C(\Omega)$. Its trace, being an element of $H^{1/2}(\partial\Omega)$, is a square-integrable function, hence can be determined by looking away from a negligible Lebesgue set. Thus we look at the trace away from the origin, where $u$ is continuous and vanishes at the boundary. Therefore the trace is $\equiv0$, that is $u\in H^1_0(\Omega)$. Yet, it is not a bounded function.<|endoftext|> TITLE: What makes Gaussian distributions special? Local field version? QUESTION [7 upvotes]: This question is inspired by the recent one about Gaussian measures over the reals: What makes Gaussian distributions special? I would be interested in a similar list of characterizations for the probability measure, on the field of $p$-adic numbers $\mathbb{Q}_p$, whose density with respect to the standard additive Haar measure is given by the indicator function of the ring of $p$-adic integers $\mathbb{Z}_p$. REPLY [4 votes]: Perhaps one answer is the appearance of the Gaussian in the oscillator (a.k.a. Weil) representation --- which makes sense even in non-zero characteristic. You're probably aware of the construction of this infinite dimensional representation for the double cover of $\mathrm{SL}_2(\Bbb{R})$. The story for $\mathrm{SL}_2 \big( \Bbb{F}_q \big)$ where $q$ a power of a prime strongly parallels the continuous story except, of course, it's finite: Choose any non-square $\delta \in \Bbb{F}_q$ and form the quadratic extension $\Bbb{F}_q(\delta) \cong \Bbb{F}_q^2$. As usual, we identify elements of $\Bbb{F}_q(\delta)$ as linear combinations $z = a + \sqrt{\delta} b$ with $a, b \in \Bbb{F}_q$ subject the usual formulae for addition and multiplication; conjugation and norm are expressed as $\bar{z} = a - \sqrt{\delta} b$ and $\mathrm{N}(z) = a^2 - \delta b^2$ respectively. We'll need to choose (any) non-trivial additive character $\psi: \Bbb{F}_q \longrightarrow \Bbb{C}^*$ together with any multiplicative character $\chi: \Bbb{F}_q(\delta)^* \longrightarrow \Bbb{C}^*$ which generates the character group of $\Bbb{F}_q(\delta)^*$. Morally $\psi$ plays the role of the exponential function $\exp: \Bbb{R} \longrightarrow \Bbb{R}^*$ in this finite context. Now set \begin{equation} W_\chi := \ \left\{ \begin{array}{l} \displaystyle \text{all functions} \ f: \Bbb{F}_q(\delta) \longrightarrow \Bbb{C} \ \ \text{such that} \\ \displaystyle f(wz) = \overline{\chi(w)} \, f(z) \ \text{whenever $\mathrm{N}(w)=1$} \end{array} \right\} \end{equation} which is $q-1$ dimensional. The oscillator representation $\varrho_\chi: \mathrm{SL}_2(\Bbb{F}_q) \longrightarrow \mathrm{GL}(W_\chi)$ is determined by the action of the follow elements, which generate $\mathrm{SL}_2(\Bbb{F}_q)$: \begin{equation} \begin{array}{ll} \displaystyle \varrho_\chi { \scriptstyle \begin{pmatrix} x & 0 \\ 0 & x^{-1} \end{pmatrix}} f(z) &\displaystyle = \ f(xz) \\ \displaystyle \displaystyle \varrho_\chi { \scriptstyle \begin{pmatrix} 1 & \ \ y \\ 0 & \ \ 1 \end{pmatrix}} f(z) &\displaystyle = \ \underbrace{\psi \big( y \, \mathrm{N}(z) \big)}_{\text{the Gaussian $G_y(z)$}} \cdot f(z) \\ \displaystyle \displaystyle \varrho_\chi { \scriptstyle \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}} f(z) &\displaystyle = \ -q \widehat{f}(-\bar{z}) \end{array} \end{equation} where $\widehat{f}$ is the Fourier transform of $f$ with respect to the additive structure of $\Bbb{F}_q(\sqrt{\delta})$. Like the continuous Gaussian, $G_y(z)$ is an eigenfunction of this Fourier transform. As far as I know, the construction that I've outlined (and which I learned from Amritanshu Prasad's online notes) can be carried out for $\mathrm{SL}_2\big( \Bbb{Z}/q\Bbb{Z} \big)$ where $q= p^N$ is still a power of a fixed prime $p$. Furthermore a coherent choice of additive and multiplicative characters $\psi_N$ and $\chi_N$ can be made for each $N \geq 1$ so that these oscillator representations agree with the inverse system \begin{equation} 1 \longleftarrow \Bbb{Z}/p\Bbb{Z} \stackrel{\ \text{mod $p$}}{\longleftarrow} \Bbb{Z}/p^2\Bbb{Z} \stackrel{\ \text{mod $p^2$}}{\longleftarrow} \Bbb{Z}/p^3\Bbb{Z} \longleftarrow \, \cdots \end{equation} thus allowing the oscillator representations $\varrho_{\chi_N}$ to be exported to the $p$-adic integers $\Bbb{Z}_p$, which will inherent some kind of Gaussian-like function $\varprojlim G_{\bf y}$ where ${\bf y}= (y_1,y_2,y_3, \dots)$ and $y_N = y_{N+1} \, \text{mod $p^N$}$ for each $N \geq 1$.<|endoftext|> TITLE: Why does not a closed 3-manifold modelled on SL(2,R) admit a metric of nonpositive curvature? QUESTION [8 upvotes]: I was reading the paper `actions of discrete groups on nonpositively curved spaces' written by Kapovich and Leeb. In this paper, they proved that generic mapping class groups are not Hadamard groups, i.e. no discrete actions on CAT(0) spaces by semi-simple isometries. In their proof, they said that since the unit tangent bundle of a hyperbolic surface is modelled on SL(2,R)-geometry, it does not admit a metric of nonpositive curvature. How can I prove the above statement? Can you suggest me any references? REPLY [10 votes]: If you read our paper a bit further, you will find that on page 348 we mention that this result is due to Eberlein and give a reference to his 1982 paper. More precisely, he proves a more general theorem that a nonpositively curved compact Riemannian manifold whose fundamental group has nontrivial center has a finite-sheeted covering space which splits smoothly as a product of a torus with another factor. For Seifert manifolds, this is equivalent to the types $E^3$ and $H^2\times R$.<|endoftext|> TITLE: Canonical lift of the deformation of an ordinary abelian variety QUESTION [11 upvotes]: If $A/k$ is a principally polarised ordinary abelian variety ($k$ a perfect field of characteristic $p$, we may assume it is finite for simplicity), we have a canonical lift $\hat{A}/W(k)$. Now if I take a deformation $A_{\epsilon}/k[\epsilon]$, does there still exist a canonical lift of this deformation to a deformation of (the generic fiber of) $\hat{A}$? By the Kodaira-Spencer mapping deformations are essentially encoded by differentials of $A$, so the question boils down to whether differentials on $A$ lift canonically to differentials on $\hat{A}$. By Katz, Serre-Tate local moduli, Section 3, differentials on any lift $\tilde{A}$ correspond to points in $T_p(A^\vee)(k)$, and his main theorem 3.7.1 describes the compatibility of this identification with the Kodaira-Spencer map. Is there a way to use this to lift differentials canonically on the canonical lift? REPLY [14 votes]: No. The picture over a general base is this: let $A_0\to S_0$ be an ordinary abelian variety over a characteristic $p$ scheme $S_0$, and let $S_n$ ($n\geq 0$) be compatible flat liftings of $S_0$ over $\mathbf{Z}/p^{n+1}$. Let $F\colon S_0\to S_0$ be the Frobenius. Then the pull-back $(F^n)^* A_0$ has a canonical lifting $A_{n,\rm can}$ to $S_n$. Thus if $S_0$ is perfect (in which case automatically $S_n = W_{n+1}(S_0)$), then we can undo the Frobenius twist $(F^n)^*$ above, and we obtain a compatible system of lifts $A_n = ((F^{-n})^* A_0)_{n, \rm can}$ to $S_n$, i.e. a formal abelian variety over the formal scheme $S_\infty = \varinjlim_n\, S_n$. If you prefer, you can also say that $A_0$ has a canonical lifting $\tilde A_n$ to $W_{n+1}(S_0)$ for all $n$ (this is an equivalent point of view taken in a paper by Borger and Guerney). If $S_n$ is as above, there is a canonical map $\phi_n \colon S_n\to W_{n+1}(S_0)$ such that $\phi_n\circ i = j\circ F^n$ where $i\colon S_0\hookrightarrow S_n$ and $j\colon S_0\to W_{n+1}(S_0)$ are the canonical closed immersions. Then $\phi_n^* \tilde A_n \simeq A_{n, \rm can}$ as abelian schemes over $S_n$. Unfortunately, in your case $S_0 = \operatorname{Spec} k[\varepsilon]/(\varepsilon^2)$ the Frobenius factors through $\operatorname{Spec} k$, so the above gives you nothing. But it does give you something e.g. for $S_0 = \operatorname{Spec} k[\![t]\!]$. EDIT. Coming back to the heart of your question. Let $D=\operatorname{Def}_{A_0/W(k)}$ be the formal deformation space of $A_0$ over $W(k)$. The essence of Serre-Tate theory is that $D$ carries a canonical lifting of Frobenius $F\colon D\to D$, which in particularly chosen multiplicative coordinates $q_{ij}$ ($1\leq i,j\leq g$), i.e. $D\simeq \operatorname{Spf} W(k)[\![q_{ij}-1]\!]$, has the particularly simple form $F^*(q_{ij})=q_{ij}^p$. (These coordinates may only exist over a finite separable extension of $k$, so let us assume that $k$ is algebraically closed for simplicity.) The "canonical coordinates" $q_{ij}$ can be recovered back from $F^*$ once we fix a basis of the $p$-adic Tate module of $A_0$, and then the Hodge $F$-crystal of the universal formal abelian variety over $D$ is explicitly described as in Katz's article or the accompanying article by Deligne and Illusie (with an appendix by Katz) "Cristaux ordinaires et coordonnees canoniques". So how does the Frobenius $F\colon D\to D$ give us canonical liftings? It is easy to check (see section 1 in Katz "Travaux de Dwork") that for every lifting of Frobenius $F$ on $R\simeq W(k)[\![x_1, \ldots, x_n]\!]$ there exists a unique homomorphism $\sigma_F\colon R\to W(k)$ over $W(k)$ which commutes with the Frobenius lifts, i.e. $\sigma_F F=F_{W(k)}\sigma_F$ where the second $F_{W(k)}$ is the unique Frobenius lift $F_{W(k)} = W(F_k)\colon W(k)\to W(k)$ on the Witt vectors. If $\operatorname{Def}_{A_0/W(k)} \simeq \operatorname{Spf} R$ and $F$ is the Serre-Tate lift of Frobenius, then $\sigma_F$ corresponds to an element $\tilde A_{\rm can}$ of $\operatorname{Def}_{A_0/W(k)}(W(k))$, i.e. a (formal) lifting of $A_0$ over $W(k)$. What happens if we replace $W(k)$ with some $p$-adic $W(k)$-algebra $V$ (a quotient of $W(k)[\![x_1, \ldots, x_n]\!]$ for some $n$)? Do we get some canonical lift of $A_0$ over $\operatorname{Spf} V$? For this to make sense, we must assume that we are given some deformation $A_{V_0}$ of $V_0$ over $\operatorname{Spf} V_0$ where $V_0 = V/pV$, and that $V$ is endowed with a Frobenius lifting $F_V\colon V\to V$. Thus the question is whether there exists a natural (possibly unique) $\sigma\colon R\to V$ with $\sigma F=F_V\sigma$ lifting the $R_0\to V_0$ induced by $A_{V_0}$. The answer to this is no. For example, let $V=W_2(k)[\varepsilon]/(\varepsilon^2)$, so $V_0 = k[\varepsilon]/(\varepsilon^2)$. Let $F_V\colon V\to V$ be a lifting of Frobenius, so $F_V(\varepsilon)=\varepsilon^p + p\beta = p\beta$ for some $\beta\in V_0$. Let $R \simeq W(k)[\![q_{ij}-1]\!]$ as in Serre--Tate theory, so $F(q_{ij})=q_{ij}^p$. Let $R_0\to V_0$ send $q_{ij}$ to $\varepsilon + 1$ for all $i$ and $j$ (just so that the tangent direction does not lie in any of the hyperplanes/subtori $q_{ij}=1$). Suppose that $\sigma\colon R\to V$ lifts $R_0\to V_0$, so $\sigma(q_{ij}) = \varepsilon + 1 + pg_{ij}$ for some $g_{ij}\in V_0$. The condition $\sigma F = F_V\sigma$ means that $(\varepsilon+1+pg_{ij})^p = p\beta + 1 + pg_{ij}^p$. But the left hand side is $(1+pg_{ij})^p + p\varepsilon(1+pg_{ij})^{p-1} = 1 + p\varepsilon$. We get $\beta + g_{ij}^p = \varepsilon$. Thus $\varepsilon - \beta$ has to be a $p$-th power in $V_0$, which is a non-trivial condition on $\beta$. (Perhaps this can be simplified) What is true though is this. We have a canonical homomorphism $W_{n+1}(V_0)\to V_n$ ($V_n=V/p^{n+1}$), taking $(x_0, \ldots, x_n)$ to $y_0^{p^n} + py_1^{p^{n-1}} + \cdots + p^ny_n$, where $y_i\in V_n$ are lifts of $x_i\in V_0$. Moreover, the lifting of Frobenius $F$ on $R$ induces a homomorphism $t_F \colon R_n\to W_{n+1}(R_0)$ (called "Cartier's arrow" in Illusie "Complexe de de Rham-Witt...", chapitre 0). We can now form the composition $$ \tilde\sigma_0\colon R_n \to W_{n+1}(R_0)\xrightarrow{W_{n+1}(\sigma_0)} W_{n+1}(V_0)\to V_n $$ where $\sigma_0\colon R_0\to V_0$ is the given map. This map turns out to satisfy $\tilde\sigma_0 F = F_{V_n}\tilde\sigma_0$ for every lifting of Frobenius $F_{V_n}\colon V_n\to V_n$. However, modulo $p$ this map reduces to $F^n\sigma_0 = \sigma_0 F^n$. Therefore we obtain a lifting of $(F^n)^*A_{V_0}$, not of $A_{V_0}$, over $V_n$. This recovers what I said in the first paragraph.<|endoftext|> TITLE: Elementary theory of the category of groupoids? QUESTION [11 upvotes]: One axiomatisation of set theory, the Elementary Theory of the Category of Sets, or ETCS for short, comes from category theory and states that sets and functions form a locally cartesian-closed, finitely complete and co-complete Heyting pretopos with a subobject classifier and a natural numbers object, whose generating object is the terminal object and whose epimorphisms are split. Is there a corresponding axiomatic charactersation of the category of groupoids, in which one could do groupoid theory in, that does not involve first defining the concept of $\infty$-groupoid or homotopy types or other infinity categorical structures, and if so, what are the axioms? And just as ETCS could be considered as a foundation of mathematics that only require sets and propositions, could this Elementary Theory of the Category of Groupoids serve as a more general foundation of mathematics that includes everything that could be done in ETCS as well as providing proper foundations for $1$-category theory? The collection of objects $Ob(C)$ in a category $C$ tend to be a groupoid; when $Ob(C)$ is a set, the category is called strict, so category theory as defined in ETCS or another set theory like ZFC could only speak of strict categories rather than general categories. Edit: In a finitely complete category, finite limits are saturated under the terminal object and pullbacks. Does this still remain true when one moves to (2,1)-terminal objects and (2,1)-pullbacks and (2,1)-limits in finitely (co)complete (2,1)-categories? For ($\infty$,1)-categories, it doesn't seem to be the case that finite ($\infty$,1)-limits are saturated under ($\infty$,1)-terminal objects and ($\infty$,1)-pullbacks, if I am reading the nLab article on Lurie's ($\infty$,1)-pretopos correctly. In a (2,2)-category, (2,2)-terminal objects and (2,2)-pullbacks are known not to be enough for all finite (2,2)-limits; (2,2)-powers with the interval category are also needed. Edit 2: I commented somewhere below that this theory as a foundational theory should be expressed in first order logic with isomorphisms, rather than first order logic with equality. I don't think this is the case anymore; ETCS should be the theory expressed in first order logic with isomorphism, as sets in ETCS are only isomorphic rather than equal. Rather, ETCG should be expressed in first order logic with equivalence of groupoids. It is only models of ETCG internal to ETCG that are expressed in first order logic with isomorphism, in the same way that models of ETCS internal to ETCS are expeessed in first order logic with equality. REPLY [6 votes]: I'm showing up a bit late to this party, but maybe I still have something to add. As Andrej and others have pointed out, one can obtain a type theory for 1-groupoids by starting with any form of HoTT and adding a 1-truncation axiom. (It's amusing (and perhaps deep) that in the type-theoretic context, it's easier to start from a type theory for $\infty$-groupoids and then assert an extra axiom that cuts down to 1-groupoids.) The paper of Harper and Licata is about a version of such a theory that satisfies the nice technical property of "canonicity", but if all you want is a formal system you can just add the 1-truncation axiom to Book HoTT --- and modify the univalence axiom so it doesn't contradict that. Basically you should assert only that there is a univalent universe of sets (0-truncated types). Now, it sounds from the question like you're more interested in a "category-theoretic" phrasing of this, looking more like the phrasing of ETCS as "the category of sets is an elementary topos such that blah". I don't think this has been written down as such; for higher values of $n$ it's more common to discuss Grothendieck $n$-toposes than elementary ones. But similar to the type-theoretic case, one could start from a proposed definition of elementary $(\infty,1)$-topos and cut it back down to a (2,1)-category. This would produce something like: an elementary (2,1)-topos is a (2,1)-category (i.e. a category enriched in groupoids, or a 2-category or bicategory whose hom-categories are groupoids) such that It has finite limits and colimits, in the bicategorical sense. It is locally cartesian closed, in a bicategorical sense: pullback along any morphism has a right bicategorical adjoint. It has a subobject classifier, i.e. the functor sending each object to the poset of fully faithful inclusions into it is representable. For any faithful morphism, there is a generic faithful morphism classifying it and such that the morphisms it classifies are closed under composition, finite fiberwise limits and colimits, and dependent products. The last axiom is the "object classifier", and there's some room for discussion in how it should be phrased. The above phrasing will allow you to use it to reason about arbitrary 0-truncated objects, but it requires some large cardinals to model. I expect that with the object classifier, one can prove that the category is "(2,1)-exact" in the sense that any "internal groupoid" in a suitable sense is the "kernel" of some quotient. A while ago I made some study of exactness conditions for 2-categories, the remnants of which can be found here. If one wants to avoid any universes, one could take this exactness as part of the definition instead, although exactness isn't the only purpose of the universe so this would impoverish the definition a bit. Finally one would want to add some axioms analogous to the blah in ETCS. A (2,1)-category should be "well-pointed" if the terminal object is a "generator" in a suitable sense. Certainly the functor $E(1,-) : E \to Gpd$ should be locally faithful. Probably it's too much to ask it to be locally fully faithful. One might need to add some more "projectivity and indecomposablity" conditions analogous to the constructive notion of well-pointedness for 1-categories, if they don't follow automatically the way they do for classical 1-categories; I haven't thought too much about this. Then if one wants a "classical" version analogous to ETCS, one can add the axiom of choice. The only thing to be aware of here is that it can pertain only to 0-truncated objects, as discussed in type theory in the HoTT Book. In a (2,1)-category, one way to state the internal axiom of choice is that for any (not necessarily truncated) object $U$ and 0-truncated (i.e. representably faithful) morphisms $Y\to X\to U$ such that $Y\to X$ is surjective, there exists a surjective map $V\to U$ such that the pullback of $Y\to X$ to $V$ has a section. Note that $V$ need not be 0-truncated either. In the presence of well-pointedness, this might simplify to a more "external" axiom of choice analogous to ETCS's "all epimorphisms split", but I haven't thought much about that either.<|endoftext|> TITLE: De Bruijn's sequence is odd iff $n=2^m-1$: Part I QUESTION [8 upvotes]: Among the families of sequences studied by Nicolaas de Bruijn (Asymptotic Methods in Analysis, 1958), let's focus on the (modified) $$\hat{S}(4,n)=\frac1{n+1}\sum_{k=0}^{2n}(-1)^{n+k}\binom{2n}k^4.$$ An all-familiar fact states: the Catalan number $C_n=\frac1{n+1}\binom{2n}n$ is odd iff $n=2^m-1$. In the same tradition, I ask: QUESTION. Is this true? $\hat{S}(4,n)$ is odd iff $n=2^m-1$ for some $m\in\mathbb{Z_{\geq0}}$. POSTSCRIPT. This question has been extended to my other MO question. REPLY [5 votes]: I just wished to suggest an alternative (mechanical) proof whose details can be provided (using the WZ-method) for the identity $$\sum_{k=0}^{2n}(-1)^{n+k}\binom{2n}k^4=\binom{2n}n\sum_{k=0}^n(-1)^k\binom{2n+k}k^2\binom{2n}{n+k}.\qquad \qquad (*)$$ The idea is: verify that both sides of $(*)$ satisfy the 2nd-order recurrence \begin{align*} A(n)a(n+2)- B(b)a(n+1) +C(n)a(n)=0 \end{align*} with initial conditions $a(0)=1$ and $a(1)=14$; where \begin{align*} A(n)&=(2n + 3)(48n^2 + 66n + 23)(n + 2)^3 \\ B(n)&=13056n^6+ 96288n^5+ 289600n^4+ 453428n^3 + 388698n^2 + 172598n+31030 \\ C(n)&=4(n + 1)(48n^2 + 162n + 137)(2n + 1)^3. \end{align*}<|endoftext|> TITLE: Integral homology of braid groups as a ring QUESTION [5 upvotes]: Let $Br_k$ denote the braid group on $k$ strands. In Corollary A.4 of "Homology of Iterated Loop Spaces" (Page 348), Cohen-Lada-May compute $H_i(Br_k;\mathbb Z)$ as an abelian group for each $i$ and $k$. There is a ring structure on $\bigoplus_{k,i} H_i(Br_k;\mathbb Z)$ induced by the maps $Br_k \times Br_j \to Br_{k+j}$. Is the ring structure on $\bigoplus_{k,i} H_i(Br_k;\mathbb Z)$ known? Cohen-Lada-May compute it with field coefficients. Equivalently, is the ring structure on $H_*(\Omega^2 S^2;\mathbb Z)$ known? REPLY [8 votes]: Using the Hopf fibration you can show that $\Omega^2S^2\simeq\mathbb{Z}\times\Omega^2S^3$. Also $\pi_1(\Omega^2S^3)=\pi_3(S^3)=\mathbb{Z}$, so the Universal Coefficient Theorem gives $[\Omega^2S^3,S^1]=H^1(\Omega^2S^3)=\text{Hom}(\pi_1(\Omega^2S^3),\mathbb{Z})$. Using this we see that $\Omega^2S^3=S^1\times W$ for some $W$. We then find that the homotopy groups of $W$ are all finite, and thus that the homology groups of $W$ are all finite. Thus, the reduced integral homology of $W$ is the direct sum of the $p$-local homologies for all $p$, and it is easy to see that the product of any two summands is zero in the ring structure. The Cohen-Lada-May calculation shows that $H_*(W;\mathbb{Z}/p)=P\otimes E$, where $E$ is exterior on generators $x_k$ of degree $2p^k-1$ for $k>0$, and $P$ is polynomial on generators $y_k=\beta(x_k)$ of degree $2p^k-2$. It follows that the Bockstein homology $\ker(\beta)/\text{image}(\beta)$ is just $\mathbb{Z}/p$ in degree zero (by decomposing the whole ring as the tensor product of subrings generated by a single pair $\{x_k,y_k\}$, and applying the Kunneth isomorphism). This means that in positive degrees we have $\ker(\beta)=\text{image}(\beta)$. Let $\rho\colon H_*(X;\mathbb{Z})\to H_*(X;\mathbb{Z}/p)$ be the reduction map. There is a lifted Bockstein operation $\widetilde{\beta}\colon H_k(X;\mathbb{Z}/p)\to H_{k-1}(X;\mathbb{Z}_{(p)})$ with $\beta=\rho\widetilde{\beta}$ and $\widetilde{\beta}\rho=0$. Using this one can check that $\rho$ gives an isomorphism $H_*(W;\mathbb{Z}_{(p)})\to\ker(\beta)$ in positive degrees. I don't think you can expect to get a nice presentation of this ring by generators and relations. In this kind of context, you should usually avoid integral homology if you possibly can. Most things that you might want to do with integral homology can be done by combining rational and mod $p$ homology. In some cases Brown-Peterson homology is more tractable than integral homology.<|endoftext|> TITLE: Finite order elements of $\mathrm{GL}_d(\mathbb{Z})$ that are conjugate to powers of themselves QUESTION [8 upvotes]: $\DeclareMathOperator\GL{GL}$Let $A\in \GL_d(\mathbb{Z})$ have finite order $n.$ Suppose that $k\in \mathbb{Z}$ is relatively prime to $n.$ Is $A^k$ conjugate to $A$ in $\GL_d(\mathbb{Z})$? For $d\leq 4$ the answer is yes. Indeed the papers "On the finite subgroups of $\GL(3,\mathbb{Z})$" by K. Tahara, 1971 and "Conjugacy Classes of Torsion in $\GL_n(\mathbb{Z})$", by Q. Yang, 2015 list all torsion elements up to conjugacy for the cases $d=2,3$ and $d=4$ respectively. There are not many cases that need to be checked, so I checked each case and the answer turns out to be "yes" in all of these cases. But I have no general argument for why things work out when $d\leq 4$, only computations. REPLY [14 votes]: The answer is "no" in general. There may be an elementary way of seeing this, but I will frame this in representation theoretic terms and will describe a general construction. The question is equivalent to the following question: let $C_n$ be a cyclic group of order $n$, let $g$ be a generator, and let $\rho\colon C_n\to {\rm GL}_d(\mathbb{Z})$ be an integral representation of $C_n$, defined by $g\mapsto A$. Let $\sigma\in {\rm Aut}(C_n)$ be defined by $g\mapsto g^k$. Then is the integral representation $\rho\circ\sigma$ of $C_n$ isomorphic to $\rho$? Let me now explain a general construction of representations for which the answer is "no". Set $d=\phi(n)$, the Euler phi function. For example if $n$ is prime, then we will have $d=n-1$. Fix a primitive $n$-th root of unity $\zeta_n$, and let $I$ be an ideal in $\mathbb{Z}[\zeta_n]$, the ring of integers of the $n$-th cyclotomic field $\mathbb{Q}(\zeta_n)$. Let $C_n$ act on $I$ by letting $g$ act by multiplication by $\zeta_n$. Since $I$ is an ideal, multiplication by $\zeta_n$ preserves it as a set, so this defines a representation $\rho_I\colon C_n\to {\rm GL}_d(\mathbb{Z})$. Moreover, it is not hard to see that if $I$ and $J$ are two such ideals, then $\rho_I$ is isomorphic to $\rho_J$ if and only if $I$ and $J$ represent the same class in the ideal class group of $\mathbb{Q}(\zeta_n)$. Finally, given $k$ coprime to $n$, there exists an element $\sigma$ of the Galois group of $\mathbb{Q}(\zeta_n)$ such that $\sigma\zeta_n = \zeta_n^k$. This $\sigma$ also defines an automorphism of $C_n$ via $g\mapsto g^{k}$. It is easy to see that $\rho_I\circ \sigma=\rho_{\sigma^{-1}I}$. Therefore, to produce an example of the type you desire, you can take a cyclotomic field in which the Galois group acts non-trivially on the class group and let $I$ be a representative of any class in the class group that is not preserved by the Galois group. The smallest example that you can produce this way has $n=23$, $d=22$. Indeed, the cyclotomic field $\mathbb{Q}(\zeta_{23})$ is the first cyclotomic field with non-trivial class group, which has order $3$, and complex conjugation acts by $-1$ on it. So take any non-principal ideal $I$ in $\mathbb{Z}[\zeta_{23}]$, write down a $\mathbb{Z}$-basis for it, and write out the matrix $A$ of multiplication by $\zeta_{23}$ on this basis. Then $A$ will not be conjugate to $A^{-1}=A^{22}$ in ${\rm GL}_{22}(\mathbb{Z})$.<|endoftext|> TITLE: Is there a metamathematical $V$? QUESTION [22 upvotes]: As with many of you, I've been following Peter Scholze's recent question about universes with great interest. In ring theory, we don't often have to deal with proper classes, but they occasionally pop their heads. For instance, one cannot take the product of all countable rings---but one can replace that proper class with the set of isomorphism types of countable rings and get something that encodes a lot of the important information that would have existed in the proper class product (if it existed). I've similarly viewed the axiom of universes as an elegent way to avoid Russell's paradox when wanting to talk about something "close" to the proper class of all rings, groups, sets, etc.; the discussion at Peter Scholze's question has been very enlightening in that regard. This brings to mind a related question I've been struggling with for quite a while. I view ZFC as a formalization of those mathematical techniques that I take for granted when working with collections. As I've delved deeper into first order logic, and formal languages, the importance (in my mind) of metamathematical assumptions has increased. One of the questions that I'm told was historically important was whether or not there was a "completed" infinity, or in other words whether or not the natural numbers actually form a completed whole collection. I'd say that modern mathematics comes down strongly on the side of a completed infinity. In particular, I've understood that the axiom of infinity in ZFC is supposed to reflect a similar metamathematical assumption. One does not think of the axiom of infinity merely as a formal rule, interpretable in a finitistic sense (even if, technically, it could be). On a more concrete level, I'd guess most mathematicians think of questions about the infinite behavior of Turing machines as actually having answers (metamathematically and/or Platonically). (This is not to dismiss those who prefer to think of the natural numbers as an uncompleted collection. These issues definitely raise interesting philosophical question as well as mathematical questions, such as how much these metamathematical assumptions affect standard mathematics.) So, my first question is whether or not, metamathematically speaking, we should take $V$ (the proper class of all sets) as a completed whole or not. If that is too philosophical, the second is much more mathematical: Does this matter for formalizing mathematics? If not, does taking $\mathbb{N}$ as a completed whole matter for formalizing/doing mathematics, or can we easily do without that assumption? If we believe $\mathbb{N}$ exists as a completed whole, and work by analogy, then the answer to the first question seems to be obviously yes. But if we take Russell's paradox at face value, then the answer seems to be obviously no. Moreover, the different standard options available to formalize mathematics seems to leave this question open-ended: A ZFC-formalist might say no, there is no meta-$V$ because there are not even any formal proper classes, they are just an elegant informal way to talk about uncompleted infinities. An NBG-formalist might say yes, and our meta-$V$ is only somewhat like sets. An MK-formalist might say yes, and our meta-$V$ is really nearly a set. A believer in ZFC+Universes, might say yes or no. Yes, if we think of our meta-$V$ varying as we change our universe of expression, but maybe no if we want it to have all the properties of being the full (meta-)universe (on pain of Russell's paradox). The reason I'm asking this question is that I would guess most mathematicians would say something like "No, it doesn't matter whether or not we take $V$ as meta-mathematically existing." But I'd guess those same mathematicians would say that it does matter that we take $\mathbb{N}$ as meta-mathematically existing, else we can't even begin to formalize how to construct a (completed) language, etc. REPLY [7 votes]: Let me offer a different perspective, informed by my recent conversion from ZFC to ETCSR, i.e. Lawvere's elementary theory of the category of sets (with replacement). Recall that the two theories are "equivalent" in that each interprets the other, and models naturally correspond; the difference between them is essentially linguistic. In ETCSR, one formalizes the category of all sets. So there is a (meta?)mathematical category of all sets $\mathrm{Set}$. I'd like to make the following analogy: Mathematics probably starts by counting, leading to the integers $0,1,2,\ldots$. This quickly leads one to wonder how far this goes -- is there a "largest integer"? Of course, that's an obviously self-contradictory concept. Then there was a conceptual leap in mathematics, from mere numbers to the concept of sets. We allowed ourselves to contemplate taking all the integers, forming a new kind of object, a set. Then just like Peano arithmetic formalizes the integers on some inductive idea of the formation of integers, we formalize sets in ZFC based on some inductive idea of the formation of sets. In particular, we can form larger and larger sets. But just like for the integers, it turns out that the idea of "largest set" is self-contradictory. In the ETCSR point of view, what we do now is to make another conceptual leap in mathematics, from sets to categories. In this conception, we allow ourselves to contemplate the category of all sets, and then also "completed" categories of groups, of rings, etc. So somehow just like for the integers, the correct question was now that of "a largest integer larger than all the others", but that of collecting all the integers into a new kind of object -- a set -- the correct question here is not that of "a largest set containing all the other sets", but that of collecting all the sets in to a new kind of object -- a category. I realize that the last 'correct' is up for debate, and that this perspective puts me on a (slippery?) slope that would ultimately want to formalize the category of all categories, ..., leading one to $\infty$-categories and homotopy type theory. Maybe that's the way to go, but for now I'd only take one step at a time.<|endoftext|> TITLE: What is the lowest complexity definition of $\mathbb{Z}$ in an infinite extension of $\mathbb{Q}$ QUESTION [8 upvotes]: In 2009, Jochen Koenigsmann showed that $\mathbb{Z}$ is universally definable in the field $\mathbb{Q}$. And in 2012, Jennifer Park proved a result which implies that $\mathbb{Z}$ is $\exists\forall$-definable in many if not all finite extensions of $\mathbb{Q}$. But I’m wondering about infinite extensions. My question is, is there a low-complexity definitions of $\mathbb{Z}$ in some infinite extension of $\mathbb{Q}$? What is the lowest-known complexity for a definition of $\mathbb{Z}$ in some infinite extension of $\mathbb{Q}$? REPLY [10 votes]: There is an existential definition of $\mathbb{Z}$ in the rational function field $\mathbb{R}(t)$ by a beautiful result of Denef using elliptic curves (Proposition 2 of The diophantine problem for polynomial rings and fields of rational functions, Trans. Amer. Math. Soc. 242 (1978), 391-399 doi:10.1090/S0002-9947-1978-0491583-7). Regarding algebraic infinite extensions of $\mathbb{Q}$ I'm not aware of any such result. In this related discussion, Tom Scanlon mentions "natural infinite algebraic extensions in which $\mathbb{Z}$ is definable", but I'm not sure what he is referring to, and what the complexity of the definition would be.<|endoftext|> TITLE: Tensor triangulated categories associated to schemes and their families QUESTION [6 upvotes]: For any essentially small, rigid and idempotent-complete tensor triangulated (TT for short) category $\mathcal{T}$ Balmer (The spectrum of prime ideals in tensor triangulated categories) constructs a locally ringed topological space Spec($\mathcal{T}$). When $\mathcal{T}$ is the derived category of perfect complexes, $D^o(S)$, over a quasi-compact quasi-separated (qcqs) scheme $S$, the construction recovers $S$ as a locally ringed space together with its structure sheaf from the tensor triangular structure of $D^o(S)$. My questions are: (1) Given a TT category, can we tell when it is equivalent to $D^o(X)$ for a qcqs scheme $X$? Let's call the TT categories that fulfill the criterion schematic. (2) If we take an essentially small additive category and put different schematic tensor triangulated structures on it, we'd get different schemes via Balmer's construction. Can all flat families of schemes of a certain type (e.g., smooth curves of genus $g$ over a field $k$) be obtained by varying the tensor structure alone? (3) What are some natural discrete invariants of TT-categories and how can we build their moduli spaces after fixing values of the discrete invariants? REPLY [7 votes]: I don't have an answer to this, but this is too long for a comment. As for your fist question, one thing to check is whether the resulting locally ringed space is a scheme. This doesn't happen always, as Balmer showed in his paper Spectra, spectra, spectra Proposition 9.7. He concretely shows that the locally ringed space associated to $SH^{fin}_{(p)}$, the topological stable homotopy category of finite spectra localized at p with smash product as the monoidal structure, is not a scheme. He does so by checking that if it were a scheme it would be the spec of the ring of global sections, which he shows is local and then compares the number of points of the underlying space. I believe that you could abstract a bit how this counterexample goes and arrive to a statement with a somewhat contrived test to check whether the space is a scheme. It is my impression that the general suspicion is that triangulated categories of topological nature would hardly give you a scheme, but then again I don't know of very general theory on this idea. However even if the space is a scheme I think it wont guarantee you the TT-category is a derived category. A semisimple abelian category is triangulated, so the usual tensor product on something like $k-Mod$ will yield $Spec(k)$ under Balmer's reconstruction, but $k-Mod$ is not (I don't think?) equivalent to a derived category of a scheme. But this example looks a bit extreme. As for your second question this seems even tougher, the extreme case I can think of is when the derived category already determines the scheme as the Bondal-Orlov reconstruction setting, some discussion on the tensor products that arise in this situation was discussed in this MO question although this doesn't address your question per se.<|endoftext|> TITLE: What is dispersive estimate? QUESTION [7 upvotes]: Consider free NLS: $i\partial_tu+\Delta u=0, \quad u(0, x)=u_0$ The solution of this IVP, can be written as $$u(x,t)=e^{it\Delta}u_0(x)$$ It is clear to me that how to prove following estimate: $$ \|e^{it\Delta} f \|_{L^{p'}} \leq |t|^{-\frac{d}{2} \left( \frac{1}{p}-\frac{1}{p'} \right)} \|f\|_{L^{p}} \quad (1\leq p \leq 2)$$ My question is: Why this estimate is known as dispersive estimates? REPLY [5 votes]: To complete Piero's answer, let me give a definition of dispersiveness. To stay at a rather simple level, let me consider an linear evolution PDE with constant coefficients, $$P(\partial_t,\nabla_x)u=0.$$ Hereabove, $P$ is a polynomial. Let me assume a property of homogeneity, $$P(\lambda^\alpha\tau,\lambda\xi)\equiv\lambda^m P\tau,\xi).$$ It is met by the wave equation and the Schrödinger operator. When studying the wave propagation, one faces the algebraic equation $$P(\tau,\xi)=0,\qquad\xi\in{\mathbb R}^n,\quad\tau\in{\mathbb C}.$$ This defines roots $\tau_1(\xi),\ldots,\tau_k(\xi)$ - when $\tau_j(\xi)$ is real, $\nabla_\xi\tau_j$ is a group velocity. In the best situation, they have constant multiplicities for $\xi\ne0$, and then are smooth functions. Then the equation is dispersive if none of these functions is linear, meaning that the velocities $\nabla\tau_j$ do vary with $\xi$. Because $\tau(\lambda\xi)=\lambda^\alpha\tau(\xi)$ for $\lambda>0$, every equation whose order in time differs from the order in space ($\alpha\ne1$) is dispersive - mind however that if $\alpha<1$, then the Cauchy problem is not even well-posed. But even if $\alpha=1$, the equation can be dispersive, as shown by $\Box u=0$. Other examples come from linearized gas dynamics: the acoustic waves satisfy $\tau(\xi)=v\cdot\xi\pm c|\xi|$ ($v$ the fluid velocity, $c$ the sound speed) and thus are dispersive, though the entropy waves satisfy $\tau(\xi)=v\cdot\xi$, hence are not dispersive. The fact that dispersive equations satisfy $L^p-L^q$ decay estimates is an extension of so-called Strichartz estimates. REPLY [4 votes]: The intutitive picture is the following: there is a certain amount of "mass", which at time $t=0$ you might imagine concentrated in a heap near a location $x=0$. This heap evolves with time according to a differential equation. If the equation is a transport equation or a 1D wave equation (or the solution is a soliton for some suitable nonlinear equation), the heap moves in some direction but remains concentrated near some point $x(t)$. This behaviour is non dispersive. But if the equation is of dispersive type, the heap also spreads as time evolves: the total mass will remain the same, by conservation of mass, but it will spread on a larger and larger region of space. This is a dispersive behaviour. In your example, if you take $p=2$ you have conservation of mass, measured in $L^2$ norm. However, if you take $p=\infty$, you see that the "peak" of the heap becomes lower, at a rate of $t^{-n/2}$. If you take as initial data a wave packet, you can actually see this effect: the packet moves at a speed proportional to the frequency (as postulated by quantum mechanics), it spreads (this is called decoherence in physics), and its height decreases precisely at the stated rate.<|endoftext|> TITLE: Translation of "The joy of learning" by Hironaka QUESTION [8 upvotes]: According to this Quanta article about June Huh, there exists a memoir by Heisuke Hironaka called The Joy of Learning. It seems to be this short article: Heisuke Hironaka, The joy of learning, SEIBUTSU BUTSURI KAGAKU 44 Issue 2 (2000) pp. 53-57, doi:10.2198/sbk.44.53 which is in Japanese. On the other hand, this page refers to an English translation by Bang Seung-yang. Does anyone know where this translation can be found? Supposedly some generations of Japanese and Korean children have been greatly inspired by it, so it could be very interesting to read. REPLY [12 votes]: It seems like that there is no such translation. Moreover, the essay you linked and the book you linked are different material, although they have the same title. I searched the book whose author is Heisuke Hironaka, but I cannot find any essays written by Heisuke Hironaka, except for those in Japanese. Does it mean the book referred by June Huh and the linked catalogue does not exist? No. It is common to change the title of a book, movie, etc., during translation. An example would be Castle in the Sky, whose original title is Tenkū no Shiro Rapyuta (天空の城ラピュタ, whose meaning is Laputa: a castle in the sky.) We may guess that the same happens for Hironaka's autobiography, but neither June Huh nor the translator of the catalog has not double-checked the original title of the book. The above scenario is in part supported by your linked catagolue on Korea University. It mentioned the translator is Bang Seung-Yang. I can find that 방승양 (Bang Seung-Yang) is the translator of Hironaka's book Hakmun ui Jeulgeoum (학문의 즐거움, The joy of Learning.) Korea University is, as the name suggests, a university in Korea. Hence we may guess the catalogue is originally written in Korean, then translated into English later. But the translater was likely not doing any double-check the title of the books. Then what is the original title of Hironaka's autobiography? After a quick search, I found that a thread on Reddit suggested Gakumon no Hakken (学問の発見, The discovery of academics.) Hironaka wrote other books for non-mathematicians, like Ikiru koto Manabu Koto (生きること学ぶこと, About Living, about learning), and Hironaka Heisuke no Sūgaku Kyōshitsu - Dare demo Sūgaku ga Suki ni nareru (広中平祐の数学教室―誰でも数学が好きになれる, Heisuke Hironaka's Mathematics class - Everyone can love mathematics.) I compared the contents of the books with the Korean one, and I can see that Reddit's suggestion was correct: the original title of The Joy of Learning should be The discovery of academics.<|endoftext|> TITLE: Subgroups of Mod(S) generated by Dehn twists depend only on intersection numbers? QUESTION [5 upvotes]: $\DeclareMathOperator\Mod{Mod}$Let $S$ be a closed surface and $\Mod(S)$ be its mapping class group. It is a well known fact, proved in the Primer on Mapping class groups for example, that the subgroup of $\Mod(S)$ generated by two Dehn twists $T_a$ and $T_b$ depends only on the geometric intersection number $i(a,b)$. Indeed, if $i(a,b) \geq 2$ a ping pong lemma argument on the curve graph shows that $\langle T_a, T_b \rangle$ is a free group. Also, if $i(a,b) = 0$, then $\langle T_a, T_b \rangle$ is clearly free abelian. Lastly, if $i(a,b) = 1$, it can be shown that $\langle T_a, T_b \rangle$ is a braid group on 3 strands, unless $S$ is a torus. I am wondering whether such a statement can be proven in general, even without knowing a full classification of the groups that can appear. More precisely, if $\{a_1, \cdots, a_n\}$ and $\{b_1, \cdots, b_n\}$ are two collections of curves such that $i(a_i,a_j) = i(b_i,b_j)$ for all $i,j$, is it true that $\langle T_{a_1}, \cdots, T_{a_n} \rangle \simeq \langle T_{b_1}, \cdots, T_{b_n} \rangle$ as subgroups of $\Mod(S)$?. It seems that the group $\langle T_{a_1}, \cdots, T_{a_n} \rangle$ depends only on a regular neighborhood $N_a$ of the curves $a_i$, which is then homeomorphic to any regular neighborhood $N_b$ of the curves $b_i$. But in general the injection $i \colon N_a \to S$ does not induce an injective map on the mapping class group levels, so this does not seem to give a proper argument. Any help is greatly appreciated! REPLY [6 votes]: This is not true and I don't see the "right" side-conditions to make it true. Here is an example. Suppose that $a_1, a_2, a_3$ all lie in a single handle (surface of genus one, with one boundary component). and all meet exactly once, pairwise. Then the twist about $a_3$ lies in the group generated by the others, giving the usual three-strand braid group. Suppose that $b_1, b_2, b_3$ all meet exactly once, pairwise. Let $N$ be a regular neighbourhood of their union. Assume that $N$ has genus one and three boundary components. Assume that all boundary components are essential in the ambient surface $S$. Then the group generated is some Artin group with three generators.<|endoftext|> TITLE: Groups with a unique lonely element QUESTION [18 upvotes]: Does there exist a finite group $G$ of order greater than two containing a unique element $g$ such that $$ g\notin\langle x\rangle \hbox{ for all $x\in G\setminus\{g\}$ ?} $$ Or we have another fantastic property of the order-two group? Clearly, such a group must be a 2-group, and the unique lonely element must be central and of order two. REPLY [23 votes]: I think that the nontrivial semidirect product of a cyclic group of order 4 $\langle x\rangle$ acting on another cyclic group of order 4 $\langle y\rangle$ is an example of such a group. The center of this group is $\langle x^2,y^2\rangle$ and $x^2y^2$ is not a square. I came up with this trying to prove that no such group existed. In an example the center cannot be cyclic and then a minimal example had to be like this.<|endoftext|> TITLE: A finitely presented group whose rational cohomology is not nilpotent QUESTION [7 upvotes]: Does there exist a finitely presented (preferably $\text{FP}_{\infty}$) group $\Gamma$ and an element $\alpha \in \text{H}^{\ast>0}(B\Gamma;\mathbf{Q})$ that is not nilpotent? If non-discrete groups were allowed, the Euler class $e \in \text{H}^2(BS^1;\mathbf{Q})$ would do the trick, and there are corresponding classes in $\text{H}^2(BC_p;\mathbf{F})$ for $\mathbf{F}$ a finite field of chracteristic $p$, and $C_p$ cyclic of order $p$. One approach I thought of is to apply the Kan–Thurston theorem. But the (uncountable) groups appearing in their construction cannot be easily replaced by finitely generated ones, unless the complex one starts with is of low dimension. See the second half of sub-section 2.2 in their paper. REPLY [7 votes]: Let me compile the comments into an official answer: yes, such a group exists. As @dodd predicted in a comment, Thompson's group $F$ does the trick. Brown's computation of the cohomology ring (http://pi.math.cornell.edu/~kbrown/papers/homology.pdf) reveals non-nilpotent elements, e.g., the element he denotes $u$, which lives in degree 2 and generates a divided polynomial ring. (Thanks to @BenjaminSteinberg for the comment making me realize things were a lot more straightforward than I initially thought.)<|endoftext|> TITLE: Algebraic and rational parts of a real number QUESTION [6 upvotes]: Let $\alpha$ be a positive real number. Does it make sense to define the closest rational to $\alpha$ as the number $R(\alpha)=\frac{p_1}{p_2}$ such that $p_1,p_2$ are positive co-prime integers minimizing $p_2 \cdot |p_2\alpha - p_1|$? Clearly, there are going to be some irrational numbers for which this makes no sense, for instance $\alpha=\frac{1+\sqrt{5}}{2}$. My guess is that such numbers are rare: there are countably infinitely many of them, or their set has Lebesgue measure 0. Do we have $R(\pi) = \frac{355}{113}$? I looked at the fist 12 convergents of $\pi$, and $\frac{355}{113}$ achieves the minimum. This is related to approximations of irrational numbers, continued fractions, and the irrationality measure of a number. More generally, we could define the closest algebraic number of degree $d$, as the number $R_d(\alpha)$ defined as follows. $$P^*= \arg\min_P\Big(H(P)\Big)^{\mu(d)}\cdot |P(\alpha)|,\\ R_d(\alpha) = \Big(H(P^*)\Big)^{\mu(d)}\cdot |P^*(\alpha)| $$ where $P$ is any polynomial of degree $d$ with integer coefficients, with highest and lowest coefficients not equal to $0$, and $H(P)$ is the height of $P$, that is, its highest coefficient in absolute value. To make this work for most $\alpha$, how should we choose $\mu(d)$? Does $\mu(d)=d$ work? It seems to work if $d=1$. As of now, as far as I know, all results involving $d>1$ are conjectures. Related material includes the Wirsing conjecture. See also the "Generalizations" section in the Wikipedia article on Roth's theorem, here. Update on Feb 8, 2021: It is possible that the best approximation, if $d$ is an even integer, may be a complex number. Also, see my new question here, about approximations by dyadic fractions. The plan is to look at approximations using the first $n$ digits of $\alpha\in [0, 1]$ in base $b$, where (say) $b=\sqrt{2}$, focusing on values of $n$ where a long run of zeros start, leading to approximations of transcendental numbers by quadratic irrationals. This is briefly discussed in the comments in my new question. REPLY [10 votes]: Let $\alpha$ be an irrational. We shall consider its continued fraction $[a_0;a_1,a_2,\dots]$. Recall some basic results about convergents of continued fractions (see e.g. here): letting $p_n,q_n$ be the sequence of numerators and denominators of convergents, for any $n>1$ we have $q_{n+1}=a_{n+1}q_n+q_{n-1}>a_{n+1}q_n$ and $$\left|\alpha-\frac{p_n}{q_n}\right|<\frac{1}{q_nq_{n+1}}<\frac{1}{a_{n+1}q_n^2},$$ hence $q_n|q_n\alpha-p_n|<\frac{1}{a_{n+1}}$. Therefore if $a_{n+1}$ are unbounded, then $q_n|q_n\alpha-p_n|$ does not attain a minimum. Therefore $R(\alpha)$ does not exist for those $\alpha$. Therefore $R(\alpha)$ can only exist for badly approximable numbers. It is known that those numbers form a set of measure zero, and most natural constants besides quadratic irrationalities, including $\pi$ and all higher degree algebraic irrationals, are conjectured to not lie in it. Therefore $R(\pi)$ probably doesn't exist. On the other hand, if $\alpha$ is badly approximable, then this still doesn't necessarily mean $R(\alpha)$ necessarily exists, as you note with $\alpha=\frac{1+\sqrt{5}}{2}$. In fact I believe it won't exist for any quadratic irrational. However, using the bound $$\left|\alpha-\frac{p_n}{q_n}\right|>\frac{1}{q_n(q_{n+1}+q_n)}>\frac{1}{(a_{n+1}+2)q_n^2},$$ we at the very least get that for those numbers the quantity $q_n|q_n\alpha-p_n|$ is bounded away from zero (note that $q|q\alpha-p|$ can only be smaller than $1/2$ if $p/q$ is a convergent, so we don't lose much from looking at just looking at convergents). Last remark I have is that there are uncountably many $\alpha$ for which $R(\alpha)$ exists. Indeed, from the above considerations it follows easily that this is the case if for some $N$ we have that the continued fraction of $\alpha$ contains a partial denominator $N$, but from some point on all denominators are at most $N-2$.<|endoftext|> TITLE: Action of symmetric matrices under $\mathrm{O}(n)$ QUESTION [8 upvotes]: $\DeclareMathOperator\Sym{Sym}\DeclareMathOperator\O{O}\DeclareMathOperator\GL{GL}$Let $k$ be an algebraically closed field of characteristic 0 (it can even be $\mathbb{C}$ if you like), and let $n\in\mathbb{N}$. Let $\Sym(n)$ denote the space of $n\times n$ symmetric matrices with entries in $k$. Let $\O(n)\subseteq \GL(n)$ denote the orthogonal group, that is all $n\times n$ matrices $A$ with entries in $k$ such that $AA^t=A^tA=I_n$. Then $\O(n)$ acts on $\Sym(n)$ by conjugation. I would like nice representatives for the orbits of this action. Let $B\in \Sym(n)$. If $B$ is diagonalizable, then it is a theorem that it can be diagonalized by an orthogonal matrix, so all the 'semisimple orbits' have a diagonal matrix as a representative. So my question is about when $B$ is not diagonalizable — what is a 'nice' form we can put it in under orthogonal conjugation? (For an example of a non-diagonalizable symmetric matrix, see https://math.stackexchange.com/questions/1658393/why-a-complex-symmetric-matrix-is-not-diagonalizible.) Seeking explanations or references. Thank you! REPLY [7 votes]: The adjoint action of $SO(n,\mathbb C)$ on $\mathrm{Sym}_0(n,\mathbb C)$ (we can replace $O(n,\mathbb C)$ by $SO(n,\mathbb C)$, and $\mathrm{Sym}(n,\mathbb C)$ by the subspace of traceless matrices, since the action splits off the scalar multiples of the identity) is the isotropy representation of the complex symmetric space $SL(n,\mathbb C)/SO(n,\mathbb C)$. Kostant and Rallis (1971) investigated such representations. This was later generalized by Vinberg (1976) to the case of a $\theta$-group (the $\mathfrak g_0$-action on $\mathfrak g_1$, where we consider the decomposition of a complex Lie algebra $\mathfrak g=\mathfrak g_0 +\mathfrak g_1+\cdots + \mathfrak g_m$ into eigenspaces of an automorphism $\theta$ of order $m$ ($1$-parameter group of automorphisms in case $m=\infty$). Among other things, they found that for such a representation $(G_0,V)$: There is a Jordan decomposition: every $x\in V$ can be uniquely written as $x_s+ x_n$, where $x_s$ is a semisimple element (this means that $ad_{x_s}$ is a semisimple transformation) , $x_n$ is a nilpotent element (this means that $\mathrm{ad}_{x_n}$ is a nilpotent transformation), and they commute. A maximal Abelian subspace $c$ of $V$ consisting of semisimple elements is called a Cartan subspace. Any two Cartan subspaces are $G_0$-conjugate. Semisimple elements are characterized by the fact that their $G^0$-orbits are closed, or they lie in a Cartan subspace, or they are $G^0$-conjugate to an element in a given Cartan subspace. In your example, a Cartan subspace is given by the diagonal matrices, and the semisimple elements are exactly the diagonalizable symmetric matrices. Nilpotent elements are characterized by the fact that $0$ lies in the closure of their orbits. Nilpotent elements of course come in $G^0$-orbits, and there are only finitely many nilpotent orbits. In the case of a complex symmetric space $\mathfrak g=\mathfrak k + \mathfrak p$ (under $\theta$), the Kostant-Sekiguchi correspondence establishes a bijection between the set of nilpotent $K$-orbits on $\mathfrak p$ and the set of nilpotent $G_{\mathbb R}$-orbits on $\mathfrak g_{\mathbb R}$, where $\mathfrak g_{\mathbb R}$ is the noncompact real form of $\mathfrak g$ whose Cartan involution induces $\theta$ on $\mathfrak g$. In turn, the nilpotent $G_{\mathbb R}$-orbits on $\mathfrak g_{\mathbb R}$ are parametrized by partitions. More explicitly, in our case we have the symmetric space $\mathfrak{sl}(n,\mathbb C)=\mathfrak{so}(n,\mathbb C)+\mathrm{Sym}_0(n,\mathbb C)$ and the nilpotent $SO(n,\mathbb C)$-orbits in $\mathrm{Sym}(n,\mathbb C)$ are in bijection with the nilpotent $SL(n,\mathbb R)$-orbits in $\mathfrak{sl}(n,\mathbb R)$. The latter can be understood in terms of the Jordan canonical form. As an explicit example of the Kostant-Sekiguchi correspondence, the nilpositive element $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ of an $\mathfrak{sl}_2$-triple in $\mathfrak{sl}(2,\mathbb R)$ corresponds under the Cayley transform to the nilpositive element $\frac12\begin{pmatrix}i&1\\1&-i\end{pmatrix}$ of a normal $\mathfrak{sl}_2$-triple in $\mathfrak{sl}(2,\mathbb C)$; that is essentially the example given by @JohannesEbert. One can work the details of the correspondence more generally, see e.g. the book https://www.amazon.com/Nilpotent-Orbits-Semisimple-Lie-Algebra/dp/0534188346<|endoftext|> TITLE: Mathematical construction of $\phi^4$ Euclidean field theory QUESTION [11 upvotes]: One possible approach to constructive field theory is to define it on a lattice and take the scaling limit, and there are famous results stating that in $d\geq4$ this cannot lead to a non-trivial theory. What is the status of approaches using the Gaussian free field? What I mean is this: let $\Omega\subseteq\mathbb{R}^d$ be a smooth domain and consider the Gaussian free field $\{\langle f,\varphi\rangle\}_{f\in H_0^1(\Omega)}$ in $\Omega$ with zero boundary conditions. Let $F$ be a nice functional of $\varphi$, e.g., a mollified two-point function between $x,y\in\Omega$. Why doesn't the following definition of $\varphi^4$ theory work? $$ \langle F \rangle_\lambda := \lim_{\Omega\to\mathbb{R}^d} \frac{\mathbb{E}\left[F(\varphi)\exp(-\lambda\int_\Omega\varphi^4)\right]}{\mathbb{E}\left[\exp(-\lambda\int_\Omega\varphi^4)\right]} \,. $$ I suppose the first question one should ask about this is how to make sense of $\int_\Omega\varphi^4$, and then, if the infinite volume limit exists. Do people in constructive field theory work on this approach? What are some of the major hurdles in such or similar frameworks? REPLY [11 votes]: When $d=2$, this works fine and this is precisely how Nelson originally constructed the $\Phi^4$ measure (in finite volume). Already for $d = 3$, the $\Phi^4$ measure is singular with respect to the free field, even in finite volume, so this approach is bound to fail. The reason why it is singular is subtle, but you can (kind of) convince yourself that this should be the case from the fact that the correct renormalisation in $d=3$ has an additional logarithmic correction on top of Wick renormalisation.<|endoftext|> TITLE: Is $\iiint_{[0, 1]^3} \lvert f(x)+f(y)+f(z)\rvert\, dx\, dy\, dz \ge \int_0^1 \lvert f(x)\rvert\, dx$? QUESTION [24 upvotes]: $\newcommand\abs[1]{\lvert#1\rvert}\newcommand\Abs[1]{\left\lvert#1\right\rvert}$Question: Let $f(x)$ be a continuous real-valued function defined on the interval $[0, 1]$. Does the following inequality hold? $$\int_{0}^{1}\int_{0}^{1}\dotsi\int_0^1\int_0^1\abs{f(x_{1})+f(x_{2})+\dotsb+f(x_{n})}dx_1 \; dx_{2}\dotsm\;dx_{n} \ge \int_0^1 \abs{f(x)}dx.$$ Even the case $n=3$ would be interesting. This is inspired by the case $n=2$, which was a Putnam problem from 2003, as follows. Theorem. Let $f(x)$ be a continuous real-valued function defined on the interval $[0, 1]$. Then $$\int_0^1\int_0^1\abs{f(x)+f(y)}dx \; dy \ge \int_0^1 \abs{f(x)}dx.$$ Proof by Kent Merryfield: Let $P$ be the subset of $[0, 1]$ on which $f\ge 0$ and $N$ the set on which $f < 0$. As is conventional, define $f^+(x) = \max(f(x), 0)$ and $f^-(x) = \max(-f(x), 0)$. Thus, $f = f^+ - f^-$, $|f| = f^+ + f^-$, and $f^+$ equals $0$ everywhere on $N$ while $f^-$ equals zero everywhere on $P$. Then \begin{align*}\int_0^1 \int_0^1 \abs{f(x) + f(y)}\,dx\,dy &= \int_P \int_P \abs{f(x) + f(y)}\,dx,dy + \int_P \int_N \abs{f(x) + f(y)} \,dx\, dy \\ &+ \int_N \int_P \abs{f(x) + f(y)}\, dx \,dy+ \int_N \int_N \abs{f(x) + f(y)}\, dx\, dy.\end{align*} We tackle these terms one at a time. \begin{align*}\int_P \int_P \abs{f(x) + f(y)}\, dx\, dy &= \int_P \int_P (f(x) + f(y))\,dx \,dy\\ &= \abs P \int_P f(x)\,dx + \abs P \int_P f(y) \,dy = 2\abs P\int_P f^+(x) \,dx\end{align*} where we use the notation $\abs P$ to mean the measure (total net length) of the set $P$. Similarly, $\int_N \int_N \abs{f(x) + f(y)} \,dx \,dy = 2\abs N\int_N f^-(x) \,dx$. The other two terms are equal to each other (as shown by interchanging $x$ and $y$). \begin{align*}\int_P \int_N \abs{f(x) + f(y)} \,dx \,dy &= \int_P \int_N \abs{f^+(x) - f^-(y)} \,dx \,dy\\ &\ge \Abs{\int_P \int_N f^+(x) - f^-(y) \,dx \,dy}\\ &= \Abs{ \abs N\int_P f^+(x) \,dx - \abs P\int_N f^-(y) \,dy}\end{align*} If we let $A = \int_P f^+(x) \,dx$, $B = \int_N f^-(x) \,dx$, and $I = \int_0^1 \int_0^1 \abs{f(x) + f(y)} \,dx \,dy$, then we have found that $I \ge 2\abs P A + 2\abs N B + 2\abs{(\abs N A - \abs P B)}$. For convenience, we now square this: \begin{align*}I^2 &\ge 4\left[(\abs P A + \abs N B)^2 + (\abs N A - \abs P B)^2 + (\text{other positive terms})\right]\\ &\ge 4(\abs P^2A^2 + \abs N^2B^2 + \abs N^2A^2 + \abs P^2B^2)\\ &= 4(\abs P^2 + \abs N^2)(A^2 + B^2).\end{align*} But for real $a$ and $b$, $(a + b)^2 \le 2(a^2 + b^2)$ since $2(a^2 + b^2) - (a + b)^2 = (a - b)^2$. Hence, $2(\abs P^2 + \abs N^2) \ge (\abs P + \abs N)^2 = 1^2$, since $\abs P + \abs N$ is the measure of the interval $[0, 1]$. Also, $2(A^2 + B^2) \ge (A + B)^2 = \left(\int_0^1 \abs{f(x)} \,dx\right)^2$. $\square$ REPLY [20 votes]: Here is the proof using the alternative route. Let $X$, $Y$ be two independent real-valued random variables such that $EX,EY\ge 0$ and $\min(E|X|,E|Y|)=I$. We want to prove that $E|X+Y|\ge I$. Again, as in both the OP and Iosif's post, we can consider only the case when $X$ is $A$ with probability $P$ and $-B$ with probability $Q$, while $Y$ is $a$ with probability $p$ and $-b$ with probability $q$, where $A,B,a,b\ge 0$. Then we need to show that the inequality $$ (A+a)Pp+|a-B|pQ+|b-A|Pq+(B+b)Qq0, \alpha+\beta=1$, and we are done again. It would be interesting to see what factor on the right hand side can be put in place of $1$ for $n\ge 3$. I suspect that the worst case for even $n$ is addition of $n$ Rademacher independent random variables ($\pm 1$ with probability $\frac 12$ each) even if we allow them to be different (but with expectations of the same sign) but I have no proof of it at the moment. The odd case may have a rather ugly answer even for $n=3$.<|endoftext|> TITLE: Does $G_2(\mathbb{Z})$ depend on the choice of an integral model? QUESTION [8 upvotes]: I am trying to understand constructions of exceptional groups of type $G_2$ (over rings). In this post, by a model (of type $G_2$) I mean an affine smooth group scheme over $\mathbb{Z}$ such that the fibres are connected simple algebraic groups of type $G_2$. In Gross' paper Groups over Z (see Page 272) a model $\mathbb{G}$ is given using Coxeter's integral octonions, and it is mentioned there that $\mathbb{G}(\mathbb{Z})$ is isomorphic to $G_2(\mathbb{F}_2)$ as abstract groups. Models are not unique: For example, this $\mathbb{G}$ is a non-split model, and there is also a split one as given in Appendix B of Conrad's paper Non-split reductive groups over Z. Question: Do we have $\mathbb{G}'(\mathbb{Z})\cong G_2(\mathbb{F}_2)$ for every model $\mathbb{G}'$ of type $G_2$? REPLY [12 votes]: The short answer is no, because a Chevalley group has infinitely many $\mathbb{Z}$-points (even Zariski dense by the Borel density theorem). For the long answer, let me first completely describe all $\mathbb{Q}$-models and $\mathbb{Z}$-models of groups of type $G_2$. Let $G_0/\mathbb{Q}$ be the split reductive group of type $G_2$. (It is both simply connected and adjoint since the $G_2$ Cartan matrix has determinant $1$.) This group has a reductive $\mathbb{Z}$-model $\underline{G}_0$ which is unique up to isomorphism by the theory of Chevalley groups. Now the $\mathbb{Q}$-forms of $G_0$ are classified by $\mathrm{H}^1(\mathbb{Q},G_0)$, bearing in mind that $G_0$ is adjoint and has no outer automorphisms. By the work of many people, the restriction map $\mathrm{H}^1(\mathbb{Q},G_0) \rightarrow \mathrm{H}^1(\mathbb{R},G_0)$ is an isomorphism (for references see Theorems 5.12.24 and 5.12.31 of Poonen's rational points book). By table 1.3 in the Gross' paper you mention, $\mathrm{H}^1(\mathbb{R},G_0)$ has size $2$, where the nontrivial element is represented by the compact form. Let's call the corresponding $\mathbb{Q}$-form $G$. It can be constructed as the automorphism group of (a $\mathbb{Q}$-form of) the octonions (as done in the Gross' paper) and its real points $G(\mathbb{R})$ are compact. Gross constructs a $\mathbb{Z}$-model $\underline{G}$ of $G$, again using octonions. It has the property that $\underline{G}(\mathbb{Z})\simeq \underline{G}_0(\mathbb{F}_2)$ as abstract groups. He shows using the mass formula that $\underline{G}$ is the only $\mathbb{Z}$-model of $G$. Conclusion Up to isomorphim, there exist exactly two semisimple groups of type $G_2$ over $\mathbb{Q}$. Both admit unique models over $\mathbb{Z}$. This is a very special property of $G_2$! Usually a semisimple group has infinitely many $\mathbb{Q}$-forms, most of them will not have integral models, and if such integral models exist they might not be unique. (Considering $\mathrm{SL}_2$ should already give you an idea.) Back to your question: you are asking whether we have an isomorphism $\underline{G}_0(\mathbb{Z}) \simeq \underline{G}_0(\mathbb{F}_2)$. But the left hand side is not even finite: by a theorem of Borel and Harish-Chandra, $\underline{G}_0(\mathbb{Z})$ is a lattice in $G_0(\mathbb{R})$. By the Borel density theorem, this implies that $\underline{G}_0(\mathbb{Z})$ is Zariski dense in $G_0$, so in particular $\underline{G}_0(\mathbb{Z})$ is infinite.<|endoftext|> TITLE: How to shrink a square with minimal distortion? QUESTION [10 upvotes]: $\newcommand{\CO}{\text{CO}_2}$ $\newcommand{\euc}{\mathfrak{e}}$ $\newcommand{\SO}{\text{SO}_2}$ $\newcommand{\al}{\alpha}$ $\newcommand{\dist}{\text{dist}}$ $\newcommand{\Lip}{\text{Lip}_{\text{inj}}}$ $\newcommand{\sAverage}[1]{\langle#1\rangle}$ $\newcommand{\IP}[2]{\sAverage{#1,#2}}$ $\newcommand{\pl}{\partial}$ $\newcommand{\Sone}{\mathbb{S}^1}$ Let $0<\sigma_1<\sigma_2$ satisfy $\sigma_1\sigma_2=1$, and let $D=[-1,1]^2$. Does there exist a Lipschitz injective a.e. map $\phi:D \to D$ such that $d\phi$ has almost everywhere the fixed singular values $\sigma_1,\sigma_2$? Does there exist such a $\phi$ that maps the boundary onto itself? By injective a.e. I mean that $|\phi^{-1}(y)| \le 1$ for a.e. $y \in D$. Edit If $[-1,1]^2$ is replaced by a disk, then each $ \phi_t:(r,\theta) \to (r,\theta+t \log r)$ has constant singular values. Unfortunately, an analogous approach doesn't work for the square. Here is why: Let $\alpha(\theta)$ be an arc length parametrization of $\partial D$, $D=[-1,1]^2$, i.e. $\alpha:\frac{8}{2\pi}\mathbb{S}^1 \to \partial D$, $|\dot \alpha|=1$. Let $h:(0,1) \times \frac{8}{2\pi}\mathbb{S}^1 \to \frac{8}{2\pi}\mathbb{S}^1$, and assume that for every $r \in (0,1)$, $h(r,\cdot)$ is a diffeomorphism of $\frac{8}{2\pi}\mathbb{S}^1$. Define $$ f\big(r\alpha(\theta)\big)=r\alpha(h(r,\theta)). $$ $f$ maps each scaled copy $r\text{Image}(\alpha)$ diffeomorphically onto itself. The case of the "logarithmic" map $\phi_t$ above corresponds to choosing $h(r,\theta)=\theta+t \log r$. Thinking of $f$ as a map $$ (r,\theta) \mapsto (r,h(r,\theta)), $$ one gets that $$ [df]_{E_1,E_2}=\begin{pmatrix} h_{\theta} & \bigg(rh_r(r,\theta)+b(h(r,\theta))-b(\theta) h_{\theta}(r,\theta)\bigg)/s(\theta) \\0 & s(h(r,\theta))/s(\theta) \end{pmatrix}, \tag{1} $$ where $ s:=|\al \times \dot \al|, b:=\IP{\al}{\dot \al}, $ and $(E_1,E_2)$ is an orthonormal frame in the domain of parametrization $(0,1] \times \frac{8}{2\pi}\Sone$, given by $$ (E_1,E_2)_{r,\theta}=\big(\frac{1}{r} \pl_{\theta}, \frac{1}{|\al \times \dot \al|} (\pl_r-\frac{\IP{\al}{\dot \al}}{r} \pl_{\theta})\big). $$ More explicitly, we define $\Phi:(0,1] \times \frac{8}{2\pi}\Sone \to (D,\euc) $, where $\euc$ is the standard Euclidean metric, by setting $\Phi(r,\theta):=r\al(\theta)$. $(E_1,E_2)$ is an orthonormal frame in $(0,1] \times \frac{8}{2\pi}\Sone$ w.r.t the pullback metric $\Phi^*\euc$. For the case where the domain $D=[-1,1]^2$ is a square, $s=|\al \times \dot \al|$ is constant, so Equation $(1)$ reduces to $$ [df]_{E_1,E_2}=\begin{pmatrix} h_{\theta} & \bigg(rh_r(r,\theta)+b(h(r,\theta))-b(\theta) h_{\theta}(r,\theta)\bigg)/s \\0 & 1 \end{pmatrix}, \tag{2} $$ so $f$ is area-preserving if and only if $h_{\theta}=1$, i.e. $h(r,\theta)=\theta+g(r)$. Then Equation $(2)$ further reduces to $$ [df]_{E_1,E_2}=\begin{pmatrix} 1 & \big(rg'(r)+b(\theta+g(r))-b(\theta) \big)/s \\0 & 1 \end{pmatrix}. $$ Thus $df$ has constant singular values if and only if $$ A(r,\theta):=rg'(r)+b(\theta+g(r))-b(\theta) $$ is constant. $b(\theta)$ is piecewise-affine in $\theta$, with jumps at the corners, so the locations of jumps in the difference $b(\theta+g(r))-b(\theta)$ depend on both $r,\theta$ and cannot be cancelled by the remaining term $rg'(r)$. Variational motivation for the question: Such maps $\phi$ correspond to distortion minimizing maps from the square into a smaller square. Let $0<\lambda \le 1/2$, and let $\Lip(D,\lambda D)$ be the space of injective a.e. Lipschitz maps $D \to \lambda D$ having nonnegative Jacobian. Set $E:\Lip(D,\lambda D) \to \mathbb{R}$ by $$E(\phi)=\int_{D} \dist^p( d\phi,\SO )\,\,dx.$$ I proved (Theorem 1.1 here) that for $p>2$, $E(\phi) \ge (1-2\lambda)^{p/2}$, and that equality holds if and only if $\sigma_1(d\phi)+\sigma_2(d\phi)=1, \sigma_1(d\phi)\cdot \sigma_2(d\phi)=\lambda^2$. $\frac{\phi}{\lambda}:D \to D$ has constant distinct singular values. The motivation for choosing specifically the square, is that we can tile other shapes in the plane up to an arbitrary accuracy, so if we have a solution on a square, we can glue these solutions to approach an infimal distortion energy for other shapes as well. REPLY [5 votes]: Update. I'll show in the end of this post in the proof that the answer to the question is positive for all $\sigma_1\in (0,1)$. The square is denoted by $\square$. Set up. Let us fix $x,y\in (0,1)$ with $x^2-2x+y^2<0$ and denote by $\square_{x,y}$ the square with vertices $$(x,y), (y,-x), (-x,-y), (-y,x)$$ Now let us start do define the map $\varphi_{x,y}:\square\to \square$. The map will be identical on the boundary of the square. First we define it on $\square\setminus \square_{x,y}$. The map will send the boundary of $\square_{x,y}$ to the boundary of $\square_{y,x}$. On vertices it acts as follows: $$(x,y)\to (x,-y),\, (y,-x)\to (-y-x),\, (-x,-y)\to (-x,y),\, (-y,x)\to (y,x)$$ This is a is depicted on the figure. The set $\square\setminus \square_{x,y}$ is cut into $8$ triangles (with green boundary) and the map $\varphi_{x,y}$ is linear on them. Now we extend this map to the $\square_{x,y}\setminus \square_{x',y'}$ and so on. Easy claim. The constructed map is area preserving and $\sigma_1$ takes only two values - for half of triangles it is one and for the other half it is different. Finally, we should choose $(x,y)$ so that $\sigma_1$ takes just one value. By simple continuity it is clear that such $(x,y)$ will form a non-empty (algebraic) curve. The equation of the curve is given below. In the above figure the brown triangle on the left goes to the brown triangle on the right. Added. So, for this map to have constant $\sigma_1$ we need an equality on $x,y$ to hold. To write down the equality, denote by $A$ the matrix $$A=\begin{pmatrix} y+1 & 1-x\\ 1-x & 1-y \end{pmatrix}$$ Then we should have $$2+\frac{4y^2}{(1-x)^2}=tr\left( A\cdot\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \cdot A^{-2}\cdot \begin{pmatrix} 0& 1\\ 1 & 0 \end{pmatrix} \cdot A\right) $$ This is the equality between the traces of matrices $BB^T$ and $CC^T$, where $B$ corresponds to the linear map $\varphi_{x,y}$ induced on a triangle adjacent to an edge of $\square$ and $C$ corresponds to a triangle adjacent to an edge of $\square_{x,y}$. Proof. So let us denote by $D$ the domain $$y\ge 0,\, x<1,\, x^2-2x+y^2<0$$ where the vertex $(x,y)$ of $\square_{x,y}$ can lie, so that the map $\varphi_{x,y}$ is well defined. On the figure below the domain is in light blue. Let us introduce two functions on $D$: $\sigma_2^A(x,y)$ and $\sigma_2^B(x,y)$. The first function is equal to $\sigma_2$ of the map $\varphi_{x,y}$ restricted to triangle $A$ and the second is $\sigma_2$ of the map $\varphi_{x,y}$ restricted to triangle $B$. Now, we have the following claims: $\sigma_2^A(x,y)$ and $\sigma_2^B(x,y)$ are smooth on $D$. On the segment $y=0$ both functions are equal to $1$ When the point $(x,y)$ approaches the circular arc $x^2-2x+y^2=0$ the function $\sigma_2^B(x,y)$ tends to $+\infty$, while $\sigma_2^B(x,y)$ stays bounded. When the point $(x,y)$ approaches the segment $x=1$ the function $\sigma_2^A(x,y)$ tends to infinity, while $\sigma_2^A(x,y)$ stays bounded. Now if one stares a bit more on the figure on trajectories in pink, one sees that the function $\sigma_2^A(x,y)$ goes from $1$ to $+\infty$ when we move along any such trajectory from $(0,0)$ to $x=1$. At the same time $\sigma_2^B(x,y)$ stays bounded. Moreover, on trajectories that are close to $y=0$ it is close to $1$, while on trajectories close to $x^2-2x+y^2<0$ it is large. This proves that on any trajectory there is a point where $\sigma_2^A(x,y)=\sigma_2^B(x,y)$. By varying further the pink trajectory from the segment $y=0$ to the circular arc we get any $\sigma_2\ge 1$ that we want. PPS I add here one more picture. It is the previous one rotated by 90 degrees, it is a bit easier to see it this way. In pink I draw the level sets $\sigma_2^B=const$, in green $\sigma_2^A=const$. For $const=1$ these curve coincide with the vertical green-pink segment. Otherwise they are different and they always intersect.<|endoftext|> TITLE: Commutation of limits and colimits: Is there a choice diagram? QUESTION [8 upvotes]: I was looking at this question about a "soft proof" of the fact that finite limits (shape $I$) commute with filtered colimits (shape $J$) in Set, using only the fact that the diagonal $J \to J^I$ is final. If we consider the simple case of intersection and unions of sets $A_{i,j}$, we have the formula $$\bigcap_{i \in I}\bigcup_{j \in J} A_{i,j} = \bigcup_{f : I \to J} \bigcap_{i \in I} A_{i,f(i)}$$ where $f : I \to J$ is a "skolemization" of the existential quantifier on the left hand side (This requires AC for infinite $I$). I am wondering if in the same vein, a limit-of-colimits can be replaced by a colimit-of-limits over a diagram category. That is, do we have for every diagram $A: I \times J \to \mathbf{Set}$ ($I$ possibly finite) that $$\mathrm{lim}_{i \in I}\mathrm{colim}_{j \in J}\; A(i,j) \cong \mathrm{colim}_{F \in J^I} \mathrm{lim}_{i \in I}\; A(i,F(i))$$ In this case, using the finality of $I \to J^I$ we'd obtain the desired commutation $$\mathrm{lim}_{i \in I}\mathrm{colim}_{j \in J}\; A(i,j) \cong \mathrm{colim}_{j \in J} \mathrm{lim}_{i \in I}\; A(i,j)$$ So the formula should hold for $I$ finite $J$ filtered, and I've checked it for $J$ discrete as well if I'm not mistaken. In general, going from the lhs to the rhs requires to build up a skolemizing diagram $F$ not only on objects but on morphisms as well, which seems to involve some complicated choice. Is a formula like the above known, or is my intuition off somewhere? REPLY [4 votes]: This isn't true in general. Take $I = BG$ and $J = BH$ to be one-object groupoids, so that $A(i, j)$ becomes a set $A$ with commuting actions of $G$ and $H$. The left hand side is obtained by taking the $H$-orbits of $A$, and then the $G$-fixed points of the result. The right hand side isn't as easy to describe precisely, but it's some quotient of the coproduct over all group homomorphisms $f : G \to H$ of the elements of $A$ fixed by the action of $(g, f(g))$ for every $g \in G$. Take $G = \mathbb{Z}/2$, $H = \mathbb{Z}$ and let $A = \mathbb{Z}/2$ with action of both $G$ and $H$ given by addition (mod 2). Then $A$ has a single $H$-orbit, which is (obviously) fixed by $G$, so the left hand side has a single element. On the other hand, the only group homomorphism $f : \mathbb{Z}/2 \to \mathbb{Z}$ is the zero map, and there aren't any elements of $A$ fixed by the action of $(g, 0)$ for $g = 1 \in \mathbb{Z}/2$, so the right hand side is empty. What went wrong in this example? Well, let's try to construct an isomorphism between the two sides anyways. Given an element $x$ on the left, we should try to construct a "corresponding" group homomorphism $f : G \to H$ (not necessarily uniquely determined). The element $x$ is an $H$-orbit of $A$ which is fixed by $G$. Let's pick an element $a \in A$ of this orbit. Then for each $g \in G$, since $ga$ generates the same orbit as $a$, there must be some $h \in H$ with $(g, h)a = a$. We would like to send $x$ to the rule $f$ which maps $g$ to such an $h$. However, $h$ is only well-defined as an element of some quotient of $H$, namely the quotient by the subgroup which fixes $a$ (or $ga$, since the actions commute). So we obtain a group homomorphism from $G$ to a quotient of $H$, but it might not lift to $H$ itself. One thing I'm not sure about is what happens in the $\infty$-category of spaces, which generally has better exactness properties since there is no set-truncation in the construction of colimits. If the isomorphism holds there, then one could recover the version for sets when $J$ is filtered, since filtered colimits of sets are also homotopy colimits.<|endoftext|> TITLE: Mathematical predictions of AdS/CFT QUESTION [21 upvotes]: What sorts of mathematical statements are predicted by the AdS/CFT correspondence? My "understanding" (term used very loosely) is that this correspondence isn't a mathematically rigorous statement, but I can still imagine that certain baby cases might spit out interesting well-defined identities or relationships. I'm very happy for this question to be broadly interpreted. REPLY [4 votes]: New exact solutions in classical general relativity have been found using AdS/CFT methods. It may sound trivial since there are already encyclopedias of exact solutions, but those new ones have quite unexpected properties. Some references to those works are found e.g. in sec. 4.2 of Hubeny's notes (in the same document you encounter an extensive catalog of applications of AdS/CFT to problems in mathematical physics).<|endoftext|> TITLE: About an argument in Olsson's book QUESTION [5 upvotes]: The following picture is from Algebraic spaces and stacks (p.54) by Martin Olsson. I don't understand how to conclude that $\alpha$ is induced by a nonzero class in the end. It seems that there might be an exact sequence consists of edge maps like $E_2^{n0}\to H^n\to E_2^{0n}$ if there is a convergent first quadrant spectral sequence $E_2^{rs}\Longrightarrow H^{r+s}$, but this doesn't fall into the usual assumption for such an exact sequence to exist. Any ideas? Another way to explain Olsson's argument is also fine. Edit. To summarize (by Minseon Shin's answer), the conclusion is: If in addition $E_{2}^{n-i,i} = 0$ for $0 TITLE: Dealing cards numbered $1$ to $n$ into piles QUESTION [5 upvotes]: Is anything known about the following? I hold in my hand a shuffled pack of cards numbered $1$ to $n$. One by one, I place them all, face up, on a table in piles. For each card I deal from my hand, say card numbered $k$, I open a new pile only if the top card of an existing pile is not $k + 1$; otherwise I place that card on top of that pile. If at any stage (before dealing a new card) the bottom card of an existing pile is one less than the top card of another pile, I combine the two into a new pile in the obvious way. What is the most likely maximum number of piles that will be formed at some stage while so dealing the $n$ cards? What is the expected maximum number of piles that will be formed at some stage while so dealing the $n$ cards? In both cases, count piles only if no two piles can be combined into one. REPLY [3 votes]: At Timothy Chow's request, here is a table of $A(n,m)$, the number of permutations in $\mathfrak{S}_n$ with maximum number of piles equal to $m$. Note that clearly $A(n,m)=0$ if $m > \lceil n/2 \rceil$: $ \begin{array}{c|c c c c} n/m & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & 1 \\ 2 & 2 \\ 3 & 4 & 2 \\ 4 & 8 & 16 \\ 5 & 16 & 92 & 12 \\ 6 & 32 & 472 & 216 \\ 7 & 64 & 2312 & 2520 & 144 \\ 8 & 128 & 11104 & 24480 & 4608 \\ 9 & 256 & 52880 & 216432 & 90432 & 2880 \\ 10 & 512 & 250912 & 1815264 & 1418112 & 144000 \end{array}$ There are certainly some patterns visible, but the whole triangle does not appear to be in the OEIS.<|endoftext|> TITLE: Is this theory the complete theory of the real ordered field? QUESTION [24 upvotes]: We know that the real ordered field can be characterized up to isomorphism as a complete ordered field. However this is a second order characterization. That raises the following question. Consider the following theory. We take as axioms the axioms for ordered fields, and then add an axiom schema that states that every nonempty set that is definable without parameters that is bounded above has a least upper bound. Is that theory the complete first order theory of the real ordered field? And if it is not, can someone exhibit a model of that theory that is not elementarily equivalent to the real ordered field? I asked this question on math stack exchange, but I did not receive an answer. REPLY [5 votes]: Edit: As Emil Jeřábek has pointed out in the comments, there is an error in the paper linked below, and the example does not work. I'll leave the answer here to document the error and add to the interesting list of attempted answers gathered here. An erratum has now appeared: Corrigendum to "[..]" RML, referencing this discussion. Let me add that Dmytro Taranovsky has given a correct example, but I would still be very interested to see a "forcing-free" and especially an "algebraically natural" example of a $\emptyset$-definably complete ordered field which is not definably complete. This question was answered negatively by Mojtaba Moniri in his recent paper On definable completeness for ordered fields, Reports on Mathematical Logic, Number 54 (2019), pp. 95-100. An explicit example is given: $\mathbb{R}((t^\Gamma))$, where $\Gamma$ is the ordered additive group of dyadic rationals. Here $\mathbb{R}((t^\Gamma))$ is the Hahn series field over $\mathbb{R}$ with value group $\Gamma$: its elements are formal sums $\sum_{e\in \Gamma}c_et^e$ with $c_e\in \mathbb{R}$ such that the support $\{e\in \Gamma\mid c_e\neq 0\}$ is well-ordered. This field is complete for cuts definable without parameters, but not real closed: $t$ has no cube root.<|endoftext|> TITLE: Is there a simple proof that $Ax^3+By^3=C$ has only finitely many integer solutions QUESTION [5 upvotes]: One can use Thue's 1909 result to show that the Diophantine equation $Ax^3 + By^3 = C$ ($A,B$ not perfect cubes, $C\neq 0$) has finitely many integer solutions $(x,y)$. But does there exist a simple way to prove this equation (because I think this is a special equation? REPLY [3 votes]: I think both of the answers given avoid the modern treatment of Thue equations, which despite being newer is probably simpler in many ways. As my doctoral advisor is a pioneer in this area I feel obligated to explain these ideas. Let us first emphasize that Thue's theorem, extending to the theorems of Siegel, Dyson, and Roth, is completely ineffective in the height aspect (i.e., it gives no way at all to control the size of solutions) and performs very badly in the count aspect (i.e., it is in principle possible to extract a bound for the number of solutions, but it is terrible). The height of the solutions can be bounded using effective bounds on linear forms of logarithms, due to Baker. However these height bounds are terrible (super-polynomial in the coefficients of the polynomial and the integer $h$ in the equation $F(x,y) = h$) and are of little practical use. This is still the best general method for height bounds of solutions as far as I know. In terms of bounding the number of solutions, there has been significant progress. The key idea comes from a paper of Bombieri and Schmidt: one can obtain a fairly cheap way to reduce a Thue equation $F(x,y) = h$ into a family of Thue equations $G_a(x,y) = 1$, where $a \in A$ lies in some finite indexing set. The forms $G_a$ have the same degree as $F$. One then obtains a uniform bound, depending only on the degree $d = \deg F = \deg G_a$, for the number of solutions of Thue equations of the shape $G(x,y) = 1$ which is valid for any binary form $G$ of degree $d$ with non-zero discriminant. Thus, one obtains a bound of the shape $C_d \cdot |A|$, where $C_d$ is the bound for the number of solutions to the unit equation $G(x,y) = 1$ for forms of degree $d$. This approach exploits the fact that for any integer $m$, the set of solutions to the congruence $F(x,y) \equiv 0 \pmod{m}$ lies in a finite number of lattices of $\mathbb{Z}^2$ (that number could be 0, if the congruence is not solvable). This is because $F$ splits into a product of linear forms over an algebraically closed field. One then writes $h = p_1^{a_1} \cdots p_k^{a_k}$, and since $F(x,y) = h$ has a solution the congruence $F(x,y) \equiv 0 \pmod{p_i^{a_i}}$ is solvable for each $1 \leq i \leq k$. This implies that $F$ has a linear factor over $\mathbb{F}_{p_i}$. Each such factor gives rise to a lattice $\Lambda_j(p_i)$. For simplicity, let us assume that $F$ is unramified at $p_i$. Then we may use Hensel's lemma to show that each of the $\Lambda_j(p_i)$'s can be lifted uniquely to a lattice $\Lambda_j(p_i^{a_i})$ whose elements correspond to solutions of the congruence $F(x,y) \equiv 0 \pmod{p_i^{a_i}}$. The ramified case is a bit more delicate, but is handled in full generality by this paper of Cam Stewart. Thus, for each $p_i^{a_i}$ we obtain a finite number of lattices (at most $d = \deg F$) $\Lambda_j(p_i^{a_i})$ and we apply a transformation sending $\mathbb{Z}^2$ to $\Lambda_j(p_i^{a_i})$, thus obtain forms $F_{\Lambda_j(p_i^{a_i})}(x,y)$. We have thus exchanged our original Thue equation $F(x,y) = h$ for at most $d$ equations of the form $$\displaystyle F_{\Lambda_j(p_i^{a_i})}(x,y) = h p_i^{-a_i}.$$ We may then repeat this process, and eventually obtain a finite number $A$ (roughly $d^k = d^{\omega(h)}$ many) of lattices and auxiliary forms $G_a$ with $a \leq A$, and the unit equation $G_a(x,y) = 1$. Assuming we can obtain a bound for these latter equations which depends only on $d$, then we will get a bound roughly of the shape $C_d \cdot d^{\omega(h)}$, which is $O_{d,\epsilon}(h^\epsilon)$ for any $\epsilon > 0$. How are such uniform bounds achieved? The key observation is that binary forms with integer coefficients and non-zero discriminant has discriminant at least one in absolute value. In particular, we are now in a situation where the discriminant is comparably large compared to the integer $h$. This means that we can rely on the Mahler measure of the form (really the corresponding polynomial $F(x,1)$) and the discriminant alone to bound the number of large solutions, using the Thue-Siegel principle, which is an effective (and very simple, i.e., does not require the construction of auxiliary polynomials as in Thue's method) but fairly weak way to push apart large solutions. One then bounds the small solutions suitably, and then optimize the counting by controlling various parameters. I should mention that Stewart makes a critical observation in his 1991 JAMS paper linked above, where he exploits this principle further by noting that it is really the property that $h$ is small compared to the discriminant of $F$ that matters. Here one observes that as one replaces the forms $F$ with the forms $F_{\Lambda_j(p_i^{a_i})}$ the discriminant actually increases substantially, thus one can often move into a favourable situation where one obtains a finite number of equations of the shape $G_a(x,y) = g$ with $g$ very small (but not necessarily equal to one) compared to the discriminant of $G_a$ without using all of the prime factors of $h$. This is still essentially the best known result, despite the passage of nearly 30 years (since Stewart's 1991 paper). Indeed, it is conjectured that there should be a uniform bound depending only on the degree $d$ such that the number of solutions to the Thue equation $F(x,y) = h$ is bounded by $O_d(1)$. This is a special case of the uniform boundedness conjecture, which is known to be true assuming the Bombieri-Lang conjecture (this result is due to Caporaso, Harris, and Mazur) Stewart makes an even stronger conjecture in his aforementioned paper. Indeed, the Bombieri-Schmidt theorem gives unconditionally the following special case of the uniform boundedness conjecture, which I don't believe can be proved any other way: we consider the family $\mathcal{C}(d)$ of affine curves defined by two objects: a binary form $F$ with non-zero discriminant, integer coefficients, and degree $d$, and a prime number $p$ - then the affine curves $$C(F,p) : = \{F(x,y) = p\}$$ have a uniformly bounded number of integral points (i.e., primitive solutions to the Thue equation $F(x,y) = p$). Indeed, one can get something very explicit... for instance $2800 d^2$ works. This result does not require any input from the Mordell-Weil rank of the Jacobian of the curve, or any other similar quantity (i.e., Minhyong Kim's notion of global Selmer variety).<|endoftext|> TITLE: How many reals can we construct by iteratively writing down truth tables for ZFC? QUESTION [8 upvotes]: For the purposes of this question, pretend that there is a platonic, "standard" model $V$ of ZFC - so that every set theoretic statement has a meaningful truth value. Fix a bijection between $\mathbb{N}$ and the set of formulas $\varphi(x)$ with one free variable $x$ in the language of set theory, and let $\varphi_i(x)$ be the $i$th formula according to this bijection. For every set $X \in V$, let $T(X) \in 2^\mathbb{N}$ be the truth table for statements involving the set $X$, that is, the $i$ bit of $T(X)$ is set to $1$ iff $\varphi_i(X)$ is true in $V$. Note that we can't prove that the truth table $T(X)$ is a set or even define $T(X)$ within the formal framework of ZFC - assume a strong enough metatheory that we can talk about $T(X)$ as representing a well-defined thing. Additionally, if we take the idea that $V$ is standard seriously, then we should assume that $T(X)$ is a set in $V$, for every $X \in V$. Starting from the truth table $T(\emptyset)$ which encodes all true statements of $V$, we can then build the truth table $T(T(\emptyset))$ which encodes all the true statements about the nature of truth in $V$. The new set $T(T(\emptyset))$ is really new: it is not computable from $T(\emptyset)$, since $T(T(\emptyset))$ can be used to solve the halting problem for Turing machines with oracle access to $T(\emptyset)$. In particular, $T(T(\emptyset)) \ne T(\emptyset)$. I want to iterate this construction as boldly as possible. To this end, inductively define a class function $\mathcal{T} : \operatorname{Ord} \rightarrow 2^\mathbb{N}$ as follows: for every ordinal $\alpha$, if $V$ is truly standard then the restriction $\mathcal{T}|_\alpha : \alpha \rightarrow 2^\mathbb{N}$ should be a set in $V$, so we can define $\mathcal{T}(\alpha) = T(\mathcal{T}|_\alpha)$. By the same argument as the one showing that $T(T(\emptyset)) \ne T(\emptyset)$, if $\alpha < \beta \in \operatorname{Ord}$ and if $\alpha$ is definable given $\beta$ as a parameter, then $\mathcal{T}(\beta)$ is not computable from $\mathcal{T}(\alpha)$. Call an ordinal $\alpha$ "fresh" if $\mathcal{T}(\alpha) \ne \mathcal{T}(\beta)$ for all $\beta < \alpha$, and let $F \subseteq \operatorname{Ord}$ be the set of fresh ordinals. The restriction of $\mathcal{T}$ to $F$ gives an injection $F \hookrightarrow 2^\mathbb{N}$. Question(s) 1: What can we say about the order type of $F$ (with the induced ordering from $\operatorname{Ord}$)? Is $F$ uncountable? Could $|F|$ be strictly larger than $\aleph_1$ (assuming that the continuum hypothesis is false)? Can we determine whether the order type of $F$ is a limit ordinal? What can we say about the least ordinal which is not contained in $F$ (i.e., the first "stale" ordinal) - could it be definable? Are these questions independent of large cardinal hypotheses/forcing axioms? We can also ask interesting questions about the long-term behavior of the function $\mathcal{T}$. For every $\alpha \in F$, let $S(\alpha)$ be the collection of ordinals $\beta$ such that $\mathcal{T}(\beta) = \mathcal{T}(\alpha)$ (so all but one element of $S(\alpha)$ is stale). Note that at least one $S(\alpha)$ must be a proper class, and that for any definable $\alpha \in F$, we have $S(\alpha) = \{\alpha\}$. Question(s) 2: For how many $\alpha \in F$ is the collection $S(\alpha)$ a proper class? Is there an undefinable $\alpha \in F$ such that $S(\alpha)$ is a singleton? How many distinct cardinalities show up among the $S(\alpha)$s? Generalizing the problem, we can also build iterated truth tables involving any set $X \in V$ as a parameter. Define the function $\mathcal{T}_X : \operatorname{Ord} \rightarrow 2^\mathbb{N}$ inductively by $\mathcal{T}_X(\alpha) = T((\mathcal{T}_X|_\alpha, X))$, where $(\mathcal{T}_X|_\alpha, X)$ is the Kuratowski ordered pair. Call an ordinal $\alpha$ "fresh for $X$" if $\mathcal{T}_X(\alpha) \ne \mathcal{T}_X(\beta)$ for all $\beta < \alpha$, and let $F_X$ be the set of ordinals which are fresh for $X$. If we choose $X$ such that for every countable ordinal $\alpha$ there is a bijection from $\mathbb{N}$ to $\alpha$ which can be defined from $X$ and $\alpha$, then every countable ordinal will be fresh, so we will have $\omega_1 \subseteq F_X$. Question 3: How large can $F_X$ be, if we choose the set $X \in V$ to make $F_X$ as large as possible? Apologies for asking so many questions at once. I'll be happy with any nontrivial results about the nature of the function $\mathcal{T}$ or the set of fresh ordinals $F$. Edit: I just realized that Question 3 isn't nearly as interesting as I had thought it was - if we take $X$ to be an injection from an ordinal $\alpha$ into $2^\mathbb{N}$, then it is easy to prove that $\alpha \subseteq F_X$, so Question 3 is really just asking how big the continuum is. REPLY [5 votes]: I think it is easier to think about the question generalized to an arbitrary transitive model of ZFC, resisting the natural urge to grasp towards the Absolute. So fix such a model $M$, and let $\mathcal T^M(\alpha)$, $F^M$, and $S^M(\alpha)$ be as you defined them, replacing $V$ with $M$. Let's omit the superscripts, though. The ordertype of $F$ is a limit ordinal: if $\alpha$ is fresh, so is $\alpha+1$, since in general, one can recover $\mathcal T(\beta)$ from $\mathcal T(\beta+1)$. Note that if $\alpha$ is definable in $M$, $\mathcal T(\alpha)$ is fresh since it is the unique value of the function $\mathcal T$ containing the index of the formula $\varphi(\text{dom}(x))$ where $\varphi$ defines $\alpha$. In particular, an ordinal that is not fresh is not definable, and so your least stale ordinal is not definable. Note that if every ordinal of $M$ is definable in $M$, then every ordinal is fresh. More interestingly, even models with undefinable ordinals can have only fresh ordinals. (I think Con(ZFC) does not suffice to build a model of ZFC whose fresh ordinals form a set.) To avoid these examples, assume from now on that $(M,S)$ satisfies the Axiom of Replacement where $S$ is the satisfaction predicate of $M$. The fresh ordinals $F$ then form a set in $M$. Moreover, $(\mathcal T(\alpha))_{\alpha \in F}$ is definable in $M$ from any sufficiently large ordinal $\kappa$ such that $V_\kappa^M\prec M$ (and there are cofinally many). It follows that $(\mathcal T(\alpha))_{\alpha \in F}$ belongs to $\text{HOD}^M$, which I will denote by $H$. Therefore $F$, has cardinality at most $\mathfrak c$ in $H$. For any ordinal $\alpha < \mathfrak{c}^H$, the $\alpha$-th real in the canonical wellorder of $H$ is $\Sigma_2$-definable from $\alpha$ and hence is recoverable from $\mathcal T(\alpha)$. It follows that $\alpha$ is fresh. This shows $\mathfrak{c}^H\subseteq F$. Also $\mathfrak{c}^H\in F$, being definable in $M$. So the ordertype of $F$ is strictly between $\mathfrak{c}^H$ and $\mathfrak{c}^{+H}$. Now it is clear that $F$ is uncountable if and only if $M$ thinks $H$ contains uncountably many ordinal definable reals, which of course depends on the choice of $M$. The cardinality of $F$ is larger than $\aleph_1$ if and only if $H$ contains at least $\aleph_2$-many reals, which is independent of $\neg\text{CH}$. Assuming Martin's Maximum, every subset of $\omega$ is coded into the pattern of stationary reflection for any infinite stationary partition of $\{\alpha < \kappa : \text{cf}(\alpha) = \omega\}$ where $\kappa\geq \omega_2$. (This is part of the proof of Theorem 10 in Foreman-Magidor-Shelah's Martin's Maximum, saturated ideals, and nonregular ultrafilters.) So assuming $M$ satisfies MM, large cardinals, and Woodin's HOD Hypothesis, every real is in $H$. Therefore it is conceivable that your question decided by the conjunction of forcing axioms and large cardinals, but this is open... For the second question, note that by the pigeonhole principle there is an unbounded class $C\subseteq M$ of ordinals such that for all $\kappa_0,\kappa_1\in C$, the theory of $\kappa_0$ in $(M,S)$ with real parameters is the same as that of $\kappa_1$. Let $\langle-,-\rangle$ be an $M$-definable pairing function. It follows that $\mathcal T\langle\kappa_0,\alpha\rangle$ is equal to $\mathcal T\langle\kappa_1,\alpha\rangle$ for all $\alpha < \mathfrak{c}^H$, since $\alpha$ is interdefinable over $M$ with the $\alpha$-th real in the canonical wellorder of $H$ and $\mathcal T\langle\kappa_0,\alpha\rangle$ depends only on the theory of $\kappa_0$ and $\alpha$ in $(M,S)$. But similarly, for each $\kappa\in C$, the $\mathcal T\langle\kappa,\alpha\rangle$ are distinct for $\alpha < \mathfrak{c}^{H}$; let $\xi_{\alpha}$ be the least ordinal $\xi$ such that $\mathcal T(\xi) = \mathcal T\langle\kappa,\alpha\rangle$. Letting $E\subseteq F$ be the set of ordinals $\alpha$ such that $S(\alpha)$ a proper class in $M$, it follows that $\{\xi_\alpha : \alpha < \mathfrak{c}^{H}\}\subseteq E$, so $E$ has cardinality $\mathfrak{c}$ in $H$. So much of what was said above applies to $E$ as well.<|endoftext|> TITLE: Computing the $p$-adic gamma function $\Gamma_p$ QUESTION [5 upvotes]: Let $p>2$ be a prime. For $n \in \mathbb{Z}^+$ we can define \begin{equation} F(n) = (-1)^n \prod_{1 TITLE: Irrationality of $e^{x/y}$ QUESTION [5 upvotes]: How to prove the following continued fraction of $e^{x/y}$ $${\displaystyle e^{x/y}=1+{\cfrac {2x}{2y-x+{\cfrac {x^{2}}{6y+{\cfrac {x^{2}}{10y+{\cfrac {x^{2}}{14y+{\cfrac {x^{2}}{18y+\ddots }}}}}}}}}}}$$ Since $a_i \geq b_i$ for all $i \geq 1$. By the condition of irrationality of generalized continued fraction, proving this directly proves that $e^{x/y}$ will be an irrational number! REPLY [7 votes]: I think this might be a solution. The Continued Fraction Expansion of the hyperbolic tanh function discovered by Gauss is $$\tanh z = \frac{z}{1 + \frac{z^2}{3 + \frac{z^2}{5 + \frac{z^2}{...}}}} \\\\$$ We also know that the hyperbolic tanh function is related to the exponential function with the following formula $$\tanh z=\frac{e^z-e^{-z}}{e^z+e^{-z}}$$ Putting $\frac{x}{y}$ in place of $z$ in the previous equation we get $$\frac{e^{\frac{x}{y}}-e^{-{\frac{x}{y}}}}{e^{\frac{x}{y}}+e^{-{\frac{x}{y}}}}= \frac{{\frac{x}{y}}}{1 + \frac{{\frac{x}{y}}^2}{3 + \frac{{\frac{x}{y}}^2}{5 + \frac{{\frac{x}{y}}^2}{...}}}} \\\\$$ This continued fraction can be simplified into $$\frac{e^{\frac{x}{y}}-e^{-{\frac{x}{y}}}}{e^{\frac{x}{y}}+e^{-{\frac{x}{y}}}}= \frac{x}{y + \frac{x^2}{3y + \frac{x^2}{5y + \frac{x^2}{...}}}} \\\\$$ This equation can be further be simplified as $$1+\frac{2}{e^{\frac{2x}{y}}-1}=y + \frac{x^2}{3y + \frac{x^2}{5y + \frac{x^2}{...}}} \\\\$$ $$\frac{e^{\frac{2x}{y}}-1}{2}=\frac{1}{(\frac{y}{x}-1) + \frac{x}{3y + \frac{x^2}{5y + \frac{x^2}{...}}}}$$ From this equation after some algebraic manipulation, we finally get the continued fraction expansion of $e^{x/y}$ as $${\displaystyle e^{x/y}=1+{\frac {2x}{2y-x+{\frac {x^{2}}{6y+{\frac {x^{2}}{10y+{\frac {x^{2}}{14y+{\frac {x^{2}}{18y+\ddots }}}}}}}}}}}$$ This is an infinite generalized continued fraction. We will now state the necessary and sufficient condition for the continued fraction proved by Lagrange given in Corollary 3, on page 495, in chapter XXXIV on "General Continued Fractions" of Chrystal's Algebra Vol.II to converge into an irrational number. \begin{theorem}The necessary and sufficient condition that the continued fraction $$\frac{b_1}{a_1 + \frac{b_2}{a_2 + \frac{b_3}{a_3 + \dots}}}$$ is irrational is that the values $a_{i}, b_{i}$ are all positive integers, and if we have $|a_i| > |b_i|$ for all $i$ greater than some $n$ \end{theorem} In the continued fraction of $e^{x/y}-1$ as we have derived $a_{i}, b_{i}$ are equals to $2(2i-1),x^2$ except $i=1$.Therefore we have $|a_i| > |b_i|$ for all $i>\frac{\frac{x^2}{2}+1}{2}$. Hence we have proved that $e^{x/y}-1$ is irrational which in turn means $e^{x/y}$ is irrational.<|endoftext|> TITLE: Why are compactly generated $\infty$-categories closed under limits in $\operatorname{Cat}_{\infty}$? QUESTION [5 upvotes]: In Proposition 5.5.7.6 of Lurie's Higher Topos Theory, Lurie states that the $\infty$-category $\operatorname{Pr}^R_{\omega}$ of compactly generated $\infty$-categories and filtered-colimit-preserving right adjoints is closed under limits in $\operatorname{Cat}_\infty$. I would like to understand some detail of the proof. Lurie says that most of the proposition follows from Theorem 5.5.3.18, which states that the inclusion $\operatorname{Pr}^R \subseteq \operatorname{Cat}_\infty$ is closed under limits. However, as $\operatorname{Pr}^R_{\omega}$ does not seem to be a full subcategory of $\operatorname{Pr}^R$, I don't understand why the result of 5.5.3.18 gives us that the adjunctions $F_\alpha: \mathcal{C}_\alpha \rightleftarrows \mathcal{C}: G_{\alpha}$ appearing in the proof of 5.5.7.6 correspond to morphisms in $\operatorname{Pr}^R_{\omega}$, i.e. why $F_{\alpha}$ preserves compact objects/$G_\alpha$ filtered colimits. Trying to trace back through the results leading up to 5.5.3.18, I get the impression that the fact that $G_{\alpha}$ is accessible in the proof of 5.5.3.18 indirectly relies on Proposition 5.4.7.7, which says that any right adjoint is accessible. However, this proposition uses that we may choose the regular kardinal $\kappa$ as high as we wish before showing that $G_{\alpha}$ is $\kappa$-continuous, and in particular cannot be applied if we wish to show this for $\kappa = \omega$. Question: Why do the maps $G_{\alpha}: \mathcal{C} \to \mathcal{C}_{\alpha}$ appearing in the proof of 5.5.7.6 preserve filtered colimits? REPLY [7 votes]: By lemma 5.4.5.5., the forgetful functor $Pr^R_\omega\to Cat_\infty$ preserves pullbacks : the projection functors in the pullback preserve filtered colimits. It's easy to prove that the same thing holds for arbitrary products ( a colimit in $\prod_\alpha C_\alpha$ is computed pointwise, which means precisely that each projection preserves it), and so by 4.4.2.7, the forgetful functor preserves all limits. The main point being the following : Given a diagram $F: I\to Cat_\infty$ where each $F(i)$ has $K$-indexed colimits and each $i\to j$ induces a $K$-indexed colimit preserving functor $F(i)\to F(j)$, then $\lim_{i\in I}F(i)$ has $K$-indexed colimits and each projection preserves them. You prove this the same way, using 5.4.5.5. and 4.4.2.7 to the category of $\infty$-categories that have $K$-indexed colimits, and $K$-indexed colimit-preserving functors.<|endoftext|> TITLE: How to show this symmetric function inequality QUESTION [5 upvotes]: Question: let $x_{i}>0$ $(i=1,2,\cdots,n)$, such that $x_{i}\neq x_{j},\forall i\neq j$, find all real numbers $p$ that satisfy the following inequality $$\sum_{i=1}^{n}\dfrac{x^p_{i}}{\prod_{j\neq i}(x_{i}-x_{j})}\ge 0$$ when $n=3$, I have solved it as $p\le 0$ or $p\ge 1$, because we only use this $$x^p\ge y^p+p(x-y)y^{p-1},x>y>0$$so WLOG $x>y>z$ we have $$\sum_{cyc}\dfrac{x^p}{(x-y)(x-z)}\ge\dfrac{y^p+p(x-y)y^{p-1}}{(x-y)(y-z)}+\dfrac{y^p}{(y-z)(y-x)}+\dfrac{y^p+p(z-y)z^{p-1}}{(z-x)(z-y)}=0$$ For $n=4$ I tried some values, Now I conjecture $p\ge 2$ or $0\le p\le 1$, and I can't prove it in general, now I can't find the $p$. Thanks REPLY [20 votes]: Your expression is secretly a divided difference for the function $x \mapsto x^p$. In general, divided differences for a function $f$ are defined inductively by $f[x_0] = f(x_0)$ and $f[x_0, ..., x_n] = \frac{f[x_0, ..., x_{n-1}] - f[x_0, ..., x_{n-2}, x_n]}{x_{n-1} - x_{n}}$. This is a symmetric function of $x_0, ..., x_n$, and it can be expressed explicitly as $f[x_0, ..., x_n] = \sum_{i=0}^n \frac{f(x_i)}{\prod_{j \ne i} (x_i - x_j)}.$ The main useful fact about divided differences is the mean value theorem for divided differences: if $f$ is $n$ times differentiable and $x_0, ..., x_n$ are distinct real numbers, then there is some $\xi$ in the smallest interval containing the $x_i$ such that $f[x_0, ..., x_n] = \frac{f^{(n)}(\xi)}{n!}.$ So to answer your question, you just need to determine for which $p$ the $n$th derivative of the function $x \mapsto x^p$ is positive (keeping in mind that your $n$ is off by one from my $n$), which is a fairly easy exercise. REPLY [11 votes]: The conjecture for $n=4$ is correct. For general $n$ and any $x_1,\ldots,x_n$, $$ f(p) := \sum_{i=1}^{n}\dfrac{x^p_i}{\prod_{j\neq i}(x_i-x_j)} $$ is positive for $p > n-2$ and switches sign at $p=0,1,2,\ldots,n-2$ and nowhere else. For example: If $n=5$ then $f(p) \geq 0$ iff $p \geq 3$ or $1 \leq p \leq 2$ or $p \leq 0$; if $n=6$ then $f(p) \geq 0$ for $p \in [0,1]$, $[2,3]$, or $[4,\infty)$; if $n=7$ then $f(p) \geq 0$ for $p \in (-\infty,0]$, $[1,2]$, $[3,4]$, or $[5,\infty)$; etc. The key fact is the following exponential analogue of "Descartes' law of sines": Proposition. Let $x_1,\ldots,x_n$ be pairwise distinct positive real numbers, and $c_i$ any real numbers not all zero. Then the function $f(p) := \sum_{i=1}^n c_i^{\phantom.} x_i^p$ has at most $n-1$ real zeros, counted with multiplicity. Proof: Induction on $n$. For $n=1$ the function $f$ is a nonzero multiple of $x_1^p$, which is never zero; this establishes the base case. Now let $n>1$, and assume we have proved the case $n-1$ of the Proposition. Let $f_1(p) = x_n^{-p} f(p) = \sum_{i=1}^n c_i (x_i/x_n)^p$, which is of the same form as $f$ with $x_n=1$ and has the same zeros and multiplicities. Then the inductive hypothesis applies to the derivative $$ f'_1(p) = \sum_{i=1}^{n-1} (c_i \log (x_i/x_n)) \, (x_i/x_n)^p; $$ so $f'_1$ has at most $n-2$ zeros, counted with multiplicity. By Rolle's Theorem it follows that $f_1$ has at most $n-1$ zeros, counted with multiplicity. This completes the induction step and the proof. QED It remains to check that our $f$ vanishes at $p=0,1,\ldots,n-2$; then we can use our Proposition to deduce that each of these $n-1$ zeros is simple and there are no others. This can be done by recognizing $f(p)$ as the sum of the residues of the differential $x^p \, dx \left/ \prod_{j=1}^n (x-x_j) \right.$ (which is regular at $x=\infty$ for integers $p \leq n-2$). Alternatively we can relate $f(p)$ to a Schur-Vandermonde determinant with a repeated row. EDIT here's a more self-contained argument that $f(0)=f(1)=\cdots=f(n-2)=0$. Let $P(x)$ be any polynomial with $\deg P < n$. Then $P(x) \left/ \prod_{j=1}^n (x-x_j) \right.$ has a partial-fraction decomposition $\sum_{i=1}^n c_i / (x-x_i)$. Taking $x \to x_i$ (or multiplying through by $\prod_{i=1}^n (x-x_i)$ and setting $x = x_i$), we see that $c_i = P(x_i) \left/ \prod_{j\neq i}(x_i - x_j) \right.$. On the other hand, taking $x \to \infty$ (or multiplying through by $\prod_{i=1}^n (x-x_i)$ and comparing $x^{n-1}$ coefficients), we see that the $x^{n-1}$ coefficient of $P$ is $\sum_{i=1}^n c_i$. In particular if $P(x) = x^p$ for some nonnegative integer $p \leq n-2$ then $f(p) = \sum_{i=1}^n c_i = 0$. $\Box$<|endoftext|> TITLE: Looking for a combinatorial proof for a Catalan identity QUESTION [13 upvotes]: Let $C_n=\frac1{n+1}\binom{2n}n$ be the familiar Catalan numbers. QUESTION. Is there a combinatorial or conceptual justification for this identity? $$\sum_{k=1}^n\left[\frac{k}n\binom{2n}{n-k}\right]^2=C_{2n-1}.$$ REPLY [12 votes]: More generally, $$\sum_{k\ge1} \frac{k}{m}\binom{2m}{m-k}\cdot\frac{k}{n} \binom{2n}{n-k} = C_{m+n-1}.$$ This can be proved by the same reasoning as in Timothy Budd's answer. This formula gives the LDU (or in this case, LU) factorization of the Hankel matrix for Catalan numbers $(C_{m+n-1})_{m,n\ge1}$. There is a similar formula for the Hankel matrix $(C_{m+n-2})_{m,n\ge1}$, involving the remaining ballot numbers. More generally, there are explicit formulas for LDU factorizations of Hankel matrices of moments of other orthogonal polynomials. (The Catalan numbers are moments of Chebyshev polynomials.)<|endoftext|> TITLE: Equivalence of definitions of equivalence of étale Lie groupoids QUESTION [5 upvotes]: I've come across two definitions of an equivalence of étale Lie groupoids, and I'd like to know whether they are equivalent. Let $\mathcal{G}$ be an étale Lie groupoid with space of objects $\mathcal{G}_0$ and space of arrows $\mathcal{G}_1$. Write $\alpha$ and $\omega\colon \mathcal{G}_1 \to \mathcal{G}_0$ for the maps sending an arrow to its source and target, respectively. Definition 1. In Metric Spaces of Nonpositive Curvature, Bridson–Haefliger define an equivalence of étale (Lie) groupoids $f\colon \mathcal{G} \to \mathcal{H}$ to be a smooth functor such that $f\colon \mathcal{G}_0 \to \mathcal{H}_0$ is an étale map (a local diffeomorphism) and such that for all $x \in \mathcal{G}_0$, the map $f$ induces an isomorphism of isotropy groups $\mathcal{G}_x \to \mathcal{H}_{f(x)}$, and the map $f$ induces a bijection of orbit spaces $\mathcal{G}_1\backslash\mathcal{G}_0 \to \mathcal{H}_1\backslash\mathcal{H}_0$. Definition 2. In "Orbifolds as Groupoids: An Introduction", Moerdijk defines an equivalence of (étale) Lie groupoids $f\colon\mathcal{G} \to \mathcal{H}$ to be a smooth functor such that the map $\omega\pi_1 \colon \mathcal{H}_1\times_{\mathcal{H}_0}\mathcal{G}_0 \to \mathcal{H}_0$ defined on the fiber product $\{(g,y) \in \mathcal{H}_1\times\mathcal{G}_0 : \alpha(g) = f(y)\}$ sending a pair $(g,y)$ to $\omega(g)$ is a surjective submersion, and the following diagram is a pullback square $\require{AMScd}$ $$\begin{CD} \mathcal{G}_1 @>f>> \mathcal{H}_1 \\ @VV{\alpha\times\omega}V @VV{\alpha\times\omega}V \\ \mathcal{G}_0\times\mathcal{G}_0 @>{f\times f}>> \mathcal{H}_0\times\mathcal{H}_0, \end{CD}$$ i.e. the map $g \mapsto (\alpha(g),\omega(g),f(g))$ is a diffeomorphism from $\mathcal{G}_1$ to $\{ (x,y,g) \in \mathcal{G}_0\times\mathcal{G}_0\times\mathcal{H}_1 : \alpha(g) = f(x),\ \omega(g) = f(y)\}$. Remark. Moerdijk notes that the first condition says that each $x \in \mathcal{H}_0$ is isomorphic to some $f(y)$, so the map of orbit spaces is a surjection. He also notes that the second condition implies $f$ induces a diffeomorphism from the space of arrows $g \colon x \to y$ in $\mathcal{G}$ to the space of arrows $g' \colon f(x) \to f(y)$ in $\mathcal{H}$, so the map of orbit spaces is an injection, and for each $x \in \mathcal{G}_0$, the map $f$ induces an isomorphism of isotropy groups $\mathcal{G}_x \to \mathcal{H}_{f(x)}$. Thus Definition 2 implies all parts of Definition 1 except possibly the requirement that $f$ is an étale map. REPLY [2 votes]: Notice that because $\omega$ is étale and $\omega\pi_1$ is a submersion, $\pi_1\colon \mathcal{H}_1\times_{\mathcal{H}_0}\mathcal{G}_0 \to \mathcal{H}_1$ is a submersion, as is $\alpha\pi_1$. Since $\alpha$ is étale, $\pi_2\colon \mathcal{H}_1\times_{\mathcal{H}_0}\mathcal{G}_0 \to \mathcal{G}_0$ is étale. Therefore because the composition $\alpha\pi_1 = \pi_2f$ is a submersion, so is $f \colon \mathcal{G}_0 \to \mathcal{H}_0$. Because $f$ induces a bijection of orbit spaces, only points in the same orbit may be identified by $f$. Since $\mathcal{G}$ is étale, its orbits are discrete. Therefore the dimension of $\mathcal{G}_0$ is equal to the dimension of $\mathcal{H}_0$ and we conclude that $f\colon \mathcal{G}_0 \to \mathcal{H}_0$ is étale. (Actually, $f\colon \mathcal{G}_1 \to \mathcal{H}_1$ is also étale). Therefore Definition 2 implies Definition 1. Edit: Definition 1 implies Definition 2. Consider the fiber product $$\begin{CD} \mathcal{H}_1\times_{\mathcal{H}_0}\mathcal{G}_0 @>{\pi_2}>> \mathcal{G}_0 \\ @V{\pi_1}VV @VfVV \\ \mathcal{H}_1 @>\alpha>> \mathcal{H}_0. \end{CD}$$ Because $f$ is étale, $\pi_1$ is étale; therefore $\omega\pi_1$ is a fortiori a submersion. It is surjective as soon as, for every $x \in \mathcal{H}_0$, there exists $y\in \mathcal{G}_0$ and an arrow $f(y) \to x$. This is true because $f$ induces a bijection of orbit sets. Now consider the fiber product $$\begin{CD} K = (\mathcal{G}_0\times\mathcal{G}_0)\times_{\mathcal{H}_0\times\mathcal{H}_0}\mathcal{H}_1 @>{\pi_2}>> \mathcal{H}_1 \\ @V{\pi_1}VV @V{\alpha\times\omega}VV \\ \mathcal{G}_0 \times\mathcal{G}_0 @>{f\times f}>> \mathcal{H}_0\times\mathcal{H}_0. \end{CD}$$ We want to show that the map $F\colon \mathcal{G}_1 \to K$ defined by $g \mapsto (\alpha(g),\omega(g),f(g))$ is a diffeomorphism. Notice that because $f\times f$ is étale and $\pi_2F = f$ is étale, $F$ is étale. We need therefore only show that it is a bijection. It is easy to argue that because $f$ induces a bijection of orbit sets and isomorphisms on isotropy groups, for each $x$ and $y \in \mathcal{G}_0$, the map $f$ induces a bijection from the set $\{g\colon x \to y\}$ to the set $\{g' \colon f(x) \to f(y)\}$. Therefore $F$ is a diffeomorphism.<|endoftext|> TITLE: Example of non-saturated (co)fibration category QUESTION [6 upvotes]: A cofibration category is saturated if it satisfies the following equivalent conditions: Every map which becomes an isomorphism in the homotopy category is already a weak equivalence. The weak equivalences are closed under retracts. The weak equivalences satisfy the two-out-of-six property. The second condition is one of the axioms of a model category, and most categories with weak equivalences that arise in practice satisfy the last two properties more or less by construction. In fact, I realized I don't know any examples of cofibration categories which are not saturated. So the question is: What is an example of a non-saturated cofibration category? And, what are its maps which become isomorphisms in the homotopy category that are not already weak equivalences? I think that "finite spaces" (in a suitable sense) with simple homotopy equivalences as the weak equivalences is supposed to form such an example; but I don't know where to find the details. Edit: By a "cofibration category" I mean any of a "catégorie dérivable à droite" in the sense of Cisinski a "precofibration category" in the sense of Radulescu-Banu a "cofibration category" in the sense of Baues, without the axiom on fibrant models (though I don't see why this axiom should force the category to be saturated) I'm not picky about exactly which axioms assume which objects are cofibrant, but I do want the weak equivalences to satisfy two-out-of-three. REPLY [6 votes]: In chapter 2 of "Spaces of PL Manifolds and Categories of Simple Maps" Waldhausen, Jahren, and Rognes show that finite simplicial sets with injective maps as cofibrations and surjective maps with contractible point inverses as weak equivalence, forms a cofibration category. This category is not saturated. This is because all homotopy equivalences are inverted in the homotopy category since the projection $X \times I \rightarrow X$ is simple, and this forces the two inclusions $X \rightarrow X \times I$ to be equal in the homotopy category. Since there are many homotopy equivalences that are not surjective (not even taking into account Whitehead torsion), there are many maps inverted that are not weak equivalences.<|endoftext|> TITLE: Dictionary of arithmetic symmetries and Langlands QUESTION [7 upvotes]: To a number theorist automorphic forms appear to be adelic point-counting generating functions for arithmetic schemes. This is what the conjectured equality of their $L$ functions tells us. The fact that these generating functions are highly symmetric (Hecke action) seems to correspond to the fact that arithmetic schemes are highly symmetric (Galois action). That is why we try to match up Hecke symmetries with Galois symmetries when proving their relationship. If we think of algebraic groups together with their arithmetic subgroups as comprising all the arithmetic symmetries there are, it seems reasonable that whenever these symmetries show up in nature, for example in the form of Galois action on an arithmetic scheme, we want to look up this comprehensive dictionary of arithmetic symmetries to see where it occurs in order to understand it. Is this a clue to the awesome mystery that is the arithemetic Langlands correspondence? Does this count as a moral justification for the conjecture, Langlands' achievement being not only intuiting that auto-morphic actions of adelic algebraic groups are all the arithmetic symmetries there are, but also providing an indexing for the dictionary for this purpose by boldly extrapolating from class field theory after recasting it as the $GL_1$ case of a vast conjecture based on scant evidence at the time? Even if the intuition is valid as far as it goes, it leaves big questions to think about, starting with "why only reductive groups?" and "what are the qualifiers that need to go in front of arithmetic symmetries to make it more precise?", which I'd love to understand. REPLY [8 votes]: One way to interpret this question: Should we be able to guess or morally justify the philosophy behind the Langlands conjecture without analyzing important examples and special cases? I'm going to answer this one "no, we shouldn't". The Langlands conjectures emerged from Langlands' construction of the $L$-function of an automorphic form and his prior knowledge of Artin $L$-functions of a Galois representation, class field theory, and other topics. Wiping this away, I don't see how one could have guessed it. It's true that both sides can be described as "arithmetic symmetry", sort of (maybe one would better be described as "symmetric arithmetic"), but lots of things can be described as "arithmetic symmetry" at this level of detail, and they're not all in correspondence. Some examples are automorphisms of arithmetic schemes, fundamental groups of arithmetic schemes, and actions of fundamental groups on things other than vector spaces. It's true that we are more likely to see a correspondence between two classes of highly symmetric, arithmetic objects than between one such class and one class of non-symmetric or non-arithmetic objects. For example, we are unlikely to find a deep correspondence between Galois representations and finite graphs. Thus I think your approach is somewhat backwards. The way to guess the Langlands correspondence is to first notice that special cases of the two sides correspond, and that concrete features appear similar on both sides, using easier steps like the Satake isomorphism, class field theory, and so on, and then generalize. Trying say "I believe a priori there should be some correspondence between these two types of objects, now I just have to figure out which ones match up with which ones and what that means in concrete terms" is trying to make the hard parts easy and the easy parts hard. The closest thing I can imagine to this is to take the physical perspective on geometric Langlands, derive it from physical ideas (electric-magnetic duality in $N=4$ supersymmetric Yang-Mills), and deduce that it should have some arithmetic analogue or consequence. This faces two challenges, one is that making the translation is very hard (even having access to both the physical and arithmetic-geometric sides, it took Kapustin and Witten a lot of effort to see how they match up) and the physical ideas are not "morally obvious" in themselves. This perspective gets the closest to seeing the two sides as "fundamentally the same". We think of representations of the Galois group into $\hat{G}$ as $\hat{G}$-bundles with flat connection, and we think of points of $G(F)\backslash G(\mathbb A_F)/ G( \hat{\mathcal O}_F)$ as bundles without flat connection (where the flat connection reappears when we take the cotangent bundle of the moduli space of bundles), and we compare the two types of bundles. But the comparison is not anything simple - note that $G$ is not $\hat{G}$ - it's roughly supposed to be an advanced type of Fourier transform. One reason it's hard to imagine understanding this without getting into the weeds is that any philosophical description of how the Galois group of $\mathbb Q$ relates to representations of arithmetic groups over $\mathbb Q$ seems like it would apply equally well to how the Galois group of $\mathbb Q(x)$ relates to representations of arithmetic groups over $\mathbb Q(x)$. However, the first step of the Langlands program, class field theory, fails in that setting. There are notions of higher class field theory, but they don't just involve comparing representations and representations, and they don't seem likely to generalize to non-abelian groups. So we're saying something special about number fields that doesn't hold for higher-dimensional fields. (From my arithmetic-geometric perspective, I would say class field theory is a manifestation of Poincare duality, and the statement of Poincare duality depends heavily on the dimension. From the physical perspective, Langlands comes from a duality in four-dimensional field theory, and you don't get the same duality in higher dimensions.) Why reductive groups and not general groups? The rule of thumb you should always use in these types of things is that before conjecturing something for reductive groups you should test it for $\mathbb G_m$, and before conjecturing something for unipotent groups, or non-reductive groups, you should test it for $\mathbb G_a$. Characters of $\mathbb G_m(\mathbb A_F)/\mathbb G_m(F)$ correspond to characters of the Galois group by class field theory. Characters of $\mathbb G_a(\mathbb A_F)/\mathbb G_a(F)$ correspond to elements of $F$, not anything Galois-theoretic. What more is there to say?<|endoftext|> TITLE: Quadratic Diophantine equations with all values prime QUESTION [5 upvotes]: Given a quadratic Diophantine equation over the integers in two variables, can we say much about when it has only finitely many solutions with the additional assumption that both variables are prime? That is, given integers $a$, $b$, $c$, $d$, and $e$, $a$ and $c$ both non-zero. Then it seems reasonable to conjecture that there are only finitely many primes $x$ and $y$ with $$ax^2 +bxy +cy^2 + dx +ey + f=0,$$ barring the exceptional case where the left-hand side itself factors. The obvious heuristic here is that solutions to the equation (whether or not they are prime) should grow roughly exponentially based on standard methods to solve quadratic Diophantine equation. Thus, if we call the $n$th solution, $(x_n,y_n)$ then the chance they are both prime should be $O(\frac{1}{\log k^n})=O(\frac{1}{n})$ (where $k$ is some constant). So the chance they are both prime is about $O(\frac{1}{n^2})$, and so the relevant series converges, since $\sum_{n \geq 1} \frac{1}{n^2}$ converges. There are some cases where it is not hard to prove this sort of conjecture, using completely elementary methods. For example, it is not hard to show the following: Proposition: Suppose that $p$ is prime, $b \geq 1$, $m \geq 2$, $a \geq 1$, and $$p^2 + bp + ma^2 = mq^2.$$ Then $p \leq b + 4am$. Proof sketch: The equation can be written as $$p(p+b)=m(q-a)(q+a)$$ and so $p$ needs to divide one of the terms on the right-hand side. One interesting thing here is that in this case, we only need that $p$ is prime, which is stronger than what the above heuristic would suggest, since that heuristic uses both variables being prime, whereas here no assumption about the primality of $q$. (One can in this case also tighten the bound if one does do a little more work or if one also assumes that $q$ is prime.) My guess is that answering this question in complete generality is going to be tough. So my question then is what broad sets of cases can we prove this for? REPLY [2 votes]: here's one that puts up a good fight. The table can be extended readily with $w_{n+4} = 6 w_{n+2} - w_n$ and same for $v$ jagy@phobeusjunior:~$ ./Pell_Target_Fundamental 2 3 2 23 120 Automorphism matrix: 3 4 2 3 Automorphism backwards: 3 -4 -2 3 3^2 - 2 2^2 = 1 w^2 - 2 v^2 = 23 = 23 Sun Feb 7 13:49:34 PST 2021 1. w: 5 w= 5 v: 1 SEED KEEP +- v = 1 2. w: 11 w= 11 v: 7 SEED BACK ONE STEP 5 , -1 v = 7 3. w: 19 w= 19 v: 13 v = 13 4. w: 61 w= 61 v: 43 v = 43 5. w: 109 w= 109 v: 77 v = 7 11 6. w: 355 w= 5 71 v: 251 v = 251 7. w: 635 w= 5 127 v: 449 v = 449 8. w: 2069 w= 2069 v: 1463 v = 7 11 19 9. w: 3701 w= 3701 v: 2617 v = 2617 10. w: 12059 w= 31 389 v: 8527 v = 8527 11. w: 21571 w= 11 37 53 v: 15253 v = 7 2179 12. w: 70285 w= 5 14057 v: 49699 v = 13 3823 13. w: 125725 w= 5^2 47 107 v: 88901 v = 19 4679 14. w: 409651 w= 11 167 223 v: 289667 v = 7 41381 15. w: 732779 w= 67 10937 v: 518153 v = 518153 16. w: 2387621 w= 2387621 v: 1688303 v = 83 20341 17. w: 4270949 w= 4270949 v: 3020017 v = 7^2 11 13 431 18. w: 13916075 w= 5^2 19 29297 v: 9840151 v = 9840151 19. w: 24892915 w= 5 43 115781 v: 17601949 v = 17601949 Sun Feb 7 13:51:34 PST 2021 w^2 - 2 v^2 = 23 = 23 5, 11, 19, 61, 109, 355, 635, 2069, 3701, 12059, 21571, 70285, 125725, 409651, 732779, 2387621, 4270949, 13916075, 24892915, 1, 7, 13, 43, 77, 251, 449, 1463, 2617, 8527, 15253, 49699, 88901, 289667, 518153, 1688303, 3020017, 9840151, 17601949, jagy@phobeusjunior:~$<|endoftext|> TITLE: Paris-Harrington principles parametrized by functions $f:\mathbb N \to \mathbb N$ QUESTION [8 upvotes]: Recall that the Paris-Harrington Principle, $\mathsf{PH}$, is the statement that for each $e, r, k < \omega$ there is an $N < \omega$ so that given any coloring $c:[N]^e \to r$ there is an $H \subseteq N$ so that $c \upharpoonright [H]^e$ is constant, and the cardinality of $H$ is larger than both $k$ and ${\rm min}(H)$. This last clause separates $\mathsf{PH}$ from finite Ramsey's theorem. Famously Paris and Harrington showed that $\mathsf{PH}$ is not provable in $\mathsf{PA}$ but it is true in the standard model. Of course in general the interest from a logical point of view is the non-provability result but in this question I want to focus on the fact that it's true in the standard model. The standard proof of this fact mimics the proof of finite Ramsey's theorem as an application of Konig's Lemma and infinite Ramsey theorem noting that, in the infinite homogeneous set one gets, we can choose $H$ as big a finite segment as we like. If people are unclear what I mean, I can write this down. The same proof actually shows that for $any$ $f:\omega \to \omega$ we can actually arrange, in the definition of $\mathsf{PH}$ that not only $|H| > {\rm min}(H)$ but actually $|H| > f({\rm min}(H))$. Let us denote this (seemingly stronger) statement by $\mathsf{PH}_f$. So, for example, $\mathsf{PH}$ is $\mathsf{PH}_{identity}$. My question is: When is it the case that for $f, g:\omega \to \omega$ do we have $\mathsf{PH}_f$ implies $\mathsf{PH}_g$ over $\mathsf{PA}$ as a base theory? For example, one particular instance of this question that I'm interested in is: Does $\mathsf{PH}$ imply $\mathsf{PH}_f$ for all computable $f$? What about all arithmetic $f$? REPLY [4 votes]: The following paper by my supervisor Andreas Weiermann studies the principle PH$_f$ you mention: A. Weiermann, A classification of rapidly growing Ramsey functions Proc. Amer. Math. Soc. 132 (2004), no. 2, 553–561. There are many follow-up papers with plenty of variations by Andreas and his collaborators. As to your question, any answer here depends greatly on the exact technical details, so do consult the original paper if possible. Intuitively, the answer is two-fold as follows: PH$_f$ is not provable in PA for $f$ equal to log$^d$ for any fixed $d\in \mathbb{N}$. One can obtain sharper result as follows: $d$ can be replaced with an unbounded (provable say in ZFC) function that grows so slow that there are models of PA in which it is bounded above. An example of the latter is $H_{\epsilon_0}^{-1}$, i.e. the inverse of the Hardy function with ordinal $\epsilon_0$. Note that $H_{\epsilon_0}^{-1}$ may seem weird, but PA actually proves that it is total.<|endoftext|> TITLE: Is there a condensed / pyknotic refinement of the shape of an $\infty$-topos? QUESTION [14 upvotes]: Let $\mathcal E$ be an $\infty$-topos. Recall that Lurie defines the shape of $\mathcal E$ as the left-exact, accessible functor $\Gamma \Delta: Spaces \to Spaces$ where $\Delta: Spaces^\to_\leftarrow \mathcal E: \Gamma$ is the the constant / global sections adjunction, the unique geometric morphism from $\mathcal E \to Spaces$. The $\infty$-category of left-exact, accessible endofunctors of $Spaces$ is better-known as the $\infty$-category of Pro-spaces. As a pro-space, the shape of $\mathcal E$ is like having a space, but even better, because its homotopy groups are not just groups, but pro-discrete groups. For instance, the shape of the etale site of a scheme $X$ is essentially the etale homotopy type of $X$, and its fundamental groupoid is essentially the etale fundamental groupoid, complete with its usual profinite topology (I am glossing over some additional finiteness considerations; to be honest this is largely because I don't fully understand them and have even less idea how they change the picture conceptually.) This situation -- of homotopy groups which come with a topology -- is (from some perspective) exactly what the condensed mathematics of Clausen and Scholze / pyknotic mathematics of Barwick and Haine was developed for, and my understanding is that the example of the etale homotopy type is particularly germane, as Bhatt and Scholze's first application of the theory is to study the pro-etale site and its associated fundamental group. So I think the answer to the following questions must be yes, at least modulo the finiteness conditions I haven't understood: Question 1: Is there a condensed / pyknotic "re-interpretation" of the pro-discrete topology on the homotopy groups of the shape of an arbitrary $\infty$-topos? Question 2: Is there a condensed / pyknotic refinement of the shape of an arbitrary $\infty$-topos, perhaps containing more information than the usual pro-space? I think if I read Bhatt and Scholze, it should all be in there. Unfortunately, I'm hampered by the fact that the etale fundamental group is usually approached differently by algebraic geometers, in a way which circumvents considering all of the data of the etale homotopy type. This approach is doubtless more practical for algebraic geometry (and of course, my way of putting it is completely ahistorical, as Grothendieck constructed the etale fundamental group long before Artin and Mazur constructed the etale homotopy type), but unfortunately I am not an algebraic geometer and I've never dug into it to understand. It appears that Bhatt and Scholze use a similar approach, which is an obstacle for me to understand it. So I'm wondering if somebody can translate these formal aspects of the situation into a language which I personally happen to understand better. REPLY [8 votes]: In subsection 13.8.10. of version 7 of the "Exodromy" paper by Barwick, Glasman and Haine (arXiv:1807.03281v7), the authors define the "pyknotic étale homotopy type". They say that it should recover the pro-étale fundamental group of Bhatt–Scholze and will explore this in a future work.<|endoftext|> TITLE: Initiation to constructive mathematics QUESTION [8 upvotes]: What are some good introductory references to constructive mathematics for non-specialist mathematicians? I would like to learn more about constructive mathematics, just to improve my general mathematical knowledge, but I have some difficulties in finding good references for non-specialists. Are there introductory texts including: a rough description of the main motivations and of the history of the field; a discussion of the different meanings given to "constructive"; explicit examples of constructive vs. non-constructive proofs; a discussion about what one looses compared to "classical" mathematics? Of course, I do not necessarily expect a single text satisfying all these conditions, a collection of a few references would do the job. REPLY [14 votes]: I have a soft spot for Constructivism in Mathematics: An Introduction (2 volumes) by Troelstra and van Dalen as an overview. And for what one loses compared to classical mathematics (as well as its historical importance), it's probably worth going straight to Bishop's Foundations of constructive analysis. However, an important thing to be aware of when reading older and historical works is that they may give a misleading impression of constructivism as currently practiced. Historical overviews such as Troelstra-van Dalen or Bridges-Richman will often spend time contrasting schools of constructivism such as Brouwer's intuitionism, Markov's constructive recursive mathematics, and Bishop's constructive mathematics. This is certainly interesting and historically important, but in my (humble) opinion, the trend of the future in constructivism is towards "neutral" constructive mathematics, which historically was a bit of a latecomer. The three historical "big schools" all assume various principles that deviate from neutral constructivism, some of which are even classically contradictory. While some of those principles (notably ones like Markov's principle and countable choice, which are weak "computable" forms of LEM and AC and in particular are consistent with classical mathematics) are still in wide use today, it's become more fashionable to at least notice when they are used and attempt to avoid them if possible. One important reason for this is that neutral constructivism is valid internal to any elementary topos, whereas these additional principles are not. Topos theory has also led to important insights into the best way to do mathematics constructively, such as the use of locales as a good notion of "space". Thus, I would recommend supplementing your historical reading with some more modern work. Bauer's Five stages of accepting constructive mathematics cited in the comments is an excellent start. But unfortunately, I don't know of a good book-length work to follow it. One of my own early introductions to constructivism was a course on constructive locale theory given by Peter Johnstone, much of which became Part C of Sketches of an Elephant; but that may be a bit difficult to follow without some background in category theory and topos theory (such as obtained from Parts A and B). The HoTT Book also includes some explanation of modern approaches to doing mathematics constructively, particularly chapters 10 and 11 (plus a few remarks in the introduction); but, again, it may be hard to get there without slogging through chapters 1-9. And of course there is Taylor's Practical foundations of mathematics, which has many nice features but which Peter Johnstone famously called "the book of which Mathematics Made Difficult was a parody". So overall I think there is room here for someone to write a good book.<|endoftext|> TITLE: Maximal (minimal) value of $S=x_1^2x_2+x_2^2x_3+\cdots+x_{n-1}^2x_n+x_n^2x_1$ on condition that $x_1^2+x_2^2+\cdots+x_n^2=1$ QUESTION [6 upvotes]: since $x_1^2+x_2^2+\cdots+x_n^2=1$ is sphere,a compact set,so $S$ has a maximal(minimal) value. But when I try to solve it using the Lagrangian multiplier method, I don't know how to solve these equations. Clearly $x_1=x_2=x_3=\cdots = x_n=\frac{1}{\sqrt{n}}$ is an extremal point, but I don't know if it's the maximal(minimal) value. I want to know how to solve the problem. Also, could this problem be solved by elementary methods, like some inequality techniques? REPLY [6 votes]: For $n=3$ we need to find $$\max_{a^2+b^2+c^2=1}(a^2b+b^2c+c^2a).$$ Indeed, let $\{|a|,|b|,|c|\}=\{x,y,z\}$, where $x\geq y\geq z\geq0$. Thus, by Rearrangement and AM-GM we obtain: $$\sum_{cyc}a^2b\leq|a|\cdot(|a||b|)+|b|\cdot(|b||c|)+|c|\cdot(|c||a|)\leq$$ $$\leq x\cdot xy+y\cdot xz+z\cdot yz=y(x^2+xz+z^2)\leq y\left(x^2+\frac{x^2+z^2}{2}+z^2\right)=$$ $$=\frac{3}{2}y(1-y^2)=\frac{3}{2\sqrt2}\sqrt{2y^2(1-y^2)^2}\leq\frac{3}{2\sqrt2}\sqrt{\left(\frac{2y^2+2-2y^2}{3}\right)^3}=\frac{1}{\sqrt3}.$$ The equality occurs for $a=b=c=\frac{1}{\sqrt3}$, which says that we got a maximal value. For $n=4$ we need to find $$\max_{\sum\limits_{cyc}a^2=1}\sum_{cyc}a^2b.$$ Indeed, by C-S and AM-GM we obtain: $$\sum_{cyc}a^2b\leq\sqrt{\sum_{cyc}a^2\sum_{cyc}a^2b^2}=\sqrt{(a^2+c^2)(b^2+d^2)}\leq\frac{1}{2}(a^2+c^2+b^2+d^2)=\frac{1}{2}.$$ The equality occurs for $a=b=c=d=\frac{1}{2},$ which says that we got a maximal value. For $n\geq5$ we can use the Lagrange Multipliers method, but it does not give nice numbers. For example, for $n=5$ the maximum occurs, when $(x_1,x_2,x_3,x_4,x_5)||(0.79...,3.24...,3.78...,2.48...,1),$ which gives a value $0.45...$ The following inequality is also true. Let $a$, $b$ and $c$ be real numbers such that $a^2+b^2+c^2=1$. Prove that: $$a^3b^2+b^3c^2+c^3a^2\leq\frac{1}{3\sqrt3}.$$<|endoftext|> TITLE: Did André Bloch or any other mathematician receive the Becquerel Prize? QUESTION [19 upvotes]: On this biography page of André Bloch, it is said that The Académie des Sciences awarded him the Becquerel Prize just before his death. This claim is also repeated in PlanetMath, Wikiversity and also Hersh & John-Steiner's book: Loving and Hating Mathematics: Challenging the Myths of Mathematical Life (+link). There are several other resources backing this claim. But on the other hand, the official website of the Becquerel Prize says: (emphasis mine) The Alexandre Edmond Becquerel Prize was established in 1989 by the European Commission at the occasion of the 150th anniversary of Becquerel’s classical experiment in which he discovered the photovoltaic effect. Its purpose is to honour scientific, technical or managerial merit in the development of photovoltaic solar energy, attained over a long period of continuous achievements, or very exceptionally, for some extraordinary invention or discovery. So clearly, this Becquerel Prize has nothing to do with mathematics. I suspected that there was another Becquerel Prize in 1930s or 1940s, which was awarded by the Académie des Sciences. But I found no record of that except the Wikipedia page of Yvette Cauchois, who was a physicist known for her contributions to x-ray spectroscopy and x-ray optics: Henri Becquerel Prize from the French Academy of Sciences (1935) Therefore this limited sample just strengthens my suspicion that there might no Becquerel Prize for mathematicians. And this part of André Bloch's biography also adds up to the other myths surrounding him. So my questions, as the title says, are: Was there a Becquerel Prize for mathematicians during that era? Did André Bloch receive it? REPLY [4 votes]: I was looking for the same claim about another mathematician (namely Vazgain Avanissian) that I came to this question. Following the clues left by the previous answer and the comments, I found a mention of Pierre Fatou in Project Euclid. Fatou has been mentioned in the OP's comment. However, the piece below includes added information. It neither answer OP's questions asked in the body of the question, nor any information about the mathematician whom I am looking for. But it gives an answer to the "any" part of the title: "any other mathematician ...". I hope, this in turns leads someone else to a better picture of the situation.<|endoftext|> TITLE: For what topological groups $G$ can we take $EG \rightarrow BG$ to be of the form $S^{\infty} \rightarrow BG$? QUESTION [17 upvotes]: Title. For what topological groups $G$ can we take $EG \rightarrow BG$ to be of the form $S^{\infty} \rightarrow BG$? If $G$ is a subgroup of either $S^0,S^1,S^3$ or $S^7$ this induces a free action on $S^{\infty}$ and thus a $G-$principal bundle $S^{\infty} \rightarrow BG$. Does the reverse direction hold? That is, if $G$ acts freely on $S^{\infty}$ is it contained in some of the $S^0,S^1,...,S^7$ as a subgroup such that the induced action on $S^{\infty}$ is the same? REPLY [19 votes]: I like to think of $EG$ and $BG$ in terms of configuration spaces. The space $BG$ can be identified with the following configuration space. It consists of configurations of finitely many points in the open interval $(0,1)$ with the points having labels in the topological group $G$. It is topologized so that when points collide, the labels are multiplied (using orientation of the interval to determine the order); points labeled by the identity element of $G$ can always be added or removed; Points can "disappear" by sliding off either end of the interval. It is a nice exercise to see that this agrees with the usual definition of $BG$ as the geometric realization of a simplicial space. $EG$ has a similar description as configurations of points in the half-open interval $[0, 1)$. In this case points cannot slide off the closed end, and can only "disappear" by sliding off the open end. Sliding everything off the open end gives a contraction onto the empty configuration, whence $EG$ is contractible. The map $EG \to BG$ is just the restriction of configurations. The action of $G$ on $EG$ is the following. Each configuration in $EG$ may be view as having the point $0 \in [0,1)$ as part of it - either it is already labeled or we give it the label $e \in G$. The action of $G$ just multiplies the label of the point $0$ on the left. From these descriptions (or the usual simplicial ones) you can realize $EG$ as a certain colimit of simple spaces which consist of products of intervals (open and half-open) and copies of $G$. If $G$ is a finite dimensional Lie group with countably many components, then from this colimit description it is possible to see that locally $EG$ is of the form $K \times \mathbb{R}^\infty$ where $K$ is a neighborhood retract of $\mathbb{R}^\infty$ (which might be different at different points - we do not care). If $G$ has finitely many components $K$ will even be a locally finite CW-complex. If there are countably many components $K$ will look like a finite dimensional countable CW-complex, which can still be embedded nicely in $\mathbb{R}^\infty$ as a neighborhood retract. From the results cited in this excellent MO answer: https://mathoverflow.net/a/293409/184 we deduce the following surprising facts (1) $EG$ is actually locally modeled on $\mathbb{R}^\infty$ and (2) for spaces locally modeled on $\mathbb{R}^\infty$, homotopy equivalence implies homeomorphism. Since both $EG$ and $S^\infty$ are contractible spaces locally modeled on $\mathbb{R}^\infty$, it follows that we have a homeomorphism $EG \cong S^\infty$. So in summary: For any finite dimensional Lie group $G$ with countably many components you may take $EG \cong S^\infty$. For example $G$ can be any countable discrete group. However the free action of $G$ on $S^\infty$ is realized through a possibly strange homeomorphism and likely has nothing to do with $G$ acting on finite dimensional spheres.<|endoftext|> TITLE: Grothendieck group generated by classes of invertible sheaves QUESTION [7 upvotes]: Given a smooth, projective (complex) varieties $X$, is it true that the grothendieck group $K_0(X)$ of equivalence classes of coherent sheaves on $X$, is generated by clases of invertible sheaves i.e., any class of a coherent sheaf on $X$ can be written as a linear combination of classes of invertible sheaves on $X$? Any reference/idea will be most welcome. REPLY [12 votes]: Here is a different, perhaps more elementary example. Consider the Grassmannian $\mathrm{Gr}(2,4)$ of lines in $\mathbf{P}^3$, and let $\mathscr{Q}$ be the universal quotient bundle. The line bundle $\mathrm{det}(\mathscr{Q})$ generates the Picard group of $\mathrm{Gr}(2,4)$. If $[\mathscr{Q}]$ were in the subgroup of $\mathrm{K}^0$ generated by line bundles, then we would have $[\mathscr{Q}]=\sum_i n_i [\mathrm{det}(\mathscr{Q})^{\otimes r_{i}}]$ for some integers $n_i, r_i$. Taking Chern classes implies $$c_2(\mathscr{Q})= m c_1(\mathscr{Q})^2$$ for some integer $m$, which cannot be true. Indeed, fix a point $p$ and a line $L$ in $\mathbf{P}^3$; the Chern class $c_1(\mathscr{Q})$ is geometrically represented by the Schubert cycle $\Sigma_1(L)=\{L' \ | \ L\cap L'\neq\emptyset\}$, while $c_2(\mathscr{Q})$ is represented by $\Sigma_2(p)=\{L' \ | \ p\in L'\}$. Then $$1=\int_{\mathrm{Gr}(2,4)} c_2(\mathscr{Q})^2\neq m^2\int_{\mathrm{Gr}(2,4)} c_1(\mathscr{Q})^4=2m^2.$$ The integral on the left is the number of lines through two distinct points in $\mathbf{P}^3$, while the integral on the right is the number of lines meeting four given (general) lines in $\mathbf{P}^3$. I think only few varieties have the property which you want. It is certainly true for curves: if you have a locally free sheaf $\mathscr{E}$ on a curve, then you can find an exact sequence of locally free sheaves $$0\rightarrow\mathscr{E}'\rightarrow\mathscr{E}\rightarrow\mathscr{E}''\rightarrow 0,$$ where the ranks of $\mathscr{E}'$ and $\mathscr{E}''$ are strictly smaller than $\mathscr{E}$ (unless, of course, $\mathscr{E}$ is already of rank $1$). The same property holds if $\mathscr{E}$ is a locally free sheaf of rank $>2$ on a surface. Indeed, after twisting with an invertible sheaf, we may assume that $\mathscr{E}$ is globally generated; by a standard result (e.g. Exercise 8.2 in Chapter 2 of Hartshorne) we can then find a nonzero section $s:\mathscr{O}\rightarrow\mathscr{E}$ whose cokernel is locally free. The K-theory of surfaces can thus be generated by locally free sheaves of rank $1$ and $2$, and analogous statements hold in higher dimension.<|endoftext|> TITLE: On $p$-groups with abelian automorphism group QUESTION [8 upvotes]: Let $G$ be a $p$-group of order $p^{n}\geq p^{7}$ and its automorphism group is elementary abelian $p$-group. Then, it is clear that $G$ is nilpotent of class $2$. However, the converse is not true in general (take for example the unitriangular group of $3\times 3$ upper triangular matrices over $\mathbb{F}_{p}$). In fact, Marta Morigi proves that there is a $p$-group of order $p^{7}$ and class $2$ and its automorphism group is elementary abelian $p$-subgroup of order $p^{12}$. Furthermore, I think the converse can be true for 2-groups $G$ with abelian direct factor but I don't know how to do this. Are there any assumption to add to p-groups of class $2$ to get the converse for arbitrarily integer $n$? Let $p$ be an odd prime and $G$ be a purely non-abelian p-group of class $2$ and order $p^{n}\geq p^{7}$. Does the automorphism group of $G$ is abelian? Are there some other class of $p$-groups with elementary abelian automorphism $p$-groups?. Any response or reference may be very helpful. Thank you in advance. REPLY [6 votes]: This is just an amplification of my comment above. Theorem 3.3 of this this survey article, A Survey on Automorphism Groups of Finite p-Groups, by Geir T. Helleloid (2006) describes a result published in U. M. Webb, The number of stem covers of an elementary abelian p-group, Math. Z. 182 (1983), no. 3, 327–337. A stem cover of a group $Q$ is a maximal stem extension of $Q$; that is, a maximal (under group extensions) group $G$ with normal subgroup $N$ such that $G/N \cong Q$ and $N \le Z(G) \cap [G,G]$. (So, in a stem cover, $N$ is isomorphic to the Schur Multiplier of $Q$.) This result says that, for $p$ an odd prime, the proportion of stem covers of an elementary abelian group of order $p^n$ for which the automorphism group is an elementary abelian $p$-group approaches 1 as $n \to \infty$. Elementary abelian gropups have lots of distinct stem covers, so this is producing plenty of examples of the type you are looking for.<|endoftext|> TITLE: What are cospectra, and why have they received so little attention? QUESTION [18 upvotes]: A cospectrum (in the context of homotopy theory) is defined to be a sequence of spaces $X_0, X_1, \ldots, X_n, \ldots, $ equipped with maps $X_{n+1}\to \Sigma X_n$, for each $n$. So cospectra are similar to spectra, except that the structure maps point in the opposite direction. Cospectra were first introduced by Elon Lima in 1959. I learned about them form Browder's classic paper on the Kervaire invariant problem. Browder seems to make rather essential use of this construction. I had never seen the concept before, so tried to search the literature on cospectra. I found virtually nothing. There are a couple of papers developing some properties of cospectra, but they did not seem to lead to any further activity. As far as I can see, practically no one else investigated cospectra or used them for anything. So we have a definition that is a more or less natural variant of a very influential one. A definition that was used once in an important paper, and nowhere else. I find this curious, therefore I want to ask Question 1 Is there a reasonable way to rewrite Browder's proof without cospectra? According to my limited understanding, the reason for introducing cospectra is that they provide an alternative approach to Spanier-Whitehead duality. Lima's motivation was to define Spanier-Whitehead dual of spaces more general than finite CW complexes. As far as I could make out, Browder's motivation was similar. He needed to have a notion of Spanier-Whitehead dual that was well-behaved for non-finite spectra, and cospectra seem to do the job for him. Therefore I wonder if the same could be accomplished using the duality between spectra and pro-spectra, or maybe by just using finite approximations to spectra. If question 1 does not have an obvious positive answer, then there is a natural follow up: Question 2 How come no one else found use for cospectra? Is there some good mathematics lying that way, waiting to be discovered? REPLY [7 votes]: Brayton Gray had a nice preprint about cospectra and unstable $v_n$-periodic homotopy. This was maybe 20 years ago, and I can't find any online version. To confirm my memory that this existed, by searching the web, I also found mention of this in notes by Neil Strickland. I likely have a paper version in my office, which I am not visiting very often right now. But maybe I, or someone else, can reconstruct his basic idea.<|endoftext|> TITLE: Internal logic of locally strongly finitely presentable categories QUESTION [8 upvotes]: There is a duality between locally strongly finitely presentable categories and (Cauchy complete) cartesian categories, i.e. multisorted algebraic theories. The internal logic of cartesian categories is well known to be first-order equational logic. What is the internal logic of locally strongly finitely presentable categories? It should at least subsume "exact logic" (i.e. have finite conjunction, existential quantification, and quotients of equivalence relations), since the category of models for any multisorted algebraic theory is exact. However, it may not be describable as fragment of first-order logic, because although there exist arbitrary disjunctions, conjunction does not distributive over disjunction. REPLY [2 votes]: The comments are getting a bit long (sorry, that is largely my fault), so I think it's worth expanding on Jiří Rosický 's point, which comes very close to completely answering the question. Adamek, Lawvere, and Rosicky introduced the notion of an algebraically exact category in How algebraic is algebra?, giving a precise definition of what it means for a category to have all of the "exactness" properties enjoyed by (possibly multisorted) varieties (= algebraic categories = locally strongly finitely-presentable) categories. The definition is conceptually illuminating: the forgetful functor $Var \to Cat$ has a left adjoint, and an algebraically exact category is defined to be a pseudoalgebra for the induced pseudomonad on $Cat$. They observed that every algebraically exact category $\mathcal C$ has the following properties: $\mathcal C$ has limits. $\mathcal C$ has sifted colimits. $\mathcal C$ is Barr-exact. finite limits commute with filtered colimits in $\mathcal C$. regular epimorphisms are stable under products in $\mathcal C$. filtered colimits distribute over products in $\mathcal C$.\ (This one must apparently follow from the rest: regular epis are stable under pullback.) They conjectured that these properties completely characterize the algebraically exact categories. This conjecture was proven under various additional assumptions by these and other authors, until it was finally proven in full generality by Garner in A characterization of algebraic exactness (see Garner for a full bibliography). Upshot: It would seem the appropriate internal logic would be "whatever you can express using the above properties". In particular, Barr-exactness gives a good fragment of logic. Stability of regular epis under pullback probably lets you do a reasonable amount more. Perhaps others can elaborate on / correct these statements. Side note: I would assume that only the "finitary" fragment of the above properties would be relevant to building an internal logic as it's usually conceived. I would be interested to be proven wrong about that though!<|endoftext|> TITLE: Functional-analytic proof of the existence of non-symmetric random variables with vanishing odd moments QUESTION [28 upvotes]: It is known that a random variable $X$ which is symmetric about $0$ (i.e $X$ and $-X$ have the same distribution) must have all its odd moments (when they exist!) equal to zero. The converse is a strong temptation which has been around for perhaps a hundred years. Question. Suppose $\mathbb E[X^n] = 0$ for all odd $n$. Is it true that $X$ is symmetric ? This question was solved in Churchill (1946). In fact, he proved something much stronger Theorem. Let $X$ be a random variable and let $(a_m)_{m \in \mathbb N}$ be a sequence of real numbers. Then for every $\epsilon > 0$, the exists a random variable $Y$ such that (1) $\mathbb E[Y^{2m+1}] = a_m$ for all $m \in \mathbb N$, and (2) The Kolmogorov distance between $X$ and $Y$ is at most $\epsilon$. Of course this theorem immediately implies a negative answer to the above question. The proof given in the paper is constructive, but somewhat mysterious. Question. Is there simple / modern way to prove the above theorem using functional-analytic tools ? After all, the theorem simply says (roughly) that the set of random variables of with odd-moments given by the sequence $(a_m)$ is dense in the space of random variables equipped with Komolgorov distance. Note. The expected advantage of a general functional-analytic solution is that it would perhaps extend to constraints which are more general than those implied by prescribed odd moments. REPLY [12 votes]: This doesn't really require modern functional analytic tools, but we can prove a statement (due originally to Edelheit, according to Jochen Wengenroth in the comments) like Let $V$ be a Frechet space, complete with respect to a family of norms $||x||_i$, $i=0,1,\dots$. Let $L_1,L_2,\dots$ be linear forms on $V$, with $L_i$ bounded with respect to $||x||_i$ but unbounded with respect to any linear combination of $||x||_j$ for $j0$, and $L_m(f) = \int f x^{2m+1} dx$. linear forms obtained by integrating against $x^{2n+1}$. This gives us a function $f(x)$ with desired odd moments and norm $\epsilon$, which we can make into a nonnegative function with the same moments and integral $\epsilon$ by the trick $g(x) =f^+(x) + f^-(-x)$. Then take $(1-\epsilon)$ of any measure plus $\epsilon$ times $g$ for suitable $g$. To prove this: We fix a bunch of small constants $\epsilon_{ij}$, for $i,j \in \mathbb N$, to be chosen later. After rescaling $L_1$, we may assume $|L_1(\cdot)| \leq ||\cdot||_1$. Choose $x_1$ with $L_1(x_1)=1$ and $||x_1||_0 < \epsilon_{01}$ . Rescale $||\cdot||_2$ and $L_2$ so that $||x_1||_2<\epsilon_{21}$ and $|L_2( \cdot) |\leq ||\cdot||_2$ and then find $x_2$ with $||x_2||_0 < \epsilon_{02}$, $||x_2||_1< \epsilon_{12}$, and $L_2(x_2)=1$. Rescale $||\cdot ||_3$ and $L_3$ so that $||x_1||_3<\epsilon_{31}$, $||x_2||_3<\epsilon_{32}$, and $L_3(\cdot) \leq ||\cdot||_3$, and iterate this process. Let $M$ be the $\mathbb N \times \mathbb N$ matrix with entries $M_{ij} = L_i (x_j)$. We have $|M_{ij} |\leq \epsilon ij$ if $i \neq j$ and $1$ if $i=j$. Let $N = I + (I-M) + (I-M)^2 + (I-M)^3+ \dots $ be the inverse matrix, choosing $\epsilon_{ij}$ small enough that this sum converges. Taking $\epsilon_{ij}$ as small as we want, we can make the off-diagonal entries of $N$ as small as desired and the diagonal entries as close to $1$ as desired. Then letting $$x =\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} N_{ij} x_i a_j $$ we have $L_i(x) = a_i$ and $$ || x||_0 \leq \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} |N_{ij}| ||x_i||_0 |a_j| \leq \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} |N_{ij}| \epsilon_{0i} |a_j|$$ which we can make as small as desired, and for any $k>0$ $$ || x||_k \leq \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} |N_{ij}| ||x_i||_k |a_j| \leq \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} |N_{ij}| \epsilon_{ki} |a_j| +\sum_{j=1}^{\infty} |N_{kj}| ||x_k||_k |a_j|$$, where the first term can be made as small as desired and the second term can, at least, be made convergent. For example $\epsilon_{ij} = 2^{-i-j}/ (|a_j|+1)$ for $i \neq j$ and $\epsilon_{0i} = 2^{-i} \cdot \epsilon / (|a_i|+1) $ will ensure $(N- I)_{ij}$ is bounded by $(3/2) 2^{-i -j} / (|a_j|+1)$ and so $||x||_0 = O(\epsilon)$ and $||x||_k = O ( \epsilon \cdot (1 + ||x_k||_k ) (1+ |a_k|) ) . $ The sign trick to make the function nonnegative is not strictly necessary. We can prove a similar statement like Let $f_1,f_2,\dots$ be functions on a measure space. Assume that $$ \inf \frac{ f_i(x)}{1 + \sum_{j TITLE: Rational maps from $\mathbb CP^n$ to $\mathbb CP^{n-1}$, fixing $\mathbb CP^{n-1}$ QUESTION [6 upvotes]: Consider $\mathbb CP^n$ and let $H\subset \mathbb CP^n$ be a hyperplane. Suppose $\varphi: \mathbb CP^n\to H$ is a rational map that fixes $H$ pointwise. I believe that $\varphi$ must be a projection from a point $p\in \mathbb CP^n\setminus H$ to $H$. How to prove this? Added. F_L gave a nice counterexample to the above question. But I still hope that the answer is positive if one has the following restriction on $\varphi$: The intersection of the locus of indeterminacy of $\varphi$ with $H$ has codimension $\ge 2$ in $H$. Does the statement hold under this additional condition? (In the counter-example of F_L, $p_1\in L$ is a point of indeterminancy of $\varphi$) REPLY [6 votes]: With the additional restriction, $\varphi$ is indeed a linear projection. Applying a linear change of coordinates, $H$ is given by $x_0=0$, so $\varphi$ is $[x_0:\cdots:x_n]\mapsto [0:P_1(x_0,\ldots,x_n):\cdots: P_n(x_0,\ldots,x_n)]$ for some homogeneous polynomials $P_1,\ldots,P_n\in \mathbb{C}[x_0,\ldots,x_n]$ of the same degree $d$. You may choose them without common factor. Your assumption is that the restriction to $H$ is the identity. So this means that $[0:x_1:\cdots:x_n]\mapsto [0:P_1(0,x_1,\ldots,x_n):\cdots: P_n(0,x_1,\ldots,x_n)]$ is the identity on $H$. Hence there is a polynomial $A\in \mathbb{C}[x_1,\ldots,x_n]$, homogeneous of degree $d-1$, such that $$P_i(0,x_1,\ldots,x_n)=A\cdot x_i$$ for $i=1,\ldots,n$. The locus where $A=0$ and where $x_0=0$ is then contained in the intersection of the base-locus of $\varphi$ with $H$. As you assumed that this latter should be of codimension $2$ in $H$, it implies that $A$ is a non-zero constant. Hence, $d=1$, so $\varphi$ is of degree $1$ and is a linear projection.<|endoftext|> TITLE: Approximation of analytic function by a fixed number of monomials QUESTION [5 upvotes]: This question seems simple but I can't manage to disprove it. Let $N\in \mathbb{N}$. We know that by its analyticity that this precise linear combination of monomials $ \sum_{n=0}^K \frac1{n!} x^n $ converges uniformly to $g$ on $[0,1]$ and that for $K$ large enough $$ \max_{x \in [0,1]}\, \|\sum_{n=0}^K \frac1{n!} x^n - \exp(x)\| TITLE: Which great mathematicians were also historians of mathematics? QUESTION [20 upvotes]: As the question title suggests, which great mathematicians were also historians of mathematics? We all know plenty of great mathematicians, but not many historians of mathematics. Not to mention that the history of mathematics can be dismissed as mere sociology sometimes and isn't very popular. REPLY [5 votes]: It seems to me that Otto Toeplitz (1881-1940) has not been mentioned yet. The following paragraph from the Wikipedia article on Toeplitz gives a quick idea of his contributions to the history of mathematics: In 1929, he cofounded "Quellen und Studien zur Geschichte der Mathematik" with Otto Neugebauer and Julius Stenzel. Beginning in the 1920s, Toeplitz advocated a "genetic method" in teaching of mathematics, which he applied in writing the book Entwicklung der Infinitesimalrechnung ("The Calculus: A Genetic Approach"). The book introduces the subject by giving an idealized historical narrative to motivate the concepts, showing how they developed from classical problems of Greek mathematics. It was left unfinished, edited by Gottfried Köthe and posthumously published in German in 1946 (and translated into English in 1963). Harold M. Edwards's book on Fermat's Last Theorem was structured having in mind the genetic method. This is what Edwards himself said about the method of his book in the preface to the first edition of it: «... The basic method of the book is... the genetic method. The dictionary defines the genetic method as "the explanation or evaluation of a thing or event in terms of its origin and development."... It is important to distinguish the genetic method from history. The distinction lies in the fact that the genetic method primarily concerns itself with the subject—its "explanation or evaluation" in the definition above—whereas the primary concern of history is an accurate record of the men, ideas, and events which played a part in the evolution of the subject. In a history there is no place for detailed descriptions of the theory unless it is essential to an understanding of the events. In the genetic method there is no place for a careful study of the events unless it contributes to the appreciation of the subject. This means that the genetic method tends to present the historical record from a false perspective. Questions which were never successfully resolved are ignored. Ideas which led into into blind alleys are not pursued. Months of fruitless effort are passed over in silence and mountains of exploratory calculations are dispensed with. In order to get to the really fruitful ideas, one pretends that human reason moves in straight lines from problems to solutions. I want to emphasize as strongly as I can that this notion that reason moves in straight lines is an outrageous fiction which should not for a moment be taken seriously. Samuel Johnson once said of the writing of biography that "If nothing but the bright side of characters should be shown, we should sit down in despondency, and think it utterly impossible to imitate them in anything. The sacred writers related the vicious as well as the virtuous actions of men; which had this moral effect, that it kept mankind from despair." This book does, for the most part, show only the bright side, only the ideas that work, only the guesses that are correct. You should bear in mind that this is not a history or biography and you should not despair. You may well be interested less in the contrast between history and the genetic method than in the contrast between the genetic method and the more usual method of mathematical exposition. As the mathematician Otto Toeplitz described it, the essence of the genetic method is to look to the historical origins of an idea in order to find the best way to motivate it, to study the context in which the originator of the idea was working in order to find the "burning question" which he was striving to answer... In contrast to this, the more usual method pays no attention to the questions and presents only the answers. From a logical point of view only the answers are needed, but from a psychological point of view, learning the answers without knowing the questions is so difficult that it is almost impossible. That, at any rate, is my own experience...» It might be noteworthy that Michael Spivak implicitly referred to the genetic approach in the preface to volume one of his A comprehensive introduction to differential geometry: «... The second premise for these notes is that in order for an introduction to differential geometry to expose the geometric aspect of the subject, an historical approach is necessary... Of course, I do not think that one should follow all the intricacies of the historical process, with its inevitable duplications and false leads. What is intended, rather, is a presentation of the subject along the lines which its development might have followed; as Bernard Morin said to me, there is no reason, in mathematics any more than in biology, why ontogeny must recapitulate philogeny. When modern terminology is introduced, it should be as an outgrowth of this (mythical) historical development.»<|endoftext|> TITLE: Set of prime numbers $q$ such that $\sum\limits_{p\leq\sqrt{q}}p=\pi(q)$, where $p$ are prime numbers QUESTION [9 upvotes]: The question is: does the set of prime numbers $q$ such that $\sum\limits_{p\leq\sqrt{q}}p=\pi(q)$, where $p$ are prime numbers, contain infinitely many elements? You can find the first elements here (http://oeis.org/A329403). Any insight on this would be welcomed. Thanks in advance for your time and effort! REPLY [14 votes]: Nice question! The answer is affirmative. If $\sigma(x)$ denotes the sum of primes up to $\sqrt{x}$, then it suffices to show that $\pi(x)-\sigma(x)$ changes sign infinitely often, because $$\pi(x)-\sigma(x)<0<\pi(x+1)-\sigma(x+1)$$ never holds. I thank Juan Moreno and Will Sawin for this simple but crucial observation. My streamlined argument below owes to reuns' ideas as well. Let us introduce the notation $$\pi(x)=\mathrm{li}(x)-\mathrm{li}(1)+\rho(x),$$ then we get \begin{align*}\sigma(x) &=\int_1^\sqrt{x}t\,d\pi(t)\\ &=\int_1^\sqrt{x}\frac{t\,dt}{\log t}+\int_1^\sqrt{x} t\ d\rho(t)\\ &=\int_1^x\frac{du}{\log u}+\int_1^\sqrt{x} t\ d\rho(t)\\ &=\mathrm{li}(x)-\mathrm{li}(1)+\int_1^\sqrt{x} t\ d\rho(t). \end{align*} So the difference $\pi(x)-\sigma(x)$ can be directly estimated by the error term in the prime number theorem. We can analyze this difference further by considering the Mellin transform $$\int_1^\infty x^{-s}\ d(\pi(x)-\sigma(x))=\int_1^\infty(x^{-s}-x^{1-2s})\ d\rho(x).$$ We shall only need to look at the left-hand side. Integrating by parts, we see that it is holomorphic in a region containing the half-plane $\Re(s)>1$ and the half-line $s>2/3$. In fact, in this region, the left-hand side equals $$\log\zeta(s)-\log\zeta(2s-1)+\frac{1}{2}\log\zeta(4s-2)+f(s),\tag{$\ast$}$$ where $f(s)$ is holomorphic for $\Re(s)>2/3$. As in the proof of Theorem 15.2 in Montgomery-Vaughan: Multiplicative number theory I, this eventually yields that $$\pi(x)-\sigma(x)=\Omega_\pm(x^c)\quad\text{for any}\quad c<3/4.$$ Indeed, let us assume that this bound fails for some $c<3/4$. Without loss of generality, $c>2/3$. Then, by Landau's lemma (which is Lemma 15.1 in the same book), $(\ast)$ is holomorphic in the half-plane $\Re(s)>c$. However, this is easily seen to be false, and we are done.<|endoftext|> TITLE: When are indiscrete reflection groups Coxeter groups? QUESTION [7 upvotes]: A well-known theorem of Coxeter states that any discrete group $W$ which is generated by reflections across (possibly affine) hyperplanes in Euclidean space is a Coxeter group: it has a presentation with generators $\{r_i, i \in I\}$ and relations $r_i^2=e, \forall i$ and $(r_ir_j)^{m_{ij}}=e$ for some numbers $m_{ij} \in \{2,3,\ldots\} \sqcup \{\infty\}$. I am interested in when the discreteness hypothesis can be discarded. For example, take the simplest indiscrete case: let $W$ be generated by reflections across two central hyperplanes meeting at a $\pi$-irrational angle. Then $W$ is not discrete since it contains a dense set of rotations, but $W$ is still a Coxeter group with generators $\{r_1,r_2\}$ corresponding to the two given reflections and $m_{12}=\infty$. Note that I do not require $W$ to be generated by finitely many reflections, nor my Coxeter groups to have $|I|$ finite. However I would still be interested to know about any results particular to the finite case. Questions: Is any such group $W$ a Coxeter group? I suspect not. As a particular example: is the orthogonal group a Coxeter group? Again, probably not, but I don't see how to prove this. Is there some nice condition, more general than discreteness, which guarantees that $W$ is a Coxeter group? This other question also asks for reflection groups that aren't Coxeter groups, but without focusing on discreteness. It also doesn't insist on working in Euclidean space and does assume finite $|I|$. It got no answers. REPLY [2 votes]: Over on math.SE, I pointed out that there are three reflections $a$, $b$ and $c$ in $O(3)$ such that $\langle a, b \rangle$, $\langle a, c \rangle$ and $\langle b, c \rangle$ are infinite dihedral, but $(abc)^6 = 1$. However, I didn't actually prove that it wasn't a Coxeter group. Maybe this question will motivate someone to finish off the proof.<|endoftext|> TITLE: Holomorphic Urysohn Lemma QUESTION [11 upvotes]: Let $M,N$ be two disjoint closed holomorphic submanifolds of $\mathbb{C}^n$. Is there a holomorphic map $f:\mathbb{C}^n\to \mathbb{C}$ with $f(M)=0,\;f(N)=1$. REPLY [12 votes]: Yes, since the morphism of sheaves $\mathcal{O}_{\mathbb{C}^n} \to \mathcal{O}_{M \cup N}$ is surjective, it is surjective on global sections by Cartan's theorem B. Thus, any holomorphic function on $M \cup N$ has a holomorphic extension to $\mathbb{C}^n$, in particular, this holds for the function $f$ which is $\equiv 0$ on $M$ and $\equiv 1$ on $N$. Edit: To expand on the use of Cartan's theorem B, note that we have a short exact sequence of coherent sheaves $0 \to \mathcal{J} \to \mathcal{O}_{\mathbb{C}^n} \to \mathcal{O}_{M \cup N} \to 0$, where $\mathcal{J}$ is the kernel of $\mathcal{O}_{\mathbb{C}^n} \to \mathcal{O}_{M \cup N}$ (i.e., the ideal defining $M \cup N$), and thus we get the exact sequence of sheaf cohomology $H^0(\mathbb{C}^n,\mathcal{O}_{\mathbb{C}^n}) \to H^0(\mathbb{C}^n,\mathcal{O}_{M \cup N}) \to H^1(\mathbb{C}^n,\mathcal{J})$, and $H^1(\mathbb{C}^n,\mathcal{J})=0$ by Cartan's theorem B, which means that there is a surjective morphism on global sections.<|endoftext|> TITLE: Groups with maximal element order 6 QUESTION [9 upvotes]: I'm dealing with finite groups $G$ in which the maximal order of an element is $6$. With GAP I found out that for all groups with order $<1000$ the number of elements of order 6 $k := |\{x\in G : ord(x)=6\}|$ is $0$ or $2$ $\mod 6$. I'm trying to understand, why $k \not\equiv 4 \mod 6$. I'm collecting some facts about such a group $G$: $|G|=2^\alpha 3^\beta 5^\gamma$, with $\alpha,\beta,\gamma$ being natural numbers. $k$ must be even, as elements of order $6$ occur in pairs $x,x^{-1}$. Let $n_a := |\{x\in G : x^a=e\}$ for a divisor $a$ of $|G|$. Frobenius theorem says $a$ divides $n_a$. So $n_3 \equiv 3 \mod 6$, as elements of order $3$ occur pairwise. Can anyone help me out or suggest some tools to tackle this problem. Thanks! Edit: I again computed with GAP in groups with small order: $n_4 \not\equiv 0 \mod 6$ if $4$ divides $|G|$. And $n_2 \not\equiv 0 \mod 6$ if $4$ does not divide $|G|$. If $5$ does not divide $|G|$ then $k=n_6 - n_4 -n_3 + 1\equiv n_4 + 4 \mod 6$. This means it suffices to prove $n_4 \not\equiv 0 \mod 6$ if $4$ divides $|G|$ or analogous for $n_2$. REPLY [12 votes]: Let $P$ be a Sylow $3$-subgroup of $G$ (which we may assume to be nontrivial) and consider the conjugation action of $P$ on the set $X$ of all elements of order six in $G$. Remove from $X$ all of its $P$-fixed points. The resulting set $Y$ is a union of non-trivial $P$-orbits and therefore $|Y|$ divisible by three. As $y^{-1} \in Y$ whenever $y \in Y$, we see that $|Y|$ is even and so divisible by six. It remains to count the $P$-fixed points in $X$, that is, the elements of order six in $C_G(P)$. Your conditions on $G$ force that $C_G(P)$ is the direct product of $Z(P)$ (an elementary abelian $3$-group) and a (possibly trivial) elementary abelian $2$-group. If $|C_G(P)|=2^a3^b$, then the number of elements of order six in $C_G(P)$ is $(2^a-1)(3^b-1)$, and your claim follows.<|endoftext|> TITLE: Can $E_8$ be enlarged? QUESTION [9 upvotes]: Is there any finite 8-dimensional point group which contains the $E_8$ Coxeter group as a subgroup other than $E_8$ itself? REPLY [18 votes]: No. Let $G$ be a finite subgroup of $GL_n(\mathbb R)$ containing $W(E_8)$ as a subgroup. Because $G$ is compact, $G$ must preserve a symmetric positive definite form on $\mathbb R^8$. Since $W(E_8)$ preserves a unique such form, it must be that one. Let $H$ be the largest subgroup of $G$ generated by reflections. Then $H$ contains $W(E_8)$ and thus is an irreducible reflection group, hence a Coxeter group. Examining the table of Coxeter groups and looking for entries in dimension $8$, there are four possibilities: $W(A_8), W(B_8), W(D_8), W(E_8)$. Because $W(E_8)$ has the highest order of these, we must have $H = W(E_8)$. Now, by construction, $H$ is a normal subgroup of $G$, so $G$ normalizes $W(E_8)$, and hence $G$ is contained in the automorphism group of the $E_8$ root system. Because the Coxeter-Dynkin diagram $E_8$ has no nontrivial automorphisms, this is $W(E_8)$ and so $G= W(E_8)$, as desired.<|endoftext|> TITLE: Diagonalizing quaternionic unitary matrices QUESTION [9 upvotes]: The quaternionic unitary group $\mathrm{U}(n,\mathbb{H})$, also called the compact symplectic group $\mathrm{Sp}(n)$, consists of $n \times n$ quaternionic matrices $g$ such that $gg^\ast = 1$, where $$ (g^\ast)_{ij} = \overline{g}_{ji}$$ and the overline denotes quaternionic conjugation. My question: is every quaternionic unitary matrix conjugate to a diagonal one? That is: given a quaternionic unitary matrix $g$, is there a quaternionic unitary matrix $h$ such that $h g h^{-1}$ is diagonal, meaning that $(h g h^{-1})_{ij} = 0$ when $i \ne j$? I feel pretty sure this is true. If it is, what I really want is a reference to a proof, or a proof. REPLY [11 votes]: This follows from the general fact that, in a compact connected Lie group, every element is conjugate to an element in a maximal torus (and all maximal tori are conjugate). This result is proved in just about every book that treats compact Lie groups. For example, see Helgason's "Differential Geometry, Lie Groups, and Symmetric Spaces" or Bröcker and tom Dieck's "Representations of Compact Lie Groups". The diagonal elements of $\mathrm{U}(n)\subset\mathrm{Sp}(n)$ form a maximal torus in $\mathrm{Sp}(n)$, so every element in $\mathrm{Sp}(n)$ is conjugate in $\mathrm{Sp}(n)$ to a diagonal unitary matrix.<|endoftext|> TITLE: Motivation for the definition of complex orientable cohomology theory QUESTION [7 upvotes]: PRELIMINARY DEFINITIONS: Let $E^*$ be a multiplicative generalized cohomology theory. By the suspension isomorphism we have: $$ \tilde{E^2}(S^2)\cong\tilde{E^0}(S^0)=E^0(pt) $$ So there is a special element $t\in\tilde{E^2}(S^2)$ which corresponds to $1_E\in E^0(pt)$ under the above isomorphism. $E^*$ is said to be complex orientable if the inclusion map $i:S^2=\mathbb{C}P^1\to\mathbb{C}P^{\infty}$ induces a sujective morphism $i^*:\tilde{E^2}(\mathbb{C}P^{\infty})\to\tilde{E^2}(S^2)$. A pair $(E^*,x_E)$ of a complex orientable cohomology theory $E^*$ and a choice of an element $x_E\in\tilde{E^2}(\mathbb{C}P^{\infty})$ such that $i^*(x_E)=t$ is called an oriented cohomology theory. It is well known that singular cohomology, complex K-theory and complex cobordism are examples of orientable cohomology theories. QUESTION: I would like to motivate the previous definition. In particular, I would like to motivate this definition through some classical result for singular cohomology/ K-theory. So, I am searching for an answer of this kind:"since for singular cohomology/K-theory holds theorem X (and we want to generalize this stuff to generalized cohomology), then it is natural to define oriented cohomology theories as above" My attempt: according to various textbooks, for example [1, Theorem 3.10], we have the following classical result: for any complex oriented cohomology theory $(E^*,x_E)$, there is an isomorphism $$ E^*(\mathbb{C}P^{\infty})\cong E^*(pt)[[x_E]] $$ this allows you to define the first Chern class in $E^*$-cohomology and other good stuff follows, for example you can associate to each cohomology theory a formal group law (see [2]). So, the definition of complex orientable cohomology given above is the good one if you want to define Chern classes for generalized cohomologies. Any other ideas? Thank you in advance. REFERENCES: [1] A.Kono, D.Tamaki-Generalized cohomology [2] J.Lurie-Chromatic homotopy theory course http://people.math.harvard.edu/~lurie/252x.html [3] M.Hopkins-Complex oriented chomology theories and the language of staks https://people.math.rochester.edu/faculty/doug/otherpapers/coctalos.pdf REPLY [12 votes]: As you wrote, complex orientability can be characterized by the cohomology of $\mathbb{C}P^\infty$: $E$ is complex orientable if $E^*(\mathbb{C}P^\infty)$ splits according to the cell structure of $\mathbb{C}P^\infty$, i.e. $E$ doesn't "see" the attaching maps of $\mathbb{C}P^\infty$. For example, this happens whenever $E_*$ is concentrated in even degrees, for degree reasons. This is the case for singular cohomology and complex K-theory. To contrast this with a negative example: For real K-theory, we don't get such a splitting: The attaching map of the $4$-cell of $\mathbb{C}P^\infty$ is the Hopf map $\eta$, which acts nontrivially on $KO$. This kind of splitting can be expressed in terms of the Atiyah-Hirzebruch spectral sequence, and using the multiplicative structure you have on there, and the fact that singular cohomology of $\mathbb{C} P^\infty$ is polynomial, the question of whether the Atiyah-Hirzebruch spectral sequence splits boils down to the existence of a single element in $E^2(\mathbb{C}P^\infty)$ (which corresponds to the $2$-cell). The choice of such an element is precisely the usual definition of complex orientation.<|endoftext|> TITLE: Alternative approaches to topological QFTs QUESTION [10 upvotes]: A while ago I read the paper 'Quantum Field Theory and the Jones Polynomial' by Edward Witten. This article uses a lot of concepts from physics like BRST symmetry and the Chern-Simons action which are perhaps a bit mysterious to an audience of general mathematicians. One common way to get QFT into a precise language which mathematicians can comprehend is to recast everything into categorical language, so for example a mathematical interpretation of Witten's work in the 1989 paper is to take an approach with modular tensor categories. My only issue with all this is that although it is mathematically much more precise and clear, it feels like a lot of the physical intuition is lost and in some cases one can read lecture notes on QFT written by mathematicians without knowing that there is any physics involved at all. Is there some possible mathematical interpretation of Witten's work on topological QFTs which is mathematically precise but still keeps a bit more of the original physical and geometric intuition without framing everything in terms of categorical language? (As far as I am aware, this language is mostly not used by physicists except where they need it for some particular problem). REPLY [9 votes]: It sounds like what you would like is a rigorous version of Witten's original Feynman integral & Wilson loop approach. This is not a totally unreasonable thing to ask for, since QFTs have been rigorously constructed along these lines, by demonstrating the existence of moments of the Euclidean-signature Feynman integrals. It would indeed be wonderful to just run some Monte Carlo simulations and see knot invariants emerge. But there are major obstacles to this program for the particular QFT Witten was thinking about. It is not known even at a physical level of rigor how to define the SU(2) Chern-Simons theory in this fashion. It's an open research topic. In fact, this question is still under active investigation by physicists, even for the far simpler abelian Chern-Simons theory. (See Bietenholz, Nishimura, Sodano '02 and DeMarco, Wen '19.) Roughly, what one would like is a lattice gauge theory which flows to Chern-Simons of the desired level at large distances. There's a number of obstacles here. I'm only going to mention some of them. The first sign of the trouble is that, in the continuum, the Euclidean Chern-Simons action is purely imaginary and unbounded, rather than real and pleasantly bounded like the Yang-Mills action. Euclidean path integral constructions usually depend on having Gaussian-ish behavior, and there's no reason to expect that here. To make matters worse, there isn't any exact lattice version of the Chern-Simons class. You can approximate it in various ways, but the errors in the approximation will make the lattice theory non-topological. Even the abelian Chern-Simons theory (which has a quadratic classical action) has problems on the lattice. There's a natural discretization of the classical action, due to Frohlich & Marchetti, but it has a fatal flaw: It has zero modes, which means that the exponential $e^{-S(A)}$ is not integrable, even on a finite lattice, where the integration measure $dA$ is well-defined. There's a theorem due to Berruto, Diamanti, & Sidano which says, in effect, that any lattice action for a gauge field which is gauge-invariant, linear in derivatives, and odd under parity reversal (as the cubic Chern-Simons action should be) will share these problems. You can't fix this by screwing around with the discretization, and there's not much reason to expect better behavior in the case of non-abelian gauge groups. There is a reasonable way out of this problem: One can add to the lattice Chern-Simons action a parity-even term, which lifts the degeneracies. The leading order parity-even term you can add is the Maxwell action. The abelian theory is basically OK at this point. The Maxwell term's effect on long-distance physics is negligible, so you basically have a definition of abelian Chern-Simons. I think one could even construct it along the lines of Driver and Gross. (Frohlich & King sketch the idea.) But the analogue of the Jones polynomials here are just linking numbers. The nonabelian case is probably worse: The Maxwell term that showed up in regularization gets replaced by the Yang-Mills Lagrangian. The basic story isn't expected to change: the long distance scaling limit should be dominated by Chern-Simons. So that's good. But now you've got to show that 3d Yang-Mills theory makes sense. This is expected to be easier than the Millenium Prize problem, but it shares a lot of the same features. So as not to end on a down-note, let me mention that Yang-Mills theory is not the only possible lattice QFT that leads to Chern-Simons. There's a quite large line of research (starting, I think, with Fradkin & Shaposnik) in which fairly accessible QFTs, like the 3d Thirring model, are shown to give rise to Chern-Simons at long distances. I don't know to what degree the constructive QFT community has taken up these challenges; I've seen it mainly from lattice and condensed matter people.<|endoftext|> TITLE: Intersections and curvature in the plane QUESTION [10 upvotes]: Let $D$ be a nonempty compact convex plane region whose boundary is a smooth curve whose radius of curvature is at most 1 everywhere. Can the boundary of $D$ intersect a circle of radius 1 in more than two points? REPLY [4 votes]: David Speyer writes "the only way this strategy could work is if we improve the $2$ in the previous paragraph to $\pi$." That improvement follows. Suppose $f < 0$ and $f + f'' \leq 0$ on $(0,c)$, with $f(0)=f(c)=0$. We show $c \geq \pi$, with equality if and only if $f(\cdot)$ is a negative multiple of $\sin(\cdot)$. Let $\alpha = \tan^{-1}(f'/f)$, so $\alpha(0) = \pi/2$ and $\alpha(c) = -\pi/2$. Note that if $f(x) = A\sin x$ then $\alpha(x) = \pi/2 - x$. In general we compute $$ \alpha' = \frac{f\!f''-{f'}^2}{f^2+{f'}^2}, $$ whence $$ \alpha' + 1 = \frac{f(f+f'')}{f^2+{f'}^2} > 0. $$ Therefore $\alpha' \leq -1$ on $(0,c)$. Since $\alpha$ decreases by $\pi$ on that interval, the interval must have length at least $\pi$. Moreover if $c=\pi$ then $f+f''=0$ on all of $(0,\pi)$, which together with $f(0)=f(\pi)=0$ makes $f$ a multiple of the sine function. $\Box$<|endoftext|> TITLE: Which reflection groups can be enlarged? QUESTION [8 upvotes]: Based on this question (which focuses on the case $E_8$) I wonder the following: Question: For each finite reflection group $\Gamma\subseteq\mathrm O(\Bbb R^d)$, what is the largest finite group $\bar \Gamma\subseteq\mathrm O(\Bbb R^d)$ that contains $\Gamma$? Some inclusions already happen among the reflection groups and their extensions: For $I_2(n)$ there is no such largest group, because $$I_2(n)\subset I_2(2n)\subseteq I_2(4n)\subset\cdots\subset I_2(2^r n)\subset \cdots.$$ In general, we have $D_d\subset B_d$. In general, $A_d\subset A_d^*$, where $A_d^*$ is the extended group that results from the additional symmetries of the Coxeter-Dynkin diagram of $A_d$. We have $A_3=D_3\subset B_3=A_3^*$. We have $A_4\subset A_4^*\subset H_4$. We have $D_4\subset B_4\subset F_4\subset F_4^*$, again, $F_4^*$ is the extended group resulting from the additional symmetries of the Coxeter-Dynkin diagram. We have $E_6\subset E_6^*$, for the same reason as above. We have $A_7\subset A_7^*\subset E_7$. We have $A_8\subset A_8^*\subset E_8$. (Thanks to Daniel for noting the extensions of $A_4^*, A_7^*$ and $A_8^*$). I believe that $H_3, H_4,B_5,E_7,E_8$ and $B_d,d\ge 9$ cannot be enlarged by the same reasoning as given in this answer (because these groups are the largest reflection groups in their respective dimension and their Coxeter-Dynkin diagrams have no additional symmetries). So we are left with the following: Question: Can we enlarge the groups $B_d$ ($d\in\{3,5,6,7\}$), $F_4^*$, $E_6^*$ and $A_d^*$ ($d\not\in\{4,7,8\}$)? Maybe the inclusion $A_d\subset A_d^*\subset\cdots$ is not the right chain leading to the largest group. And I also have not touched on the reducible groups which can also be enlarged in some cases, e.g. the Coxeter-Dynkin diagram of $I_2(n)\oplus I_2(n)$ has also additional symmetries. We also have inclusions like $I_1\oplus B_d \subset B_{d+1}$. REPLY [8 votes]: $A_1^*$ is maximal. $A_2 = I_3$ and $A_2^* = I_6 = G_2$ (for the first one, consider the Dynkin diagram, for the second one, any group containing it as an index $2$ subgroup must normalize it, hence must be the normalizer) so $A_2^*$ is not maximal. $A_3 = D_3$ and $A_3^* = B_3$. $B_3$ has order $48$, a multiple of $16$, while the only larger $3$-dimensional reflection group is $H_3$, of order $120$, not a multiple of $16$, so $B_3$ is a maximal reflection group (and thus maximal, because it has no outer automorphisms EDIT: in $O_3$). But $A_n$ doesn't embed into $B_n$ for $n\geq 4$ because the alternating group on $n+1$ letters is fine simple in this range, thus has no nontrivial homomorphism onto the symmetric group on $n$ letters, hence no nontrivial automorphism into the extension of the symmetric group on $n$ letters by an abelian group. So $A_n^*$ is maximal unless $A_n^*$ is contained in an exceptional reflection group. In particular, $A_n^*$ is maximal for $n \geq 9$ or $n=5$. Because $F_4$ has order $1152$, and $H_4$ has order $14400$, which is not a multiple of $1152$, $F_4$ does not embed into $H_4$. It also doesn't embed into a maximal $A_4$ or $B_4$ since those are smaller, so $F_4$ is a maximal reflection group, and thus $F_4^*$ is a maximal symmetry group. I'm not sure why $B_5$ was on your list as there's no exceptional group in dimension $5$. Maybe you meant $B_8$. $A_6$ can't embed into $E_6$ because the order of $E_6$ is not divisible by $7$. $B_6$ can't embed into $E_6$ because the order of $E_6$ is $72 \cdot 6!$ and thus is not divisible by $2^6 \cdot 6!$. $E_6$ can't embed into the other two since it has the largest order. So these are all maximal reflection groups and their normalizers are maximal symmetry groups. The order of $E_7$ is $72 \cdot 8!$ which is divisible by the order of $B_7= 2^6 \cdot 7! = 2^3 \cdot 8!$, with quotient $9$. But the Weyl group of $E_7$, mod the center of order $2$, is simple, and can't have a subgroup of order $9$. if it did, its order would divide $9!\cdot 2$. So B_7$ is maximal. According to Daniel Sebald, $B_8$ is not contained in $E_8$. Thus $B_8$ is maximal. In summary, $A_n^*$ is maximal for all $n$ except $2$, $4$, $7$, $8$, $B_n$ is maximal for all $n$ except $2,4$, $I_n$ is never maximal, and the normalizer of every exceptional group is maximal.<|endoftext|> TITLE: Is there a filtered splitting of product labelling spaces? QUESTION [6 upvotes]: For a well-based space $X$ denote by $C(\mathbb{R};X)$ the unordered configuration space of points on the real line with labels in $X$, and a point can vanish if its label reaches the basepoint. (Alternatively, you can think about the free $E_1$-algebra over the based space $X$.) Note that $C(\mathbb{R};X)$ is filtered by subspaces $C_{\le r}(\mathbb{R};X)$ which contain configurations of at most $r$ labelled points. Now let $X$ and $Y$ be well-based and path-connected spaces. According to Segal, we have homotopy equivalences $$C(\mathbb{R};X\times Y)\to \Omega\Sigma(X\times Y),$$ and we use that $\Sigma(X\times Y)$ splits up to homotopy into $\Sigma X\vee \Sigma Y\vee \Sigma(X\wedge Y)$. We also have a homotopy equivalence $$C(\mathbb{R};X\vee Y\vee (X\wedge Y))\to \Omega \Sigma(X\vee Y\vee (X\wedge Y)).$$ Now my question is: Can we invert the homotopy equivalences in such a way that the resulting equivalence $$C(\mathbb{R};X\vee Y\vee (X\wedge Y)) \to C(\mathbb{R};X\times Y)$$ is filtration-preserving? Or is there even an explicit, geometric description of such a map? REPLY [8 votes]: The answer to your first question is no. And this can be seen by homology considerations. Note that this equivalence induces an isomorphism of Hopf algebras $$ H_*(C(\mathbb R; X \vee Y \vee (X\wedge Y)); \mathbb F) \simeq H_*(C(\mathbb R; X \times Y); \mathbb F)$$ for any field coefficients $\mathbb F$. Also, one knows that $H_*(C(\mathbb R; Z); \mathbb F)$ is isomorphic to the free tensor algebra generated by $\tilde H_*(Z;\mathbb F)$. The isomorphism must send primitives to primitives. Examples show that this will not preserve the filtration on homology induced by the filtration you are asking about. Here is a simple example: let $X=Y=S^2$ and $\mathbb F = \mathbb Q$. Let $x \in H_2(X;\mathbb Q)$ and $y \in H_2(Y;\mathbb Q)$ be generators. Then the filtration 1 primitive one might write as $x \bar{\times} y \in H_4(X \wedge Y;\mathbb Q)$ must map to the primitive $x \times y + x * y \in H_4(X \times Y;\mathbb Q)$. But this element has filtration 2. (Here $*$ is the multiplication induced by the H-space structure.) (Of course, even more simply, filtration 1 on the one side certainly can't be mapped as an equivalence to filtration 1 on the other side!) As to your second question, just add the three maps $$X \times Y \rightarrow X \hookrightarrow C(\mathbb R; X \vee Y \vee (X\wedge Y)),$$ $$X \times Y \rightarrow Y \hookrightarrow C(\mathbb R; X \vee Y \vee (X\wedge Y)),$$ $$X \times Y \rightarrow X \wedge Y \hookrightarrow C(\mathbb R; X \vee Y \vee (X\wedge Y)),$$ and then extend to a map of $\mathbb E_1$-algebras.<|endoftext|> TITLE: Ordered logic is the internal language of which class of categories? QUESTION [7 upvotes]: Wikipedia says: The internal language of closed symmetric monoidal categories is linear logic and the type system is the linear type system. "A Fibrational Framework for Substructural and Modal Logics" says: Linear logic is ordered logic with exchange, so to model this we add a commutativity equation $$x \otimes y \equiv y \otimes x$$ Does this mean that if we take the definition of a closed symmetric monoidal category, and then remove the requirement that the tensor product $\otimes$ be commutative, we obtain the class of categories whose internal language is ordered logic? Would braided monoidal categories, where $x \otimes y \cong y \otimes x$, have an internal language that's strictly between linear logic and ordered logic in terms of restrictions? REPLY [6 votes]: Yes, ordered logic is the internal language of non-symmetric monoidal categories. As with linear and nonlinear logic, if the ordered logic contains function-types then they correspond to internal-homs making the monoidal category closed, although one has to be a bit careful since in the non-symmetric case there are two inequivalent notions of internal-hom; sometimes one speaks of "left closed" and "right closed" to distinguish, with either "closed" or "biclosed" when both are present. And yes, in principle braided monoidal categories should have an internal "braided logic". I've never seen such a thing written down, but something along those lines could be obtained by writing the mode theory of a braided monoidal category in our general framework.<|endoftext|> TITLE: Conditions under which the preimage of a submanifold in nontrivial in homology QUESTION [8 upvotes]: Let $\pi: M^{n+k} \to N^n$ be a fibre bundle with fibre $F$ between compact smooth manifolds. What are “mild” sufficient conditions on the topology of $M$, $N$ and $F$ so that given a closed $p$-submanifold $\Sigma^p \subset N$ which is nontrivial in $H_p(N; \mathbb{Z})$, the preimage $\pi^{-1}(\Sigma)$ is nontrivial in $H_{p+k}(M;\mathbb{Z})$? I guess spectral sequences may help, but I am not familiar with them. REPLY [4 votes]: As pointed out by Nicolas Tholozan in the comments, the map $\pi^{-1}:H_p(N)\to H_{p+k}(M)$ which sends $[S\subset N]$ to $[\pi^{-1}(S)\subset M]$ is Poincaré dual to the map $\pi^*:H^{n-p}(N)\to H^{n-p}(M)$ induced by $\pi$ on cohomology. (Here I'm assuming $M$ and $N$ are both oriented and using integral coefficients.) So you're really asking for conditions under which the induced map in cohomology is injective. One easy example is when $\pi$ admits a section $\sigma:N\to M$, in which case $\operatorname{Id}_{H^*(N)}=(\pi\circ \sigma)^*=\sigma^*\pi^*$.<|endoftext|> TITLE: Amenable groupoid C*-algebras satisfy the UCT in English? QUESTION [6 upvotes]: As is by now well known, Tu proved in 1998 that the C*-algebras coming from amenable groupoids satisfy the so-called UCT (universal coefficient theorem). Unfortunately, I don't speak french and I've only found this result in french. Question: does anyone happen to know of any account of the proof that is written in English? Tu's original paper is fairly long, and hence would require a lot of work to read for a non-speaker. Even talks available online, or particular cases (such as the same for amenable groups) would be helpful and very much appreciated. [1] J.L. Tu, La conjecture de Baum–Connes pour les feuilletages moyennables, K-Theory 17 (1999) pp. 215–264, (pdf) REPLY [2 votes]: For anyone that might see this in the future, in the chapter 12 of here you may find a short summary of the main ideas of Tu's proof.<|endoftext|> TITLE: When the definition of a set starts to matter in category theory QUESTION [15 upvotes]: In most introductory courses to category theory, the precise definition of a set is more-or-less ignored. The idea being that all basic results in the subject hold for any reasonable definition of a set. At some point one feels that this must cease to be the case. So what is the simplest (a simple) example of a categorical concept or result that actually relies on the details of the definition of a set that one is using? REPLY [14 votes]: I didn't have time to write up a proper answer on initially seeing your question and much of what I have to say has been said in the comments and Tim's answer, but I'll still offer some specifics on things mentioned in the comments. The definition of a set is, as Tim Campion pointed out, the axioms of whatever set theory you're working in. They do not so much define a set as describe a primitive notion that behaves like what we think sets should behave like, with different axiomatizations giving rise to differing notions of set. These differing axiomatizations do have an impact on category theory because they change the behavior of ${\bf Set}$, and consequently the behavior of presheaf categories, and they also impact our ability to manipulate large categories. As mentioned by Ingo Blechschmidt in the comments, Mike Shulman has an excellent paper surveying some of these consequences. I will summarize some of them here, but I highly recommend you check out his paper. A striking result referenced in the Shulman paper is due to Colin McLarty, establishing that the NF axiomatization of what a set is yields a ${\bf Set}$ that isn't Cartesian closed. In ZFC we really only run into issues if we want to manipulate large categories as a whole, for example ${\bf Set}$ or ${\bf Group}$, which are not actually objects in ZFC since they're proper classes. We can get around this with shenanigans about formulas the metalanguage, but anyone looking for an integrated and 'natural' treatment of large categories on level footing with small ones will be disappointed in this setting. NBG is a conservative extension of ZFC (meaning it doesn't prove anything about sets that ZFC can't) which does allow proper classes to be real objects in the theory, but we still run into some discomfort when dealing with large categories. NBG manages to be conservative over ZFC by restricting it's comprehension axiom to only apply to sets, not proper classes -- in practical terms, as Mike points out in his paper linked above, this means (for example) that we can't prove by induction that a large category $\mathcal{C}$ has an $n$-fold Cartesian product $\mathcal{C}^n$. We can get around this by constructing it directly as the category of functions from $n$ into $\mathcal{C}$, but the unavailability of canonical proof methods like induction is troubling. MK is a non-conservative extension of ZFC, essentially NBG but with full class comprehension allowed so we have access to all of the standard proof tools for large categories. This new theory can prove things ZFC can't, like the consistency of ZFC, and is thusly strictly stronger in a meaningful sense. MK also has its own serious issues when working with large categories -- we can't define the category of functors between two large categories, and this applies to NBG as well. Using full MK further suggests that we try to look at the category of classes, since they're really the category of collections we want to work with right? And bam, once again we're back to a situation where we have to play games in the metalanguage, or conservatively extend/step up the consistency strength of our theory. This leads mathematicians to situations like Grothendieck universes, where it's always possible to step up to the next universe if we need to talk about 'all the somethings' in the current universe. This is equivalent to working in ZFC plus an axiom asserting the existence of an inaccessible cardinal. All the extra baggage of universes or inaccessibles is still somewhat of a sledgehammer for the problem at hand, though; all we want is for large categories to 'be like small categories' in enough ways that we can carry out all the constructions we care about with large categories, but inaccessibles or Grothendieck universes also have a plethora of other consequences (like the need to juggle universes1). A solution to these problems comes in the form of reflection principles, which are essentially axioms asserting that proper classes look enough like sets that we don't have to soil ourselves when they appear, but don't endow them with enough independence to give rise to a whole hierarchy of universes we need to ask questions about. All of this is discussed at length in the Shulman paper referenced above, with additional references therein. First paper: Shulman, Mike. Set theory for category theory. arXiv:0810.1279v2 [math.CT] Second paper: McLarty, Colin. Failure of Cartesian Closedness in NF. J. Symbolic Logic 57 (1992), no. 2, 555--556. https://doi.org/10.2307/2275291 1As Tim points out in the comments, how many universes we have to juggle when taking this route is up to us. Skilled jugglers may use an infinite number, while those new to the approach may use only two.<|endoftext|> TITLE: Comparison: Formal Wirthmüller isomorphism of Fausk-Hu-May vs. Balmer et. al QUESTION [7 upvotes]: $\newcommand{\Cc}{\mathcal{C}}$ $\newcommand{\Dd}{\mathcal{D}}$ $\newcommand{\tensor}{\otimes}$ $\DeclareMathOperator{\Sp}{Sp}$ This question is about comparing the approaches for a formal Wirthmüller isomorphism by Fausk-Hu-May [FHM] in isomorphisms between left and right adjoints and by Balmer-Dell'Ambrogio-Sanders [BDS] in Grothendieck-Neeman duality and the Wirthmüller isomorphism. In both articles, a formal Wirthmüller isomorphism is studied: we are given a tensor-exact functor $f^*: \Dd \to \Cc$ between tensor-triangulated categories that has both a left adjoint $f_!: \Cc \to \Dd$ and a right adjoint $f_*: \Cc \to \Dd$ and we are looking for a `twisted isomorphism' between $f_!$ and $f_*$. In [FHM] this takes the form of an isomorphism $f_*(-) \overset{\cong}{\implies} f_!(- \otimes C)$ for some object $C \in \Cc$, and in [BDS] this takes the form of an isomorphism $f_*(- \tensor \omega_f) \overset{\cong}{\implies} f_!(-)$ for another object $\omega_f \in \Cc$. In [BDS] a comparison is given with the approach of [FHM]: it is stated in proposition 4.4 that if the situations of [BDS] and [FHM] both apply simultaneously, then the object $\omega_f$ is the dual in $\Cc$ of the object $C$. However, I don't understand the rather short proof: it is claimed that the formal assumptions already give an isomorphism $f_* \cong f_!(- \tensor C)$ by [FHM, Thm. 8.1]. But it seems to me that [FHM, Thm. 8.1] requires some non-formal input, namely one needs to explicitly check that $f_*(G) \to f_!(G \tensor C)$ is an isomorphism for some set of generating compact dualizable objects $\{G\}$ of $\Cc$. Indeed, it takes May in the follow-up article The Wirthmüller isomorphism revisited quite some time to prove that the latter condition is satisfied in the setting of equivariant homotopy theory. It seems that Proposition 4.4 of [BDS] is cutting this short. Question 1: Why does it follow in [BDS,Prop. 4.4] that $f_*(-) \implies f_!(- \otimes C)$ is a natural isomorphism? Question 2: If it doesn't follow, is it otherwise possible to compare $C$ and $\omega$ without checking that $f_*(-) \implies f_!(- \otimes C)$ is a natural isomorphism on all compact generators? Let's assume that we know that both isomorphisms hold and additionally that $\omega_f$ is invertible (with inverse $C$). Then both approaches give a natural isomorphism $f_*(-) \implies f_!(- \otimes C)$. Question 3: (Why) do these two isomorphisms agree? REPLY [3 votes]: Thanks for pointing this out and sorry for the confusion. Proposition 4.4 of our paper does appear to be misstated. The meaning of "Wirthmüller context in the sense of [FHM]" needs to be strengthened to include the isomorphism that [FHM, Theorem 8.1] doesn't provide for free. The Proposition could be (correctly) stated as: "Suppose the basic adjunction $f^* \dashv f_*$ satisfies a Wirthmüller isomorphism in the sense of [FHM]; i.e., suppose that $f^*$ has a left adjoint $f_!$ and that there exists an object $C \in \mathcal C$ together with an isomorphism $f_!(C\otimes -)\simeq f_*(-)$. Then its dual $\mathrm{hom}(C,1)$ is isomorphic to $\omega_f$." The point is that we can recognize the dualizing object from existing isomorphisms that have been established in specific examples by specific non-formal means. (See e.g., Remark 2.16 of my other paper, "The compactness locus of a geometric functor and the formal construction of the Adams isomorphism".) In any case, you are correct that there appears to be no reason why the choice of isomorphism $f_!(C)\simeq f_*(1)$ would imply an isomorphism $f_!(C\otimes -)\simeq f_*(-)$. Indeed, this is one of the points about these works --- the $C$ that appears in [FHM] is non-canonical and the isomorphism of Theorem 8.1 doesn't come for free; in contrast, the relative dualizing object is canonical and the isomorphism of [BDS] comes essentially for free (in our rigidly-compactly generated setting). Also, the real point of our Remark 4.3 and (misstated) Proposition 4.4 is to show that the Wirthmüller isomorphism (à la [FHM]) implies the Grothendieck-Neeman isomorphism, contrary to the discussion in [FHM], about these being distinct settings. I think this is one of the things that was clarified by our paper. Finally, as you mention (in your answer) the example from May's second paper doesn't exactly fit into our setup since the equivariant stable homotopy category over the $N$-fixed universe is not rigidly-compactly generated --- not all of the generators are dualizable. Nevertheless, one can give an example of an $f^*:\mathcal D \rightarrow \mathcal C$ between rigidly-compactly generated categories which admits an invertible object $C\in \mathcal C$ such that $f_!(C) \simeq f_*(1)$ and yet $\mathrm{hom}(C,1) \not\simeq \omega_f$. I'll give this example as an answer to your other MathOverflow question. This will thus provide an example (fitting into our setup) where the associated [FHM] map $f_* \rightarrow f_!(C\otimes -)$ is not an isomorphism.<|endoftext|> TITLE: Can the Chern-Simons invariant of a cusped hyperbolic $3$-manfiold be defined mod $\mathbb Z$? QUESTION [7 upvotes]: For a closed hyperbolic $3$-manifold $M$, the Chern-Simons invariant $CS(M)$ can be defined as an element of $\mathbb R/\mathbb Z$. When $M$ is cusped it can still be defined, but is now only defined modulo $1/2$, i.e. as an element of $\mathbb R / \frac{1}{2}\mathbb Z$. (This is mentioned here.) In the cusped case, is there some extra data that lets us refine $CS(M)$ to an element of $\mathbb R/\mathbb Z$? Maybe something like a choice of class in $H^1(M, \mathbb Z / 2\mathbb Z)$? I'm particularly interested in the case of $M = S^3 \setminus K$ a knot complement. REPLY [4 votes]: Remark 7.2 in Christian's paper suggests that for the choice of a lift of the discrete-faithful $PSL_2(\mathbb{C})$ representation to $SL_2(\mathbb{C})$, there ought to be a lift of the Chern-Simons invariant to $\mathbb{R}/\mathbb{Z}$. As he points out, this is equivalent to the choice of a spin structure (which is only a torsor over $H^1(M;\mathbb{Z}/2)$). This should be sandwiched in the diagram on p. 30: $\require{AMScd}$ $$\begin{CD} H_3(SL_2(\mathbb{C}),P) @>>> H_3(SL_2(\mathbb{C}),\pm P) @>>> H_3(PSL_2(\mathbb{C}),P) \\ @V V V @V VV @V VV \\ \mathbb{C /4\pi^2 Z} @>>> \mathbb{C /2\pi^2 Z} @>>> \mathbb{C /\pi^2 Z}. \end{CD}$$ See also the discussion on p. 2, paragraph 2 to see why I think this normalization corresponds to the normalization given in your question.<|endoftext|> TITLE: Is an open subset of a cofibration a cofibration? QUESTION [6 upvotes]: Suppose $A \to X$ is a cofibration in topological spaces, and $U \subseteq X$ is an open subset. Is $U \cap A \to U$ a cofibration? Sorry if this is rather simple, but I don't have much experience with this sort of algebraic topology. Naively, it looks as though the universal example for the homotopy extension property (see May's Concise course 6.2) implies that cofibrations are stable under all pullbacks. However, if this were true, I would have expected to see it stated somewhere. REPLY [2 votes]: The answer is yes if $X$ is metrizable. As noted after Satz 1 in the paper by Dold cited in the answer by skupers Dold, Albrecht Die Homotopieerweiterungseigenschaft (=HEP) ist eine lokale Eigenschaft. Invent. Math. 6 (1968), 185–189. such a map $\tau$ exists if $X$ is metrizable, since then $A = \bar A$ is closed in $X$ and you can take $\tau(x)$ to be the distance from $x$ to $X - U$. Closed cofibrations are closed under pullbacks along Hurewicz fibrations. This is Theorem 12 of Strøm's paper Strøm, Arne Note on cofibrations. II. Math. Scand. 22 (1968), 130–142 (1969).<|endoftext|> TITLE: Why is embedded contact homology so powerful? QUESTION [15 upvotes]: The Embedded Contact Homology (ECH), introduced by M. Hutchings, is an invariant of (contact) three-manifolds. Since its introduction, well-known conjectures in symplectic/contact topology in dimension $3$, including the Weinstein conjecture and its gen erali zations, the Arnold chord conjecture, are proved based on this invariant. Furthermore, the ECH capacities are used to establish a Weyl law, which is the foundation of the equidistribution result for generic contact forms. There are also nice applications of ECH to symplectic embedding problems in dimension $4$. The question is: why is ECH so powerful? To illustrate the question, the Symplectic Field Theory (SFT) should capture more information from pseudo-holomorphic curves than ECH but the above results were not (yet) proved using SFT. One quick answer to this question would be the celebrated ECH=SWF theorem, which allows one to use powerful results/computations from monopole Floer homology. But eventually, one might hope to stay in the world of pseudo-holomorphic curves without appealing to gauge-theoretic invariants, in order to go to higher dimensions. So the question might be phrased as: Is ECH a low-dimensional miracle? If not, what lessons should we learn from the selections of pseudo-holomorphic curves which define ECH? REPLY [10 votes]: Without elaborating much there are three key points, with the first two laying the bedrock for the third: ECH counts J-curves without caring about most information of the actual branched covers of such curves. Relatedly and more to the point, ECH counts J-curves with certain ECH index, and this picks out "the right" curves (separating itself from SFT). In dimension 4 (where the J-curves live) we have the adjunction formula. A lot of results deal with nontriviality of ECH, which comes from nontriviality of monopole Floer homology. (Ex: nontriviality of monopole Floer is what Taubes used to get existence of Reeb orbits, i.e. proof of Weinstein conjecture.) This is for the same reason that the Seiberg-Witten invariants are so powerful, because Taubes' SW = Gr result gives nontriviality results about symplectic 4-manifolds. (ECH is the "categorification" of the Gromov invariants.) Here is another crucial point disguised as an application: On a symplectic 4-manifold with (negative) contact 3-manifold boundary, the standard "ECH curve count" yields a relative invariant in the $ECH_*$ of the boundary, while the standard "SFT curve count" yields a relative invariant in the (ordinary) contact homology $CH_*$ of the boundary. But if the contact structure is overtwisted then $CH_*$ is necessarily trivial, whereas $ECH_*$ can easily be nontrivial.<|endoftext|> TITLE: Is it a reasonable way to write a research article assuming truth of a conjecture? QUESTION [18 upvotes]: I have found a conjecture in a research article (published in a good journal) on number theory, which is not well known but very reasonable. Let me be clear that, there is no counter-example that vote down the conjecture, rather, its trueness has been proved in some special situation so far. I need this conjecture to develop a tool in order to write a research article. Unfortunately, I am unable to prove the conjecture as of now. Maybe others can prove it after some time. My question: Is it a reasonable way to write a research article assuming truth of the conjecture? REPLY [2 votes]: This is not fundamentally different from the (fairly subtle) question of whether any mathematical result in general is "interesting". Publication typically requires (at least) a result that is both true and interesting. From a logical viewpoint, there is absolutely no problem with proving the truth of a theorem that includes a conjecture in its hypothesis. It is a theorem like any other. Meanwhile, a major factor in what makes a result uninteresting for publication is if it can be trivially derived from simpler results. This is not an objective characteristic but a time-dependent fact of human knowledge. In your situation, if and when the conjecture is resolved, we know it will be possible to derive your theorem trivially from a simpler result. Namely, if the conjecture is true, there will be a simpler theorem that omits the conjecture from the hypothesis, from which your theorem will follow trivially. And if the conjecture is false, your theorem will be derivable as a tautology. But I would argue that until we know which obtains, your theorem is not yet uninteresting.<|endoftext|> TITLE: Varieties where every algebra is projective? QUESTION [9 upvotes]: Is it possible to classify all varieties (in the sense of universal algebra) where every algebra is projective? Several years ago I asked a similar question, with "free" in place of "projective". It turned out it had been answered by Steve Givant in his 1975 thesis, but a variant question I threw in as an afterthought was still open -- to classify all varieties where every finitely-generated algebra is free. Keith Kearnes, Emil Kiss, and Agnes Szendrei were able to give a classification. In both cases (with or without the finite-generation condition in the hypothesis) the answer is the same: the only such varieties are sets, pointed sets, vector spaces over a division ring, and affine spaces over a division ring. So now I'm feeling a little greedier, and want to relax freeness to projectiveness. I think that now the "finitely-generated" and "infinitely-generated" cases will diverge. For instance, every finitely-generated Boolean algebra is projective, but not every Boolean algebra is projective. As pointed out by Keith Kearnes below, every finite Boolean algebra is projective except for terminal algebra (the corresponding statement under Stone duality is that every finite set is injective except for the empty set). So it's possible there's no divergence here. On the other hand, if "finitely-generated" is weakened to "finitely-presented", then as observed by Jeremy Rickard in the comments, we start seeing modules over general von Neumann regular rings, so in this direction things do start to change. There are also at least two notions of projectivity to consider -- projectivity means that an algebra lifts against all epimorphisms, whereas regular-projectivity means that it lifts against all regular epimorphisms (i.e. surjections). As pointed out by Benjamin Steinberg in the comments, it really makes most sense for us to use the regular-projective versions (and there are too many questions on this question anyway) so let's focus exclusively on the "regular-projective" versions of the question: Question 1': For which varieties is it the case that every algebra is regular-projective? Question 2': For which varieties is it the case that every finitely-generated algebra is regular-projective? Question 3: For which rings is the the case that every finitely-generated module is projective? I'm pretty sure that a ring $R$ has all modules projective if and only if $R$ is a finite product of finite-dimensional matrix algebras over division rings by the Artin-Wedderburn theorem. Projectivity and regular-projectivity coincide in the abelian setting. So the guess would be that for Question 1, the only varieties are finite products of those varieties where every algebra is free (i.e. of sets, pointed sets, algebras over a division ring, and affine spaces over a division ring) or varieties whose categories of algebras are equivalent to such (although perhaps a syntactic characterization of this condition is nontrivial?). For Question 2 and Question 3 I suspect there might be more interesting examples, which I'd like to hear about even if a classification is out of reach. Just to be sure we're on the same page (since there are equivalent definitions of "projective" in the abelian case which are inequivalent in general), I say that $P$ is projective if for every epimorphsim $A \twoheadrightarrow B$ and every homomorphism $P \to B$, there exists a map $P \to A$ making the obvious triangle commute. This implies that every epimorphism splits, but I think the converse does not hold. "Regular projective" is the same, but with $A \twoheadrightarrow B$ assumed to be surjective. REPLY [5 votes]: Tim asked about varieties where every algebra is free in 2014 on MO. I wrote a partial answer in 2015. Later in 2015, Emil Kiss, Agnes Szendrei and I wrote a paper to answer Tim's question. My email records show that Emil, Agnes and Keith also discussed the problem of determining which varieties have the property that every algebra is projective (or every finitely generated algebra is projective). We didn't solve it, but made some inital observations. Let me say that a variety has ${\bf Proj}$ if all algebras are projective and has ${\bf Proj}_{f.g.}$ if all finitely generated algebras are projective. First, some trivial observations: A variety where every algebra is free will have ${\bf Proj}$. These are the variety of sets, the variety of pointed sets, the variety of vector spaces over a division ring, or the variety of affine spaces over a division ring. A variety of modules has ${\bf Proj}$ iff every cyclic module is projective iff the associated ring is semisimple. (Jeremy mentions this in the comments.) Any variety with ${\bf Proj}_{f.g.}$ has the property that every member has a singleton subalgebra. (In particular, the variety of Boolean algebras does NOT have ${\bf Proj}_{f.g}$, since the 1-element BA is not projective.) You probably already see the reason: In the presence of ${\bf Proj}_{f.g.}$, any surjection onto the 1-element algebra must have a section. The class of varieties with ${\bf Proj}$ is closed under products, matrix powers, definitional equivalence, and localization to the range of an invertible idempotent. This includes Ben's example, since the variety of rectangular bands is the product of two varieties definitionally equivalent to sets. Here are some slightly less obvious (but still easy) observations. Let $\theta$ be a congruence on an algebra $A$ and let $S\leq A$ be a subalgebra. The $\theta$-saturation, $S^{\theta}$, of $S$ is the union of the $\theta$-classes that intersect $S$. It is a subalgebra of $A$ and it is the least subalgebra of $A$ that is a union of $\theta$-classes. Call $S$ saturated by $\theta$ (or say that $\theta$ saturates $S$) if $S^{\theta}=S$. If $\theta$ is a congruence on $A$, then a subalgebra $S\leq A$ is a complement to $\theta$ if $S^{\theta}=A$ and the restriction of $\theta$ to $S$ is the equality relation. A variety $\mathcal V$ has ${\bf Proj}$ (${\bf Proj}_{f.g.}$) iff every (f.g.) algebra is a retract of a (f.g.) free algebra. A variety $\mathcal V$ has ${\bf Proj}$ iff the following condition holds: Whenever $A\in {\mathcal V}$ has a congruence $\theta$, then $\theta$ has a complementary subalgebra. (There is a corresponding property for ${\bf Proj}_{f.g.}$. Also, if the condition holds for free algebras, then it will hold for all algebras.) A locally finite variety $\mathcal V$ has ${\bf Proj}_{f.g}$ if and only if $\mathcal V$ is ``Frattini-free''. That is, every finite algebra in $\mathcal V$ has trivial Frattini congruence. (The Frattini congruence of an algebra $A$ is the join of all congruences which saturate all maximal subalgebras of $A$.) Since the algebras in varieties with ${\bf Proj}_{f.g.}$ have at least one trivial subalgebra, we decided to examine all examples of varieties generated by a single, finite, idempotent strictly simple algebra to see if ${\bf Proj}_{f.g.}$ holds. ($A$ is idempotent if every fundamental operation satisfies an identity of the form $f(x,x,\ldots,x)\approx x$. $A$ is strictly simple if it is simple and has no nontrivial proper subalgebra.) The results were these: If ${\mathcal V} = HSP(A)$, where $A$ is a finite, idempotent, strictly simple algebra, then $\mathcal V$ has ${\bf Proj}_{f.g.}$ if and only if $A$ has types 1, 2, or 3 in the sense of tame congruence theory. (This theory allows five types of local behavior, 1--5. Type 1 simple algebras are those that look locally like a $G$-set, type 2 simple algebras consists of algebras that look locally like a vector space, type 3 simple algebras are those that look locally like a Boolean algebra, type 4 simple algebras are those that look locally like a lattice, type 5 simple algebras are those that look locally like a semilattice.) We examined infinitely many different clones to determine that the type 1,2, 3 varieties have ${\bf Proj}_{f.g}$ while the type 4, 5 cases do not. Our results seem to suggest that ordered algebras are unlikely to have ${\bf Proj}_{f.g.}$. Here we interpreted locally the following failure of ${\bf Proj}_{f.g.}$ from the variety of distributive lattices. (The distributive lattice on the right is not projective, since the map does not have a section.) EDIT 2-14-21 [From Tim] From what you say, it sounds like you may have found some examples of ${\bf Proj}_{f.g.}$ varieties which are not (or not obviously) products of varieties where all algebras are free -- is that accurate? Yes, but all examples we found "resembled" modified versions of the variety of sets, a variety of modules or the variety Boolean algebras. The simplest new example that hasn't yet been mentioned is the variety of rectangular bands with an involution. The rectangular band identities are: $x(yz)\approx (xy)z, x^2\approx x$, and $xyz\approx xz$. Add to this variety a unary function $x'$ satisfying $(x')'\approx x$ and $(xy)'\approx y'x'$. The resulting variety is not decomposable into a product of other varieties (since it has a unique simple member up to isomorphism, and products of at least 2 nontrivial varieties have at least 2 simples), however it does have the property that all members are projective. Not all are free. In fact, the variety of rectangular bands with involution is categorically equivalent to the variety of sets (which implies all algebras in this variety are projective), but it is not definitionally equivalent to the the variety of sets (since not all algebras in this variety are free). REPLY [2 votes]: Here is an example of a variety where every object is regular-projective. I am not sure what you mean by a product of varieties, so I don’t know for sure if it fits into that context but I suspect it does. First note that an algebra is regular projective if and only if it is a retract of a free object. This is more or less exactly like the standard argument for modules since regular epimorphisms are pullback stable. A rectangular band is a semigroup satisfying the identities $x^2=x$ and $xyx=x$. Every non-empty rectangular band is isomorphic to one for the form $A\times B$ where $A$ and $B$ are sets and the multiplication is of the form $(a,b)(a’,b’)=(a,b’)$. The free rectangular band on $X$ is $X\times X$ with the above multiplication where $X$ embeds diagonally to get the universal map. Clearly, if $X$ is of cardinality at least the max of the cardinality of $A,B$, then we can write $A\times B$ as a retract of $X\times X$. Notice that these are precisely direct products of left zero semigroups with right zero semigroups. Since either of these kinds of algebras are term equivalent to sets, it wouldn’t be surprising to me if this example is captured by what you mean as a product of examples. I believe the category of rectangular bands is equivalent to the product of two copies of the category of sets so that is probably what you mean by a product.<|endoftext|> TITLE: Regularly well-powered iff regularly co-well-powered? QUESTION [6 upvotes]: Let $\mathcal C$ be a finitely complete, finitely cocomplete category. Then the following are equivalent: $\mathcal C$ is regularly well-powered (i.e. every $C \in \mathcal C$ has a small set of regular subobjects); $\mathcal C$ is regularly co-well-powered (i.e. $\mathcal C^{op}$ is regularly well-powered). Proof: Freyd showed in his paper Concreteness that (1) is equivalent to $\mathcal C$ is concrete (i.e. there exists a faithful functor $\mathcal C \to Set$); which is in turn equivalent (because there exist faithful functors $Set^{op} \to Set$, e.g. the contravariant powerset functor) to $\mathcal C^{op}$ is concrete; which by the dual of Freyd's result, is equivalent to (2). Question: Is there an elementary proof that $(1) \Leftrightarrow (2)$? Can the hypotheses be relaxed? For instance, does $(1) \Leftrightarrow (2)$ hold when $\mathcal C$ is assumed to have equalizers and coequalizers, but not necessarily all finite limits and colimits? REPLY [4 votes]: Your first question is handled in Freyd’s paper, which shows that the notion of “a binary relation controlled by a set” is richer than you would think. Freyd says a relation $\square$ from $L$ to $R$ between large sets is resolvable if there is a small subset $L’\subseteq L$ such that every $l\in L$ has the same $\square$-relatives as some $l’\in L’$. Remarkably, resolvability of a relation implies that of its opposite! This is the key elementary foundation for everything in this paper. The proof is easier than thinking of the statement: for each subset $A$ of $L’$ for which this is possible, pick $r_A$ such that $A$ is the set of all $r_A$’s $\square$-relatives in $L’$. Then it’s easy to check that the (small!) set of all $r_A$s resolves $\square^{\mathrm{op}}$. The categorical application of this notion is to say that a span $A\leftarrow X\to B$ $\square$-relates to a cospan $A\to Y\leftarrow B$ if and only if the square commutes. Isbell proved that a concrete category has $\square$ resolvable for every $A,B$, and Freyd proved the converse. This is saying that a category is concrete if and only if it doesn’t have too many distinguishable maps into or equivalently out of any pair of objects, which is a hyper-general (co)wellpoweredness condition. If now the category has binary products, respectively coproducts, then it is equivalent to look at the analogous relations involving forks, respectively coforks, centered at a single object $A$, and one proves this directly. It’s actually the first lemma toward proving the concreteness theorem. Resolvability of these relations is what Freyd calls generalized regular (co)-well-poweredness, so we do know that these are equivalent for categories with binary products and coproducts; adding equalizers and coequalizers gets you the result you mentioned, and again it’s easy to prove directly that resolvability of the generalized regular subobject relation is equivalent to regular wellpoweredness when there are equalizers. So that handles your (1). Toward your (2), I’m pretty convinced you can show these results are optimal by fiddling with Isbell’s original non-concrete category, which is a union of a large number of commutative squares spanning and cospanning two fixed objects $A,B$. This is (generalized) regular well- and co-wellpowered because there aren’t any nontrivial forks or coforks. Now freely adjoin a product $A\times B$. You haven’t added any coforks but $A\times B$ is at the middle of a large number of distinguishable forks, so this is a guy which is not generalized regular well powered but is (generalized) regular cowellpowered. Now freely adjoin equalizers and then coequalizers. Since we’ve never created any new coforks and every nontrivial fork is centered on $A\times B$, I claim that all this adds is an initial object and a terminal object plus a large family of distinct regular subobjects of $A\times B$. Now this has equalizers and coequalizers, is not regular well powered but is regular well copowered. So I think the moral is that forks are less meaningful that squares in the absence of products.<|endoftext|> TITLE: On a result of Euler on pseudoprimes QUESTION [5 upvotes]: In several sources (for instance on page 58 of the first ed. of Crandall & Pomerance book on prime numbers or at the end of this paper by J. H. Jaroma), I have seen a result that goes like this: Let $p$ be an odd prime congruent to $-1$ modulo $4$. Then $2p+1$ is a prime iff $(2p+1) \mid (2^{2p}-1)$. Do you know if the hypothesis $p \equiv -1 \pmod 4$ may be removed? If the result is still valid for primes congruent to $1$ modulo $4$, I wonder why it is that it is not mentioned in the sources I referred to above... REPLY [5 votes]: Yes, the result holds for every odd prime number $p$... I certainly find it somewhat "strange" that it is only stated for primes congruent to $-1$ modulo $4$ in several places: Proposition. Let us suppose that $p$ is an odd prime number and that $2p+1$ divides $2^{2p}-1$. Then, $2p+1$ is a prime number. Proof. (I learnt it from J. I. Restrepo) For the sake of contradiction, let us suppose that $2p+1$ is not a prime number and that $q$ is a prime number dividing $2p+1$. From the hypothesis and Fermat's little theorem we have that \begin{eqnarray} 2^{2p} \equiv 1 \pmod{q}\\ 2^{q-1} \equiv 1 \pmod{q}. \end{eqnarray} It follows from these congruences that $\mathrm{ord}_{q}(2)=:\mathfrak{o}$ is a common positive divisor of $2p$ and $q-1$; since there are only four positive divisors of $2p$ and $q-1 \leq (p-\frac{1}{2}) TITLE: Law of the iterated logarithm in C([0,1]) QUESTION [6 upvotes]: Let $X_1, X_2, \ldots $ be a sequence of independent identically distributed random variables with values in the space $C([0,1])$ of real continuous functions on $[0,1$]. Assume for simplicity that the $X_i$'s are centered, in the sense that $E( X_i(t)) = 0$ for $t \in [0,1]$). Are there any necessary and sufficient conditions (on the common distribution of the $X_i$'s) in order that $$ \limsup_{n \to \infty} \frac {1}{\sqrt {2n \ln \ln (n)}} \, \| X_1 + \cdots +X_n \|_\infty < \infty $$ almost surely (where $\| \cdot \|_\infty$ is the sup norm)? What would then be the value of the limsup? REPLY [4 votes]: What you want is the so-called bounded law of the iterated logarithm (BLIL) in $C[0,1]$. A necessary and sufficient condition for the BLIL in an arbitrary real separable Banach space $B$ was given by Ledoux and Talagrand '88 -- see e.g. Theorem A in the subsequent paper by Einmahl '93. This necessary and sufficient condition involves the boundedness of $(S_n/\sqrt{n\ln\ln n})_{n\ge3}$ in probability, which may be hard to verify in $C[0,1]$. A sufficient condition (involving a certain, rather weak condition on the modulus of continuity of $X_1(\cdot)$) for the BLIL in $C[0,1]$ is implied by Theorem 2.1 in the earlier paper by Kuelbs '76.<|endoftext|> TITLE: Euler characteristic and rational Poincaré series QUESTION [5 upvotes]: $\DeclareMathOperator\len{len}\DeclareMathOperator\Tor{Tor}$Let $(A,\mathfrak{m})$ be a regular local ring, and $x \in \mathfrak{m}^2$ be a non-zero prime element. So $R:=A/(x)$ is a non-regular Cohen-Macaulay local domain. By Eisenbud's famous work "Homological algebra on a complete intersection, with an application to group representations", the (infinite) minimal free resolution of any finitely generated $R$-module $M$ becomes periodic of period at most $2$ after at most $\dim R$ steps. For two finitely generated $R$-module $M, N$ such that $\Tor_i^R(M,N)$ has finite length for all $i$, we define their Poincaré series as $P_{M,N}(t)=\sum_{i=0}^{\infty}\len_{R}(\Tor_i^R(M,N))t^{i}$. By Eisenbud's result, $P_{M,N}(t)$ is a rational function with only possible poles at $t=1$. We then define the Euler characteristic as $\chi(M,N):=P_{M,N}(-1)$. Questions: has such generalized Euler characteristic been studied before? Is it well-behaved e.g satisfying usual properties of Euler characteristic? Do we know analogs of Serre's homological conjecture for it? REPLY [2 votes]: $\DeclareMathOperator\Tor{Tor}$I studied this function in my thesis (gee, typing this answer makes me feel old!). In fact, this function can be defined even when the Tor modules have finite length eventually. One simply takes the difference $\theta^R(M,N):=\ell(\Tor_{2e}(M,N)) - \ell(\Tor_{2e-1}(M,N))$ for some $e\gg0$. When $M$ has finite projective dimension on the punctured spectrum of $R$, then $\theta^R(M,N)$ is defines for all $N$, and gives a map from the Grothendieck group of $\mathrm{mod}(R)$ to $\mathbb Z$. The idea was suggested in an old paper by Hochster, and this is now widely called Hochster's theta invariant. You can find some basic information in my paper here. The most interesting case is when $R$ has isolated singularity, then one has a pairing on the Grothendieck group. I made a number of conjectures along the line of the Serre's homological conjectures. They are too numerous to list, you can read about them in Section 3 of this paper with Kurano, available here. This function (and the Ext version, which is sometimes called the Herbrand difference, first suggested in Buchweitz's famous unpublished note) have appeared many times recently from different aspects. Here is a sample of some of my favorite: projective geometry and algebraic K-theory, (Mark Walker and his collaborators), topological K-theory (Buchweitz-Van Straten), Adams operations (Michael Brown) or Chern character on matrix factorizations (Polishchuck-Vaintrob). If you are actually doing research on this topic, feel free to send me an email and I can load you up with more specific information. PS: a version for complete intersection exists, and was written down in the second part of my thesis. Unfortunately I never got around to publish it, but you can check the papers that cite it for more update information.<|endoftext|> TITLE: Examples of Banach manifolds with function spaces as tangent spaces QUESTION [15 upvotes]: I have recently been learning the theory of Banach manifolds through Serge Lang's book on Differential Manifolds. So far the objects seem rather interesting but my intuition always comes from the finite dimensional case, which admittedly is very helpful but I certainly want to add onto this. I am wondering if there is a reference where one has examples of Banach manifolds where the tangent spaces are function spaces (ie Sobolev spaces). Obviously, I would like to avoid the trivial example of the function space being the manifold. In some sense I would like an example that embraces the infinite dimensional tangent space as well as giving me some understanding as to what the topological space/smooth structure should be in the setting. REPLY [5 votes]: Banach manifolds as a natural generalization of the finite dimensional case became “popular” in the late 60s. In Physics an obvious example would be the projective Hilbert space, the state space. Klingenberg’s book on Riemannian Geometry has some chapters on (Hilbert) Riemannian Geometry. Infinite Hamiltonian systems and symplectic geometryare treated extensively by many authors (E.G. Marsden, Ratiu, Weinstein etc.). However results by Eells/Elworthy 1970, Elworthy 1972, Henderson 1969, 1972, show that Hilbert manifolds are essentially open subsets of their tangent space and smooth Banach manifolds are more or less all diffeomorphic (Henderson 1972). Also it would appear natural to look at the diffeomorphism group of a finite dimensional manifold. The tangent space should be given by the space of vector fields. Unfortunately a result by (Omori 1970) shows Banach manifolds are not suitable in that important case, more general (topological) vector spaces are needed. See the introduction of Kriegl/Michor 1997: A Convenient Setting of Global Analysis.<|endoftext|> TITLE: Rigged Hilbert spaces and the spectral theory in quantum mechanics QUESTION [21 upvotes]: I'm trying to learn some quantum mechanics by myself, and because of my mathematics background, I'm trying to understand it in a rigorous way. Since then, I've been intrigued by the use of rigged Hilbert spaces in this context. The topic caught my attention one more time after my discussion on physics stack exchange about the Dirac formalism. Repeating a little bit what I've said in the linked question, physicists use $|x\rangle$ as eigenvectors of the position, and write every element $|\psi\rangle$ of the underlying Hilbert space as: \begin{eqnarray} |\psi\rangle = \int dx \,\psi(x) |x\rangle \tag{1}\label{1} \end{eqnarray} This is analogous to what is usually found in the theory of Hilbert spaces, in which for every $x\in \mathscr{H}$ and $\{e_{n}\}_{n\in \mathbb{N}}$ Hilbert basis: \begin{eqnarray} x = \sum_{n\in \mathbb{N}}\langle e_{n},x\rangle e_{n} \tag{2}\label{2} \end{eqnarray} As it turns out, the space of vectors $|\psi\rangle$ are usually taken to be $L^{2}(\mathbb{R}^{d})$, so that the position operator $\hat{x}$ makes sense as a multiplication operator on a dense subspace. However, it is a self-adjoint operator with continuous spectrum, so it has no eigenvectors at all in $L^{2}(\mathbb{R}^{d})$ and, thus, $|x\rangle$ does not make mathematical sense in this context. In any case, the use of (\ref{1}) demand that the coefficients $\psi(x)$ to be elements of $L^{2}(\mathbb{R}^{d})$ themselves, just as the coefficients $\alpha_{n} = \langle e_{n},x\rangle$ are elements of $\ell^{2}(\mathbb{N})$ in (\ref{2}), so we end up using the components $\psi(x)$ to do actual calculations most of the time (at least, explicit calculations such as solving the Schrödinger equation). This is what intrigues me. The above scenario is present at the very beginning of every discussion of quantum mechanics and, at least to me, these very few words are more than enough to justify the need of "some bigger space", which seems to be a rigged Hilbert space. However, a quick search on the internet and you find only a few things about rigged Hilbert spaces, in general papers discussing its connections with quantum mechanics. I've checked also my books on rigorous quantum mechanics and none of them seem to address the problem. However, all of them discuss spectral theory, which I've heard is another way to approach all this. Question: Why does it seem that rigged Hilbert spaces are still a paper subject, not fully incorporated into the usual expositions of rigorous quantum mechanics, while spectral theory is deeply rooted? Is the spectral theory formalism preferable to treat the problems mentioned above instead of rigged Hilbert spaces by any reason? And why is this? REPLY [12 votes]: You ask 'why spectral theory instead of rigged Hilbert spaces?'. There are some practical/pedagogical reasons: One is that you'll need basic spectral theory in quantum mechanics anyways, so might as well get as much mileage as you can out of it. Another is that spectral theory can be largely explained by analogy with finite-dimensional linear algebra. Rigged Hilbert spaces require nuclear vector spaces, which while quite nice are not particularly intuitive. (One can cut corners and define them via nested Hilbert spaces, but this is also clearly more complicated than just having one Hilbert space.) There are also sociological reasons: Much of the work on rigorous quantum physics follows von Neumann, and he used spectral theory. He wasn't wrong to do so; the most important thing to get right at the time was the shocking behavior of atomic spectra. Rigged Hilbert spaces came from Gel'fand and his collaborators thirty years later. By that point, much of the work in rigorous quantum physics was being done by people who'd already set everything up in terms of C*-algebras. One might also blame Reed & Simon, whose influential and otherwise fabulous analysis book rather sassily only recommends "the abstract rigged approach to readers with an emotional attachment to the Dirac formalism". Lastly, I'd argue that spectral theory and rigged Hilbert spaces aren't really different things. The theory of rigged Hilbert spaces is just a nice way of setting up spectral theory for unbounded operators. I prefer rigged Hilbert spaces for quantum physics, because they let you make explicit use of the operators you actually want to talk about, instead of inventing circumlocutions to only talk of bounded operators. But it must be acknowledged that the C*-algebra language has allowed a lot of progress to be made, and that the distinctions between these things are invisible to experiment.<|endoftext|> TITLE: Large cardinals without replacement QUESTION [9 upvotes]: Let $ZC$ be Zermelo set theory with choice, which differs from $ZFC$ in omitting the axiom scheme of replacement. EDIT: I think I want to include foundation in the axioms, which apparently isn't normally considered to be part of Zermelo set theory. As is well-known, $ZC$ is much weaker than $ZFC$; for instance $V_{\aleph_\omega}$ models $ZC$. One way to measure the difference is by interpolating between the two with the theories $ZC_n$ for $n \in \mathbb N$, where one adds an axiom or axiom scheme of $\Sigma_n$ replacement; I believe one has $ZC = ZC_0$ and informally $ZFC = ZC_\omega$. At the extreme low end, I believe that $ZC$ doesn't even prove that every well-order is isomorphic to an ordinal, so if it makes things easier in the following to replace $ZC$ with $ZC_1$, I don't think I'd object. What I'd like to know is whether $ZC$ admits a "large cardinal hierarchy" resembling in some sense the familiar hierachy of large cardinal axioms one can add to $ZFC$, and which give another way of calibrating consistency strength of such theories. When trying to adapt large-cardinal ideas from $ZFC$ to $ZC$, I imagine there are plenty of potential issues. For one thing, as usual equivalent formulations of a statement in $ZFC$ may become inequivalent in $ZC$, so one must think carefully about choosing "correct" formulations. More concerningly, the canonical example of a model of $ZC$ given by $V_{\aleph_\omega}$ suggests that perhaps one way $ZC$ differs from $ZFC$ is that in $ZC$ the universe doesn't necessarily "extend endlessly upward", and so "adding to the top" by hypothesizing "large cardinals" may simply be an ineffective way to generate stronger extensions of $ZC$ which are not extensions of $ZFC$. So I think the somewhat-more-specific questions I have are: Question 1: Is there anything analogous to the large cardinal hierarchy when it comes to theories which extend $ZC$ but not $ZFC$? Or is it rather the case that anything recognizable as a "large cardinal axiom" in $ZC$ will likely imply replacement anyway? Question 2: If so, does this hierarchy lie strictly below $ZFC$ in consistency strength? Or is it rather possible to get a theory without replacement which is stronger than $ZFC$ in consistency strength, or even incomparable? Question 3: What's an example of an analog of a large cardinal axiom in $ZC$? (Or if the answer to (1) is "no", then: what are some other interesting ways to get theories between $ZC$ and $ZFC$ other than the theories $ZC_n$?) My favorite large cardinal axiom happens to be Vopenka's principle. So for instance, is there a version of Vopenka in $ZC$, and if so, is the resulting theory weaker than $ZFC$? I'd also be interested in asking similar questions about $BZC$, the material-set-theoretic analog of ETCS, where the language is modified so that there simply aren't any unbounded quantifiers at all. But perhaps that would be too radically different a setting from $ZFC$ to really get a grip on the question. REPLY [5 votes]: Overall, the large cardinal axiom hierarchy is very similar between ZC (ZFC minus replacement; we are including regularity) and ZFC. A large cardinal axiom (unprovable in ZFC) satisfied by $κ$ typically implies in ZC that $V_κ$ satisfies ZFC + weaker large cardinal axioms. However, the axioms typically do not imply additional replacement above $κ$ even for $Σ^V_3$ axioms such as existence of a strong or a supercompact cardinal (and hence their strength is truncated accordingly); but an extendible or a proper class of strong cardinals gives $Σ_2$ replacement, and a proper class of extendibles gives $Σ_3$ replacement. Equiconsistencies also typically carry over to ZC. For example, ZC + $L(ℝ)⊨\text{AD}$ is equiconsistent with ZC + $ω$ Woodin cardinals whose supremum exists. However, there are some differences. One is notational. ZC does not prove that $ω2=ω+ω$ exists as the transitive set. However, ZC interprets $Σ_1$ replacement, and we can either add $Σ_1$-replacement, or implicitly speak of codes for ordinals and other sets. Defining HOD requires replacement. However, there are inequivalent first order definable approximations of OD in ZC, one of which is $∪_{V_κ \text{ exists}} \mathrm{OD}^{V_κ}$. The lack of singular infinite cardinals creates some strength differences. For example, ZC + $∀κ \, (κ^+)^L < κ^+$ is equiconsistent with ZC rather than ZC + $0^\#$. Still, there are other covering properties whose violation has high strength in ZC, and nonexistence of an inner model with a Woodin cardinal should still imply that the core model exists. Also, ZC($j$) + “$j$ is a nontrivial elementary embedding $V→V$” (called Wholeness Axiom; it proves ZFC without replacement for $j$-formulas) is consistent relative to the $\mathrm{I}_3$ axiom, as opposed to the Kunen inconsistency in ZFC($j$) due to the axiom of choice and existence of $j^ω(\mathrm{crit}(j))$. There are large cardinal axioms for ZC that are implied by ZFC. Borel determinacy is equivalent to $∀r∈ℝ \,∀α < ω_1 \, ∃β \, L_β(r) ⊨ \text{“} ω_α \text{ exists”}$. Also, the least $α<β$ with $\mathrm{Theory}(V_α)=\mathrm{Theory}(V_β)$ are strictly between $ω_1^L$ and $c^+$. Bounded quantifier ZC (BZC, also called Mac Lane set theory) is useful for some equiconsistency statements, and as a minimal base theory. For example, BZC + a proper class of Woodin cardinals is conservative over $\text{Z}_2 + \text{PD}$ (second order arithmetic with projective determinacy). In turn, key theorems about universally Baire sets relying on a proper class of Woodin cardinals can in fact be proved in BZC + a proper class of Woodin cardinals (even though it does not prove that the set of all universally Baire sets of reals exists). By reflection, for every consistent axiom A, ZC+A has lower consistency strength than ZFC+A, but this need not apply to schemas. For example, Vopěnka's principle over ZC (or just BZC) implies ZFC.<|endoftext|> TITLE: How to see that the determinant of this matrix is nonzero for all primes? QUESTION [21 upvotes]: I'm trying to show that $\sum_{i = 0}^{p-2} (i+1)^{-1} t^{i+n}$ where $0 \leq n \leq p-2$ spans the vector space $\mathbb{F}_p[t]/(1-t)^{p-1}$ as a rank $p-1$ module over $\mathbb{F}_p$. In other words, I would like to show that the determinant of the following matrix is a unit in $\mathbb{F}_p$. I've shown this for $p = 2, 3, 5, 7, 11, 13$. I have tried to use induction but failed. This is such a natural matrix I am hoping someone recognizes it! $$\begin{pmatrix} 1 & 0 & -1 & (p-2)^{-1} & \cdots & 4^{-1} & 3^{-1} \\ 2^{-1} & 1 & 0 & -1 & (p-2)^{-1} & \cdots & 4^{-1} \\ 3^{-1} & 2^{-1} & 1 & 0 & -1 & \ddots & \vdots \\ \vdots & 3^{-1} & 2^{-1} & 1 & 0 & \ddots & (p-2)^{-1} \\ (p-3)^{-1} & \ddots & \ddots & \ddots & \ddots & 0 & -1 \\ (p-2)^{-1} & (p-3)^{-1} & (p-4)^{-1} & \ddots & 2^{-1} & 1 & 0 \\ -1 & (p-2)^{-1} & (p-3)^{-1} & \cdots & 3^{-1} & 2^{-1} & 1 \end{pmatrix}$$ Edit: At risk of overcrowding the above question, I am adding some context below the line. Feel free to ignore it. Let us look at the action of $G := C_p \rtimes C_{p-1} \simeq (\mathbb{F}_p, +) \rtimes (\mathbb{F}_p^*, \times)$. Then, $G$ acts on $x \in X := F_p$ as follows, $(c, m)(x) = c + mx$. Let R be a $\mathbb{Z_p}$-algebra, and $A=R[C_p]$ be the permutation representation. If $x \in X$, we write $[x]$ as the corresponding element in the module. Let us fix $\sigma := (1,1)$ and $\tau := (0, a)$ to be the generators of $G$, where $a$ is a fixed primitive root of $\mathbb{Z}/p$. Now, let $B$ be the kernel of the augmentation map $R[X] \to R$. This is a free representation of rank $p-1$ over $R$. Further, as a $C_p$-module, $B$ is isomorphic to $R[X]/N$, where $N$ is generated by $(1+\sigma+\cdots+\sigma^{p-1})$. Further, $N$ is isomorphic to the trivial representation as a $C_p$ module. Let $V$ be a rank 1 $C_{p-1}$-subrepresentation of $B$. We may extend the inclusion map $V \to B$ of $C_{p-1}$ representations to a map of $G$ representations: $$f: R[G] \otimes_{R[C_{p-1}]} V \to B.$$ I wish to pick $V$ such that this map is surjective. By Nakayama's lemma, since we are working with local rings, it suffices to show this map is surjective mod $p$. To get to the below phrasing, I rewrote $R[X]$ as $R[t]/(t^p-1)$. Here, $\sigma$ acts by taking $t^{i} \mapsto t^{i+1}$. Then, if we consider $B$ mod $p$, which we call $\overline{B}$, then $\overline{B} = \mathbb{F}_p[t]/(t-1)^{p-1}$. There are $(p-1)$ 1-d $C_{p-1}$ subrepresentations of $B$, where $t$ in $\mathbb{F}_p^\times$ acts by multiplication with $t^r$, where $1 \leq r \leq p-1$. Further, from trial and error I noticed that picking $V$ to be the subrepresentation where $r = 1$, seems to make $f$ surjective. In other words, choosing $V$'s generator mod p to be $y := \sum_{j = 0}^{p-2}a^{-j}t^{a^k} = \sum_{i=0}^{p-2} i^{-1}t^i$ seems to make $f$ surjective. Note that the image of $f$ is $(y, \sigma(y), \cdots, \sigma^{p-2}(y))$. REPLY [10 votes]: As discussed in the comments, I don't see how to extract the desired matrix from the original question (about spanning the vector space). However, the matrix being nonzero IS equivalent to the following: The polynomials $\sum_{i = 0}^{p-2} (i+1)^{-1} t^{i+n}$ for $0\leq n\leq p-2$, along with the polynomial $t^{p-1}$, span the vector space $\mathbb{F}_p[t]/(1-t)^p$. To see this, the change of basis matrix from $1,t,\cdots, t^{p-1}$ to our set of polynomials is given by $\begin{pmatrix} 1 & 2^{-1} & 3^{-1} & 4^{-1} & \cdots & (p-1)^{-1} & 0 \\ 0 & 1 & 2^{-1} & 3^{-1} & 4^{-1} & \cdots & (p-1)^{-1} \\ (p-1)^{-1} & 0 & 1 & 2^{-1} & 3^{-1} & \ddots & \vdots \\ \vdots & (p-1)^{-1} & 0 & 1 & 2^{-1} & \ddots & 4^{-1} \\ 4^{-1} & \ddots & \ddots & \ddots & \ddots & 2^{-1} & 3^{-1} \\ 3^{-1} & 4^{-1} & 5^{-1} & \ddots & 0 & 1 & 2^{-1} \\ 0 & 0 & 0 & \cdots & 0 & 0 & 1 \end{pmatrix}$ (here we use that $(1-t)^p=1-t^p$.) Because of the form of the bottom row, the determinant of this matrix is equal to the determinant of the minor given by the first $(n-1)$ rows and columns, which is the transpose of the original matrix. Now let us address this question about polynomials. Set $P(t)=\sum_{i=0}^{p-2}(i+1)^{-1}t^i.$ Then the question is equivalent to showing that there is no choice of a polynomial $Q(t)$ of degree $\leq p-2$ and a constant $c\in\mathbb{F}_p$ with $P(t)Q(t)=ct^{p-1}$ in $\mathbb{F}_p[t]/(1-t)^p$. Assume otherwise. First note that, if $p>2$, $t-1$ divides $P(t)$. Indeed, $P(1)=1^{-1}+2^{-1}+\cdots+(p-1)^{-1}$, which is zero modulo $p$. If $p=2$ then the matrix trivially has nonzero determinant. Therefore, we have $0=P(1)Q(1)=c$, so we have $P(t)Q(t)=0$. This can only be possible if $P(t)$ is a multiple of $(t-1)^2$. This in turn would imply that $tP(t)=\sum_{i=0}^{p-2}(i+1)^{-1}t^{i+1}$ is a multiple of $(t-1)^2$, and hence that the derivative $(tP(t))'=\sum_{i=0}^{p-2}t^i$ is a multiple of $(t-1)$. But, by inspection, we see that its value at $1$ is $p-1$, a contradiction.<|endoftext|> TITLE: Do odd-weight cusp forms have analytic rank 0? QUESTION [8 upvotes]: Let $f(z)=\sum_{n\ge 1}a_nq^n$ be a cusp form, where $q=e^{2\pi i z}$. Let $ L(s) = \sum_{n\ge 1} a_nn^{-s} $ be its corresponding L-function. The completed L-function of $L(s)$, $\Lambda(s)$, should satisfy a functional equation $\Lambda(s)=\bar \Lambda(a-s)$ for some integer $a$. The analytic rank of $f(s)$ is defined as the order of vanishing of $L$ at $a/2$. Question: If $f$ is of odd weight, is its analytic rank $0$ (or equivalently, $L(a/2)\neq0$)? REPLY [2 votes]: The statement needs some fixing. You must mean newform, rather than cusp form, as otherwise it could be interpreted that linear combinations can be taken (although they must possess identical sign of the functional equation to be allowed). This said, I think the answer is expected to be no as discussed by François Brunault, BUT the related question allowing (nonholomorphic) Maass forms of eigenvalue 1/4 has a different answer, as noted by Armitage. That is, contrary to the $L$-function expectations otherwise, there are $L$-functions of motivic weight 1 with $\Lambda(s)=-\Lambda(1-s)$, which vanish at $s=1/2$ due to this symplectic root number. https://doi.org/10.1007/BF01404125<|endoftext|> TITLE: Is every finite $d$-dimensional matrix group generated by $d$ elements? QUESTION [13 upvotes]: The question is in the title. If $\Gamma\subset\mathrm{GL}(\Bbb R^d)$ is a finite matrix group, can it be generated by (at most) $d$ elements? I suspect that this hope is too naive, but I have no counterexamples. I would also be interested in bounds on the number of generators. REPLY [19 votes]: The answer to the question is yes. In Theorem 1.2 of this paper. the authors (Colva Roney-Dougal and myself) prove that if $G \le {\rm GL}(n,F)$ with $F$ a field, where (a) $G$ is finite; (b) either $G$ is completely reducible or ${\rm char} F = p$ and $O_p(G)=1$; and (c) $F$ does not contain a primitive fourth root of unity, then $d(G) \le n$. This is best possible when ${\rm char} F \ne 2$, because there is an obvious embedding of $C_2^n$ in ${\rm GL}(n,F)$.<|endoftext|> TITLE: Rank of $A\otimes B - B\otimes A$ QUESTION [6 upvotes]: Let $A$, $B\in\mathbb{R}^{n\times n}$ be full-rank random matrices and define the Kronecker products $P=A\otimes B$ and $Q=B\otimes A$. Through example-based examination, I have found that $\text{rank}(P-Q)=n^2-n$, but I am struggling to formulate a proof of this. Thus far, I have focused on the fact that $P$ and $Q$ are permutation-similar such $Q=ZPZ$, where $Z=Z^T=Z^{-1}$. We know that $P$ and $Q$ have the same eigenvalues and that the eigenvectors, $\Phi_P$ and $\Phi_Q$, are related by $\Phi_P=Z\Phi_Q$. Furthermore, $\Phi_P-Z\Phi_QZ=T=[T_1 \ldots T_{n^2}]$, where $T_i=0$ for $i \in \{1:n+1:n^2\}$. I presume that this somehow shows the rank condition, but I cannot figure out the coupling. I would like to ask whether I am on the right track; and if yes, can you provide some pointers on how I proceed to show the rank condition? REPLY [7 votes]: A slight variant of Fedor's answer: using a QZ (generalized Schur) factorization $A=QT_A Z, B = Q T_BZ$, you can make an orthogonal change of basis such that $A$ and $B$ are both upper triangular. Then $P$ and $Q$ are both upper triangular, too, and the eigenvalues of $P-Q$ are its diagonal elements $a_{ii}b_{jj} - b_{ii}a_{jj}$, $i,j=1,\dots,n$. This makes it evident that $n$ of them are zeros, and that the rest of the entries in the upper triangle, generically, is not. (EDIT: note that this change of variables does not preserve eigenvalues, since $Q$ and $Z$ are different in general, but it does preserve the rank.)<|endoftext|> TITLE: What is the smallest group for which Broué's abelian defect group conjecture has not yet been verified? QUESTION [6 upvotes]: Let $G$ be a finite group. Let $p$ be a prime dividing $|G|$. Let $k:=\overline{\mathbb{F}_p}$. Let $b$ be a $p$-block of $kG$ with abelian defect group $D$. Let $H:=N_G(D)$. Let $c$ be the Brauer correspondent of $b$. M. Broué conjectured in the 90's that $b$ and $c$ are derived equivalent under these assumptions. I would like to ask the following: Questions: Does there exist an up-to-date list of small groups for which this conjecture has been verified? E.g., is it true for all small groups of order less than $200$, say ? What is the smallest example (w.r.t. $|G|$) which is not yet verified? Thank you very much for the help. REPLY [5 votes]: I don't know the group of smallest order for which the conjecture has not been verified. But certainly it is known to be true for all groups of order less than 200. There are general results that deal with most small groups. For example, For $p$-soluble groups (and in particular soluble groups) it is known to be true, and in fact the derived equivalence is a Morita equivalence in this case. I think this was originally proved by Dade. For blocks with cyclic defect group it is known to be true. If it is true for two groups then it is true for their direct product. For groups of order less than 200, these facts deal with all cases except for $p=2$ and $G=A_5, S_5, \operatorname{SL}(2,5)$ or $\operatorname{PSL}(2,7)$, and all of those cases have been verified.<|endoftext|> TITLE: Modular forms and number of representations by binary quadratic forms QUESTION [5 upvotes]: Let $Q(x,y)$ be a positive definite quadratic form of discriminant $d$. Let $r_Q(n)$ be the number of solutions of $Q(x,y)=n$. It is known that the function $f_Q(\tau)=\sum_{n=0}^{\infty}r_Q(n)q^n$ is a modular form of weight $1$. Let $Q_1,\ldots,Q_{h_d}$ be representatives for nonequivalent classes of forms of discriminant $D$. Then we have the formula $$\sum_{k=1}^{h_d}r_{Q_k}(n)=\sum_{t\mid n}\chi_d(t).$$ I know how to prove this using ideals and $L$-series, or in an elementary way. Is there a proof that uses the fact that $f_Q(\tau)$ is a modular form? Can you give me a reference? REPLY [4 votes]: The coefficient $w$ in theorem 64 is usually $2,$ but is $4$ for discriminant $-4,$ then $6$ for discriminant $-3$<|endoftext|> TITLE: Condensed / pyknotic approach to orbifolds? QUESTION [9 upvotes]: Does condensed / pyknotic mathematics afford an (yet!) another approach to orbifold theory? Let me admit up-front that I don't know much about either condensed / pyknotic mathematics or about orbifold theory. But I think the following is a precise question which gets at the the above vague question. I believe there are several approaches to defining what an orbifold is, so for safety let me restrict attention to the $(2,1)$-category $\mathcal C$ of topological orbifolds $X$ such that $X$ is the orbifold quotient of a topological manifold $X_0$ by a properly discontinuous group action. I hope that even if there are differences between definitions of topological orbifolds, they should all at least agree on the category $\mathcal C$ up to equivalence of $(2,1)$-categories. On the other hand, every compactly generated weak Hausdorff space, and in particular every topological manifold, may be regarded as a condensed set in a canonical way. Moreover, there inclusion functor $Set \to Gpd$ induces an inclusion functor $Cond(Set) \to Cond(Gpd)$ from condensed sets to condensed groupoids. Let $\mathcal D$ be the full, $(2,1)$-subcategory of $Cond(Gpd)$ whose objects are $(2,1)$-categorical quotients of topological manifolds by properly discontinuous group actions. There are canonical inclusion functors $Man \to \mathcal C$ and $Man \to \mathcal D$ where $Man$ is the category of topological manifolds (where every 2-morphism is the identity). Question: Is there an equivalence of $(2,1)$-categories $\mathcal C \simeq \mathcal D$ respecting the two inclusions $Man \rightrightarrows \mathcal C, \mathcal D$? REPLY [11 votes]: Great question! I think the answer ought to be yes. But I must also make a disclaimer that I don't really know what orbifolds are, and the comments by David Roberts make be believe that what I thought they should be is not what they actually are. To me, an orbifold should be to a manifold as a Deligne--Mumford stack is to a scheme. In particular, they should naturally form a $2$-category (just like groupoids do). Recall that a Deligne--Mumford stack is a stack (=sheaf of groupoids) $X$ on the category of schemes equipped with the étale topology such that there is a surjective étale map $Y\to X$ from a scheme $Y$. Now the most general definition of an étale map of stacks is actually a bit subtle. Let me only consider the case of $0$-truncated maps (which is the only case relevant here, as $Y$ and thus $Y\to X$ is $0$-truncated). The case of separated étale maps is OK: A map $f: Y\to X$ is separated étale if and only if for all maps $X'\to X$ from a scheme $X'$, the fibre product $Y'=Y\times_X X'$ is representable by a scheme, and $Y'\to X'$ is separated étale. To verify that this is a good definition, one uses that separated étale maps of schemes satisfy fpqc descent. For general étale maps, one has to iterate this procedure a bit: One should only ask that $Y'$ is an algebraic space, and $Y'\to X'$ an étale map of algebraic spaces. To define the latter, one takes an étale cover of $Y'$ and asks that the composite map to $X$ is étale. Now the translation to (topological, say) manifolds is somewhat complicated by the fact that separatedness assumptions are much more commonly assumed by default. For example, I think $\mathbb R$ with a doubled origin (i.e. two copies of $\mathbb R$ glued along $\mathbb R\setminus\{0\}$) does not count as a topological manifold, even while its algebraic analogue has long become a honorable member of the category of schemes. I will thus assume that the "correct" notion of orbifold is the one one obtains by pretending that all schemes are separated, and all étale maps are separated étale. One then gets the following notion. Consider the category of topological manifolds $\mathrm{Man}$, and make it into a site by using open covers as covers. Then an orbifold is a stack (=sheaf of groupoids) $X$ on $\mathrm{Man}$ such that there is a surjective separated étale map $Y\to X$ for some manifold $Y$. Here, a map $Y\to X$ of stacks on $\mathrm{Man}$ is separated étale if after pullback $X'\to X$ from any manifold $X'$, the fibre product $Y'=Y\times_X X'\to X'$, $Y'$ is a manifold, and $Y'\to X'$ is a local isomorphism. Here is a different way to proceed, starting from condensed groupoids. Recall that these are sheaves of groupoids on the site $\mathrm{ProFin}$ of profinite sets, with covers given by finite families of jointly surjective maps. Again, we first define separated étale maps $f: Y\to X$ of condensed groupoids: This means that after any pullback $S\to X$ to a profinite set $S$, the fibre product $Y\times_X S$ is representable by a locally profinite set, and the map to $S$ is a local isomorphism. Any manifold $M$ defines a condensed set $\underline{M}$ via sending $S$ to the set of continuous maps from $S$ to $M$, and this embeds manifolds fully faithfully into condensed sets. Now one can define an orbifold to be a condensed groupoid $X$ that admits a surjective separated étale map $\underline{M}\to X$ from the condensed set corresponding to a manifold $M$. We claim that these definitions are equivalent. First, there is a functor: Any orbifold $X$ in the condensed sense defines a sheaf of groupoids on $\mathrm{Man}$ by sending $M$ to the groupoid of maps $\underline{M}\to X$. Proposition. This functor is an equivalence of categories. Let me just explain the key step. Assume that $M$ is a topological manifold, and $f: Y\to \underline{M}$ is a separated étale map of condensed sets. Then $Y=\underline{N}$ for some manifold $N$, where $N\to M$ is a local isomorphism. To prove this, the idea is the following. Take any $y\in Y$ mapping to $m\in M$. We want to find a neighborhood of $y$ mapping isomorphically to a neighborhood of $m$. But for separated étale maps, sections always extend from closed subsets (like $\{m\}\subset M$) to small strict neighborhoods (and uniquely so, up to further shrinking the neighborhood). By descent, this can be checked after taking surjections from profinite sets to closed neighborhoods, and after pullback to profinite sets, the claim is easy to see. Let me also advertise that in Etale cohomology of diamonds, I'm very much proceeding along such lines to define various "geometric objects" in $p$-adic geometry akin to (pretty wild forms of) orbifolds, by essentially adopting the condensed point of view (but this paper predates the introduction of the "condensed" term). In that paper, the role of profinite sets is played by "strictly totally disconnected perfectoid spaces", which are essentially profinite sets of geometric points. The difference here is that there is not just one kind of geometric point, but rather one for any complete algebraically closed nonarchimedean field $C$ (of characteristic $p$). One then studies all geometric objects by regarding them as quotients of (disjoint unions of) such strictly totally disconnected perfectoid spaces. The analogue of the key step is Lemma 15.6 (or really Lemma 11.31) there, comparing étale maps defined via descent to strictly totally disconnected spaces to étale maps defined more geometrically. That it was possible to develop the whole machinery from this point of view of descent to strictly totally disconnected spaces is what convinced me personally that it's not a completely crazy idea to regard manifolds as quotients of locally profinite sets.<|endoftext|> TITLE: Reference for the Brown representability theorem in the case of locally presentable (∞,1)-categories QUESTION [10 upvotes]: The abstract Brown representability theorem, proved by Brown in 1964 (a missing assumption of a cogroup structure on generators was added later), states that a contravariant functor F from a category C satisfying certain conditions to the category of sets is representable if and only if it sends coproducts to products and weak pushouts to weak pullbacks. This formulation lifts to the setting of (∞,1)-categories, as exemplified by model categories (Jardine) and quasicategories (Lurie, Theorem 1.4.1.2 in Higher Algebra). (In these structured settings, weak pushouts are replaced by (∞,1)-pushouts.) There is also a version for the special case of triangulated categories (Neeman, Theorem 3.1). All these formulations require C to be compactly generated. Since more than one statement can be found in the literature, here is the precise version I am interested in (Theorem 1.4.1.2 in Higher Algebra): if C is a locally presentable (∞,1)-category with a set of compact objects that admit a cogroup structure in Ho(C) and generate C under homotopy colimits, then a functor F: Ho(C)^op→Set is representable if and only if it sends coproducts in Ho(C) to products and (∞,1)-pushouts in C to weak pullbacks of sets. Looking at the proofs, it appears to me that the theorem should continue to hold in the case of locally α-presentable (∞,1)-categories (not necessarily compactly generated) for any regular cardinal α, provided the functor F satisfies one additional condition: for any α-filtered category I (taking I to be an ordinal should suffice) and any diagram D: I→C, the induced map F(colim D) → lim F∘D should be a surjective map of sets. That is, F should send α-filtered (∞,1)-colimits in C to weak α-cofiltered limits. I haven't checked all the details carefully, but if this is the case, is this written up anywhere? What is a citeable reference for the Brown representability theorem in the case of locally presentable (∞,1)-categories? A question of secondary importance is whether one can give a simple counterexample that shows the above condition involving α-filtered (co)limits to be necessary. The specific (∞,1)-category C that I am interested in is the category of (∞,1)-sheaves of spectra (or pointed spaces) on the site of smooth manifolds. This (∞,1)-category is not compactly generated. REPLY [10 votes]: The theorem is, in fact, false even for finitely locally presentable $\infty$-categories, and indeed for the $\infty$-category of spaces, as has been known since Heller's paper On the Representability of Homotopy Functors, Journal of the London Mathematical Society (2) 23 (1981) pp. 551-562, doi:10.1112/jlms/s2-23.3.551. Heller's counterexample relies on the existence of a group admitting a unique conjugacy class of injections from every group in any given small set; of course no such group exists handling all groups at once. The basic problem is that the homotopy category of spaces has no small set of objects detecting isomorphisms, a critical ingredient in Brown's original work as well as all its more specialized successors. (You need a small set of types of cells to construct your representing complex out of.) Arguably, the whole historical focus of algebraic topology (and of the study of Brown representability in particular) on pointed connected spaces, which are $\infty$-categorically not the objects of interest at all, comes down to the need to have a set of objects, in this case the spheres, detecting isomorphisms. As for your question, your formulation is more complex than needed: the only $\alpha$-filtered colimits you care about are weak ones in the homotopy category that you'll use to build up your representing cell complex out of generators, so the correct hypothesis, as it has always been since Brown, is just the preservation of coproducts and weak pushouts in the homotopy category. But you'll need the generators to appear $\alpha$-compact with respect to these weak colimits, and if $\alpha$ is uncountable then your homotopy category does not admit such constructions, see my preprint Detecting isomorphisms in the homotopy category with Christensen. EDIT: Here is an explicit counterexample to the "only if" direction of the proposed theorem in a locally presentable $\infty$-category. (Note: there are no counterexamples to the "if" direction, but there are also no known examples of functors satisfying its hypotheses, even weakened to consider only sequences, with $\alpha$ uncountable.) Generally, it suffices to find an object $Z$ and an $\alpha$-indexed sequence $X_i$ together with a cocone under $(X_i)$ in the homotopy category with legs $f_i:X_i\to Z$ such that, for any choice of homotopies $H_{ij}:X_j\otimes \Delta^1\to Z$ between $f_i|_{X_j}$ and $f_j$, it is never possible to fill the induced maps $X_k\otimes \partial \Delta^1\to Z$ consisting of $H_{ik},H_{jk},$ and $H_{ij}|_{X_k}$. In this situation $(f_i)$ is a cocone in the homotopy category not inducing any map $\mathrm{colim} X_i\to Z$, which shows the functor represented by $Z$ does not send the $(\infty,1)$-colimit of the $\alpha$-indexed diagram $(X_i)$ to a weak colimit of sets. The above framework is valid in any locally presentable $\infty$-category $\mathcal C$, and heuristically one should get counterexamples in essentially any such $\infty$-category and for essentially any choices of $Z$ and $(X_i)$. Franke's argument (introduction of On the Brown representability theorem for triangulated categories, Topology 40 (2001) pp.667-680, doi:10.1016/S0040-9383(99)00034-8) for this heuristic in the stable situation was as follows: There is a spectral sequence converging to $\mathrm{Hom}_{\mathcal C}(\mathrm{colim} X_i,Z)$ whose second page contains $\lim^q H^p(\mathrm{Hom}_{\mathcal C}(X_i,Z))$, and the functor represented by $Z$ will only send $\mathrm{colim} X_i$ to a weak limit of sets if this spectral sequence collapses. Unfortunately, nobody seems to know how to compute these derived limits well enough to show that what intuitively ought to happen actually does happen-there are isolated explicit examples of ordinal-indexed sequences of abelian groups with nontrivial derived limits, but I don't know how to cook one into this spectral sequence. So the only place I know how to follow the recipe above is in unpointed spaces, and indeed in the homotopy category of groupoids, where I can compute stuff directly. Christensen and I analyze an appropriate space $Z$ in section 3.3 of our preprint. The idea is to start with $(X_i)$ itself, then construct $Z$ as a kind of unfinished homotopy colimit that has an incoherent family of homotopies between its canonical maps from the $X_i$. The utility of working with groupoids here is that one can combinatorially show that such a cocone, with intuitively shouldn't cohere into a map from $\mathrm{colim} X_i$, actually does not. In particular, this construction uses properties of the homotopy category of spaces that are totally orthogonal to the lack of generating cogroups, so should make one more confident that the expected phenomena would appear in the stable situation if we could wrangle the spectral sequence. While the $\alpha$-filtered colimit idea does not work, Brown representability can be proved efficiently for these categories. Use the fact that an $\alpha$-presentable $\infty$-category is reflective in a finitely presentable one, and steal Brown representability from there--it's easy to prove that a reflective subcategory of a category on which every functor preserving coproducts and weak pushouts is representable enjoys the same properties. When the resulting finitely presentable $\infty$-category is generated by a set of compact cogroups, you win. Unfortunately, as discussed above, the homotopy categories of finitely presentable $\infty$-categories do not usually have Brown representability available to steal. There is a way out: the spheres do detect isomorphisms in the homotopy 2-category of the $\infty$-category of spaces, and from there it follows that the whole Brown representability story goes through perfectly for all $\alpha$-presentable $\infty$-categories--the only cost is that you have to take functors valued in groupoids, not in sets. Your mileage may vary on how "citeable" this is, but see 5.3.6 in my 2020 thesis, Some 2-categorical aspects of $\infty$-category theory.<|endoftext|> TITLE: Where was it first shown that every homotopy self-equivalence of $S^1\times S^2$ is homotopic to a homeomorphism? QUESTION [10 upvotes]: The claim in the title is proved on pp.19-20 of Topological rigidity for non-aspherical manifolds by M. Kreck and W. Lueck. Is there an earlier (classical) reference? REPLY [13 votes]: My guess is that the oldest reference might be Pontryagin's 1941 paper on the homotopy classification of maps from a 3-dimensional complex to the 2-sphere, the English version of which is in Recueil Mathématique 51 pp. 331-359. The application to homotopy equivalences of $S^1\times S^2$ is in an example on page 356 at the end of Section 4 of the paper.<|endoftext|> TITLE: Bounding size of partial difference sets given size of partial sumsets QUESTION [6 upvotes]: In this paper by Katz and Tao, the following bounds were established. Let $A,B$ be finite subsets of an abelian group, with $|A|,|B|\le N$. We fix some $G \subset A\times B$. We define $C = \{a+b:(a,b) \in G\},D=\{a+2b:(a,b)\in G\}, X = \{a-b:(a,b)\in G\}$. If $|C|\le N$, then $|X| \le N^{2-1/6}$, and it is possible for $|X| \ge N^{\log(6)/\log(3)} = N^{2-0.369\dots}$ to hold. If $|C|\le N,|D|\le N$, then $|X| \le N^{2-1/4}$, and it is possible for $|X| \ge N^{2-1/2}$ to hold. I understand these problems were originally studied to better understand Besicovitch sets and Kakeya's conjecture. I am aware of the paper of Dvir which basically settles Kakeya's conjecture for finite fields, but to my understanding, this would not say anything about what can be deduced of $|X|$. I find this problem of bounding $|X|$ given $|C|$ and $|D|$ to be quite interesting in its own right, so has this problem been studied further? If so, what are the best known bounds? Update: I looked at this survey by Katz and Tao which was written shortly after the aforementioned paper. On page 8 of the survey, they talk about this problem of understanding partial sumsets and partial difference sets. They say: There is a substantial literature on the relative sizes of sum-sets $A_0 + A_1$ and difference sets $A_0 − A_1$ (see e.g. the excellent survey [17]), but much less is known about partial sum-sets and difference sets, when one only considers a subset $G$ of pairs $A_0 \times A_1$. It has been 20 years since this survey. Can anyone comment to whether the area of partial sumsets has been studied more since then? If not, I'd be interested to hear some speculation to why this is the case. REPLY [5 votes]: Some improvements to the lower bounds appear in Lemm, Marius, New counterexamples for sums-differences, Proc. Am. Math. Soc. 143, No. 9, 3863-3868 (2015). ZBL1358.42006. In particular for 1, there are now examples with $|X| \geq N^{1.77898}$ and for 2 there are examples with $|X| \geq N^{1.61226}$. They build upon an earlier entropy-theoretic construction of Ruzsa (folklore, not sure if it was published previously to Lemm's paper) that improved on the lower bounds in my original paper with Nets, in particular producing examples to 1 with $|X| \geq N^{1.726}$. As far as I know there have been no improvement in the upper bounds (unless one also adds more slices to the hypothesis). The recent paper Green, Ben; Ruzsa, Imre Z., On the arithmetic Kakeya conjecture of Katz and Tao, Period. Math. Hung. 78, No. 2, 135-151 (2019). ZBL1438.42040. forms some further equivalent forms of the various problems here, and gives some further applications, but does not numerically improve the exponents for the upper and lower bounds. I would imagine these two papers between them would pretty much summarise the current state of the art. EDIT: There is also a "no-go" result at Katz, Nets Hawk, Elementary proofs and the sums differences problem, Collect. Math. 2006, Spec. Iss., 275-280 (2006). ZBL1213.42084. that shows that the known upper bound techniques can never improve the bound on $|X|$ below $N^{1.5}$ no matter how many slices one assumes to be bounded.<|endoftext|> TITLE: information-theoretic derivation of the prime number theorem QUESTION [18 upvotes]: Motivation: While going through a couple interesting papers on the Physics of the Riemann Hypothesis [1] and the Minimum Description Length Principle [2], a derivation(not a proof) of the Prime Number Theorem occurred to me. I thought I would share it here as I wonder whether other mathematicians have pursued this direction. It appears to have interesting implications although the arguments are elementary. An information-theoretic derivation of the prime number theorem: If we know nothing about the primes in the worst case we may assume that each prime number less than or equal to $N$ is drawn uniformly from $[1,N]$. So our source of primes is: \begin{equation} X \sim U([1,N]) \tag{1} \end{equation} where $H(X) = \ln(N)$ is the Shannon entropy of the uniform distribution. Now, given a strictly increasing integer sequence of length $N$, $U_N = \{u_i\}_{i=1}^N,$ where $u_i = i$ we may define the prime encoding of $U_N$ as the binary sequence $X_N = \{x_i\}_{i=1}^N$ where $x_i =1$ if $u_i$ is prime and $x_i=0$ otherwise. With no prior knowledge, given that each integer is either prime or not prime, we have $2^N$ possible prime encodings(i.e. arrangements of the primes) in $[1,N] \subset \mathbb{N}$. If there are $\pi(N)$ primes less than or equal to $N$ then the average number of bits per arrangement gives us the average amount of information gained from correctly identifying each prime number in $U_N$ as: \begin{equation} S_c = \frac{\log_2 (2^N)}{\pi(N)}= \frac{N}{\pi(N)} \tag{2} \end{equation} Furthermore, if we assume a maximum entropy distribution over the primes then we would expect that each prime is drawn from a uniform distribution as in (1) so we would have: \begin{equation} S_c = \frac{N}{\pi(N)} \sim \ln(N) \tag{3} \end{equation} As for why the natural logarithm appears in (3), we may first note that the base of the logarithm in the Shannon Entropy may be freely chosen without changing its properties. Moreover, given the assumptions if we define $(k,k+1] \subset [1,N]$ the average distance between consecutive primes is given by the sum of weighted distances $l$: \begin{equation} \sum_{k=1}^{N-1} \frac{1}{k} \lvert (k,k+1] \rvert = \sum_{k=1}^{N-1} \frac{1}{k} \approx \sum_{l=1}^\lambda l \cdot P_l \approx \ln(N) \tag{4} \end{equation} where $P_l = \frac{1}{l} \cdot \sum_{k= \frac{l \cdot (l-1)}{2}}^{\frac{l\cdot (l-1)}{2}+l-1} \frac{1}{k+1}$ and $\lambda = \frac{\sqrt{1+8(N+1)}-1}{2}$. This is consistent with the maximum entropy assumption in (1) as there are $k$ distinct ways to sample uniformly from $[1,k]$ and a frequency of $\frac{1}{k}$ associated with the event that a prime lies in $(k-1,k]$. The computation (4) is also consistent with Boltzmann's notion of entropy as a measure of possible arrangements. There is another useful interpretation of (4). If we break $\sum_{k=1}^{N} \frac{1}{k}$ into $\pi(N)$ disjoint blocks of size $[p_k,p_{k+1}]$ where $p_k,p_{k+1} \in \mathbb{P}$: \begin{equation} \sum_{k=1}^{N} \frac{1}{k} \approx \sum_{k=1}^{\pi(N)} \sum_{n=p_k}^{p_{k+1}} \frac{1}{n} = \sum_{k=1}^{\pi(N)} (p_{k+1}-p_k)\cdot P(p_k) \approx \ln(N) \tag{5} \end{equation} where $P(p_k)= \frac{1}{(p_{k+1}-p_k)} \sum_{b=p_k}^{p_{k+1}} \frac{1}{b}$. So we see that (4) also approximates the expected number of observations per prime where $P(p_k)$ may be interpreted as the probability of a successful observation in a frequentist sense. This is consistent with John Wheeler's it from bit interpretation of entropy [5], where the entropy measures the average number of bits(i.e. yes/no questions) per prime number. Now, we note that given (3), (4) and (5) we have: \begin{equation} \pi(N) \sim \frac{N}{\ln(N)} \tag{6} \end{equation} which happens to be equivalent to the prime number theorem. Discussion: This investigation has several consequences which may demystify the original assumptions. There are stronger motivations for maximum entropy distributions besides complete ignorance. By the Shannon source coding theorem, we may infer that $\pi(N)$ primes can't be compressed into fewer than $\pi(N) \cdot \ln(N)$ bits so this result tells us something about the incompressibility of the primes. Why might they be incompressible? By definition, all integers have non-trivial prime factorisations except for the prime numbers. In fact, there is a strong connection between maximum entropy distributions and incompressible signals. This connection may be clarified by the following insight: \begin{equation} \mathbb{E}[K(X_N)] \sim \pi(N) \cdot \ln(N) \sim N \tag{7} \end{equation} where the implicit assumption here is that for any recursive probability distribution, the expected value of the Kolmogorov Complexity equals the Shannon entropy. The distribution of the prime numbers is such a distribution as it is computable. Testable predictions: This model implies that independently of the amount of data and computational resources at their disposal, if the best machine learning model predicts the next $N$ primes to be at $\{a_i\}_{i=1}^N \in \mathbb{N}$ then for large $N$ their model's statistical performance will converge to an accuracy that is no better than: \begin{equation} \frac{1}{N}\sum_{i=1}^N \frac{1}{a_i} \tag{8} \end{equation} Moreover, it is not possible to prove that a particular object is incompressible within algorithmic information theory so the best we can do is perform rigorous experimental analysis using machine learning methods. There are in fact, two experimentally verifiable propositions: Prime encodings are algorithmically random and so they would pass the next-bit test [6]. The true positive rate for any computable primality test with bounded algorithmic information content(i.e. bounded memory) converges to zero. So far I have empirically explored both hypotheses for all prime numbers under a million(i.e. $\pi(N = 10^6$)) and I have defined a specific machine learning challenge which may be used to assess both hypotheses using state-of-the-art machine learning methods. Question: Has this line of reasoning already been developed? Note: I wrote a few more thoughts on the subject on my blog. References: Dániel Schumayer and David A. W. Hutchinson. Physics of the Riemann Hypothesis. Arxiv. 2011. Peter D. Grünwald. The Minimum Description Length Principle . MIT Press. 2007. Olivier Rioul. This is IT: A Primer on Shannon’s Entropy and Information. Séminaire Poincaré. 2018. Don Zagier. Newman’s short proof of the Prime Number Theorem. The American Mathematical Monthly, Vol. 104, No. 8 (Oct., 1997), pp. 705-708 John A. Wheeler, 1990, "Information, physics, quantum: The search for links" in W. Zurek (ed.) Complexity, Entropy, and the Physics of Information. Redwood City, CA: Addison-Wesley. Andrew Chi-Chih Yao. Theory and applications of trapdoor functions. In Proceedings of the 23rd IEEE Symposium on Foundations of Computer Science, 1982. REPLY [15 votes]: You may be interested in this arxiv paper [1], "Some information-theoretic computations related to the distribution of prime numbers", Ioannis Kontoyiannis, 2007. It discusses Chebyshev's 1852 result, $$ \sum_{p \leq n} \frac{\log p}{p} \sim \log n , $$ which is related to the prime number theorem and used in proofs of it. The paper begins by sketching Billingsley's 1973 heuristic information-theoretic argument, then makes it rigorous. The heuristic is similar to yours: Suppose we uniquely represent a number $N = \prod_{p \leq N} p^{X_p}$. Then we pick $N$ uniformly at random in $\{1,\dots,n\}$, with an entropy of $\log(n)$. The information contained in the $\{X_p\}$ variables in the same; so that collection has the same entropy. Then there is an argument that the $\{X_p\}$ are approximately geometrically distributed. This winds up giving the left side. (Note we could take all the logs to any base without changing the claim, since the change cancels out on both sides.) [1] https://arxiv.org/abs/0710.4076<|endoftext|> TITLE: Roots in Thompson's groups QUESTION [7 upvotes]: Let $G \in \{F,T,V\}$, and let $g$ be an element of Thompson's $G$. Question 1. When does $g$ have an $n$th root? If this is a complicated issue, and a description is not so easy to give, I am especially interested in the following more specific questions (for any of the $G$). Question 2. Given $(g,n)$, is it decidable whether $g$ has an $n$th root? Question 3. Can an element $g$ have roots of arbitrarily large orders? For $G = F$, if $g \neq 1_F$, looking at the slope at the leftmost point of $g$'s support we get an upper bound on the possible $n$'s, so Question 3 has a negative answer for this one. For $T$ and $V$ I don't see any trivial argument. I feel like Question 2 should have an easy "yes" answer, but I don't know how to prove that. REPLY [2 votes]: More or less to add to Jim's answer: The strand diagram material is equivalent to Brin's Revealing Pair technology (introduced in his paper Brin's Higher Dimensional Thompson groups where he introduced $2V$). We use this revealing pair technology in fashion very similar to Jim's description for $F$ elements above to describe roots of elements in $V$ in (mostly Section 7) of our paper "Centralizers in the R. Thompson group $V_n$" published in GGD in 2013 (see Centralizer's in $V$). The main point is that the centralizer of a non-torsion element is virtually $Z$ over the region of the support of the element in Cantor space. So, there is a finite set of "rootiest roots" supported there. If our element $g$ has root $r$ in this set, and $r^k=g$, then away from the support of $r$, one could put an infinitude of torsion elements of order dividing $k$. All of these created elements are also "roots" of $g$, but in a stupid way. Ewa Bieniecka in her thesis developed and generalized the revealing pair technology further, and answered the other direction of the results characterized in our centralizer's paper. I think this is a very gentle resource for learning flow graphs and revealing pairs (the crucial technologies).<|endoftext|> TITLE: The Kronecker--Hurwitz property for rings of integers in global function fields QUESTION [5 upvotes]: In Ireland and Rosen's book on number theory they give a proof of the finiteness of the class group of a number field which they attribute to Hurwitz, but which is essentially due to Kronecker (as I learnt from a comment by Franz Lemmermeyer to this answer by KConrad). The proof is based on showing that the ring of integers in a number field satisfies the following property which I here formulate for an arbitrary integral domain. Let's call an integral domain $R$ a Kronecker--Hurwitz domain if there exists a function $f:R\rightarrow\mathbb{Z}_{\geq0}$ and a finite subset $T\subset R\setminus\{0\}$ such that $f(\alpha)=0$ if and only if $\alpha=0$; for any $\alpha,\beta\in R$, $\beta\neq0$, there is a $t\in T$ and an $\omega\in R$ such that $f(t\alpha-\omega\beta)2$; then $N(\{h\gamma\}_{1})=N(\{l\gamma\}_{1})=1/q=\frac{a_{i}+1}{m}$, but $N(\{h\gamma\}_{1}-\{l\gamma\}_{1})=N(-1/t)=1/q>1/m$). REPLY [3 votes]: Let us take $f$ to be the norm (i.e. view $R$ as a finite rank module over $\mathbb F_q[t]$. Each element of $R$ acts by multiplication on this module, so its determinant lies in $\mathbb F_q[t]$. Take $f$ to be $q$ to the degree. Then it suffices to prove for $\gamma$ in the field of fractions of $R$ that there is $c \in T$ and $\omega \in R$ with $f( c\gamma - \omega)< 1$, i.e. with the degree of the determinant of $c \gamma -\omega$ negative. We can generalize this to allow $\gamma \in R \otimes_{ \mathbb F_q[t] } \mathbb F_q((t^{-1}))$, where $\mathbb F_q((t^{-1}))$ is the field of formal Laurent series. The norm makes sense for elements of this ring for the same reason - it's a free module over $\mathbb F_q((t^{-1}))$, so we can take determinants, and then look at the degree in $t$ of the leading term. The advantage of this generalization is that $(R \otimes_{ \mathbb F_q[t] } \mathbb F_q((t^{-1})) ) / R = ( \mathbb F_q((t^{-1}))/\mathbb F_q[t])^n$ is compact, and since for each $c\in R$, the set of $\gamma$ such that there exists $\omega \in R$ with $f( c\gamma -\omega)<1$ is open and invariant under translation by $R$, it suffices to check that for each $\gamma \in R \otimes_{ \mathbb F_q[t] } \mathbb F_q((t^{-1})) $ there exists $c,\omega \in R$ with $f( c\gamma - \omega)<1$, since this gives an open cover indexed by $c$ and we can find a finite subcover. Given $\gamma$, express multiplication by $\gamma$ as an $n\times n$ matrix over $\mathbb F_q((t^{-1}))$ and let $d$ be the greatest degree in $t$ of an element of that matrix. Then for $c \in R$ whose coordinates are polynomials of degree $\leq N$, and $\omega \in R$ whose coordinates are polynomials of degree $\leq N+d$, $c\gamma-\omega$ has coordinates in $\mathbb F_q((t^{-1}))$ of degree $\leq N+d$. The map $(c, \omega) \to c \gamma - \omega \mod t^{-N} \mathbb F_q[[t^{-1}]]$, where $c$ has coordinates of degree $\leq N$ and $\omega$ has coordinates of degree $\leq N+d$, is a map from an $n(N+1) + n (N+d+1)$-dimensional vector space to an $n ( 2N+d)$-dimensional vector space, and thus has nontrivial kernel, so there exist $c, \omega$ in $R$ with $c\gamma - \omega$ having all coordinates of degree $\leq -N$. Taking $N$ sufficiently large depending on the coefficients of the polynomial expressing of the determinant, we get $c \gamma -\omega$ of small norm.<|endoftext|> TITLE: If a compact convex set meets the positive orthant does it meet it at a point with a normal in the positive orthant? QUESTION [6 upvotes]: Let $C$ be a compact convex set in $\mathbb R^n$ that intersects the strictly positive orthant $\mathbb R_+^n$. Does $C\cap \mathbb R_+^n$ have to contain a point $x$ such that some vector $v\in\mathbb R_+^n$ is normal to $C$ at $x$? Here, a vector $v$ is normal to a convex set $C$ at $x$ iff for all $y\in C$, $\langle v,y \rangle\le \langle v,x\rangle$. REPLY [3 votes]: While fedja gave an answer in the comments, here is a different approach that yields a more general answer (the non-negative orthant is self-dual). Proposition: Let $C\subseteq \mathbb R^n$ be a compact convex set that meets some closed convex cone $K$. Then $C\cap K$ contains a point with a non-zero normal in the dual of $K$. If $C$ meets the interior of $K$ and the dual of $K$ has non-empty interior, then the point can be taken to have a non-zero normal in the interior of the dual of $K$. Proof: Suppose $K=\{ z : \left< x,z \right> \ge 0 \text{ for all }x\in A \}$ for a countable set $A$ (this works for any separable Banach space). Let $B$ be the basis of the largest linear space contained in $K$. Thus if $\left< x,z \right> = 0$ for all $x\in A\cup B$, then $z=0$. Enumerate $A\cup B=\{a_1,a_2,\dots\}$. Define the sequence of non-empty compact sets $C_i$ by letting $C_0 = C\cap K$ and $C_i$ be the set of points of $C_{i-1}$ that maximize $\left< a_i, \cdot \right>$. Fix $z\in\bigcup_i C_i$. I claim that $C\cap (z+K) = \{ z \}$. For suppose $w \in K$ and $z+w \in C\cap (z+K)$. The maximality conditions imply that $\left< a_i, w \right>\le 0$ for all $i$. Hence $w=0$. Let $H$ be a separating closed half-space with $z$ on its boundary that contains $C$ and the closure of whose complement contains $z+K$. The outward normal of $H$ is then in the dual of $K$. Now suppose $C$ meets the interior of $K$ at some point $z_0$ and $K^*$ has non-empty interior. Let $K_\epsilon$ be an $\epsilon$-widened closed convex cone, e.g., the cone generated by the compact set $(K\cap H)+\bar B(0;\epsilon)$ (Minkowski sum) where $H$ is a hyperplane through a point in the interior of $K^*$ and normal to the vector from the origin to that point. Then for sufficiently small $\epsilon>0$, $C\cap (z_0+K_\epsilon) \subseteq C\cap \operatorname{Int} K$. Applying the already proved part to $-z_0+C$ and $K_\epsilon$, we get a point $z$ in $C$ with a non-zero normal in $(K_\epsilon)^*$. But $\operatorname{Int} (K^*) \supseteq (K_\epsilon)^* \backslash \{ 0 \}$.<|endoftext|> TITLE: Decoupling a double integral QUESTION [17 upvotes]: I came across this question while making some calculations. QUESTION. Can you find some transformation to "decouple" the double integral as follows? $$\int_0^{\frac{\pi}2}\int_0^{\frac{\pi}2}\frac{d\alpha\,d\beta}{\sqrt{1-\sin^2\alpha\sin^2\beta}} =\frac14\int_0^{\frac{\pi}2}\frac{d\theta}{\sqrt{\cos\theta\,\sin\theta}}\int_0^{\frac{\pi}2}\frac{d\omega}{\sqrt{\cos\omega\,\sin\omega}}.$$ REPLY [10 votes]: (Thanks go to Etanche and Jandri) \begin{align}J&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-\sin^2(\theta)\sin^2 \varphi}}d\varphi d\theta\\ &\overset{z\left(\varphi\right)=\arcsin\left(\sin(\theta)\sin \varphi\right)}=\int_0^{\frac{\pi}{2}} \left(\int_0^ \theta\frac{1}{\sqrt{\sin(\theta-z)\sin(\theta+ z)}}dz\right)d\theta\tag1\\ &=\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_{u}^{\pi-u}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du \tag2\\ &=\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_{u}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du+\underbrace{\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_{\frac{\pi}{2}}^{\pi-u}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du}_{w=\pi-v}\\ &=\int_0^{\frac{\pi}{2}} \left(\int_{u}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du\\ &\overset{u\longleftrightarrow v}=\int_0^{\frac{\pi}{2}} \left(\int_{0}^{u}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du\\ &=\boxed{\frac{1}{2}\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dudv} \end{align} and to obtain the form in the OP, finally substitute $u=2\theta $, $v=2\omega $. $(1)$: $\displaystyle dz=\dfrac{\sqrt{\sin^2\theta-\sin^2 z}}{\sqrt{1-\sin^2 z}}d\varphi$, $\sin^2 a-\sin^2 b=\sin(a-b)\sin(a+b)$ $(2)$: Change of variable $u=\theta-z,v=\theta+z$<|endoftext|> TITLE: Status of a conjecture of Hirzebruch QUESTION [9 upvotes]: I was reading a paper from 1994 which claimed that the following statement was a conjecture of Hirzerbruch: If a complex surface X is homeomorphic to either $S^2 \times S^2$ or $\mathbb{C}P^2 \# \overline{\mathbb{C}P^2}$ then it is biholomorphic to one of the Hirzerbruch surfaces. I would like to know if this conjecture has been resolved or what the current status about it. Also, I would like to know the state of the art about the variant of the conjecture in which "homeomorphic" is replaced by "diffeomorphic": If a complex surface X is diffeomorphic to either $S^2 \times S^2$ or $\mathbb{C}P^2 \# \overline{\mathbb{C}P^2}$ then is it biholomorphic to one of the Hirzerbruch surfaces? If this has been proved, what is a reference? REPLY [10 votes]: Suppose $X$ is diffeomorphic to $S^2\times S^2$ or $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$. Then $X$ is biholomorphic to a Hirzebruch surface. Note that $b_1(X) = 0$, so $X$ admits a Kähler metric. It follows from Seiberg-Witten theory that any compact Kähler surface with non-negative Kodaira dimension does not admit a metric of positive scalar curvature, but $X$ does, so $X$ has Kodaira dimension $-\infty$. By the classification of surfaces, $X$ is either rational (birational to $\mathbb{CP}^2$) or a ruled surface over a curve of genus $g \geq 1$. The latter possibility cannot occur as $X$ is simply connected, so $X$ is rational and hence biholomorphic to $\mathbb{CP}^2$ or a Hirzebruch surface; see Proposition VI.3.3 of Compact Complex Surfaces (second edition) by Barth, Hulek, Peters, and Van de Ven. As $b_2(X) = 2 \neq 1 = b_2(\mathbb{CP}^2)$, we conclude that $X$ is biholomorphic to a Hirzebruch surface. If $X$ is only homeomorphic to $S^2\times S^2$ or $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$, then I believe it is still not known whether $X$ must be biholomorphic to a Hirzebruch surface. As before, $X$ must admit a Kähler metric, but $X$ need not admit a metric of positive scalar curvature a priori, so we cannot immediately determine the Kodaira dimension of $X$ as we did above. Note, if we could show that $X$ had Kodaira dimension $-\infty$, we could conclude that $X$ is biholomorphic to a Hirzebruch surface by the exact same argument. We can however use classification to show that the Kodaira dimension of $X$ is not zero or one. We can rule out Kodaira dimension zero because any Kähler surface of Kodaira dimension zero is finitely covered by a blownup torus or blownup $K3$ surface, which is not the case for $X$. The impossibility of Kodaira dimension one follows from the fact that $c_1^2 = 0$ for minimal such surfaces, so $c_1^2(X) \in \{0, -1\}$, and $c_2(X) = \chi(X) = 4$, so $\operatorname{Td}(X) = \frac{1}{12}(c_1^2(X) + c_2(X)) \not\in \mathbb{Z}$, which is impossible. Therefore, Hirzebruch's conjecture reduces to providing a negative answer to the following question: Is there a general type surface homeomorphic to $S^2\times S^2$ or $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$? It is natural to ask whether anything changes if $X$ is a compact complex surface which is only homotopy equivalent to $S^2\times S^2$ or $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$. As $X$ is simply connected, it follows from Freedman's Theorem that $X$ is actually homeomorphic to $S^2\times S^2$ or $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$, so we're again in the realm of Hirzebruch's conjecture.<|endoftext|> TITLE: What is known about the structure of finite groups admitting an automorphism where all elements have "norm" one? QUESTION [14 upvotes]: Let $G$ be a finite group admitting an automorphism $\sigma$ of prime order $p$. Define the norm map $N:G\rightarrow G$ with respect to $\sigma$ by $N(g)= g\sigma(g)\sigma^2(g)\dotsb\sigma^{p-1}(g)$. Question. If we know all elements of $G$ have norm equal to $e$, what can be said about the structure of the pair $(G,\sigma)$? In particular, must $G$ be nilpotent? For instance, if $p=2$, then $G$ must be abelian, but this need not hold for higher primes, e.g., an inner automorphism of a nonabelian $p$ group of exponent $p$. This condition on an automorphism is a natural weakening of being fixed point free, so its not unreasonable to hope that some of the structural features of that situation hold here. REPLY [8 votes]: First a note about terminology. It seems in the literature if $\sigma$ is an automorphism of $G$ such that $\sigma^n = 1$ and $g\sigma(g)\sigma^2(g) \cdots \sigma^{n-1}(g) = 1$ for all $g \in G$, we call $\sigma$ a splitting automorphism. After some searching (experts could probably give more detail), I found that the main result of Otto H. Kegel, Die Nilpotenz der $H_p$-gruppen, Math. Z. 75 (1961), 373–376 shows that if $G$ is a finite group with a splitting automorphism of prime order, then $G$ must be nilpotent. The proof by Kegel is based on results of Hughes and Thompson: D. R. Hughes and J. G. Thompson, The $H_p$-problem and the structure of $H_p$-groups, Pac. J. Math., 9, 1097–1101 (1959). For a finite group $G$ and prime $p$, let $H_p(G)$ be the subgroup generated by elements of order $\neq p$. Hughes and Thompson show that if $G$ is not a $p$-group, then either $H_p(G) = 1$, $H_p(G) = G$, or $[G:H_p(G)] = p$. In the case $[G:H_p(G)] = p$, conjugation by any element $g \in G$, $g \not\in H_p(G)$ induces a splitting automorphism of order $p$ on $H_p(G)$. Conversely, suppose that a group $H$ admits a splitting automorphism $\sigma$ of prime order $p$. Consider the subgroup $G = H \rtimes \langle \sigma \rangle$ of the holomorph of $H$. The definition of a splitting automorphism is precisely that the coset $H\sigma$ consists of elements of order $p$. Then for any integer $k$ coprime to $p$, the coset $H \sigma^k$ also consists of elements of order $p$. This follows from the fact that the map $x \mapsto x^k$ defines a bijection $H\sigma \rightarrow H \sigma^k$ which maps elements of order $p$ to elements of order $p$. Hence in this case any $g \in G$, $g \not\in H$ has order $p$, so $H_p(G) \leq H$. If $H$ is not a $p$-group, then $H_p(G) \neq 1$ and by the result of Hughes and Thompson we must have $H_p(G) = H$. So for finite groups which are not $p$-groups, the study of splitting automorphisms of order $p$ comes down to the study of what Hughes and Thompson call "$H_p$-groups", i.e. groups of the form $H_p(G)$, where $G$ is a finite group such that $[G:H_p(G)] = p$. This is seen as a generalization of Frobenius groups, which relate to fixed point free automorphisms of order $p$. Hughes and Thompson show that a $p$-Sylow of an $H_p$-group has a normal complement which is nilpotent, so in particular $H_p$-groups are solvable. In his paper Kegel continues from there, and shows that $H_p$-groups are nilpotent. I note the following about the relation between "fixed point free" and "splitting automorphism". An argument from (Gorenstein, Finite groups, Lemma 1.1 in Chapter 10): suppose that $\sigma$ is a fixed point free automorphism of $G$ of order $n$. Then the mapping $x \mapsto x^{-1}\sigma(x)$ is injective, hence also surjective. Writing $g \in G$ as $g = x^{-1}\sigma(x)$, we see that $g \sigma(g) \cdots \sigma^{n-1}(g) = e$. So $\sigma$ being fixed point free implies that $\sigma$ is a splitting automorphism. The converse fails. For $p = 2$, consider $G$ abelian of even order and not elementary abelian. Then $\sigma(g) = g^{-1}$ clearly defines a splitting automorphism of order $2$, but $\sigma$ is not fixed point free. For $p > 2$, consider $G = C_p \times C_p$ and the automorphism $\sigma$ corresponding to $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \in \operatorname{GL}_2(p)$. Then $A^p = I$ and $I + A + A^2 + \cdots + A^{p-1} = 0$. Thus $\sigma$ is a splitting automorphism of order $p$, but $\sigma$ is not fixed point free.<|endoftext|> TITLE: Probability of gaps between coordinates of a random point on the sphere QUESTION [5 upvotes]: Let $X=(X_1,\ldots,X_n)$ be a point chosen uniformly at random from the sphere $S^{n-1}\subseteq \mathbb R^n$. Given $a>0$, what is the probability that $|X_1|^2-|X_i|^2\geq a$ for all $i>1$? Is there a formula (explicit, asymptotic, etc.) for this probability in terms of $a$ and $n$? REPLY [6 votes]: $\newcommand\Z{\mathbf Z}\newcommand\ep\varepsilon\newcommand\tP{\tilde P}\newcommand\de\delta\newcommand\R{\mathbb R}$Note that the random point $\Z/|\Z|$ is uniformly distributed on the sphere $S^{n-1}$, where $\Z:=(Z_1,\dots,Z_n)$, $|\Z|:=\sqrt{Z_1^2+\dots+Z_n^2}$, and $Z,Z_1,\dots,Z_n$ are iid standard normal random variables. So, the probability in question is \begin{equation} P:=P(Z_i^2\le Z_1^2-a|\Z|^2\ \forall i\ge2). \end{equation} Let \begin{equation} a=\frac{c\ln n}n, \end{equation} where $c$ is a nonnegative real constant; we shall see that this is the right choice for $a$. Using your favorite exponential inequality, we see that eventually (that is, for all large enough $n$) \begin{equation} P(|\,|\Z|^2-n|>\sqrt n\,\ln n)\le\ep_n:=\exp\Big\{-\frac{\ln^2n}3\Big\}. \end{equation} So, \begin{equation} P=\tP+O(\ep_n), \end{equation} \begin{equation} \tP:=P(Z_i^2\le Z_1^2-b\ \forall i\ge2), \end{equation} where \begin{equation} b=na(1+\de_n)=(1+\de_n)c\ln n =c\ln n+o(1)\quad\text{and}\quad|\de_n|\le\frac{\sqrt n\,\ln n}n=o(1/\ln n). \end{equation} Next, conditioning on $Z_1$, we have \begin{equation} \tP=\int_\R dx\,f(x)(1-2G(\sqrt{(x^2-b)_+})^{n-1}, \end{equation} where $f$ is the standard normal pdf, $G(t):=P(Z>t)$, and $t_+:=\max(0,t)$. Consider here the substitution \begin{equation} e^{-u}=1-2G(\sqrt{(x^2-b)_+}), \end{equation} with $e^{-u}:=0$ if $x^2-b\le0$; note that $u\ge0$. For each real $h>0$, the integral of $f(x)(1-2G(\sqrt{(x^2-b)_+})^{n-1}=f(x)e^{-(n-1)u}$ over the set of $x>0$ such that $u>h$ is less than $e^{-(n-1)h}$. So, we can choose $h=h_n$ so that \begin{equation} h\downarrow0,\quad nh\to\infty, \end{equation} and \begin{equation} \int_{x>0\colon\, u>h} dx\,f(x)(1-2G(\sqrt{(x^2-b)_+})^{n-1} \le\ep_n \end{equation} eventually. On the other hand, if $u\le h$ and $x>0$, then $u=o(1)$ and hence \begin{equation} \sqrt{(x^2-b)_+}=\sqrt{x^2-b}=z:=G^{-1}\Big(\frac{1-e^{-u}}2\Big)\to\infty, \end{equation} \begin{equation} x=\sqrt{b+z^2}, \end{equation} \begin{equation} dx=-\frac{du\, z}{\sqrt{b+z^2}}\frac1{f(z)}\frac{e^{-u}}2. \end{equation} Next, $G(z)=\frac{1-e^{-u}}2\sim u/2$ and $G(z)=e^{-z^2/(2+o(1))}$, whence \begin{equation} z\sim\sqrt{2\ln\frac1u}, \end{equation} \begin{equation} f(z)\sim G(z)z\sim\frac u2\,\sqrt{2\ln\frac1u}, \end{equation} \begin{equation} dx\sim-\frac{du\,\sqrt{2\ln\frac1u}}{\sqrt{b+2\ln\frac1u}}\frac1{\frac u2\,\sqrt{2\ln\frac1u}}\frac{e^{-u}}2 =-\frac{du}{\sqrt{b+2\ln\frac1u}}\frac{e^{-u}}u. \end{equation} Further, \begin{equation} f(x)=e^{-b/2}f(z)\sim e^{-b/2}\frac u2\,\sqrt{2\ln\frac1u}. \end{equation} So, \begin{align*} & \int_{x>0\colon\, u\le h} dx\,f(x)(1-2G(\sqrt{(x^2-b)_+})^{n-1} \\ &\sim\int_0^h e^{-b/2}\frac u2\,\sqrt{2\ln\frac1u}\,\frac{du}{\sqrt{b+2\ln\frac1u}}\frac{e^{-u}}u e^{-(n-1)u} \\ &=\frac{e^{-b/2}}2\,\int_0^h du\,\frac{\sqrt{2\ln\frac1u}}{\sqrt{b+2\ln\frac1u}} e^{-nu} \\ &=\frac{e^{-b/2}}{2n}\,\int_0^{nh} dy\, e^{-y}\,\frac1{\sqrt{1+\frac b{2\ln\frac ny}}} \\ &=\frac{e^{-(c\ln n+o(1))/2}}{2n}\,\int_0^{nh} dy\, e^{-y}\,\frac1{\sqrt{1+\frac{c\ln n+o(1)}{2\ln\frac ny}}} \\ &\sim \frac{n^{-1-c/2}}2\,\frac1{\sqrt{1+c/2}} \end{align*} by dominated convergence. Collecting all the pieces and noting that the integrand is even in $x$, we conclude that for $a=\frac{c\ln n}n$ \begin{equation} P\sim \frac{n^{-1-c/2}}{\sqrt{1+c/2}} \end{equation} uniformly over all nonnegative $c$ in any bounded interval.<|endoftext|> TITLE: Definability of Gödel's pairing function on ordinals QUESTION [14 upvotes]: Given an infinite cardinal $\kappa$, Gödel's function is a well-known bijection $p:\kappa^2\to\kappa$. Is $p$ definable in the structure $\langle\kappa;\in\rangle$? Is $p$ definable in a bigger 2nd order structure $\langle\kappa;\mathcal P(\kappa);\in\rangle$? It looks like any typical attempt to code something like this (even + on ordinals) somehow refers to a pairing function. REPLY [11 votes]: For $\langle\kappa,{\in}\rangle$, James Hanson has already answered the question as stated. However, let me mention a generalization: it’s not just Gödel’s pairing function that’s not definable in $\langle\kappa,{\in}\rangle$—in fact, there is no injective pairing function definable in these structures. The reason for this is the computational complexity (or rather, easiness) of $\mathrm{Th}(\kappa,{\in})$. The results on quantifier elimination in [1] easily imply that if $\phi$ is an $\exists_n$ sentence, then $$\langle\kappa,{\in}\rangle\models\phi\iff\langle\omega^n,{\in}\rangle\models\phi.$$ Moreover, $\langle\omega^n,{\in}\rangle$ has an obvious $n$-dimensional interpretation in $\langle\omega,{\in}\rangle$. This provides a polynomial-time reduction of the theory of $\langle\kappa,{\in}\rangle$ to the theory of $\langle\omega,{\in}\rangle$, which is known to be decidable in PSPACE; therefore, Proposition: $\mathrm{Th}(\kappa,{\in})$ is decidable in PSPACE. On the other hand, [2] proved that no consistent theory with a definable pairing function is in ELEMENTARY (which includes PSPACE, and much, much more). Corollary: No pairing function is definable in $\langle\kappa,{\in}\rangle$. References: [1] John E. Doner, Andrzej Mostowski, Alfred Tarski: The elementary theory of well-ordering—a metamathematical study—, in: Logic Colloquium ’77 (A. Macintyre, L. Pacholski, J. Paris, eds.), Studies in Logic and the Foundations of Mathematics vol. 96, Elsevier, 1978, pp. 1–54, doi 10.1016/S0049-237X(08)71988-8. [2] Jeanne Ferrante, Charles W. Rackoff: The computational complexity of logical theories, Lecture Notes in Mathematics vol. 718, Springer-Verlag, 1979, Chapter 8, doi 10.1007/BFb0062837.<|endoftext|> TITLE: Typo in Stanley, Enumerative combinatorics II, Cor. 7.23.9? QUESTION [6 upvotes]: In Stanley, EC2, we have the following statement: I think there is a typo in the first sum after "generating function", and that $[n]_q!$ should be replaced by $(1-q)(1-q^2)\dotsb (1-q^n)$, and same for $[n]_t!$. The problem with the proof seems to be the line which is incorrect(?) as the denominator on the right hand side should be not the q/t-factorials, but the above product, according to the lemma which is referenced there. This issue (if my suspicions hold), is not listed in the Errata.. I have also checked this in Mathematica, so I am fairly confident this is a typo. REPLY [5 votes]: (In order that this question appears as answered I'm converting Richard's comment into an answer.) The (somewhat nonstandard) convention in the text, stated just before the corollary, is that $[n]!_q = (1-q)(1-q^2)\cdots(1-q^n)$. For the quantity $[n]!_q/(1-q)^n$, the notation $\mathbf{(n)!}$ is used instead.<|endoftext|> TITLE: What is Barr-Beck? QUESTION [22 upvotes]: This is a question about a naming convention. The Barr-Beck theorem (or simply Barr-Beck) is used a lot in descent theory over the past 30 years, almost invariably without a reference, like folklore. To make precise which theorem I am talking about: According to one source the Barr-Beck monadicity theorem gives necessary and sufficient conditions for a category to be equivalent to a category of algebras over a monad. There are occasionally references, e.g., to the Wikipedia page on "Beck's monadicity theorem" which attributes the theorem to Beck's thesis in 1967 under the direction of Eilenberg. This makes me wonder how Barr is related to Barr-Beck. So I did some further research and found out that there is indeed a Barr-Beck paper with a Barr-Beck theorem. But it is about triple cohomology of algebras. This seems to be different. So here are my questions: Is the Barr-Beck theorem different from Beck's theorem? If so, how? If not, how did Barr's name get attached to it? By the way, the second Beck is Jon Beck while the first is Jonathan Mock Beck with a different author number in MathSciNet. So out of curiosity: Are Jon and Jonathan Beck one and the same person? REPLY [7 votes]: It looks like Alexander's answer hits the nail on the head, but I thought I might add what Barr himself says. Barr and Wells write in Toposes, Triples, and Theories (on page 121 in the "Historical Notes on Triples" section): The next important step was the tripleableness theorem of Beck’s which in part was a generalization of Linton’s results. Variations on that theorem followed (Duskin [1969], Paré [1971]) and acquired arcane names, but they all go back to Beck and Linton. They mostly arose either because of the failure of tripleableness to be transitive or because of certain special conditions. The "arcane names" appears to be a reference to the "vulgar tripleability theorem" and the "crude tripleability theorem" which appear on page 108.<|endoftext|> TITLE: Is there a uniform solution of the Ruziewicz problem? QUESTION [9 upvotes]: For any integer $n\geq 2$ there is one and only one (up to rescaling) rotation-invariant, finitely-additive measure on the Lebesgue $\sigma$-algebra of $S^n$. The proof of this statement I'm aware of treats two cases separately: $n\geq 4$ and $n=2, 3$ (the latter in a bizarre turns of events ends up relying on the Ramanujan conjecture). I wonder if there is an argument that is uniform in $n$. REPLY [9 votes]: First, you don't need the full Ramanujan Conjecture: property $(\tau)$ is enough, so in that sense you can get a uniform proof (it's just that for $n\geq 4$ we have also property (T), which is easier to prove and stronger). Second, you can in fact induce property $(\tau)$ up the ladder (see Burger--Sarnak). Thirdly, his book [1] Sarnak makes this explicit, showing how to convert the spectral solution for the problem for $O(n)$ to a solution for $O(n+1)$. [1] Sarnak, Peter, Some applications of modular forms, Cambridge Tracts in Mathematics, 99. Cambridge: Cambridge University Press. x, 111 p. \textsterling 17.50; $ 29.95 (1990). ZBL0721.11015.<|endoftext|> TITLE: Explicit $BP_*BP$-comodule structure on $BP_*\mathbb{C}P^n$ and $BP_*\mathbb{C}P^{\infty}$ QUESTION [6 upvotes]: So as it says in the title, how can one explicitly calculate the comodule structures on $BP_*\mathbb{C}P^n$ and $BP_*\mathbb{C}P^{\infty}$ for a prime $p$? For example, $\mathbb{C}P^2$ sits in a cofiber sequence of spectra $$\Sigma^2\mathbb{S}\to \mathbb{C}P^2\to \Sigma^4\mathbb{S} $$ giving rise to a SES of $BP_*BP$-comodules $$0\to\Sigma^2BP_*\to BP_*\mathbb{C}P^2\to\Sigma^4 BP_*\to 0 $$ which defines an element in $Ext_{BP_*BP}^{1,2}(BP_*,BP_*)\cong \mathbb{Z}/2$ (this is for $p=2$). The comodule structure on $BP_*(\Sigma^2\mathbb{S}\vee\Sigma^4\mathbb{S})$ is trivial, so the comodule structure on $BP_*\mathbb{C}P^2$ should be non-trivial. I tried calculating explicitly what the comodule map should be using properties of $BP_*BP$ and the counital condition for comodule maps to get that $\Psi:BP_*\mathbb{C}P^2\to BP_*BP\otimes_{BP_*}BP_*\mathbb{C}P^2$ is given on $BP_*$-module generators by $$\Psi(g_1) = 1\otimes g_1, \Psi(g_2)= 1\otimes g_2 + t_1\otimes g_1. $$ (Here, $BP_*\mathbb{C}P^2\cong\Sigma^2BP_*\{g_1\}\oplus \Sigma^4BP_*\{g_2\}$ as $BP_*$-modules and $BP_*BP\cong BP_*[t_1, t_2,\ldots]$ with $\vert t_i\vert=2p^i-2$). I could very well be making a mistake though. I am still gaining familiarity with these $BP_*BP$ calculations. As $n$ gets bigger, doing it explicitly like this will get a lot more time-consuming it seems. Are there other ways to realize the comodule structure on $BP_*\mathbb{C}P^n$ and, more generally, on $BP_*\mathbb{C}P^{\infty}$? Has this been figured out by anyone else? If so, a reference if it exists would be great! Thanks! REPLY [9 votes]: A concise formula $$ \mu(\beta) = \beta(c(t^F)) $$ for the $BP_* BP$-coaction $\mu$ on $BP_* CP^\infty$ is given in the "Note added in proof" on page 279 of Ravenel, Douglas C.; Wilson, W. Stephen The Hopf ring for complex cobordism. J. Pure Appl. Algebra 9 (1976/77), no. 3, 241–280. A more explicit formula is given in Theorem 3.13 of Wilson, W. Stephen Brown-Peterson homology: an introduction and sampler. CBMS Regional Conference Series in Mathematics, 48. (1982) where the result is deduced from the $MU$ case, given in Theorem 1.48(e).<|endoftext|> TITLE: Character sums concerning $a^x-1$ QUESTION [7 upvotes]: Let $p$ be a prime number and $\mathbb{F}_p$ be a finite field with $p$ elements. Let $\chi$ be a multiplicative character from $\mathbb{F}_p^{\times}$ to $\mathbb{C}$, where $\mathbb{F}_p^{\times}=\{x\in\mathbb{F}_p: x\ne 0\}$. Recently, I met the following sum: $$\sum_{x=0}^{p-1}\chi(a^x-1),$$ where $a>1$ is a positive integer with $p\nmid a$. Question. Does anybody know some references concerning sums of this form? REPLY [7 votes]: The best result on the market is the one of Yu, who proves that your sum is less than $$p^{1/2} \left( \frac 2 \pi \log p + \frac 7 5 \right).$$ If $a$ is not a primitive root there is an adjustment by a factor which depends on the index of $a$. Also see Dobrowolski&Williams for essentially the same bound in the special case of Legendre symbols. Yu also proves the same bound for prime power moduli. See moreover Shparlinski for non-primitive characters, where there is an extra factor depending on the conductor of $\chi$; again from Yu, the displayed bound is in this case best possible up to a $\log q$ factor. Shparlinski and others comment that much less is studied for this kind of sums, involving multiplicative characters, than for the analogues with additive characters, especially for short sums. REPLY [4 votes]: Shparlinski has done a lot of work on problems like this, in the more general setting of linear recurrence sequences. (Your sequence of Mersenne numbers is such a sequence). I'd recommend looking at Chapter 5 of Graham Everest, Alf van der Poorten, Igor Shparlinski, Thomas Ward - Recurrence Sequences and the references therein (see e.g. p 86).<|endoftext|> TITLE: Is each squared finite group trivial? QUESTION [48 upvotes]: A semigroup $S$ is defined to be squared if there exists a subset $A\subseteq S$ such that the function $A\times A\to S$, $(x,y)\mapsto xy$, is bijective. Problem: Is each squared finite group trivial? Remarks (corrected in an Edit). I learned this problem from my former Ph.D. student Volodymyr Gavrylkiv. It can be shown that a group with two generators $a,b$ and relation $a^2=1$ is squared. So the adjective finite is essential in the above problem. Computer calculations show that no group of order $<64$ is squared. For any set $X$ the rectangular semigroup $S=X\times X$ endowed with the binary operation $(x,y)*(a,b)=(x,b)$ is squared. This follows from the observation that for the diagonal $D=\{(x,x):x\in X\}$ of $X\times X$, the map $D\times D\to S$, $(x,y)\mapsto xy$, is bijective. So restriction to groups in the formulation of the Problem is also essential. REPLY [25 votes]: I think, as implicitly suggested by Yemon Choi, it is possible to explain the proof of the answer of user49822 by making more use of idempotents. Suppose that the finite group $G$ is squared via the subset $A$. The element $ e = \frac{1}{|G|}\sum_{g \in G} g$ is a primitive idempotent of $\mathbb{C}G.$ Let $ f= \frac{1}{|A|}\sum_{a \in A} a.$ Then we have $f^{2} = ef = fe = e = e^{2}$. Thus $(f-e)^{2} = 0 = e(f-e) = (f-e)e.$ Now $f = e +(f-e)$ is the sum of commuting matrices (in the regular representation of $\mathbb{C}G$, say) with the second matrix nilpotent. Thus $f$ has trace $1$ in the regular representation of $\mathbb{C}G.$ This forces $1 \in A$ since all non-identitiy elements of $G$ have trace zero in the regular representation. But then $A = \{1 \}$, since (as in Jeremy Rickard's comment) if $a \neq 1 \in A$, then $a = 1a = a1$ gives two different expressions for $a$. Alternatively, (using traces) the fact that $1$ appears with coefficient $|G|^{-1}$ in $e$ tells us that $1$ appears with multiplicity $|G|^{-1}$ in $f$ as well, so that $\sqrt{|G|} = |A| = |G|$ and $|G| = 1.$<|endoftext|> TITLE: Non-uniqueness of $C$ with $f_!(C) = f_*(1_{\mathcal{C}})$ QUESTION [5 upvotes]: $\newcommand{\Cc}{\mathcal{C}}$ $\newcommand{\Dd}{\mathcal{D}}$ $\newcommand{\Z}{\mathbb{Z}}$ $\newcommand{\Q}{\mathbb{Q}}$ $\newcommand{\tensor}{\otimes}$ $\newcommand{\colim}{\rm colim}$ $\newcommand{\Sp}{Sp}$ $\newcommand{\iHom}{\underline{\rm Hom}}$ (This is the follow-up of this question. I have repeated the motivation for the question.) The following situation frequently arises in abstract category theory: we have symmetric closed monoidal categories $\Cc$ and $\Dd$ and a closed symmetric monoidal functor $f^*: \Dd \to \Cc$ that admits both a left adjoint $f_!: \Cc \to \Dd$ and a right adjoint $f_*: \Cc \to \Dd$, and we are interested in the relation between $f_!$ and $f_*$. This is often called a Wirthmüller context, named after the inspiring example of the Wirthmüller isomorphism in equivariant homotopy theory. (The symbol $f^*$ is just notation: there is not necessarily a morphism $f$ around.) To compare $f_!$ and $f_*$, Fausk-Hu-May [FHM] assume in the article isomorphisms between left and right adjoints the existence of an object $C \in \Cc$ satisfying $f_!(C) \simeq f_*(1_{\Cc})$, where $1_{\Cc}$ denotes the monoidal unit of $\Cc$. From this data, a natural transformation $f_*(-) \to f_!(- \tensor C)$ is constructed and conditions are given for when this is an isomorphism (the 'formal Wirthmüller isomorphism'). As remarked by Balmer-Dell'Ambrogio-Sanders [BDS] in the article Grothendieck-Neeman duality and the Wirthmüller isomorphism, the object $C$ is a priori not unique and it doesn't come with a `characterizing description'. My main question is: Question (1): Do we have counterexamples for essentially uniqueness of an object $C$ satisfying $f_!(C) = f_*(1_{\mathcal{C}})$? I actually want to follow [BDS] and stay in a more restricted setting. We assume that $\Cc$ and $\Dd$ are tensor-triangulated (or, if you prefer, stably symmetric monoidal stable $\infty$-categories) and that they are generated by a set of compact objects and admit internal homs (or equivalently: the tensor products commute with coproducts in both variables). Moreover, we assume that $\Cc$ and $\Dd$ are rigid, i.e. the compact objects are the same as the strongly dualizable objects. Finally, we assume that $f^*: \Dd \to \Cc$ is exact and strong monoidal and that it admits both a left adjoint $f_!: \Cc \to \Dd$ and a right adjoint $f_*: \Cc \to \Dd$. Examples of rigid tensor triangulated categories are spectra $\operatorname{Sp}$, genuine $G$-spectra $\operatorname{Sp}^G$ for a compact Lie group $G$ and the derived category $D(R)$ for a ring $R$. Several more advanced examples of such categories and functors are in examples 4.6 - 4.8 in the article Grothendieck-Neeman duality and the Wirthmüller isomorphism by Balmer-Dell'Ambrogio-Sanders, based on 'finite group schemes', 'motivic homotopy theory' and 'cohomology rings of classifying spaces'. However, I don't really understand them and in particular I don't know if they give counterexamples to question (1) above. Hence: Question (2): What are simple examples of functors $f^*: \Dd \to \Cc$ satisfying the above conditions? Attempts: I expect that it shouldn't be hard to come up with counterexamples for (1), but somehow I cannot make them work. For example: Consider a finite group $G$ and let $f: H \hookrightarrow G$ be a subgroup inclusion. Then the restriction functor $f^*: \Sp^{G} \to \Sp^H$ between (genuine) equivariant spectra is symmetric monoidal with left adjoint $f_!(X) = G \ltimes_H X$ and right adjoint $f_*(X) = F(G/H_+,X)$. In this case we actually have $f_! \simeq f_*$ by the Wirthmüller isomorphism. And I think that if $C \in \Sp^H$ satisfies $f_!(C) \simeq f_*(\mathbb{S}_H) \simeq f_!(\mathbb{S}_H)$, then we already have $C \simeq \mathbb{S}_H$, so this won't give a counterexample. Consider a finite group $G$ and a normal subgroup $N < G$, and let $f: G \to G/N$ be the projection. Then $f^*: \Sp^{G/N} \to \Sp^G$ between (genuine) equivariant spectra is symmetric monoidal with left adjoint $f_!(X) = X/N$ and right adjoint $f_*(X) = X^N$. Now an object $C$ with $f_!(C) \simeq f_!(\mathbb{S}_G)$ is no longer unique since $f_!(\Sigma^{\infty}_+G) \simeq f_!(\Sigma^{\infty}_+ G/N)$. But we don't have the Wirthmüller isomorphism $f_! \simeq f_*$ so it doesn't immediately produce a counterexample. I also tried a little outside the range of equivariant homotopy theory, but there I fail to even come up with functors satisfying the required criteria: Let $f: \Z \hookrightarrow \Q$ and let $f^*: \Dd = D(\Q) \to \Cc = D(\Z)$ be the forgetful functor. This is symmetric monoidal (since $\Q$ is a (derived) localization of $\Z$). However, it does not preserve compact objects: $\Q$ is not finitely generated over $\Z$. Edit: as remarked in the comments, it is actually only lax symmetric monoidal, since the monoidal unit is not sent to the monoidal unit. Let $f^*: \Dd = D(\Z/2\Z) \to \Cc = D(\Z)$ be the forgetful functor. This preserves compact objects, but it is not symmetric monoidal: we have $\Z/2Z \tensor^L_{\Z/2\Z} \Z/2\Z \simeq \Z/2\Z$, but $\Z/2\Z \tensor^L_{\Z} \Z/2\Z$ has higher Tor-groups so cannot be $\Z/2$. If $\Cc$ is cartesian closed and pointed, and $\Dd = 1$ is the trivial category, then the functor $f^*: 1 \to \Cc: * \mapsto *$ is symmetric monoidal, has both adjoints $f_!$ and $f_*$, and any object $C$ satisfies $f_!(C) \simeq f_*(1_{\Cc})$. But $\Cc$ cannot be tensor-triangulated (as $- \times -$ doesn't preserve coproducts in either variable.) REPLY [3 votes]: Consider the projective line $\mathbb{P}^1_k$ over a field $k$. The structure morphism $f:\mathbb{P}^1_k\rightarrow \mathrm{Spec}(k)$ induces a functor $f^*:D(k) \rightarrow D_{qc}(\mathbb{P}^1_k)$ which is fully faithful (equivalently, $f_*(1) \simeq 1$) and which satisfies Grothendieck--Neeman duality. Since $f_*(1)\simeq 1$ is (trivially) a rigid separable commutative algebra, there is a canonical map $\theta:1 \to \omega_f$ such that $f_*(\theta)$ an isomorphism (by Lemma 4.5 of my paper "A characterization of finite étale morphisms in tensor triangular geometry"). Thus, $f_*(1) \simeq f_*(\omega_f) \simeq f_!(1)$ where the last isomorphism is the Wirthmüller isomorphism provided by [BDS]. However, $\omega_f = \Sigma \mathcal O_{\mathbb{P}^1_k}(-2)$ is not isomorphic to $1=\mathcal O_{\mathbb{P}^1_k}$. Thus $C=\omega_f^{-1}$ and $C=1$ both satisfy $f_!(C)\simeq f_*(1)$ showing that $C$ need not be unique.<|endoftext|> TITLE: Cohesion relative to a pyknotic/condensed base QUESTION [9 upvotes]: Something that usefully emerged for me from this discussion and follow-up MO question is that rather than see cohesiveness and condensedness/pyknoticity in rivalry with one another, as my initial title proposed, that the former should be understood as a relative notion, constructions relative to a choice of base $\infty$-topos, while the latter involves the choice of a specific such base. This would suggest that a useful thing to investigate is the replacement in all things cohesive of the default base $\infty$-topos of $\infty$-groupoids by $Sh_{\infty}(ProFinSet)$. We might expect then that the package of constructions of cohesion in its various forms (infinitesimal, differential, singular (for orbifolds), elastic, solid, etc.) would be automatically available for this new base. What I would like to gain a better sense of is how important such constructions are likely to be. As a starter, then, have people considered the tangent $\infty$-topos, $T(Sh_{\infty}(ProFinSet))$, a case of infinitesimal cohesion over the new base? Presumably this involves some form of parameterised pyknotic/condensed spectra. From the above-mentioned conversations, it appears that the $\infty$-topos of $\infty$-sheaves over the pro-étale site on all schemes over a separably closed field $k$ is cohesive over $Sh_{\infty}(ProFinSet)$. Then for any spectrum object in such a cohesive setting, the associated differential cohomology follows from the relevant differential cohomology hexagon. Is this likely to be important? The article by Hisham Sati & Urs Schreiber, Proper Orbifold Cohomology, covers all forms of cohesion. REPLY [12 votes]: Let me try to cut through the jargon. One thing that confuses me are two uses of "tangent spaces" here, that I believe are quite unrelated. One is the usual notion of tangent spaces of smooth manifolds say, based on which one defines differentials, and all sorts of de Rham cohomology etc.; I believe the "differential cohomology" is of this sort. On the other hand, there is the notion of the "tangent $\infty$-topos". I vaguely understand the reason for also calling this "tangent", but it is by a series of analogies, and there seems to be no relation between these two concepts in the case at hand. More specifically, for condensed anima, the tangent $\infty$-topos is simply the $\infty$-category of pairs $(X,A)$ where $X$ is a condensed anima and $A\in \mathcal D(\mathrm{Cond}_{/X},\mathbb S)$ is a hypercomplete sheaf of spectra on the site of condensed sets over $X$. This is definitely a very interesting structure. This whole concept of $6$-functor formalisms is very much about such categories. Working with torsion coefficients, and allowing only "relatively discrete" coefficients, I've developed something along those lines in Etale cohomology of diamonds. Something even closer is in Chapter VII of Geometrization of the local Langlands correspondence (link should be active in a few days is active), where we restrict to the solid objects in $\mathcal D(\mathrm{Cond}_{/X},\mathbb Z_\ell)$. A critical role is then played by the left adjoint $f_\natural$ to pullback $f^\ast$. These do not exist in any classical setup, but have excellent formal properties. In fact, one gets a variant of a $6$-functor formalism where homology and cohomology are now on equal footing again (and arguably homology is even more primitive, again): The pullback functor $f^\ast$ admits a left adjoint $f_\natural$ ("homology") and a right adjoint $Rf_\ast$ ("cohomology"), both of which commute with any pullback. Moreover, Poincare duality holds for proper smooth maps. So yes, there's something interesting about this. On the other hand, the reference to the differential cohomology hexagon in the question confuses me. For condensed sets, there are no tangent spaces, no differential forms, etc., and you can't get them back by magic. I think the problem is that I have no idea what the term "differential cohesion" means, but my strong feeling is that to have "differential cohesion" one needs extra structure like tangent spaces on the model spaces (and that passing to the "tangent $\infty$-topos" is not at all supplying these, as this is a very different procedure). $\require{AMScd}$ Edit: Thanks to the David's for their enlightening comments! Now I understand that the part with the differentials is really only in the examples, not in the "differential cohomology hexagon". For my convenience, let me reformulate this hexagon in my own language. Say $X$ is a condensed anima, and $A\in \mathcal D(\mathrm{Cond}_{/X},\mathbb S)$ is a sheaf of spectra on condensed sets over $X$. (Or take $X$ a scheme, and $A\in \mathcal D(X_{\mathrm{proét}},\mathbb Z_\ell)$, or $X$ a small v-stack and $A\in \mathcal D(X_v,\mathbb Z_\ell)$, or...) Let $\pi$ denote the projection from the site of $X$ to the (pro-étale) site of the point. Then pullback $\pi^\ast$ has a right adjoint $R\pi_\ast$ ("cohomology") and a left adjoint that I will denote $\pi_\natural$ ("homology"). In particular, we get condensed spectra $R\pi_\ast A$ (=$R\Gamma(X,A)$), the cohomology of $A$, and $\pi_\natural A$, the homology of $A$. By adjunction, we get maps $$ \pi^\ast R\pi_\ast A\to A\to \pi^\ast \pi_\natural A. $$ Let $\overline{A}=\mathrm{cofib}(\pi^\ast R\pi_\ast A\to A)$ and $\tilde{A}=\mathrm{fib}(A\to \pi^\ast \pi_\natural A)$. Then there is a pullback square $$\begin{CD} A @>>> \overline{A}\\@VVV @VVV\\ \pi^\ast \pi_\natural A @>>> \pi^\ast \pi_\natural \overline{A} \end{CD} $$ and a pushout square $$\begin{CD} \pi^\ast R\pi_\ast \tilde{A} @>>> \pi^\ast R\pi_\ast A\\@VVV @VVV\\ \tilde{A} @>>> A \end{CD} $$ Of course, pushout squares and pullback squares are equivalent, but I want to stress that one wants to use them to recover $A$ from "simpler" information. However, to me $\tilde{A}\to A\to \overline{A}$ all feel extremely similar, and I'd regard these squares as simple statements about how to analyze the small difference between them.<|endoftext|> TITLE: Is there a nice orthogonal basis of spherical harmonics? QUESTION [9 upvotes]: Recall that a function is harmonic if its Laplacian is zero. Let $\mathrm{Harm}(n,k)$ denote the vector space of $n$-variate harmonic polynomials that are homogeneous with degree $k$. When working with spherical harmonics, we endow this vector space with the inner product $\langle\cdot,\cdot\rangle$ defined in terms of the uniform probability measure over the unit sphere $\mathbb{S}^{n-1}$. Many proofs involving spherical harmonics pass to an implicit orthogonal basis for this inner product space, but for computations, it is sometimes helpful to have an explicit basis. Question. Is there a "nice" choice of orthogonal basis for $(\mathrm{Harm}(n,k),\langle\cdot,\cdot\rangle)$? In particular, is there a choice for which there exists a fast algorithm to compute an arbitrary decomposition in the basis (à la FFT)? REPLY [5 votes]: A simple basis for $\mathcal{H}(n,k)$ is described in this question, however, not so symmetric as one may hope at a first glance, as it is shown in the answer. You may also want to check the beautiful (free online) textbook Harmonic Function Theory by Axley-Bourdon-Ramey.<|endoftext|> TITLE: Is the opposite category of commutative von Neumann algebras a topos? QUESTION [24 upvotes]: By the "category of commutative von Neumann algebras" I mean the category of all commutative von Neumann algebras with normal unital $*$-homomorphisms between them (I don't want to restrict to separable predual as I think it would prevent the existence of finite products). Of course, I don't really believe it is a topos. I would honestly be very surprised if it were but so far I can't find a clear-cut argument that would show that it is not a topos. Part of the problem is that products in this category are a bit tricky to understand. On the other hand, it is not so far-fetched to suggest this, since this category has some topos-like properties: it's definitely a "category of spaces", it has a subobject classifier given by $\mathbb{C}^2$, subobjects of each object form a complete boolean algebra, it is extensive, and it might very well be a regular category (not so clear). I don't think it is exact or cartesian closed (if it were any of these, that would make it a topos), but that's not something completely inconceivable either. Does anyone have a clean argument to show that this category is not a topos ? REPLY [22 votes]: The opposite category of commutative von Neumann algebras is not a topos because categorical products with a fixed object do not always preserve small colimits. See Theorem 6.4 in Andre Kornell's Quantum Collections.<|endoftext|> TITLE: A list of proofs of the Hasse–Minkowski theorem QUESTION [23 upvotes]: I am currently doing a project in which I intend to include the most insightful possible proof of the Hasse–Minkowski theorem (also known as the Hasse principle for quadratic forms, among other names) over $\mathbb{Q}$, as well as a separate proof of the theorem over all algebraic number fields. In order to do this, I first want to compare all the proofs given in the existing literature. So far, I have managed to compile the following list of books and papers, in which proofs of the theorem can be found: Z.I. Borevich and I.R. Shafarevich: Number Theory (1964) J.W.S. Cassels: Rational Quadratic Forms (1978) J.W.S. Cassels: Lectures on Elliptic Curves (1991) [in which the theorem is only proved in the case of three variables] H. Hasse: Über die Darstellbarkeit von Zahlen durch quadratische Formen im Körper der rationalen Zahlen (1923); Über die Äquivalenz quadraticher Formen im Körper der rationalen Zahlen (1923); Symmetrische Matrizen im Körper der rationalen Zahlen (1924); Darstellbarkeit von Zahlen durch quadratische Formen in einem beliebigen algebraischen Zahlkörper (1924) [the proof here extending through four papers and applying to all algebraic number fields] Y. Kitaoka: Arithmetic of Quadratic Forms (1993) T.Y. Lam: The Algebraic Theory of Quadratic Forms (1973) O.T. O'Meara: Introduction to Quadratic Forms (1963) [in which the theorem is proved for all global fields] J.-P. Serre: Cours d'arithmétique (1970) G. Shimura: Arithmetic of Quadratic Forms (2010) The proofs given in the above resources are all quite remarkably dissimilar in certain cases, which is encouraging, in as far as it suggests that the "canonical" proof of the theorem has yet to be established. I would hence like to ask the members of the MathOverflow community if they are aware of any other proofs of the theorem, and if they could direct me to where they can be found. I am looking especially for proofs of the theorem in the more general case over all algebraic number fields, as, apart from Hasse's original, the only full proofs of this that I have been able to find are those in the books of Lam (who proves it modulo two assumptions - the latter of which is very substantial) and O'Meara (who only proves the weak Hasse-Minkowski theorem). Proofs of the "weak" Hasse–Minkowski theorem (i.e. that pertaining to the equivalence of quadratic forms as opposed to their representing $0$) in the case of fields where the "strong" Hasse–Minkowski theorem does not hold are also especially welcome. Proofs of the theorem for a particular $n \geq 3$ are also very welcome ($n$ here denoting the number of variables of the quadratic forms). The proofs do not have to be ones found in published books: Proofs from e.g. lecture notes are also very welcome, provided they are not entirely based on a proof given in a published book. Edit: The answers to this post have helped identify the following resources, in which the theorem is either proved in its entirety or otherwise meaningfully discussed in some manner (works labelled with an asterisk are ones of which I have thus far been unable to obtain a copy): J.W.S. Cassels: Note on Quadratic Forms Over the Rational Field (1959) L.E. Dickson: Studies in the Theory of Numbers (1930)* M. Eichler: Quadratische Formen und orthogonale Gruppen (1952)* A. Gamzon: The Hasse–Minkowski Theorem (2006) [an honours thesis written at the University of Connecticut under the supervision of Keith Conrad] R. Heath-Brown: A New Form of the Circle Method, and its Application to Quadratic Forms (1996) B.W. Jones: The Arithmetical Theory of Quadratic Forms (1950) C.L. Siegel: Equivalence of Quadratic Forms (1948) Th. Skolem: Diophantische Gleichungen (1938)* T.A. Springer: Note on Quadratic Forms over Algebraic Number Fields (1957)* G.L. Watson: Integral Quadratic Forms (1960)* E. Witt: Theorie der quadratischen Formen in beliebigen Körpern (1937) I have decided to keep the works identified after the creation of this post on this separate list. I am currently in the process of reviewing all of the above resources, and will continually be amending the list with new entries and commentary on existing entries. Edit 30 July 2021: I have identified the following classical proofs of the theorem over $\mathbb{Q}$ in the case of $n=3$ (where the theorem is given in Legendre’s classical pre-$p$-adic formulation in terms of congruence conditions for the coefficients): P.G.L. Dirichlet: Vorlesungen Über Zahlentheorie, 2. Aufl. (1871) [§§156-157] C.F. Gauss: Disquisitiones Arithmeticae (1801) [§§293-298] A.-M. Legendre: Essai sur la Théorie des Nombres (1798) [§IV] For the proof of the theorem over all algebraic number fields, the $n=2$ case of course follows from the global square theorem over algebraic number fields, which was first proved in: D. Hilbert: Über die Theorie des relativquadratischen Körpers (1898) The $n=3$ case was originally proved by Furtwängler (albeit in a different formulation) in: Ph. Furtwängler: Über die Reziprozitätsgesetze für ungerade Primzahlexponenten, Parts I-III (1909, 1912 and 1913 resp.) These works were cited by Hasse in his: H. Hasse: Darstellbarkeit von Zahlen durch quadratische Formen in einem beliebigen algebraischen Zahlkörper (1924) [also listed above] For $n=4$, there is of course Hasse’s original proof, the books of Lam and O’Meara and the following paper by Springer: T.A. Springer: Note on Quadratic Forms over Algebraic Number Fields (1957) [also listed above] Edit 10 August 2021: I will here give a sketch of the proof of the Hasse-Minkowski theorem over all global fields for all $n$ except $n=3$: For $n=2$, this of course follows, as I mentioned above, from the global square theorem, which, in the case of algebraic number fields, follows from the Chebotarev density theorem. This appears as a corollary to theorem 10 of: S. Lang: Algebraic Number Theory, 2nd edition (1994) and as exercise 6.2 of: Cassels and Fröhlich (editors): Algebraic Number Theory (1967) I do not know how to prove the global square theorem for function fields over finite fields (the other kind of global fields), but then this is not relevant to my project. For $n=4$, Hasse originally proved it in a way very similar to the traditional proof of the theorem over $\mathbb{Q}$ for $n=4$, but the nicest way of proving the statement is by Springer's trick, first published in the paper listed above, and reproduced as exercise 4.4 of Cassels and Fröhlich, which reduces the statement to the $n=3$ case. For $n \geq 5$, Serre's topological argument for the theorem over $\mathbb{Q}$ from his Cours d'arithmétique generalises readily (almost verbatim) to all algebraic number fields. I will now try to discuss the only problematic case, $n=3$: It is obviously equivalent to the quadratic case of the of the Hasse norm theorem, which was, as already mentioned, originally proved by Furtwängler in part III of the above listed series of papers. I have the following to say: First, a historical remark: In the Wikipedia article on the Hasse norm theorem, there is the following: The full theorem is due to Hasse (1931). The special case when the degree $n$ of the extension is $2$ was proved by Hilbert (1897), and the special case when $n$ is prime was proved by Furtwangler (1902). This is definitely wrong - the date 1897 seems to refer to Hilbert's Zahlbericht, where the theorem was not proved. As I have already noted, the quadratic case was first proved by Furtwängler. The special case when the degree of the extension is prime was proved by Hasse in: H. Hasse: Bericht über neuere Untersuchungen und Probleme aus der Theorie der algebraischen Zahlkörper, Teil II (1930) and the case of a general cyclic extension was later proved by Hasse in: H. Hasse (1931): Beweis eines Satzes und Wiederlegung einer Vermutung über das allgemeine Normenrestsymbol (1931) I don't know how to edit Wikipedia articles, but if anyone here does, it would be good if we could correct this. As another aside: On p. 96, ch. 6 of Rational Quadratic Forms, when sketching the proof of the Hasse-Minkowski theorem over all global fields, Cassels makes the incorrect claim that the Hasse norm theorem holds for all abelian extensions - this is exactly the Vermutung that is widerlegt in Hasse's 1931 paper. - If there exists an online list of errata to Cassels' Rational Quadratic Forms, then this belongs there. The above two remarks were not directly relevant to the proof of the Hasse-Minowski theorem, but the rest of what I have to say is: The Brauer-Hasse-Noether theorem is derived as a corollary of a high-level class field theoretic result in §9.6 of ch. VII of Cassels and Fröhlich, which in turn gives us the Hasse norm theorem and hence Hasse-Minkowski for $n=3$. Moreover, the Hasse norm theorem is also proved in: G. Janusz: Algebraic Number Fields (1973) My question to the community is: Are there any other textbook proofs of the Hasse norm theorem, and are there any simpler proofs of the of theorem in the special case of quadratic extension (i.e. Furtwängler's result)? REPLY [4 votes]: Cassels also has a Note on quadratic forms over the rational field. Proc. Cambridge Philos. Soc. 55 (1959), 267–270, addendum 57 (1961), 697. https://doi.org/10.1017/S0305004100034009 There is perhaps some methodological interest in developing the theory of quadratic forms over the rational field using only the methods of elementary arithmetic. Hitherto it has appeared necessary to use theorems of a fairly deep nature, most often Dirichlet's theorem about the existence of primes in arithmetic progressions (e.g. Minkowski(1), Hasse(2), Dickson(8), Skolem(9), Burton Jones(6)). Skolem(5) uses a weaker form of Dirichlet's theorem which is rather easier to prove and Siegel(4) uses instead the machinery of the Hardy-Littlewood circle method. In this note I indicate how it is possible to develop the theory of quadratic forms over the rationals without using extraneous resources. Pall(10) states that he has also found such a development of the theory but he does not appear to have published it. In the addendum, he regrets overlooking Eichler's version. https://doi.org/10.1017/S0305004100035805<|endoftext|> TITLE: Method to evaluate an infinite sum of ratio of Gamma functions (how does Mathematica do it?) QUESTION [7 upvotes]: This question arose from Amdeberhan's question, the evaluation of a double integral, which can be reduced to the evaluation of this series: $$\sum _{n=0}^{\infty } \frac{\Gamma \left(n+\frac{1}{2}\right)^2 \Gamma \left(n+\frac{s}{2}\right)}{\Gamma (n+1)^2 \Gamma (n+s)}=\frac{\pi ^2 2^{1-s} \Gamma \left(\frac{s}{2}\right)}{\left[\Gamma \left(\frac{3}{4}\right) \Gamma \left(\frac{s}{2} +\frac{1}{4}\right)\right]^2},\;\;{\rm Re}\,s>0.$$ The evaluation of the sum is Mathematica output. Can someone enlighten me as to how this calculation proceeds? I went so far as to pay for Wolfram Alpha Pro, hoping that it would disclose the steps, but to no avail. What is even more frustrating is that for $s=1$ the right-hand-side is the square of a complete elliptic integral, which is also recognized immediately by Mathematica and was the original question in the cited post, so far without a conclusive answer. REPLY [9 votes]: This is a special case of Watson's Theorem $$\def\h{\frac{1}{2}} \def\g#1{\Gamma(#1)\,} {}_3F_2\left({a,\ b,\ c\atop\h a+\h b+\h, 2c }\biggm| 1 \right) =\frac{\g\h\g{c+\h}\g{\h a+\h b +\h}\g{c+\h -\h a -\h b}} {\g{\h a +\h} \g{\h b+\h} \g{c+\h -\h a} \g{c+\h -\h b}}. $$<|endoftext|> TITLE: Z/8Z elliptic curve with a missing generator QUESTION [11 upvotes]: We are searching for the rank $6$ elliptic curves with the torsion subgroup $\mathbb{Z}/8\mathbb{Z}$ using the families similar to Allan MacLeod's as described in A. J. MacLeod, A Simple Method for High-Rank Families of Elliptic Curves with Specified Torsion, arXiv, Number Theory [math.NT] (2014), arXiv:1410.1662v1 and came across a curve [1,0,0,-885470498073167713002317184212739304,315787360224933400897171748517120652503410335938363968] Magma Calculator, Magma V2.24-7, and mwrank (with $-b14$) return $4$ generators for this curve, e.g.: SetClassGroupBounds("GRH"); E:=EllipticCurve([1,0,0,-885470498073167713002317184212739304,315787360224933400897171748517120652503410335938363968]); MordellWeilShaInformation(E); Using model [ 1, 0, 0, -885470498073167713002317184212739304, 315787360224933400897171748517120652503410335938363968 ] Torsion Subgroup = Z/8 The 2-Selmer group has rank 6 New point of infinite order (x = 291590635879268240110317582512/87995882881) New point of infinite order (x = -780802002478160103839088529264/1152485984521) New point of infinite order (x = 526632508060857475043443664/1355049721) New point of infinite order (x = 25006056691267829443814801295058/11996130112849) After 2-descent: 4 <= Rank(E) <= 5 Sha(E)[2] <= (Z/2)^1 (Searched up to height 10000 on the 2-coverings.) Both Magma and mwrank return $5$ for the upper bound on rank of the curve and all the related 2-isogenies ($\mathbb{Z}/4\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}$, and $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ curves below) using all their 2-torsion points: E:=EllipticCurve([1,0,0,-885470498073167713002317184212739304,315787360224933400897171748517120652503410335938363968]);TwoPowerIsogenyDescentRankBound(E);print(" "); E:=EllipticCurve([1,0,0,6205127968803058885033015264909364456,-3406906009463658062905156451961488907125318013801796912]);TwoPowerIsogenyDescentRankBound(E);print(" "); E:=EllipticCurve([1,0,0,-1810312155105870840254262815766788584,-457633210238033749039428724550432643513221271485300416]);G,h:=TwoTorsionSubgroup(E);T:=[h(g):g in G];TwoPowerIsogenyDescentRankBound(E,T[2]);TwoPowerIsogenyDescentRankBound(E,T[3]);TwoPowerIsogenyDescentRankBound(E,T[4]);print(" "); E:=EllipticCurve([1,0,0,-24623218791538050601572671001307730104,-47007276920642307027416764645331974815957764878067127376]);G,h:=TwoTorsionSubgroup(E);T:=[h(g):g in G];TwoPowerIsogenyDescentRankBound(E,T[2]);TwoPowerIsogenyDescentRankBound(E,T[3]);TwoPowerIsogenyDescentRankBound(E,T[4]);print(" "); 5 [ 3, 3, 3, 3, 3 ] [ 4, 4, 4, 4, 4 ] 5 [ 7, 7, 7, 7, 7 ] [ 0, 0, 0, 0, 0 ] 5 [ 4, 4, 4, 4, 4 ] [ 3, 3, 3, 3, 3 ] 5 [ 0, 0, 0, 0, 0 ] [ 7, 7, 7, 7, 7 ] 5 [ 0, 0, 0, 0, 0 ] [ 7, 7, 7, 7, 7 ] 5 [ 0, 0, 0, 0, 0 ] [ 11, 7, 7, 7, 7 ] 5 [ 0, 0, 0, 0, 0 ] [ 9, 7, 7, 7, 7 ] 5 [ 7, 7, 7, 7, 7 ] [ 0, 0, 0, 0, 0 ] We tried implementing Jeremy Rouse's and Zev Klagsbrun's (1 and 2) approaches to employ $2$- and $4$-coverings, but could not find a rational point for the intersection of two quadrics. We used different seeds in Magma Calculator and a bound up to $3\times10^8$, e.g.: SetSeed(1); SetClassGroupBounds("GRH"); E := EllipticCurve([1,0,0,-885470498073167713002317184212739304,315787360224933400897171748517120652503410335938363968]); P1 := E![86609552610643851795318648321889070494,113996061058226165174165649863299331386757385142,41549112479447364007]; P2 := E![72445938379247589758672,7465810298311395376082337884312,117649]; P3 := E![36603887306503756637336310506125004116,9298633886669630228894900541541155002386034788,53504769840978593717]; P4 := E![29317967055802030746557715723363588500973254368,22661934910027931993184701249992172746172677836253523176,25876196859106536169778781677]; twocovers := TwoDescent(E : RemoveTorsion := true, RemoveGens := {P1,P2,P3,P4}); twocovers; fourcovers := FourDescent(twocovers[1] : RemoveTorsion := true, RemoveGensEC := {P1,P2,P3,P4}); fourcovers; _,m := AssociatedEllipticCurve(fourcovers[1] : E := E); pts := PointsQI(fourcovers[1], 3*10^8 : OnlyOne := true); pts[1]; m(pts[1]); [ Hyperelliptic Curve defined by y^2 = 4401556531432641*x^4 + 189490115768444190*x^3 + 5942548110808516561*x^2 - 11718931267663428016*x + 9974519367938837056 over Rational Field ] [ Curve over Rational Field defined by 93*$.1^2 + 505*$.1*$.2 + 715*$.1*$.3 + 2737*$.1*$.4 + 182*$.2^2 + 916*$.2*$.3 + 1233*$.2*$.4 + 690*$.3^2 - 1248*$.3*$.4 - 3453*$.4^2, 43659*$.1^2 - 100751*$.1*$.2 + 100262*$.1*$.3 + 267364*$.1*$.4 + 57470*$.2^2 + 183818*$.2*$.3 - 93739*$.2*$.4 - 647944*$.3^2 + 666862*$.3*$.4 - 133155*$.4^2 ] >> pts[1]; ^ Runtime error in '[]': Illegal null sequence Question. Considering parity, there should be one more generator on the curve. Is there a way to find it? We would greatly appreciate any hint leading to the discovery of the extra generator. If you can compute an extra generator, your name will be published at the bottom of the $\mathbb{Z}/8\mathbb{Z}$ rank 5 page. REPLY [5 votes]: Although Magma's MordellWeilShaInformation can't find the fifth generator for the $\mathbb{Z}/8\mathbb{Z}$ curve, it can find the last generator for an isogenous $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}$ curve by searching up to height $10^5$ on the $4$-coverings. SetClassGroupBounds("GRH"); E24 := EllipticCurve([1, 0, 0, -1810312155105870840254262815766788584, -457633210238033749039428724550432643513221271485300416]); MordellWeilShaInformation(E); Using model [ 1, 0, 0, -1810312155105870840254262815766788584, -457633210238033749039428724550432643513221271485300416 ] Torsion Subgroup = Z/2 x Z/4 The 2-Selmer group has rank 7 New point of infinite order (x = -443706527227992154791056873744/470078898129) New point of infinite order (x = -14367138002890543477931052222544/12054179879569) New point of infinite order (x = -269944456262034200432888/241081) New point of infinite order (x = -630597721484060507464551524086/528521730025) After 2-descent: 4 <= Rank(E) <= 5 Sha(E)[2] <= (Z/2)^1 (Searched up to height 10000 on the 2-coverings.) New point of infinite order (x = 13457909758544525530407577328967629341502974335624109249/9109819423915178769710709651398624256) After 4-descent: 5 <= Rank(E) <= 5 Sha(E)[4] is trivial (Searched up to height 10^5 on the 4-coverings.) Each found generator can be mapped back to the original $\mathbb{Z}/8\mathbb{Z}$ curve. E24 := EllipticCurve([1, 0, 0, -1810312155105870840254262815766788584, -457633210238033749039428724550432643513221271485300416]); P1_24 := Points(E24, -443706527227992154791056873744/470078898129)[1]; P2_24 := Points(E24, -14367138002890543477931052222544/12054179879569)[1]; P3_24 := Points(E24, -269944456262034200432888/241081)[1]; P4_24 := Points(E24, -630597721484060507464551524086/528521730025)[1]; P5_24 := Points(E24, 13457909758544525530407577328967629341502974335624109249/9109819423915178769710709651398624256)[1]; E8 := IsogenousCurves(E24)[1]; Coefficients(E8); b, m := IsIsogenous(E24, E8); P1 := m(P1_24); P2 := m(P2_24); P3 := m(P3_24); P4 := m(P4_24); P5 := m(P5_24); S := [P1, P2, P3, P4, P5]; T := Points(E8, Saturation(S, 0)[1][1])[1]; P1 := P1+7*T; P2 := P2+3*T; P4 := P4+2*T; P5 := P5+5*T; P1; P2; P3; P4; P5; IsLinearlyIndependent(S); [ 1, 0, 0, -885470498073167713002317184212739304, 315787360224933400897171748517120652503410335938363968 ] (19736503637532263590768972383399786896/686322068333254991401 : -9688380190239124552471385562273840631362670461480086958584/17980093484565761049363798777899 : 1) (-106392015003516680180229247909564/262444244417161 : 3315524468078731464225753019750095494243972085244/4251631139754026848091 : 1) (197824762218747711268509656464/24841227321 : -87398782145351238928127905114416134835854072/3915250679290131 : 1) (104082169729287819444025464178104869085812/94320856150181594751169 : 23929183871232548960786573517019430438238066845311306934645336/28967539063327699053963692450740447 : 1) (904856066797346401279199736603573928704209102734767139366416265493676502088160396747547111952/1980903839660916800788793856501687874959261129995118904651263730123078358161 : 7176580364279680654978748955564315110684742734823615940436244933109959252724557389297041035780397662180996315752417825806484247739820234728/88164772399749168946841346769450081472444911489955953664661823265263508750316122576099738571727015984657590089159 : 1) true<|endoftext|> TITLE: Is there a semantics for intuitionistic logic that is meta-theoretically "self-hosting"? QUESTION [15 upvotes]: One can study the standard semantics of classical propositional logic within classical logic set theory, so we can say that the semantics of classical logic is meta-theoretically "self-hosting". This property is probably a big part of why classical logic is so easy to accept as the default/implicit background/foundational logic for mathematics. In contrast, classical logic set theory is also used in the presentation of the mainstream semantic interpretations of propositional intuitionistic logic such as Kripke semantics and Heyting algebra. For example, in Kripke semantics our structures are triples $M = \langle W, \leq, v\rangle$ where $W \neq \varnothing$, $\leq$ is a preorder on $W$, and $v$ is a set function from the set of atomic propositions to the powerset of $W$, where $v$ must satisfy: for every $w_1,w_2 \in W$, for every atomic proposition $p$, if $w_1 \leq w_2$ and $w_1 \in v(p)$, then $w_2 \in v(p)$. So far, it seems that we could assume that we are working in some kind of intuitionistic set theory setting, however when we get to the definition of interpretation of formulas, it seems like there are some problems with using an intuitionistic meta-theory: \begin{array}{lll} M,w \vDash p &\text{ iff } &w \in v(p)\\ M,w \vDash \top\\ M,w \nvDash \bot\\ M,w \vDash \phi_1 \to \phi_2 &\text{ iff } &\text{for every } w'\in W.\text{ if } w \leq w' \text{ and } M,w' \vDash \phi_1 \text{ then } M,w' \vDash \phi_2\\ M,w \vDash \phi_1 \lor \phi_2 &\text{ iff } & M,w \vDash \phi_1 \text{ or } M,w \vDash \phi_2\\ M,w \vDash \phi_1 \land \phi_2 &\text{ iff } & M,w \vDash \phi_1 \text{ and } M,w \vDash \phi_2 \end{array} First off: now that we are interpreting this presentation in some kind of intuitionistic meta-theory, we will no longer necessarily have $w \vDash \phi$ or $w \nvDash \phi$, whereas before this would be true of every formula $\phi$, thanks to the law of excluded middle in classical logic. It seems like this may cause this semantics to determine a different logic than intuitionistic logic, when the meta-theory is taken to be intuitionistic. Furthermore, in my experience with this semantic interpretation, there are many times where a formula is found to be semantically valid based on some use of the law of excluded middle. For example, you might observe that there must exist some future world $w$ such that $w \vDash \phi$, or else for every future world $w$ we must have $w \nvDash \phi$. This also indicates to me that using an intuitionistic meta-theory may cause the resulting logic to be different. I haven't been able to find an example of a formula that would be intuitionistically valid, but not valid in the logic determined by Kripke semantics as interpreted within an intuitionistic meta-theory, but it seems like such a thing might exist. Can anyone point me to any work that studies this issue in depth? Or point me to work that studies the more general issue of meta-theoretic "leakage" with respect to semantics? Finally, if this is indeed a problem with Kripke/Heyting semantics, has anyone discovered a "self-hosting" intuitionistic semantics? Specifically: a semantic interpretation of intuitionistic propositional logic that assumes a meta-theory based on some kind of intuitionistic set theory, and is sound and complete with respect to a standard intuitionistic proof system. Edit: If I understand correctly, Mike Shulman's answer claims that something like the following will be a semantic interpretation of propositional intuitionistic logic, as long as the meta-theory is intuitionistic. Let $M$ be a function from the set of atomic propositions to the set $\{0,1\}$, and then interpret formulas recursively by: \begin{array}{lll} M \vDash p &= &M(p) = 1\\ M \vDash \top &= &\text{Triv.}\\ M \vDash \bot &= &\text{Abs.}\\ M \vDash \phi_1 \to \phi_2 &= & M \vDash \phi_1 \text{ implies } M \vDash \phi_2\\ M \vDash \phi_1 \lor \phi_2 &= & M \vDash \phi_1 \text{ or } M \vDash \phi_2\\ M \vDash \phi_1 \land \phi_2 &= & M \vDash \phi_1 \text{ and } M \vDash \phi_2 \end{array} This is my best attempt to adapt simple 2-value "classical" logic semantics to the situation of an intuitionistic meta-theory. The only changes I had to make was to avoid careless references to $\nvDash$, and to specify the meaning of implication separately, instead of defining it as $\neg \phi_1 \lor \phi_2$. I also changed the use of "iff" to equality to emphasize that I think we now need to think of the statements in a way that is closer to type theory. For example, for the meaning of an atomic proposition I think we are essentially saying "a proof of $M\vDash p$ is equivalent to a proof that $M(p) = 1$". For the meaning of $\bot$ we are saying "a proof of $M\vDash \bot$ is equivalent to a proof of absurdity". These two together mean that (defining $\neg \phi \equiv \phi \to \bot$) the meaning of $\neg p$ is essentially "a proof of $M \vDash \neg p$ is equivalent to a proof that $M \vDash p$ implies (meta-theoretic) absurdity". Now finally, since we are in intuitionistic meta-theory, we know that it is not given that for any atomic proposition $p$ we must have a proof of $M(p) = 1$ or a proof of $M(p) \neq 1$. This means that we can't conclude that there is always a proof of $M \vDash p \lor \neg p$ like we could in a classical meta-theory context. REPLY [9 votes]: See Palmgren, Constructive Sheaf Semantics for a completeness proof for sheaf semantics within a constructive (and predicative) metatheory. The introduction also mentions several references to earlier approaches that are generalised by sheaf semantics. I recommend particularly Troelstra and Van Dalen, Constructivism in mathematics, vol 2 as a standard reference containing various results about the semantics of constructive mathematics.<|endoftext|> TITLE: Prove that two functions are equal only when $s \equiv \pm r^{\pm 1} \pmod{q}$ QUESTION [7 upvotes]: Let us fix a positive integer $q$, and let us define a functions $P: \mathbb{Z}\times \mathbb{N} \to \mathbb{Z}$ as follows: $$ P(s,t) := \sum_{j=1}^t \left\lfloor \frac{j (s-1) + t}{q} \right\rfloor$$ If we define the function $A_s : \mathbb{N} \to \mathbb{Z}$ by: $$ A_s(t) := P(s,t) + P(-s,t). $$ I claim that $A_s = A_r$ (as functions of $t$) if and only if one of the following is true: $s \equiv r \pmod{q}$ $s \equiv -r \pmod{q}$ $sr \equiv 1 \pmod{q}$ $sr \equiv -1 \pmod{q}$. I do have a proof of both implications, but they are rather involved and a bit too technical. The hardest part is to show that $A_s = A_r$ implies one of the four bullets. I am wondering if this is indeed a hard problem and technical stuff has to play a role in a proof or if I am missing a simpler proof. Even a simple proof of the fact that each of the third and fourth bullets implies $A_s = A_r$ would be nice, since what I have is lengthy and ugly. REPLY [2 votes]: Let $\mathcal P_{q,s}$ denote the parallelogram with vertices $(0,\pm\frac{1}{q}), (1,\frac{s}{q}), (-1,-\frac{s}{q})$. The function $$I_{q,s}(t)=2A_s(t)+2\left\lfloor\frac{t}{q}\right\rfloor+2t+1$$ counts the number of lattice points in the dilation $t\cdot\mathcal P_{q,s}$. In fact this is exactly the expression you get by counting the lattice points according to their x-coordinates and then summing everything up. Now let's apply the map $(x,y)\to (x,sx-qy)$. The parallelogram becomes the square with vertices $(0,\pm1), (\pm1, 0)$ and lattice points are sent to lattice points $(x,y)$ satisfying $y-sx= 0\pmod q$. In this setting symmetry makes it easy to see why $I_{q,s}=I_{q,s'}$ for the four bullet points. In fact, $s'=-s$ corresponds to reflecting everything on the y axis, and $s'=s^{-1}$ corresponds to reflecting everything on the line $x=y$. The converse is much harder and has a really fascinating story to it. First let me mention why people care about this problem. It turns out that this Ehrhart function encodes the multiplicities that $t(t+2), t\in \mathbb N$ appears as an eigenvalue of the Laplace-Beltrami operator on the lens space $L(q,s)$. Knowing that $I(q,s)=I(q,s')$ implies that the two lens spaces $L(q,s)$ and $L(q,s')$ are isospectral (have the same eigenvalues and respective multiplicities). What is easy to see is that the four bullet points are exactly the conditions which specify lens spaces up to isometry. Of course we have $$\text{isometric} \implies \text{homeomorphic} \implies \text{isospectral}$$ but for 3-dimensional lens spaces it turns out that the opposite implication holds as well. Already the fact that the first arrow is invertible required introducing the notion of Reidemeister torsion. From the above discussion your question is completely equivalent to the statement that isospectral lens spaces in dimension 3 are actually isometric (this is known to be false in higher dimensions). So this is some evidence that the proof will probably have to be a little difficult. One proof is given in A. Ikeda, Y. Yamamoto, "On the spectra of 3-dimensional lens spaces", Osaka J. Math. 16(2) (1979), 447-469 for the special case where $q$ is a prime power or twice a prime power. It is proven in full generality in Y. Yamamoto, "On the number of lattice points in the square |x| + |y| ≤ u with a certain congruence condition", Osaka J. Math. 17(1) (1980), 9–21 The main number theoretic ingredient in Yamomoto's proof is the fact that the numbers $\cot \frac{k\pi}{q}$ for $1\le k\le \frac{q}{2}, \gcd(k,q)=1$ are linearly independent over $\mathbb Q$. If your proof is simpler than Yamamoto's it might be worth publishing for that reason alone.<|endoftext|> TITLE: Is $441$ the only square of the form $\frac{397\cdot 10^n-1}{9}$? QUESTION [5 upvotes]: Is $441$ the only square of the form $\frac{397\cdot 10^n-1}{9}$? Can it be proven? REPLY [11 votes]: If $\frac{397\cdot 10^n - 1}9$ is a square then so is $397\cdot 10^n - 1$. Let $y^2=397\cdot 10^n - 1$. Denoting $x:=10^{\lfloor n/3\rfloor}$, we get that $$y^2 = 397\cdot 10^r\cdot x^3 - 1$$ or $$(397\cdot 10^{r}y)^2 = (397\cdot 10^r\cdot x)^3 - 397^2\cdot 10^{2r}$$ where $r:=n\bmod 3\in\{0,1,2\}$. These are Mordell equations with many known solutions, which in general can be solved by finding integral points on the elliptic curve. I've solved this equation, and confirm your conjecture.<|endoftext|> TITLE: Characteristic class that cannot be represented by disjoint tori QUESTION [6 upvotes]: Is there a simply-connected smooth closed 4-manifold with a characteristic class $x \in H_2(X; \mathbb{Z})$ such that $x$ can not be represented by a disjoint union of tori in $X$? I would not know how to prove this without the characteristic hypothesis either so any thoughts on that would also be appreciated. REPLY [15 votes]: In $H_2(CP^2)$, every class $nH$ where $H$ is a generator and n is odd is characteristic. However, if $n >3$, then such a class is not represented by a torus. It is not represented by a disjoint union of tori, either. For non-zero classes in $H_2(CP^2)$ have non-zero intersection numbers. So if you had a disjoint union of tori, then at most one would be non-trivial in homology (and that one would represent $nH$, contradicting the previous step.)<|endoftext|> TITLE: Decomposing square of side length $n$ into $n$ squares in a certain "maximal" way QUESTION [10 upvotes]: I was wondering if anything is known about this problem. We are given a square of side length $n$ and we wish to embed $n$ smaller (integer) squares inside it such that the sum of the side-lengths of the squares inside is maximised. Let $f(n)$ be the maximum sum of side-lengths one can produce in this way. I have computed some values of $f(n)$, and have displayed them below. Apologies for the handwritten drawing, but I think the picture does help explain what I'm trying to get at. The inequalities I have highlighted in red are written so because I am unsure if this is the maximum that can be achieved. In particular, for $n=8$ we don't even use $8$ squares, but I cannot figure out a configuration of $8$ squares that beats $20$. (I have reason to believe that $f(8)\leq 21$, and it would be nice if this could be attained with equality.) The values I have written as equality I am more sure about, based on exhaustive computer simulation. Is anything known about this function $f$? Even basic upper and lower bounds would be really helpful. It certainly seems to be roughly $n^{3/2}$. Update. I made my simulations a bit more efficient, and they seem to indicate that $f(8) = 20$ and $f(10)=30$ are indeed equalities. We also have $f(11)\geq 35$ (and this is likely equality, since my program is taking a while trying to find an example with sum $36$.). Edit. I appreciate all of the comments! Since the constraints of original question may be a bit too difficult (and possibly artificially so), I would like to weaken the requirements slightly. In this new version of the problem, we have a rectangle of integer side lengths $a$ and $b$ and wish to fill it with $c$ squares in order to maximise the sum of side lengths of the inner squares. Thus the original problem was when $a = b = c = n$. The weaker assumptions may make it easier to form an inductive argument. I am still working on this, but would appreciate any help! REPLY [8 votes]: $f(8)=21$: The other reported values up through $n=11$ are optimal. You can solve the problem via integer linear programming as follows. For square with top left corner $(i,j)$ and side length $s$, let binary decision variable $x_{i,j,s}$ indicate whether that square is used. The problem is to maximize $\sum_{i,j,s} s x_{i,j,s}$ subject to \begin{align} \sum_{i,j,s} x_{i,j,s} &= n \tag1 \\ \sum_{\substack{i,j,s:\\ i \le i_0 \le i+s-1 \\ j \le j_0 \le j+s-1}} x_{i,j,s} &\le 1 &&\text{for $(i_0,j_0)\in \{1,\dots,n\}\times \{1,\dots,n\}$} \tag2 \\ \end{align} Constraint $(1)$ forces $n$ squares to be used. Constraint $(2)$ prevents covering cell $(i_0,j_0)$ more than once. Here are results for $n \le 50$, where $u_1(n)$ is the one-dimensional integer knapsack upper bound and $u_2(n)$ is the one-dimensional continuous knapsack upper bound: \begin{matrix} n & \lfloor \sqrt{n}\rfloor^3 & f(n) & u_1(n) & \lfloor u_2(n) \rfloor & u_2(n) & \lfloor n^{3/2} \rfloor \\ \hline 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 2 & 1 & 2 & 2 & 2 & 2.666666667 & 2 \\ 3 & 1 & 4 & 5 & 5 & 5 & 5 \\ 4 & 8 & 8 & 8 & 8 & 8 & 8 \\ 5 & 8 & 10 & 11 & 11 & 11 & 11 \\ 6 & 8 & 14 & 14 & 14 & 14.4 & 14 \\ 7 & 8 & 17 & 18 & 18 & 18.2 & 18 \\ 8 & 8 & 21 & 22 & 22 & 22.4 & 22 \\ 9 & 27 & 27 & 27 & 27 & 27 & 27 \\ 10 & 27 & 30 & 31 & 31 & 31.42857143 & 31 \\ 11 & 27 & 35 & 36 & 36 & 36.14285714 & 36 \\ 12 & 27 & 40 & 41 & 41 & 41.14285714 & 41 \\ 13 & 27 & 45 & 46 & 46 & 46.42857143 & 46 \\ 14 & 27 & 51 & 52 & 52 & 52 & 52 \\ 15 & 27 & 56 & 57 & 57 & 57.85714286 & 58 \\ 16 & 64 & 64 & 64 & 64 & 64 & 64 \\ 17 & 64 & 68 & 69 & 69 & 69.88888889 & 70 \\ 18 & 64 & 75 & 76 & 76 & 76 & 76 \\ 19 & 64 & 80 & 82 & 82 & 82.33333333 & 82 \\ 20 & 64 & 88 & 88 & 88 & 88.88888889 & 89 \\ 21 & 64 & 94 & 95 & 95 & 95.66666667 & 96 \\ 22 & 64 & 100 & 102 & 102 & 102.6666667 & 103 \\ 23 & 64 & 108 & 109 & 109 & 109.8888889 & 110 \\ 24 & 64 & 115 & 117 & 117 & 117.3333333 & 117 \\ 25 & 125 & 125 & 125 & 125 & 125 & 125 \\ 26 & 125 & 130 & 132 & 132 & 132.3636364 & 132 \\ 27 & 125 & 138 & 139 & 139 & 139.9090909 & 140 \\ 28 & 125 & 145 & 147 & 147 & 147.6363636 & 148 \\ 29 & 125 & 154 & 155 & 155 & 155.5454546 & 156 \\ 30 & 125 & 162 & 163 & 163 & 163.6363636 & 164 \\ 31 & 125 & 170 & 171 & 171 & 171.9090909 & 172 \\ 32 & 125 & 178 & 180 & 180 & 180.3636364 & 181 \\ 33 & 125 & 186 & 189 & 189 & 189 & 189 \\ 34 & 125 & 195 & 197 & 197 & 197.8181818 & 198 \\ 35 & 125 & 204 & 206 & 206 & 206.8181818 & 207 \\ 36 & 216 & 216 & 216 & 216 & 216 & 216 \\ 37 & 216 & 222 & 224 & 224 & 224.8461539 & 225 \\ 38 & 216 & 232 & 233 & 233 & 233.8461539 & 234 \\ 39 & 216 & 240 & 243 & 243 & 243 & 243 \\ 40 & 216 & 250 & 252 & 252 & 252.3076923 & 252 \\ 41 & 216 & 259 & 261 & 261 & 261.7692308 & 262 \\ 42 & 216 & 270 & 271 & 271 & 271.3846154 & 272 \\ 43 & 216 & 278 & 281 & 281 & 281.1538462 & 281 \\ 44 & 216 & 288 & 291 & 291 & 291.0769231 & 291 \\ 45 & 216 & 299 & 301 & 301 & 301.1538462 & 301 \\ 46 & 216 & 308 & 311 & 311 & 311.3846154 & 311 \\ 47 & 216 & 319 & 321 & 321 & 321.7692308 & 322 \\ 48 & 216 & 329 & 332 & 332 & 332.3076923 & 332 \\ 49 & 343 & 343 & 343 & 343 & 343 & 343 \\ 50 & 343 & 350 & 353 & 353 & 353.3333333 & 353 \\ \end{matrix}<|endoftext|> TITLE: On a curious map from the complex projective plane into $S^5$ QUESTION [24 upvotes]: I have heavily edited the post (including the title), based on a comment by @GregoryArone that my map $f$ is not injective. In an earlier version of this post, I had thought to have constructed a smooth map from $\mathrm{P}^2_\mathbb{C}$ into $S^5$, which I thought was a topological embedding. Removing a point from $S^5$ and using stereographic projection, I had thought to have found a topological embedding of $\mathrm{P}^2_\mathbb{C}$ inside $S^5$, but I was mistaken. My map was actually not injective. However, this raised an interesting question. Question set 1: does there exist a topological embedding of $\mathrm{P}^2_\mathbb{C}$ inside $\mathbb{R}^5$? Or is there maybe a topological obstruction to that? In the second part of this post, I will describe a smooth map $f$ from $\mathrm{P}^2_{\mathbb{C}}$ into $S^5$ which is a double cover that is branched over a real slice of $\mathrm{P}^2_{\mathbb{C}}$, with respect to a real structure $\sigma$ on $\mathrm{P}^2$, defined by $$ \sigma([z_0:z_1:z_2]) = [\bar{z}_2:-\bar{z}_1:\bar{z}_0].$$ Note that $\sigma$ is the real structure which is induced by the "antipodal map" $j$ on $P^1_\mathbb{C}$, defined by $$j([u_0:u_1]) = [-\bar{u}_1:\bar{u}_0].$$ I will now describe how the map $f$ is defined. First, define the map $g: \mathrm{P}^1_{\mathbb{C}} \times \mathrm{P}^1_{\mathbb{C}} \to \mathrm{P}^2_{\mathbb{C}}$: $$ ([u_0:u_1], [v_0:v_1]) \mapsto [2u_0v_0: u_0 v_1 + u_1 v_0: 2u_1v_1].$$ Then $g$ is holomorphic and onto. The symmetric group $S_2$ acts on the domain of $g$ by permuting the two factors, namely the $u$-point with the $v$-point, so to speak. The fibers of $g$ are actually the $S_2$ orbits in the domain of $g$. The (extended) Hopf map $h$ is a smooth map from $\mathbb{C}^2$ onto $\mathbb{R}^3$, defined by $$h(u_0,u_1) = \left( 2 \operatorname{Re}(u_0 \bar{u}_1), 2 \operatorname{Im}(u_0 \bar{u}_1), |u_0|^2 - |u_1|^2 \right).$$ The group $U(1)$ acts on the domain of $h$ by scalar multiplication, and the fibers of $h$ are the $U(1)$-orbits in the domain of $h$. Then the map $$ h \times h: \mathbb{C}^2 \times \mathbb{C}^2 \to \mathbb{R}^3 \times \mathbb{R}^3$$ followed by the map $\operatorname{Sym}: \mathbb{R}^3 \times \mathbb{R}^3 \to S^2(\mathbb{R}^3)$ which maps $(x,y)$ to $x \odot y$, gives a map $$k: \mathbb{C}^2 \times \mathbb{C}^2 \to S^2(\mathbb{R}^3),$$ where $k = \operatorname{Sym} \circ (h \times h)$. In turn, $k$ induces a smooth map $$\tilde{k}: \mathrm{P}^1_\mathbb{C} \times \mathrm{P}^1_\mathbb{C} \to S^5,$$ where the latter is the unit sphere in $S^2(\mathbb{R}^3) \simeq \mathbb{R}^6$. Indeed, $k$ maps $$(\mathbb{C}^2 \setminus \{ \mathbf{0} \}) \times (\mathbb{C}^2 \setminus \{ \mathbf{0} \}) \to S^2(\mathbb{R}^3) \setminus \{ \mathbf{0} \},$$ and the latter maps onto $S^5$ by the normalization map, with respect to the inner product on $S^2(\mathbb{R}^3)$ induced by the Euclidean inner product on $\mathbb{R}^3$. Note that this composed map $$(\mathbb{C}^2 \setminus \{ \mathbf{0} \}) \times (\mathbb{C}^2 \setminus \{ \mathbf{0} \}) \to S^5$$ is invariant under rescaling each of the $2$ factors of its domain individually, and so induce a smooth map which we are denoting by $\tilde{k}$, from $\mathrm{P}^1_\mathbb{C} \times \mathrm{P}^1_\mathbb{C}$ into $S^5$. The fibers of $\tilde{k}$ are actually of the form $$(\mathbf{u}, \mathbf{v}), (j\mathbf{u}, j\mathbf{v}), (\mathbf{v}, \mathbf{u}), (j\mathbf{v}, j\mathbf{u})$$ where $\mathbf{u} = [u_0:u_1]$, $\mathbf{v} = [v_0:v_1]$ are points on $\mathrm{P}^1_\mathbb{C}$ and $j$ is the "antipodal map" which was previously defined. Note that $h(j\mathbf{u}) = -h(\mathbf{u})$. We are now ready to define our map $$f: \mathrm{P}^2_\mathbb{C} \to S^5.$$ Given a point $p \in \mathrm{P}^2_\mathbb{C}$, let $w \in g^{-1}(p)$ and define $$f(p) = \tilde{k}(w).$$ Then $f$ is a well defined smooth map from $\mathrm{P}^2_\mathbb{C}$ into $S^5$, which is invariant under the real structure $\sigma$, which was previously defined. In fact, $f$ is a double cover onto its image (a codimension $1$ subset of $S^5$) which is branched over the real slice of $\mathrm{P}^2_\mathbb{C}$ with respect to $\sigma$. A generic fiber of $f$ is a pair of $\sigma$-conjugate points in $\mathrm{P}^2_\mathbb{C}$. Note that if we think of the coordinates of $S^2(\mathbb{R}^3)$ as the components of a real $3$-by-$3$ symmetric matrix $A$, then it is not too difficult to see that $\tilde{k}$ maps $\mathrm{P}^1_\mathbb{C} \times \mathrm{P}^1_\mathbb{C}$ into the real quasi-affine variety $$V = \{ \det(A) = 0 \} \cap \{ \operatorname{tr}(A^2) = 1 \} \cap \{ \operatorname{tr}(A)^2 \leq 1 \}.$$ In other words, these conditions ensure that the eigenvalues of $A$, which must be real, are of the form: $0$, $\lambda$, $\mu$ with $\lambda \mu \leq 0$ and $\lambda^2 + \mu^2 = 1$ (note that $\lambda$, or $\mu$, may be $0$). Hence the image of $f$ is contained in $V$. Question 2: is the image of $f$ equal to $V$? Edit: I think the image of $f$ is indeed $V$. Just note that it suffices to diagonalize $A$, and show that a diagonal matrix having $0$, $\lambda$ and $\mu$ as (real) eigenvalues and satisfying the previous conditions is in the image of $f$. And this is straightforward. Finally, I suspect I am just rediscovering that the complex projective plane modulo complex conjugation is the $4$-sphere, except that instead of complex conjugation, I am using a different real structure. Indeed, this "folklore" result is proved and discussed for instance in Michael Atiyah, Jurgen Berndt, Projective planes, Severi varieties and spheres, Surveys in Differential Geometry VIII, Papers in Honor of Calabi, Lawson, Siu and Uhlenbeck, International Press (2003) pp.1-27. doi:10.4310/SDG.2003.v8.n1.a1, arXiv:math/0206135. Question 3: is the image of $f$ diffeomorphic to $S^4$? If so, then it would provide yet another proof of the previous folklore result. A related question is whether or not $V$ is diffeomorphic to $S^4$. Edit: I think I can build a diffeomorphism from $S^4$ onto $V$. Think of $S^4$ as the set $$W = \{ B \,|\, \text{$B$ real symmetric $3$-by-$3$, } \operatorname{tr}(B) = 0 \text{ and } \operatorname{tr}(B^2) = 1 \}.$$ It is not too difficult to see that $W$ is diffeomorphic to $S^4$. Define a map from $W$ into $V$, by $$ B \mapsto \frac{B - \lambda_2(B) I}{\lVert B - \lambda_2(B)I \rVert},$$ where $\lambda_1(B) \leq \lambda_2(B) \leq \lambda_3(B)$ are the $3$ eigenvalues of $B$. I think that this map is perhaps a diffeomorphism from $W$ onto $V$. However, I am not sure about its smoothness when $2$ eigenvalues of $B$ collide. Can someone comment on that please? REPLY [28 votes]: I think I can prove the following Claim There is no topological embedding of $\mathbb CP^2$ into $\mathbb R^6$. The proof uses the van Kampen obstruction. Let me review the idea. Suppose there is a topological embedding $f\colon \mathbb CP^2\hookrightarrow\mathbb R^6$. Then $f$ induces a $\Sigma_2$-equivariant map of deleted squares $$ f^2_\Delta\colon \mathbb CP^2\times \mathbb CP^2\setminus \mathbb CP^2 \to \mathbb R^6\times \mathbb R^6\setminus \mathbb R^6. $$ Let $\widetilde S^5$ denote the $5$-dimensional sphere with the antipodal action of $\Sigma_2$. There is a $\Sigma_2$-equivariant map (in fact a homotopy equivalence) $$ \mathbb R^6\times \mathbb R^6\setminus \mathbb R^6 \xrightarrow{\simeq} \widetilde S^5. $$ It follows that a topological embedding $f$ would induce a $\Sigma_2$-eqivariant map $$ \mathbb CP^2\times \mathbb CP^2\setminus \mathbb CP^2 \to \widetilde S^5. $$ So to prove that there is no topological embedding, it is enough to prove that there is no such map. An equivariant map like this is essentially the same things as a nowhere vanishing section of the vector bundle $$ (\mathbb CP^2\times \mathbb CP^2\setminus \mathbb CP^2)\times_{\Sigma_2} {\widehat {\mathbb R}}^6 \to (\mathbb CP^2\times \mathbb CP^2\setminus \mathbb CP^2)_{\Sigma_2}. $$ Here $\widehat {\mathbb R}^6$ is the $6$-dimensional sign representation of $\Sigma_2$. The Euler class of this vector bundle is an obstruction to the existence of a section, and therefore to the existence of a topological embedding. This is the van Kampen obstruction. It remains to prove that the Euler class is non-zero. All cohomology groups will be taken with mod 2 coefficients. The Euler class is an element of $H^6\left((\mathbb CP^2\times \mathbb CP^2\setminus \mathbb CP^2)_{\Sigma_2}\right)$. The cohomology ring of $\left(\mathbb CP^2\times \mathbb CP^2\setminus \mathbb CP^2\right)_{\Sigma_2}$ was calculated in the following paper Samuel Feder, The reduced symmetric product of projective spaces and the generalized Whitney theorem, Illinois J. Math. 16 (1972), 323–329 https://doi.org/10.1215/ijm/1256052288 If I am parsing the result of this paper correctly, the cohomology ring is generated by two elements $u_1, x_2$, subject to just the relations $x_2^3=0$ and $u_1^3=u_1x_2$. It follows that as a vector space, the cohomology has following basis $$ 1, u_1, u_1^2, x_2, u_1x_2=u_1^3, u_1^2x_2=u_1^4, x_2^2, u_1x_2^2=u_1^5, u_1^2x_2^2=u_1^6. $$ The main point is that $u_1^6\ne 0$. Clearly $u_1$ is the Euler class of the $1$-dimensional sign representation, so $u_1^6$ is the Euler class of the $6$-dimensional sign representation.<|endoftext|> TITLE: Expected distance between two uniform points in distinct rectangles QUESTION [8 upvotes]: Are there any good approximations (especially upper bounds) for the quantity $E(\|X_1-X_2\|$), where each $X_i$ is uniformly distributed in a rectangle $[a_i,b_i]\times[c_i,d_i]$? It does not appear that I can do this analytically, but I am in a situation where I need to compute hundreds of thousands of these. Obviously a Manhattan norm approximation would be tractable, but I'd like something tighter (or even better, something that can be made arbitrarily tight by tuning parameters). REPLY [4 votes]: I tried to implement my proposal in a C-code. That is a mixture of analytic and numeric integration. It does $10^6$ rectangles with half-percent relative precision in about 16 seconds, which is a bit better than the corresponding Iosif's 30 minutes. You can play with parameters to trade speed for precision and vice versa too. The code should be self-explanatory but feel free to ask questions if something is unclear. Edit: This is the best and the fastest version. $n$ is gone now and the guaranteed relative precision is $1/N^2$ (the constant $1$ is correct, so if you want $10^{-3}$ accuracy (to compare with Mathematica time), just set $N=34$ and get $10^6$ pairs in under 10 seconds. The time is essentially proportional to $N$. For $10^{-5}$ accuracy $N=340$ and 83 seconds suffice. I'll explain the algorithm a bit later; now it makes sense :-) Edit 2: The outline of the algorithm. We shall use the averaging over the projections. If we take the discrete set of $N$ equally spaced lines $L_j$, then the approximate formula is $$ |z|\approx \frac{\pi}{2}\frac 1N\sum_{j=1}^N |P_j z| $$ where $P_j$ is the orthogonal projection operator to the line $L_j$. The relative accuracy of this approximation can be easily computed and is, as I said, $1\pm N^{-2}$. The computation of the average projection is going to be exact. For each projection, we need to evaluate the convolution of $A(z)=|z|$ with four normalized characteristic functions $F_j$ of intervals $[-U_j,U_j]$ at some point $x$. We arrange $U_j$ in the increasing order, so that $U_0 #include #include #include const double pi=3.141592653589, ppi=pi/57.6, pi2=pi/2, dl=sqrt(0.6)/2; double gghh(double a, double b, double c, double d, double x, double t) { double y=fabs(t), g=y*a, h=2*d; if(y<=a-b) g=(a*a+y*y+b*b/3)*0.5; else if(yc-d) h-=(y-c+d); return g*h; } double F(double a, double b, double c, double d, double x, double aa, double bb) { double t2=(aa+bb)*0.5, bbaa=bb-aa, dt=dl*bbaa; return bbaa*(gghh(a,b,c,d,x,t2-dt)+gghh(a,b,c,d,x,t2+dt)+1.6*gghh(a,b,c,d,x,t2))/(a*c*d); } double D(double a1,double b1, double c1, double d1, double a2,double b2, double c2, double d2, int N) { double s=0.0; double X1=b1-a1, Y1=d1-c1, X2=b2-a2, Y2=d2-c2, S1=(a2+b2-a1-b1), S2=(c2+d2-c1-d1); double t0=pi2/N, cs=cos(t0), ss=sin(t0), dcs=2*cs*cs-1, dss=2*cs*ss; double SS=fabs(S1)+fabs(S2)+fabs(X1)+fabs(X2)+fabs(Y1)+fabs(Y2); SS*=0.00000001; for(int k=0; kU[j+1]) {double u=U[j]; U[j]=U[j+1]; U[j+1]=u;} } double U0=U[0], U1=U[1], U2=U[2], U3=U[3]; double V[4]={-U3-U2,-U3+U2,U3-U2,U3+U2}, VV[4]={x-U1-U0,x-U1+U0,x+U1-U0,x+U1+U0}; double W[8]; int i=0, ii=0, kstart=-1, kfinish=-1; while(ii<4) { ++kfinish; if(V[i]M) M=r; } } printf("\n%.12f",D(1,2,3,5,4,6,7,8,4000)); printf("\n%.12f",D(1,2,3,5,4,6,7,8,N)/D(1,2,3,5,4,6,7,8,2000)-1); printf("\n%.12f",D(0,2,0,2,0,2,0,2,N)/D(0,2,0,2,0,2,0,2,2000)-1); printf("\n%.12f",D(0,3,0,0.0001,0,3,0,0.0001,N)-1); printf("\n%.12f",D(0,2,0,0,0,0,0,2,N)/D(0,2,0,0,0,0,0,2,2000)-1); printf("\n%.12f",D(0,0,0,0,3,3,4,4,N)/5-1); return 0; }<|endoftext|> TITLE: Are equivariant perverse sheaves constructible with respect to the orbit stratification? QUESTION [5 upvotes]: [Moved here from MSE] Consider a variety $X$ over a field $k$ (complex numbers is fine) with the action of a group scheme $G$, and a $G$-equivariant perverse sheaf $F$ over $X$. Question. Is it true that there exists a stratification $\tau$ of $X$ which is $G$-equivariant and such that $F$ is $\tau$-constructible? For example, one could inspect the orbit stratification. I am trying to use the characterization of invariant perverse sheaves as those perverse sheaves such that $act^* F\simeq pr_2^*F$ where $act \colon G\times_k X\to X$ is the action and $pr_2:G\times X\to X$ is the second projection. But I cannot find the solution. REPLY [8 votes]: Every complex of sheaves has a unique maximal open subset on which it is locally constant, because if it is locally constant on two open sets, it is locally constant on their union. Let $U$ be then maximal open subset of $X$ where $F$ is locally constant. Then $U$ is $G$-invariant, because $gU$ is also the maximally open set on which $F$ is locally constant and so $U= gU$. The complement $X - U$ of $X$ is also $G$-invariant, and $F$ remains $G$-equivariant on restriction to $X - U$. Now we can induct - take the maximal open subset of $X - U$ on which $G$ is locally constant, check it is $G$-invariant, and restrict $F$ to its complement, and so on.<|endoftext|> TITLE: Examples of categories cofibered in groupoids QUESTION [5 upvotes]: In Chapter 2 of Lurie's Higher Topos Theory, the first main theorem establishes a connection between categories cofibered in groupoids and left fibrations and asserts the importance of studying left fibrations in $\infty$-category theory. However, I lack understanding of the importance of categories cofibered in groupoids in ordinary category theory. Are there any good examples that illustrate the power of the notion of a category cofibered in groupoids? REPLY [8 votes]: Before getting into high-falutin' stacky considerations, I think there's something much more basic to say. Let $C$ be a 1-category. There is an equivalence between discrete fibrations over $C$ and functors $C^{op} \to Set$, i.e. presheaves. Here, "discrete fibration over $C$" is another name for "category fibered in sets over $C$", and the $\infty$-categorical analog is called a "right fibration" by Lurie (or maybe it's "left fibration" -- I get confused). Dually, there is an equivalence between discrete opfibrations over $C$ and functors $C \to Set$, i.e. copresheaves. One direction of this equivalence is usually called the Grothendieck construction, or the category of elements in 1-category theory. The $\infty$-categorical version was re-christened as straightening / unstraightening by Lurie. That is, fibered categories provide a language which is equivalent to the language of presheaves / copresheaves. The central importance of presheaves to category theory is perhaps more familiar, starting from their role in the Yoneda lemma. Now, the jump from categories cofibered in sets to categories cofibered in groupoids is essentially the jump from functors $C \to Set$ to functors $C \to Gpd$. Technically, in the world of 1-categories, this is not quite true because $Gpd$ is really a $(2,1)$-category, so talking about functors to the "underlying" $(1,1)$-category is usually a mistake -- such functors are "too strict". As a result, the correct statement is that categories cofibered in groupoids over $C$ are equivalent to pseudofunctors $C \to Gpd$. This equivalence is also referred to as the Grothendieck construction. In $\infty$-categories, we simply don't have a corresponding notion of strict functor here, so the analog of a pseudofunctor is just called a "functor". In $\infty$-category land, $\infty$-groupoids are playing the role previously played by sets in 1-category land. One manifestation of this is that a set is just a 0-truncated $\infty$-groupoid. But similarly, a groupoid is just a 1-truncated $\infty$-groupoid. So there are things in 1-category land which you have to do with groupoids rather than sets (and some of the other answers contain good examples of this), whereas when you do similar things in $\infty$-category land, you just use $\infty$-groupoids in both roles. So the moral of this story is that when Lurie talks about the importance of categories (co)fibered in groupoids, an extremely important special case of this is the importance of categories (co)fibered in sets, i.e. the importance of (co)presheaves. All of this generalizes to categories (co)fibered in categories. In 1-category land, this is the general notion of a Grothendieck fibration / Grothendieck opfibration over $C$. Such things correspond, via the Grothendieck construction, to pseudofunctors $C^{op} \to Cat$ (respectively $C \to Cat$). In $\infty$-category land, this construction is again called straightening / unstraightening by Lurie, and he calls Grothendieck fibrations cartesian fibrations and he calls Grothendieck opfibrations cocartesian fibrations. The advantage of discrete fiberations over prehseaves of sets is minimal; fibered categoies were introduced by Grothendieck precisely to avoid the annoying technical coherence conditions associated with working with pseudofunctors valued in $Cat$ or $Gpd$. Lurie uses fibrations in a similar way, to avoid having to constantly write down higher coherence data for functors valued on $Gpd_\infty$ or $Cat_\infty$ (after all, in order to get a model for either of these $\infty$-categories, one usually takes a big homotopy coherent nerve of a simplicial category, so writing functors into them is a chore).<|endoftext|> TITLE: Status of $x^3+y^3+z^3=6xyz$ QUESTION [8 upvotes]: In Erik Dofs, Solutions of $x^3 + y^3 + z^3 = nxyz$, Acta Arithmetica 73 (1995) pp. 201–213, doi:10.4064/aa-73-3-201-213, EuDML the author has studied the Diophantine equation \begin{equation} x^3+y^3+z^3=nxyz\tag{1} \end{equation} where $n$ is an integer. I am interested in the specific case $n=6$. The author mentions: For a fixed $n$-value, (1) can be transformed into an elliptic curve with a recursive solution structure derived by the "chord and tangent process". How does this work for the case $n=6$? Does this generate all or infinitely many solutions? What progress has been made in this specific case? REPLY [2 votes]: We show the integer solutions of $x^3+y^3+z^3 = nxyz$ using elliptic curve. $x^3+y^3+z^3 = nxyz\tag{1}$ Let $s=x+y+z$, then equation $(1)$ reduces to $(-3x+3s+nx)y^2+(-3s^2+6sx+nx^2-3x^2-nxs)y+s^3-3s^2x+3sx^2=0.$ This quadratic in y must have rational solutions, so the discriminant must be square number. Hence we obtain $v^2 = (-6n+9+n^2)x^4+(-2n^2s+6sn)x^3+(-18s^2+n^2s^2-6ns^2)x^2+(12s^3+2ns^3)x-3s^4.$ Furthermore, let $U=s/x$ then we obtain $$V^2 = -3U^4+(12+2n)U^3+(-18-6n+n^2)U^2+(6n-2n^2)U+9+n^2-6n\tag{2}.$$ This quartic equation is birationally equivalent to the elliptic curve below. $$Y^2-2nYX+(-72+12n+4n^2)Y = X^3+(-6n-18)X^2+(108+12n^2-72n)X-1944+216n^2+648n-72n^3\tag{3}$$ $U = \frac{\large{-12n^2+2nX-6X+108}}{\large{Y}}$ $V = \frac{\large{-5832+60n^3X+324n^2X-972nX-8n^3Y+72n^4+3888n-1620X+216Y-3X^3+162X^2-432n^3-18n^2X^2+X^3n}}{\large{Y^2}}$ $X = \frac{\large{-6V+18-12n+2nV+2n^2+6Un-2Un^2}}{\large{U^2}}$ $Y = \frac{\large{-4Un^3+4n^3-12U^2n^2+4n^2V+24Un^2-36n^2-36Un-24nV+108n+108U^2-108+36V}}{\large{U^3}}$ If the rank of elliptic curve $(3)$ is positive, we can obtain the infinitely many rational points. Hence we obtain the infinitely many integer solutions of equation $(1)$. The rank and generator are obtainded using mwrank(Cremona). Example of $n=6$. Minimal Weierstrass form is $Y^2+Y=X^3-54X-88$. Rank=1 and generator=$P(-2,3)$. The rational points $(X,Y)$ of $Y^2+Y=X^3-54X-88$ are obtained by group law using Online Magma Calculator as follows. P:=PolynomialRing(RationalField()); E := EllipticCurve([0, 0, 1, -54, -88]); P :=E![-2,3]; 2*P; 3*P; 4*P; 5*P 6*P; 7*P; The rational points $(X,Y)$ of $Y^2+Y=X^3-54X-88$. $1P:[ -2, 3]$ $2P:[ 40, 248]$ $3P:[-143/36, 1621/216]$ $4P:[56404/5041, 9350169/357911]$ $5P:[-15323534/2505889, 12752626540/3966822287]$ $6P:[3494518273/430479504, -31008773293919/8931588748992]$ $7P:[-43215340027190/7570240479649, -132725251679433577707/20828812647389616143]$ In this way, we can obtain the infinitely many rational points of $Y^2+Y=X^3-54X-88$. Hence there are infinitely many integer solutions for equation $(1).$ We can get the solutions using group law as follows. $mP: [ x, y, z]$ $1P:[ 1, 2, 3]$ $2P:[ 52, -21, -19]$ $3P:[ 1817, 3258, 5275]$ $4P:[-2847511, 3096807, -124904]$ $5P:[10840875082, 4904676969, 15051171563]$ $6P:[-150777667094725, 458665691607396, -203863624933571]$ $7P:[81821352777652044467, 29381282043563909553, 46875396961726681714]$ The positive solutions are the case of $P,3P,5P,7P$. We show only some solutions. [n] [rank] [ x y z] [ 3][ ][ 1, 1, 1] [ 5][0][ 1, 2, 1] [ 6][1][ 1, 3, 2] [ 9][1][ 3, 7, 2] [ 10][1][ 5, 18, 7] [ 13][1][ 9, 38, 13] [ 14][1][ 2, 13, 7] [ 15][1][ 7, -1, -3] [ 16][1][ 70, -9, -31] [ 17][1][ 5, 37, 18] [ 18][1][ 95, 42, 13] [ 19][1][ 9, 5, 1] [ 20][1][ 61, -13, -14] [ 21][1][ 2, 21, 13] [ 26][1][ 91, 38, 9] [ 29][1][ 43, 182, 27] [ 30][1][ 31, 21, 2] [ 31][1][ 37, -1, -27] [ 35][1][ 97, -14, -19] [ 36][1][ 151, -7, -78] [ 38][1][ 70, 629, 151] [ 40][1][ 9, -1, -2] [ 41][2][ 1, 9, 2] [ 44][1][ 819, -19, -554] [ 47][1][ 845, -38, -367] [ 51][1][ 9, 77, 13] [ 53][1][ 2, 27, 7] [ 54][1][ 2, 57, 43] [ 57][1][ 91, 310, 19] [ 62][1][13559153, -1513300, -1950953] [ 63][1][ 3775, -247, -903] [ 64][1][1338039, -119479, -232736] [ 66][1][ 3, 14, 1] [ 67][1][ 1133, 23517, 7525] [ 69][2][ 2, 73, 57] [ 70][1][1478979, -27083, -896668] [ 71][1][ 67, -7, -9] [ 72][1][-404512675962, 5450170263655, -1012930784383] [ 73][1][89200900157319, 2848691279889518, 1391526622949983] [ 74][1][ 133, 4607, 2502] [ 76][2][ 45, -2, -13] [ 77][2][ 52, -5, -7] [ 83][1][ 5, 61, 9] [ 84][1][ 56, -1, -31] [ 86][1][ 2, 91, 73] [ 87][1][ 21, -1, -5] [ 92][1][-20446843218005, 35661385544981, -548624531286] [ 94][2][ 19, 945, 746] [ 96][1][ 38, -3, -5] [ 98][1][-2559169, 59978401, -14154192] [ 99][1][14466072543, -1832602198, -1150522313] [101][1][ 1271, 3078, 79] [102][1][459338480695732254, 3816006884967068935, 13212742329826830581] [103][1][ 61, -4, -9] [103][1][58383, -1159, -26024] [105][2][ 2, 111, 91] [106][2][ 1, 54, 35] [107][1][-197624310994, 1329876450605, -83341950251] [108][1][ 39, -2, -7] [109][1][-99054267227, 7254524660292, -4035385003297] [110][1][ 2745, 18578, 1147] [112][1][81634675793306734523552997069865, -66756829882613387041310733449793, -403909122691328588518061393542] [113][1][345842, 6313383, 15170275] [116][1][ 1204, -13, -739] [117][1][ 545, 10318, 1677] [119][1][ 49, -1, -19] [120][1][ 8869, -496, -1317] [122][1][25590382918388481967217, 407249928739620845890, 23848086141138276680923] [123][1][-45191178833837, 10554611259665663, -8723981310176706] [124][1][-1882858151, 4389003335, -75992904] [126][2][ 2, 133, 111] [127][1][ 931, -45, -151] [128][1][1158179, -422318, -23611] [129][2][ 70, 2361, 629] [130][1][16177096946436536530, 639905104499493910311, 1046599292363750394389] [132][1][ 2234, -39, -905] [133][1][691440137111968428652609, 8149000233894575265542178, 27006382877335430051053793] [136][2][ 1118, -45, -203] [142][1][6587432496387235561093636933115859813174, 53881756527432415186060525094013536917351, 222932371699623861287567763383948430761525] [143][1][1636453, -1520593, -2435] [145][1][157591646586434781, 44634584148027469, 1007950541819512850] [147][1][ 21, 1529, 925] [148][1][1418519131294563, -188778746384314, -71841303293459] [149][2][ 45, 14, 1] [151][2][ 133, -9, -13] [154][2][ 2, 63, 13] [155][1][466371, -458, -442729] [156][1][-57378032205801587, 151742066509610694, -2433329851945933] [158][1][5642215349875, 7336556299898, 80828288788977] [159][1][ 31, -1, -6] [160][2][ 3691, -43, -1764] [161][2][ 95, -7, -8] [162][1][ 35, 2881, 1854] [164][1][-2187625242203395484208435, 9967112990856026231233891, -273965892543545308964986] [166][1][ 790, 611, 9] [167][1][3641058343253213, -868179733296745, -90197542563908] [172][1][23593229783585883, -16590668075015195, -127237919517328] [174][2][ 78, 7, 5] [175][1][-12984427575, 55614086497, -1343816497] [177][1][11586299300246645065650175011667633294528995894742493608006903, 499128047096078689216614212030144552460525030444760940918765730, 944421945175253160922633489847006529687358187395396664036033027] [178][1][ 2, 97, 27] [181][1][201705586625136962, 10672860536839861, 21088064331923949] [185][1][ 379, -4, -175] [186][2][ 252, -5, -67] [187][1][492233837876182300994422946725623025365, -50621375726791887196233101919521352691, -25564297137318411424451907466482829394] [189][1][18396, -209, -7891] [190][1][-297115335963207388794859858793856411, 12449186314350611078793751630598133592, -2716813894306138435285959241731689173] [191][1][56059, -1399, -11655] [192][1][ 3345, -38, -1417] [195][2][ 7, 143, 15] [196][2][ 259, -18, -19] [197][1][ 127, 11655, 6278]<|endoftext|> TITLE: Diophantine equation: $\prod_{i=1}^n (i^{i+1}+1)=k^2$ beyond $(n,k) = (4,1230)$? QUESTION [17 upvotes]: The integer $(1^2+1)(2^3+1)(3^4+1)(4^5+1)$ is a square, namely $2^23^25^241^2$. Question. What will be the next occurrence, or is there an occurrence of $$\prod_{i=1}^n (i^{i+1}+1)=k^2?$$ REPLY [22 votes]: The following criterion will most likely cover all large $n$, but actually proving this is out of reach of current technology. Proposition. Let $p$ be a Sophie Germain prime (so that $q := 2p+1$ is also prime) with $p = 11 \hbox{ mod } 12$, such that $(p-1)^p + 1$ is not divisible by $q^2$. Then $\prod_{i=1}^n (i^{i+1}+1)$ is not a square for any $p-1 \leq n < 2p$. Proof. Suppose that one of the factors $i^{i+1}+1$ is divisible by the prime $q$ for some $i \leq n$. Then $i^{i+1} = -1 \hbox{ mod } q$. Since the multiplicative group of ${\bf F}_q$ is a cyclic group of order $q-1 = 2p$, we conclude that either $i+1$ is equal to $p$ mod $2p$, or $i = -1 \hbox{ mod } q$ and $i+1$ is odd. Since $i \leq n < 2p \leq 3p-1$, the first case only occurs at $i=p-1$; since $i \leq n < 2p = q-1$, the second case does not occur at all. Hence the only factor in $\prod_{i=1}^n (i^{i+1}+1)$ that could be divisible by $q$ is $(p-1)^p + 1$. Modulo $q$, $p-1$ is equal to $-3/2$, hence $(p-1)^p + 1$ is equal to $-(3/2)^p + 1$ mod $q$. By quadratic reciprocity, $3/2$ is a square mod $q$ when $p = 11 \hbox{ mod } 12$, so $(p-1)^p + 1$ is divisible by $q$, but by hypothesis it is not divisible by $q^2$. Hence the product is not a perfect square. $\Box$ (A slight generalisation of) the Hardy-Littlewood conjecture implies that there are infinitely many Sophie Germain primes $p$ with $p = 11 \hbox{ mod } 12$ (indeed their density in any large interval $[x,2x]$ would conjecturally be of the order of $1/\log^2 x$). The condition $(2p+1)^2 \not | (p-1)^p + 1$ is reminiscent of a non-Wieferich prime condition and it is likely that only very few Sophie Germain primes fail this condition, but this is challenging to prove rigorously. (For instance, even with the ABC conjecture, the number of non-Wieferich primes is only known to grow logarithmically, a result of Silverman.) Nevertheless, this strongly suggests that the above Proposition is sufficient to rule out squares for all sufficiently large $n$. Each prime $p$ obeying the above conditions clears out a dyadic range of $n$, for instance $p=11$ obeys the conditions and so clears out the range $10 \leq n < 22$ (for all $n$ in this range the product is divisible by precisely one factor of the prime $23$). An algorithm to search for primes obeying these conditions should then keep essentially doubling the range for which no further squares are guaranteed to exist rather rapidly and numerically establish quite a large range of $n$ devoid of any additional squares. EDIT: Here is a justification of why $q^2 | (p-1)^p + 1$ is a Wieferich type condition. If this holds, then on squaring we have $$ (p-1)^{2p} = 1 \hbox{ mod } q^2$$ hence on multiplying by $p-1$ $$ (p-1)^q = p-1 \hbox{ mod } q^2$$ and then multiplying by $2^q$ $$ (q-3)^q = 2^{q-1} (q-3) \hbox{ mod } q^2.$$ By the binomial theorem and Fermat's little theorem we have $(q-3)^q = -3^q \hbox{ mod } q^2$ and $2^{q-1} q = q \hbox{ mod } q^2$, hence $$ -3^q = q - 3 \times 2^{q-1} \hbox{ mod } q^2$$ which simplifies to $$ q^2 | 3 \times 3^{q-1} - 3 \times 2^{q-1} + q$$ which can be viewed as a variant of the Wieferich condition $$ q^2 | 2^{q-1} - 1.$$<|endoftext|> TITLE: Is there any Lie groupoid structure on $Hom(\mathcal{G}, \mathcal{H})$ where $\mathcal{G}$ and $\mathcal{H}$ are Lie groupoids? QUESTION [6 upvotes]: We know that in general, there is no smooth manifold structure on $Hom(X,\, Y)$ where $X$ and $Y$ are smooth manifolds, but under certain nice conditions (see https://ncatlab.org/nlab/show/manifold+structure+of+mapping+spaces) we can give a smooth structure on $Hom(X, \, Y)$. Let $\mathcal{G}$ and $\mathcal{H}$ be two Lie groupoids. Now let us consider the category $Hom(\mathcal{G}, \, \mathcal{H})$ whose objects are homomorphisms of Lie groupoids and the morphisms are smooth natural isomorphisms. Question 1. Under what conditions on $\mathcal{G}$ and $\mathcal{H}$, we have a (canonical) Lie groupoid structure on $Hom(\mathcal{G}, \, \mathcal{H})$? Question 2. Is $Hom(\mathcal{G}, \, \mathcal{H})$ always a diffelogical groupoid in general? It would be also great if someone can suggest some literature in this direction. REPLY [2 votes]: As Dmitri points out, given a cartesian closed category $S$, the groupoid of functors and natural transformations between fixed internal groupoids $X$ and $Y$ is again an internal groupoid: this result goes back to Charles Ehresmann, but it is not difficult to write down this construction directly. However, since you mentioned the infinite dimensional manifolds of smooth maps between manifolds, then I'd like to push back against Dmitri's claim of 'almost never', since you clearly aren't just thinking of manifolds as being finite dimensional. In DMR, Raymond Vozzo, Smooth loop stacks of differentiable stacks and gerbes, Cahiers de Topologie et Géométrie Différentielle Catégoriques, Vol LIX no 2 (2018) pp 95-141 journal version, arXiv:1602.07973 we show that given a finite open cover $\{U_i\}$ of $I$, or of $S^1$ with the property that triple intersections are empty, the hom-groupoid $\mathbf{LieGpd}(\check{C}(U),X)$ is a Fréchet Lie groupoid for any finite-dimensional Lie groupoid $X$. Here $\check{C}(U)$ is the Lie groupoid with objects $\bigsqcup_i U_i$ and morphisms $\bigsqcup_{i,j} U_i\cap U_j$. A priori this is just a diffeological groupoid but we show the spaces of objects and morphisms are Fréchet manifolds and the source and target maps are submersions (in the strong sense that there are submersion charts, not that tangent spaces map surjectively). In the short announcement paper DMR, Raymond Vozzo, The smooth Hom-stack of an orbifold, In: Wood D., de Gier J., Praeger C., Tao T. (eds) 2016 MATRIX Annals. MATRIX Book Series, vol 1 (2018) doi:10.1007/978-3-319-72299-3_3, arXiv:1610.05904, MATRIX hosted version we make the more general claim that given a compact manifold $M$, and a finite open cover satisfying a certain minimality condition (and a topological condition on finite intersections of their closures), the analogous hom-groupoid is also a Fréchet Lie groupoid. The longer paper containing the more delicate proofs for this case is still in preparation, but halted due to other commitments by its authors. I suspect that these results might be able to be pushed a tiny bit further, say to the case where $\check{C}(U)$ is replaced by the analogous thing that arises from a finite open cover of a compact orbifold, but that is just intuition, we aren't pursuing that line of inquiry. Added I should have said, given a compact manifold $M$, the groupoid $\mathbf{LieGpd}(M,X)$ is Fréchet–Lie as well. If one is wiling to have more general smooth manifolds, then taking $M$ to be non-compact this is a Lie groupoid modelled on merely locally convex spaces. The topology has to be chosen carefully, it's the sort of thing my co-author Alexander Schmeding works on.<|endoftext|> TITLE: With a linear representation, how does the continuity of $G \to \mathrm{GL}(V)$ relate to that of $G \times V \to V$? QUESTION [6 upvotes]: I'm currently reading Traces of Hecke Operators by Knightly and Li, while simultaneously revisiting the adelic/representation-theoretic point of view on automorphic forms. In Knightly and Li, they give a familiar definition of a representation. That is, for a locally compact group $G$ and a normed vector space $V$, they note that a representation is a homomorphism $$ \pi: G \longrightarrow \mathrm{GL}(V) $$ such that the map $$\begin{align} G \times V &\longrightarrow V \\ (g,v) &\mapsto \pi(g)v \end{align}$$ is continuous. Sometimes I've seen this stated first in terms of the continuity of $g \mapsto \pi(g)v$. But I note that both $G$ and $\mathrm{GL}(V)$ are topological groups, so it would make sense to consider the continuity of $\pi$ directly as a function $G \longrightarrow \mathrm{GL}(V)$. I don't know how the continuity of $G \to \mathrm{GL}(V)$ relates to the continuity of $G \times V \to V$. Intuitively, I suspect that the continuity of $G \times V \to V$ implies continuity of the representation map $G \to \mathrm{GL}(V)$, but not the converse. Is this right? REPLY [9 votes]: For locally compact groups continuity of $\pi$, joint continuity of the action map $G\times V\to V$ and separate continuity in both variables are equivalent. See Theorem 2.3 of Karl H. Hofmann, Sidney A. Morris The Structure of Compact Groups (edition 3).<|endoftext|> TITLE: Random permutations without double rises (avoiding consecutive pattern $\underline{123}$) QUESTION [7 upvotes]: A permutation avoiding a consecutive pattern $\underline{123}$ is permutation $\pi = \pi_1 \pi_2 \ldots \pi_n$ with the property that there does not exists $i \in [1, n-2]$ such that $\pi_i < \pi_{i+1} < \pi_{i+2}$. Example: $53241$ is $\underline{123}$-avoiding; while $314562$ is not $\underline{123}$ avoiding, as it contains $145$. Let $a_n$ be the number of $\underline{123}$ avoiding permutations, it is known, in corresponding OEIS entry A049774, that $$ \frac{a_n}{n!} \sim e^{\frac{\pi}{3\sqrt{3}}} \left(\frac{3\sqrt{3}}{{2\pi}}\right)^{n+1}.$$ So, the simple rejection sampling is inefficient even for moderate values of $n$. How does one generate such permutations uniformly at random? I'm aware of Boltzmann sampling. But maybe you are aware of more simple and/or faster algorithms as in the case of alternating permutations. REPLY [2 votes]: I have a sampling algorithm for you written in sage / python. Instead of $\underline{123}$ avoiding permutations of length $n$ I consider words $w_1 w_2 \dots w_n$ satisfying the following properties: $0\le w_k \le n-k$ the word is weakly $\underline{123}$ avoiding, which means there is no index $i\in [1,n-2]$ such that $w_i\le w_{i+1} \le w_{i+2}$. I call these words $\underline{123}$ avoiding indexwords. They are in bijection with $\underline{123}$ avoiding permutations using the following bijection: Let $S=\{1,2,\dots,n\}$ then for a given indexword $w_1 w_2 \dots w_n$ we define $\pi_i = w_i\text{-th smallest entry of } S\setminus\{\pi_1,\dots,\pi_{i-1}\}$, where I start indexing the entries of the set with $0$. Put differently: If we keep track of the unused letters to build up the permutation in an ordered list, then $\pi_i$ is precisely the $w_i\text{-th}$ entry in that list (explaining my name for these words). These words will simplify the following algorithm for random uniform sampling. The algorithm comes in two steps: First count the number of $\underline{123}$ avoiding indexwords with a given prefix. Then create a random word iteratively by appending letters to the already constructed word using certain probabilities using the precomputed values. For this let $\alpha_n^{w_1 w_2 \dots w_l}$ be the number of $\underline{123}$ avoiding indexwords of length $n$ starting with $w_1 w_2 \dots w_l$. For calculating these values, we have two rules: Let $w_1 w_2 \dots w_l$ be weakly $\underline{123}$ avoiding, then Expansion rule $$ \alpha_n^{w_1 w_2 \dots w_l} = \sum_{0\le x\le n-l-1 \text{ and } \neg(w_{l-1}\le w_l\le x)} \alpha_n^{w_1 w_2 \dots w_l x} $$ Reduction rule $$ \alpha_n^{w_1 w_2 \dots w_l} = \alpha_{n-l+2}^{w_{l-1}w_l} $$ Start cases $$ \alpha_2^{00}=\alpha_2^{10}=1 $$ Combining the rules we obtain $$ \alpha_n^{xy}=\sum_{0\le z\le n-3 \text{ and } \neg(x\le y\le z)}\alpha_{n-1}^{yz} $$ Note that you can obtain $a_n$ by summing $\alpha_n^{xy}$ over all pairs $(x,y)\in [0,n-1]\times [0,n-2]$ Now for the sampling part. Randomly choose a pair $(x,y)\in [0,n-1]\times [0,n-2]$ with probability $\frac{\alpha_n^{xy}}{a_n}$. Assume now, we have already created a sequence $w_1 \dots w_l$ with $l1, "n needs to be at least 2" if n == 2: return 1 return sum(alpha(n-1,y,z) for z in range(n-2) if not (x<=y<=z)) def random_indexword(n): assert n>=2, "Algorithm only works for length n>=2" pairs = [(x,y) for x in range(n) for y in range(n-1)] P = [alpha(n,x,y) for x,y in pairs] s1,s2 = pairs[GeneralDiscreteDistribution(P).get_random_element()] indexword = [s1,s2] while len(indexword) TITLE: Invertible elements of the Hopf algebra quantum $SU(2)$ QUESTION [5 upvotes]: Let $SU_q(2)$ be the (polynomial) Hopf algebra introduced by Woronocicz called the quantum special unitary group. For details see https://en.wikipedia.org/wiki/Compact_quantum_group (Note that on the Wikipedia page it is the $C^*$-algebra that is discussed, but this question is about the dense Hopf algebra of the $C^*$-algebra.) Does $SU_q(2)$ contain any (non-unital) invertible elements? REPLY [5 votes]: Yes, the nonzero multiples of the identity $1$ are the only invertible elements in this algebra. I am sure that someone with more expertise in Hopf algebras than I, can provide a 'high level' proof of the result. The following is a 'low level' direct argument. For the proof, I am using some notations and basic results from Woronowicz's original paper [W]. For $\nu \in (0,1) \cup (-1,0)$, we are considering the unital $*$-algebra $A$ generated by the elements $a,b$ subject to the relations $$a^*a + b^* b = 1 \;\; , \;\; aa^* + \nu^2 b^* b = 1 \;\; , \;\; bb^* = b^*b \;\; , \;\; ab = \nu ba \;\; , \;\; ab^* = \nu b^* a \; .$$ In [W] it is proven that the elements $a^k b^n (b^*)^m$ and $(a^*)^k b^n (b^*)^m$ form a vector space basis of $A$. Define for $k \geq 0$, the subspace $A(k) \subset A$ as the linear span of $a^k b^n (b^*)^m$ with $n,m \geq 0$. For $k \leq 0$, define $A(k)$ as the linear span of $(a^*)^k b^n (b^*)^m$ with $n,m \geq 0$. One has $A(k_1) A(k_2) \subset A(k_1 + k_2)$ for all $k_1,k_2 \in \mathbb{Z}$. Another ingredient is the $*$-representation $\pi$ of $A$ on the Hilbert space $H = \ell^2(\mathbb{N} \times \mathbb{Z})$, also defined in [W], and given by $$\pi(a) e_{i,j} = \sqrt{1-\nu^2} e_{i-1,j} \;\; , \;\; \pi(b) e_{i,j} = \nu^n e_{i,j+1} .$$ Write $B = A(0)$, so that $B$ is the unital and abelian $*$-subalgebra of $A$ generated by $b$. We can view $B$ as the algebra of polynomials in two variables. We will use that $B$ has no zero divisors and that the multiples of $1$ are the only invertible elements in $B$. For every $k \in \mathbb{N}$, denote by $H_k$ the closed linear span of $e_{i,j}$ with $i \geq k$, $j \in \mathbb{Z}$. We use the following observations: for $k \geq 0$, the range of $\pi(a^k)$ is dense in $H$, while the range of $\pi((a^*)^k)$ is dense in $H_k$. Also, every $H_k$ is an invariant subspace for $\pi(B)$ and the resulting representation of $B$ on $H_k$ is faithful for every $k \geq 0$. Assume that $x,y \in A$ and $xy = 1$. Write $x = \sum_n x_n$ and $y = \sum_n y_n$ with $x_n,y_n \in A(n)$. Let $n_0$, $m_0$ be the largest integers with $x_{n_0} \neq 0$ and $y_{m_0} \neq 0$. Since $xy = 1$, we must have that $n_0 + m_0 \geq 0$. The component of $xy$ in $A(n_0 + m_0)$ is given by $x_{n_0} y_{m_0}$. We claim that $n_0 + m_0 = 0$. Assume that $n_0 + m_0 > 0$. Then, $x_{n_0} y_{m_0} = 0$. Using the notation $a_k = a^k$ for $k \geq 0$ and $a_k = (a^*)^{-k}$ for $k \leq 0$, we can uniquely write $x_{n_0} = a_{n_0} P$ and $y_{m_0} = Q a_{m_0}$ with $P,Q \in B$. We get that $$\pi(a_{n_0}) \pi(PQ) \pi(a_{m_0}) = 0 .$$ It follows that $\pi(PQ)$ is zero on $H_k$ for $k$ large enough. Hence, $PQ = 0$. This forces $P = Q = 0$, contradicting our choice of $n_0$ and $m_0$. So the claim is proven. Similarly define the smallest integers $n_1$, $m_1$ such that $x_{n_1} \neq 0$ and $y_{m_1} \neq 0$. The same reasoning leads to $n_1 + m_1 = 0$. It follows that $n_1 = n_0$ and $m_1 = m_0$. Exchanging the roles of $x$ and $y$ if necessary, we may assume that $m_0 \geq 0$ and $n_0 = -m_0$. This means that $x = (a^*)^{m_0} P$ and $y = Q a^{m_0}$ for some $P,Q \in B$. It follows that $$\pi(a^*)^{m_0} \pi(PQ) \pi(a^{m_0}) = 1 .$$ If $m_0 \geq 1$, the left hand side has a nontrivial kernel. Thus, $m_0 = 0$. It then follows that $PQ = 1$, which implies that $P$ and $Q$ are a multiple of $1$. We have proven that $x$ and $y$ are multiples of $1$.<|endoftext|> TITLE: Convergence of $\exp(tQ)$ in operator norm as $t\rightarrow\infty$ QUESTION [7 upvotes]: This is not a homework problem, so I am not sure whether this has a "good" answer or not. I came up with this question when I am now learning functional analysis and wonder whether my "freshman's intuition" for exponential works. If $Q$ is some bounded linear operator on some Banach space(and maps to the same space), then since it is bounded, we can define the exponential operator $$P(t)=\exp(tQ)$$ for every $t>0$. It is again a bounded linear operator for every fixed $t$. Question. What is a sufficient condition on $Q$ (if necessary, better!) for $\{P(t)\}$ to converge in operator norm to some operator as $t\rightarrow\infty$? My intuition tells me there should be something of $Q$ being negative--at least non-positive, resulting in $P(t)\rightarrow 0$ or some other operator. If the spectrum of $Q$ is contained in $\{z\in\mathbb{C}\, :\, \operatorname{Re}(z)<0\}$, will this convergence happen? However, I failed to link this convergence to $Q$'s spectrum. REPLY [7 votes]: Convergence to $0$ is simple: Proposition 1. The following are equivalent: (i) The operator $e^{tQ}$ converges to $0$ with respect to the operator norm as $t \to \infty$. (ii) The spectrum of $Q$ is contained in the open left halfplane $\{\lambda \in \mathbb{C}: \, \operatorname{Re} \lambda < 0\}$. Sketch of proof. As mentioned by Pietro Majer in the comments, this follows from the spectral mapping theorem for the operator exponential function, along with the spectral radius formula. $\square$ Convergence to a non-zero operator is a bit more involved. Here are the details: Theorem 2. The following are equivalent: (i) The operator $e^{tQ}$ converges with respect to the operator norm to a non-zero operator as $t \to \infty$. (ii) The spectrum of $Q$ is contained in the set $\{\lambda \in \mathbb{C}: \, \operatorname{Re} \lambda < 0\} \cup \{0\}$, and $0$ is an isolated point in the spectrum and a first order pole of the resolvent of $Q$. If the equivalent assertions are satisfied, then the limit $\lim_{t \to \infty} e^{tQ}$ equals the spectral projection of $Q$ associated with the isolated spectral value $0$. Sketch of proof. "(i) $\Rightarrow$ (ii)" Let $P$ denote the limit operator; it commutes with $e^{tQ}$ for each $t$. It's easy to check that the range of $P$ consists precisely of the fixed points of the operator semigroup $(e^{tQ})_{t \in [0,\infty)}$ and consequently, we can see that $P$ is a projection. Since the projection commutes with the semigroup, both the range and the kernel of $P$ are invariant under the semigroup. On the kernel of $P$, we have that $e^{tQ}$ converges in operator norm to $0$ as $t \to \infty$, so, according to Proposition 1, the restriction $Q|_{\ker P}$ has spectrum in the open left halfplane. On the other hand, $e^{tQ}$ acts as the identity operator on the range $\operatorname{rg}P$. Since the space $\operatorname{rg}P$ is non-zero, the number $0$ is a spectral value of the restriction $Q|_{\operatorname{rg}P} = 0$, and a first order pole of its resolvent. "(ii) $\Rightarrow$ (i)" Let $P$ denote the spectral projection of $Q$ associated with the isolated spectral value $0$. Since $0$ is a first order pole of the resolvent, it follows that the restriction $Q|_{\operatorname{Rg}P}$ is the $0$ operator, so $e^{tQ}$ acts as the identity operator on $\operatorname{Rg}P$. On the other hand, the restriction $Q|_{\ker P}$ has spectrum in the open left halfplane, so Proposition 1 tells us that $e^{tQ}$ norm converges to $0$ on $\ker P$ as $t \to \infty$. So to sum up, on the whole space we have convergence of $e^{tQ}$ to $P$ as $t \to \infty$. $\square$ Remarks 3. (a) In the proof of the implication "(ii) $\Rightarrow$ (i)" we used several results about spectral projections and poles of the resolvents. These results can be found in various classical books about functional analysis; but they are a bit scattered through several books where they are, in my experience, not so easy to digest at first glance. I thus wrote a brief summary about spectral projections and properties of poles of the resolvent, along with detailed references to various books, in the Appendices A.1 - A.3 of my PhD thesis. (b) Proposition 1 und Theorem 2 can be seen as very elementary special cases of the topic "long term behaviour of $C_0$-semigroups". The point is that, if we replace the bounded operator $Q$ with an unbounded closed operator that generates a so-called $C_0$-semigroup, the question whether one has operator norm convergence as $t \to \infty$ becomes suddenly much more involved. (c) The aforementioned topic becomes even subtler if one replaces operator norm convergence with strong convergence. For this topology, even for bounded operators $Q$ the long-term behaviour of $e^{tQ}$ becomes quite non-trivial (and even more so for $C_0$-semigroups with unbounded generators). (d) An introduction to the long-term behaviour of $C_0$-semigroups with many useful theorems can, for instance, be found in Chapter V of the book "Engel and Nagel: One-Parameter Semigroups for Linear Evolution Equations (2000)" (link to zbMATH).<|endoftext|> TITLE: Can there exist such a sequence of elementary embeddings of the universe to itself? QUESTION [6 upvotes]: Working in ZfC + Wholeness: Can we have a countable sequence of non-trivial elementary embeddings of the universe to itself, such that the range of each embedding is a subclass of the range of its successor, and such that the union of all the ranges of those embeddings is the universe itself? Formally can we have a proper class sequence $j`\mathbb N$ such that: $\forall n \in \mathbb N : j_n: V \prec V \land rng(j_{n+1}) \supset rng(j_n) \\ \forall x \exists n \in \mathbb N : x \in rng(j_n)$ Where $\mathbb N$ is the set of all finite von Neumann ordinals. Note: ZfC means ZFC but with replacement restricted to the pure language of set theory, i.e. it doesn't use the symbol $j$; however, $j$ is allowed in all instances of separation. REPLY [7 votes]: It was pointed out by Monroe that if $\lambda$ is a limit and $j:V_\lambda\to V_\lambda$ is elementary and $j_n$ is the $n$th iterate, i.e. $j_0=j$ and $j_{n+1}=j_n(j_n)$, and $\kappa_n=\mathrm{crit}(j_n)$, then since $\lim_{n<\omega}\kappa_n=\lambda$, we get every $x\in V_\lambda$ is in $\mathrm{rng}(j_n)$ for some $n<\omega$. Given this, we can arrange the increasing range condition as follows. Choose a strictly increasing sequence $\left_{i<\omega}$ such that for each $\alpha<\lambda$ there is $i<\omega$ such that $j_{n_{i-1}}\circ j_{n_{i-2}}\circ\ldots\circ j_{n_0}(\alpha)<\mathrm{crit}(j_{n_i})$. (Just construct the sequence recursively on $i$, and at stage $i$, make it work for $\alpha=\kappa_i$.) Now define $k_i:V_\lambda\to V_\lambda$ as the direct limit \begin{equation}\ldots \circ j_{n_{i+2}}\circ j_{n_{i+1}}\circ j_{n_i},\end{equation} i.e. for each $x\in V_\lambda$ set $k_i(x)=$ the common value of \begin{equation}j_{n_{i+\ell}}\circ j_{n_{i+\ell-1}}\circ\ldots\circ j_{n_i}(x)\end{equation} for large $\ell<\omega$, noting that by the choice of the $n_i$'s, for every $x\in V_\lambda$, $k_i(x)\in V_\lambda$ is well-defined; and $k_i$ is easily elementary. Note that $k_{i}=k_{i+1}\circ j_{n_i}$, and therefore $\mathrm{rng}(k_i)\subseteq\mathrm{rng}(k_{i+1})$. And note $\lim_{i<\omega}\mathrm{crit}(k_i)=\lambda$. So we get the desired conditions.<|endoftext|> TITLE: On the determinant $\det[\gcd(i-j,n)]_{1\le i,j\le n}$ QUESTION [8 upvotes]: In Sept. 2013, I investigated the determinant $$D_n=\det[\gcd(i-j,n)]_{1\le i,j\le n}$$ and computed the values $D_1,\ldots,D_{100}$ (cf. http://oeis.org/A228884). To my surprise, they are all positive! Question. Does $D_n>0$ hold for all $n=1,2,3,\ldots$? I believe that $D_n$ is always positive. How to prove this? It is easy to see that $D_n$ is divisible by $\sum_{k=1}^n\gcd(k,n)=\sum_{d\mid n}\varphi(d)\frac nd$. It seems that $\varphi(n)^{\varphi(n)}\sum_{k=1}^n\gcd(k,n)$ divides $D_n$. Maybe there is a simple explanation for this. Your comments are welcome! REPLY [13 votes]: Denote $f(k)={\rm gcd}(k,n)$. Clearly $f$ is an $n$-periodic function. $D_n$ is the circulant $D_n=\det(f(i-j):0\leqslant i,j\leqslant n-1)$ which equals $\prod_{k=0}^{n-1}h(\omega^k)$ where $\omega=e^{2\pi i/n}$ and $h(t)=f(0)+f(1)t+\ldots+f(n-1)t^{n-1}$. We have $$ h(t)=\sum_{d|n} \varphi(d)(1+t^d+t^{2d}+\ldots+t^{(n/d-1)d}). $$ If $t=\omega^k$, the sum $1+t^d+t^{2d}+\ldots+t^{(n/d-1)d}$ either equals $n/d$ (if $n$ divides $kd$) or equals $(1-\omega^{kn})/(1-\omega^{kd})=0$ otherwise. In any case it is a non-negative integer, and for $d=n$ it is strictly positive.<|endoftext|> TITLE: What kid-friendly math riddles are too often spoiled for mathematicians? QUESTION [46 upvotes]: Some math riddles tend to be spoiled for mathematicians before they get a chance to solve them. Three examples: What is $1+2+\cdots+100$? Is it possible to tile a mutilated chess board with dominoes? Given a line $\ell$ in the plane and two points $p$ and $q$ on the same side of $\ell$, what is the shortest path from $p$ to $\ell$ to $q$? I would like to give my children the opportunity to solve these riddles before the spoilers inevitably arrive. Question: What are other examples of kid-friendly math riddles that are frequently spoiled for mathematicians? Notes: There is no shortage of kid-friendly math riddles. I am specifically asking for riddles that are frequently spoiled for mathematicians because they capture a bigger idea that is useful in math, especially research-level math. As such, the types of riddles I am asking for are most readily supplied by research mathematicians. In case it is not clear whether MO is an appropriate forum for this question, see the following noteworthy precedent: Mathematical games interesting to both you and a 5+-year-old child REPLY [8 votes]: The book "1000 Play Thinks" by Ivan Moscovich contains up to 1000 of these, depending on your background. It is an absolute delight - large pages, full-coloured and playfully illustrated by Tim Robinson. Puzzles are grouped by mathematical categories (Geometry, Graphs and Networks, Numbers, Probability, Topology...), show essential examples, structures and ideas from those fields, and each has a difficulty rating and solution. Between puzzles are short introductions to subjects and historical notes of the mathematicians involved in their development. It also includes 89 references to other mathematical puzzle books. Flipping through various sections, here are a few examples: 38: Will a $70$ cm sword fit into a $30\times 40 \times 50$ cm chest? 179: Euler's Problem: "to trace a pattern without picking up your pencil or backtracking over sections." Along with $11$ images and the question "which ones do you find impossible to solve?" 186: Utilities I: Can you connect three house to three utilities without allowing any of the lines to intersect? Followed up by three Play Thinks on multipartite graphs (including the terminology, and phrased as connecting animals of various colours). 528: A description of perfect numbers, the example of 6, and the question: what is the second perfect number? Also notes that 38 perfect numbers are known, so the book is dated between 1999 and 2001. 687: You need to roll a double 6 in at least one of twenty-four throws. Are the odds in your favor? 703: Mars Colony (Gerhard Ringel's "Empire-Colony puzzle" of colouring two maps with 11 numbered regions so that both regions with the same number have the same colour.) 715: Topology of the Alphabet. Can you find the letters that are topologically equivalent to E in the given font? 859: A steel washer is heated until the metal expands by 1%. Will the hole get larger or smaller or remain unchanged? 995: Seven birds live in a nest, and send out three each day in search of food. After 7 days, every pair of birds has been one one foraging mission together. Can you work out how?<|endoftext|> TITLE: Are weighted projective spaces cut out by quadrics? QUESTION [5 upvotes]: Let $X=\mathbb{P}(a_0,\ldots, a_n)$ be a well-formed weighted projective space, and let $a=\mathrm{lcm}(a_0,\ldots,a_n)$. Then $\mathcal{O}(a)$ embeds $X$ in projective space $\mathbb{P}^N$. Question. Is $X$ cut out by quadrics in $\mathbb{P}^N$? As far as I can tell, in general this is non-obvious for knapsack problem related reasons, though it is true for surfaces by Koelman's paper A criterion for the ideal of a projectively embedded toric surface to be generated by quadrics. EDIT 3-3: Mateusz Michaelek has pointed out that $\mathcal{O}(a)$ does not generally embed X in projective space; for instance $\mathbb{P}(1,6,10,15)$ is not embedded by $\mathcal{O}(30)$, as the degree 60 monomial $x_0 x_1^4 x^2_2 x_3$ cannot be factored into degree 30 monomials. This example is example 2.17 of Toric geometry of path signature varieties. So in the question above, $\mathcal{O}(a)$ should be replaced with a very ample bundle $\mathcal{L}$, in which case I still don't know how to answer it. REPLY [3 votes]: The following code in Sage gives information about the embedding of $\mathbb{P}(1,1,2,3)$ in $\mathbb{P}^{22}$ using $\mathcal{O}_{\mathbb{P}(1,1,2,3)}(6)$. $\texttt{toric_varieties.WP(1,1,2,3).divisor([6,0,0,0]).Kodaira_map()}$ It produces the following output: Scheme morphism: From: 3-d toric variety covered by 4 affine patches To: Closed subscheme of Projective Space of dimension 22 over Rational Field defined by: -z16^2 + z0*z22, -z1*z20 + z0*z21, z21^2 - z14*z22, -z7*z16 + z0*z20, z20*z21 - z13*z22, z20^2 - z12*z22, -z1*z18 + z0*z19, z19*z21 - z11*z22, z19*z20 - z10*z22, z19^2 - z6*z22, -z1*z17 + z0*z18, z18*z21 - z10*z22, z18*z20 - z9*z22, z18*z19 - z5*z22, z18^2 - z4*z22, -z1*z16 + z0*z17, z17*z21 - z9*z22, z17*z20 - z8*z22, z17*z19 - z4*z22, z17*z18 - z3*z22, z17^2 - z2*z22, z16*z21 - z8*z22, z16*z20 - z7*z22, z16*z19 - z3*z22, z16*z18 - z2*z22, z16*z17 - z1*z22, -z7*z12 + z0*z15, z15*z19 - z14*z21, z15*z18 - z13*z21, z15*z17 - z12*z21, z15*z16 - z12*z20, -z1*z13 + z0*z14, z14*z20 - z13*z21, z14*z19 - z11*z21, z14*z18 - z10*z21, z14*z17 - z9*z21, z14*z16 - z8*z21, z14^2 - z11*z15, -z1*z12 + z0*z13, z13*z20 - z12*z21, z13*z19 - z10*z21, z13*z18 - z9*z21, z13*z17 - z8*z21, z13*z16 - z7*z21, z13*z14 - z10*z15, z13^2 - z9*z15, -z7^2 + z0*z12, z12*z19 - z9*z21, z12*z18 - z8*z21, z12*z17 - z7*z21, z12*z16 - z7*z20, z12*z14 - z9*z15, z12*z13 - z8*z15, z12^2 - z7*z15, -z1*z10 + z0*z11, z11*z20 - z10*z21, z11*z19 - z6*z21, z11*z18 - z5*z21, z11*z17 - z4*z21, z11*z16 - z3*z21, z11*z14 - z6*z15, z11*z13 - z5*z15, z11*z12 - z4*z15, z11^2 - z6*z14, -z1*z9 + z0*z10, z10*z20 - z9*z21, z10*z19 - z5*z21, z10*z18 - z4*z21, z10*z17 - z3*z21, z10*z16 - z2*z21, z10*z14 - z5*z15, z10*z13 - z4*z15, z10*z12 - z3*z15, z10*z11 - z5*z14, z10^2 - z4*z14, -z1*z8 + z0*z9, z9*z20 - z8*z21, z9*z19 - z4*z21, z9*z18 - z3*z21, z9*z17 - z2*z21, z9*z16 - z1*z21, z9*z14 - z4*z15, z9*z13 - z3*z15, z9*z12 - z2*z15, z9*z11 - z4*z14, z9*z10 - z3*z14, z9^2 - z2*z14, -z1*z7 + z0*z8, z8*z20 - z7*z21, z8*z19 - z3*z21, z8*z18 - z2*z21, z8*z17 - z1*z21, z8*z16 - z1*z20, z8*z14 - z3*z15, z8*z13 - z2*z15, z8*z12 - z1*z15, z8*z11 - z3*z14, z8*z10 - z2*z14, z8*z9 - z1*z14, z8^2 - z1*z13, z7*z19 - z2*z21, z7*z18 - z1*z21, z7*z17 - z1*z20, z7*z14 - z2*z15, z7*z13 - z1*z15, -z1*z15*z20 + z7*z12*z21, z7*z11 - z2*z14, z7*z10 - z1*z14, z7*z9 - z1*z13, z7*z8 - z1*z12, -z1*z12*z20 + z7^2*z21, -z1*z5 + z0*z6, z6*z20 - z5*z21, z6*z18 - z5*z19, z6*z17 - z4*z19, z6*z16 - z3*z19, z6*z13 - z5*z14, z6*z12 - z4*z14, z6*z10 - z5*z11, z6*z9 - z4*z11, z6*z8 - z3*z11, z6*z7 - z2*z11, -z1*z4 + z0*z5, z5*z20 - z4*z21, z5*z18 - z4*z19, z5*z17 - z3*z19, z5*z16 - z2*z19, z5*z13 - z4*z14, z5*z12 - z3*z14, z5*z10 - z4*z11, z5*z9 - z3*z11, z5*z8 - z2*z11, z5*z7 - z1*z11, z5^2 - z4*z6, -z1*z3 + z0*z4, z4*z20 - z3*z21, z4*z18 - z3*z19, z4*z17 - z2*z19, z4*z16 - z1*z19, z4*z13 - z3*z14, z4*z12 - z2*z14, z4*z10 - z3*z11, z4*z9 - z2*z11, z4*z8 - z1*z11, z4*z7 - z1*z10, z4*z5 - z3*z6, z4^2 - z2*z6, -z1*z2 + z0*z3, z3*z20 - z2*z21, z3*z18 - z2*z19, z3*z17 - z1*z19, z3*z16 - z1*z18, z3*z13 - z2*z14, z3*z12 - z1*z14, z3*z10 - z2*z11, z3*z9 - z1*z11, z3*z8 - z1*z10, z3*z7 - z1*z9, z3*z5 - z2*z6, z3*z4 - z1*z6, z3^2 - z1*z5, -z1^2 + z0*z2, z2*z20 - z1*z21, z2*z18 - z1*z19, z2*z17 - z1*z18, z2*z16 - z1*z17, z2*z13 - z1*z14, z2*z12 - z1*z13, z2*z10 - z1*z11, z2*z9 - z1*z10, z2*z8 - z1*z9, z2*z7 - z1*z8, z2*z5 - z1*z6, z2*z4 - z1*z5, z2*z3 - z1*z4, z2^2 - z1*z3 Defn: Defined on coordinates by sending [z0 : z1 : z2 : z3] to (z0^6 : z0^5*z1 : z0^4*z1^2 : z0^3*z1^3 : z0^2*z1^4 : z0*z1^5 : z1^6 : z0^4*z2 : z0^3*z1*z2 : z0^2*z1^2*z2 : z0*z1^3*z2 : z1^4*z2 : z0^2*z2^2 : z0*z1*z2^2 : z1^2*z2^2 : z2^3 : z0^3*z3 : z0^2*z1*z3 : z0*z1^2*z3 : z1^3*z3 : z0*z2*z3 : z1*z2*z3 : z3^2) Which you can see contains cubics such as $-z_1 z_{15} z_{20} + z_7 z_{12} z_{21}$.<|endoftext|> TITLE: A quantity measuring the separability of Banach spaces QUESTION [10 upvotes]: Let $X$ be a Banach space. It is natural for us to introduce a quantity measuring the separability of sets as follows: for a subset $A$ of $X$, we set $\textrm{sep}(A)=\inf\{\epsilon>0: A\subseteq K+\epsilon B_{X}$ for some countable subset $K$ of $X\}$. Clearly, $A$ is separable if and only if $\textrm{sep}(A)=0$. It is elementary that a Banach space $X$ is separable if $X^{*}$ is separable. My question is about a quantitative version of this result. Question. Does there exist a universal constant $C$ such that $$\textrm{sep}(B_{X})\leq C\cdot \textrm{sep}(B_{X^{*}}) \, ?$$ REPLY [20 votes]: For the unit ball $B_X$ of the Banach space there are only two possibilities: sep$(B_X)= 1$, if $B_X$ is not separable, and sep$(B_X)=0$ if $B_X$ is separable. Indeed, if sep$(B_X)<1$ there are $\varepsilon <1$ and a countable subset $K\subseteq B_X$ with $B_X\subseteq K + \varepsilon B_X$. But this can be iterated, i.e., $$ B_X\subseteq K +\varepsilon (K+\varepsilon B_X) \subseteq K_1+\varepsilon^2 B_X $$ where $K_1=K+\varepsilon K$ is again countable. Inductively, this implies $B_X\subseteq K_n +\varepsilon^n B_X$ for a countable set $K_n$ and hence sep$(B_X)=0$. Therefore, $C=1$ satisfies the desired inequality sep$(B_X)\le$ sep$(B_{X^*})$.<|endoftext|> TITLE: What are Koszul dualities? QUESTION [23 upvotes]: I am bewildered by the number of things I've heard referred to as "Koszul duality", and I would like to sort it out. At various different times, I believe I've seen any of the following phenomena referred to as "Koszul duality": Let $O$ be an operad (in your favorite symmetric monoidal $\infty$-category $\mathcal V$), and apply the operadic bar construction to $O$ (i.e. regard $O$ as a monoid in symmetric sequences equipped with Kelly's convolution monoidal product $\circ$, and take the the geometric realization of the simplicial symmetric sequence $[n] \mapsto O^{\circ n}$) to obtain $BO$. Then (under certain conditions?) $BO$ is a cooperad, called the Koszul dual of $O$. Dually, if $C$ is a cooperad, then applying the operadic cobar construction to $C$ yields an operad called the Koszul dual of $C$. Under certain conditions, these constructions are adjoint to one another, and under certain further conditions they are inverse to one another. There is a version of (1) called Koszul duality carrying $O$-modules to $BO$-comodules (and maybe a variant carrying $O$-algebras to $BO$-coalgebras?). Let $O$ be the $E_k$ operad, and let $A$ be an $O$-algebra. Then there is an iterated bar-construction carrying $A$ to an $E_k$-coalgebra, which is called its Koszul dual. Let $O$ be a quadratic operad in the sense of Ginzburg and Kapranov (quadratic-ness only makes sense in certain $\mathcal V$ -- basically $V$ must be chain complexes). Then there is another quadratic operad Koszul dual to $O$, and if $O$ is Koszul, then the duality operation is self-inverse at $O$. There is also Koszul duality for algebras for the associative operad. I think I've been told that the $E_k$-operad is Koszul self-dual. Since the $E_k$-operad makes sense in spaces, where quadraticness isn't even defined, I don't know what this means (I only know what it means for an operad to be dual to a coopeard in this generality, following (1)). I think what's going on is that the fundamental duality is the bar/cobar adjunction between operads $O$ and cooperads $BO$; duality between $O$-modules and $BO$-comodules, and between $O$-algebras and $BO$-coalgebras then comes along for the ride. Then some other form of duality sometimes allows one to relate cooperads back to operads, and we start talking about duality between operads, and between the modules / algebras for an operad and its dual operad from there. But I've never really seen this spelled out in this general a context. Question: What is the relationship between the above things called Koszul duality (and other things called Koszul duality which I'm missing)? I hope it's clear that this question has a different focus from this classic MO question. REPLY [4 votes]: I have finally found a source which puts together the pieces in a satisfactory way, at least in the stable setting, here: Amabel, Araminta. "Poincaré/Koszul Duality for General Operads." arXiv preprint arXiv:1910.09076 (2019). latest arxiv version. Amabel discusses (see Thm 2.20) Koszul duality of operads and cooperads in spectra (Ching has shown that the bar/cobar adjunction for spectral operads / cooperads is an equivalence here with no conditions!!! This is very surprising to me given that most authors assume very very restrictive conditions like being a quadratic operad to get similar results.) Amabel also discusses (see Thm 2.21) the relationship between Koszul duality for operads and cooperads, and the induced bar/cobar adjunctions for the (ind,nilpotent,with divided powers co)/algebras of the corresponding co/operads, with reference to Francis-Gaitsgory and Ching's thesis. According to Ching and Harper, this bar adjunction is also an equivalence if you assume only that the operad is connective -- which again seems like a much weaker assumption than I expected! Lots more good stuff in here...<|endoftext|> TITLE: Computations of divisor class monoids QUESTION [5 upvotes]: Let me first recall some definitions from the very first pages of Bourbaki, Commutative Algebra, Chapter 7, "Divisors". Let $A$ be a (commutative) domain, $K$ its field of fractions. A fractional ideal of $A$ is a finitely generated $A$-submodule of $K$. The set of all non-zero fractional ideals of $A$ is called $I(A)$. On $I(A)$ there is a natural equivalence $\sim$ relation: for two fractional ideals $\mathfrak a$ and $\mathfrak b$ we write ${\mathfrak a} \sim {\mathfrak b}$ if every principal fractional ideal containing $\mathfrak a$ also contains $\mathfrak b$ and vice-versa. The set of equivalence classes in $I(A)$ for this equivalence relation $\sim$ is called $D(A)$; its elements are called divisors of $A$. The multiplication of fractional ideals induces a multiplication on $D(A)$, which makes it a monoid. So $D(A)$ isthe divisor monoid of $A$. Now on $D(A)$ we define a second equivalence relation, where two elements $d$ and $d'$ of $D(A)$ are equivalent if for some (or equivalently any) representative $\mathfrak a$ of $d$ and for some (or equivalently any) representative $\mathfrak a$ of $d$, one has $$\mathfrak a =\mathfrak a' x \text{ for some }x \in K^\ast.$$ The quotient of $D(A)$ by this equivalence relation clearly inherits the monoid structure of $D(A)$ and is called the divisor class monoid of $A$. Bourbaki doesn't introduce a special notation for it but let us denote it by $DC(A)$. Bourbaki proves that $D(A)$ is a group (hence also $DC(A)$) if and only if $A$ is totally integrally closed (Theorem 1 of chapter 7). But I am interested in the cases where $A$ is not integrally closed, especially to the cases where $A$ is a noetherian complete domain of Krull dimension 1, or even more especially to the case where $A$ is the completed local ring at a singular point of an algebraic curve over $\mathbb C$. My question is: Has there been any systematic attempt to compute the divisor class monoid $DC(A)$ for $A$ the completed local ring at a singular point of an algebraic curve? Or at least some example of non trivial computations of such $DC(A)$? It seems to me that $DC(A)$ is a very natural invariant of a singularity of an algebraic curve. People working in the theory of singularities of algebraic or analytic curves (a vast subject) have certainly met this invariant, but I can't find any reference in the literature. Any pointers, or any suggestion to attack the problem is very welcome. Remark: I know how to compute $DC(A)$ in simple special cases, for example the case where $A$ is the complete local ring of a cusp, i.e $A=\{f \in \mathbb C[[T]], f'(0)=0\}$. This is Exercise 1 in the exercises of chapter 7, \S1 of Bourbaki. In this case $DC(A)$ is the monoid $\{1,x\}$, where $x$ satisfies $x^2=x$. (Here $x$ can be the class of the ideal $(T^2,T^3)$ of $A$, for instance). But I'd like to know the answer for more general situations. REPLY [5 votes]: Here are a few remarks about $DC(A)$ (assuming $A$ is a complete Noetherian local domain of dimension $1$). The equivalence relation in $D(A)$ is just isomorphism as $A$-modules. So you can view $DC(A)$ as the monoid of isomorphism classes of nonzero ideals $I$ in $A$ under multiplication. For any $x\in DC(A)$, $x^{n+1}=x^n$ for $n$ large enough. That is because $aI^n = I^{n+1}$ if $n$ is large enough for any minimal reduction $a$ of $I$ (here one must first enlarge the residue field, but this is safe). From above, it follows immediately that if $DC(A)$ is cancellative if and only if it is trivial if and only if $A$ is regular. This generalizes the fact in Bourbaki about being a group. Over the complex numbers, $DC(A)$ is finite if and only if $A$ has finitely many Cohen-Macaulay modules up to isomorphisms (for reference see the books on this topic by Yoshino or Leuschke-Wiegand). For instance, if $A$ is a simple singularity (ADE singularity) then this holds. In such case, you can work out the monoid explicitly with a bit of effort. Here is an example that contains what you mentioned in the last paragrach. Consider $A_n= k[[t^2,t^{2n+1}]]$. Up to isomorphisms, the only ideals are $I_i=(x^{2i}, x^{2n+1})$ with $i=0,1,...,n$. One can check that $I_iI_j \cong I_{\max\{i,j\}}$. So the monoid is $\{x_0=1,...,x_n\}$ with $x_ix_j = x_{\max\{i,j\}}$.<|endoftext|> TITLE: The symmetric square of a sphere QUESTION [6 upvotes]: $\DeclareMathOperator\Sym{Sym}$Recently, I have been thinking about the space $\Sym^2(S^n)$ of pairs of points in the sphere (including the repeated pairs $(p,p)$ for each $p\in S^n$) and possible ways to describe it. This is topologized as the quotient of $S^n\times S^n$ under the involution that switches components. I am looking (somewhat broadly and ill-definedly) for descriptions of this space and related results. I will describe below some of the results that I have encountered and I would really appreciate any pointers towards descriptions that I have missed in my literature search. (Side note: If the case of $n=1$ is something you have thought about / want to think about / enjoy, also consider taking a look at this related question that I asked on StackExchange about different ways to prove that $\Sym^2(S^1)$ is a Möbius strip.) Primarily, I am interested in the homeomorphism type of $\Sym^2(S^n)$, of which I have found two descriptions: On the topology of cyclic products of spheres by Liao cites a description due to Steenrod, where $\Sym^2(S^n)$ is formed by attaching a $2n$-cell to the (unreduced) suspension of $\Sym^{2}(S^{n-1})$. This seems quite useful, since it gives an inductive view of $\Sym^2(S^n)$ as a CW-complex. On the symmetric square of a sphere by James, Thomas, Toda and Whitehead gives a second description, as the mapping cone of a certain map $\Sigma^n(\mathbb R\mathbb P^{n-1})\rightarrow S^n$ (here $\Sigma$ denotes the unreduced suspension). This seems more specialized than the above (their calculations actually use both descriptions), but it also seems to be the more-cited description (perhaps this is because it appears in Hatcher's Algebraic Topology, on page 482). These are the only two ways I know of to describe the homeomorphism type and I would be very interested if there are others that I have missed. In terms of computations, it seems that a lot is known about the homology, homotopy and $K$-theory of the space $X_n=\Sym^2(S^n)$. However, these results are widely scattered throughout the literature, so there's a very reasonable chance that I have missed things. Here is what I have been able to find: It seems that the homology groups $H_i(X_n;\mathbb Z)$ are well-understood and should follow from Dold's purely algebraic recipe in Homology of symmetric products and other functors of complexes (though it was known before this too). I have no idea what is known about the cup product for this space. In Homotopy of two-fold symmetric products of spheres, Nakaoka describes the stable homotopy groups $\pi_{n+i}(X_n)$ for $01$, the space $X_n$ is simply-connected, so we also get $\pi_n(X_n)=\mathbb Z$ and $\pi_i(X_n)=0$ for $0 TITLE: Minimal bridging sets in infinite connected graphs QUESTION [5 upvotes]: Let $G=(V, E)$ be a connected, simple, undirected graph. We say that $B\subseteq V$ is a bridging set if $B\neq V$ and removing $B$ makes the graph disconnected, or more formally: $$G \setminus B := (V\setminus B\;, \; \{e\in E: e\cap B = \varnothing\})$$ is not connected any more. Is there an infinite connected graph $G=(V,E)$ such that for every bridging set $B\subseteq V$ there is a bridging set $B_1$ with $B_1\subseteq B$ and $B_1\neq B$? REPLY [6 votes]: Yes. Take for $V = S \coprod T$ with bijections $s : \mathbb N \to S$ and $t : \mathbb N \to T$ ; and have the edge set $K_T \cup \{(s(i),t(j)) \quad\mathtt{ iff }\quad i \le j\}$. First, see that $\forall n$, $t(\mathbb N + n)$ is a bridging set. Indeed, $s(k)$ for $k \ge n$ is isolated. This gives an infinite strict chain of bridging sets, whose limit is empty (so not a bridging set). Let $B$ be a bridging set. Let's prove that $B \supset t(\mathbb N + k)$ for a $k$. Suppose it's not the case, that is : $\forall k \in \mathbb N, \exists N(k), t(N(k)) \notin B$. Then take two nodes $u,v$, those nodes are in $T \cup s(\mathbb N + n)$, so both have $t(N(n))$ as a neighbourg.<|endoftext|> TITLE: Is the pseudoinverse the same as least squares with regularization? QUESTION [20 upvotes]: Given a linear system $Ax=b$, the pseudoinverse of $A$ is found as the matrix $A^+$ such that $x=A^+ b$ where $x$ solves the least squares problem $\min \| Ax - b \|^2 $ and $x \perp \mathcal{N}(A)$. That is, $x$ is the shortest vector in the solution space. That is, find $x$ : \begin{eqnarray} \min \| x \| \text{ such that } x \text{ minimizes } \| Ax - b \|^2 . \end{eqnarray} This is similar to the regularization problem of minimizing \begin{eqnarray} \| Ax - b \|^2 + \lambda \| x \|^2 \end{eqnarray} I do not quite can get from one to the other? Are these two things equivalent? Is there a way to post this as the same problem? Is there a connection? REPLY [3 votes]: As usual with such problems, it is most insightful to forget about matrices for a while and think about abstract vector spaces instead. Let $V$ and $W$ vector spaces and $A: V\to W$ linear†. Furthermore let $d_V$ and $d_W$ be metrics on each of the spaces. Now, given $b\in W$, we have two convex functions $\zeta_V, \zeta_b: V\to\mathbb{R}^+$: $$\begin{align} \zeta_V(v) =& d_V(v,\vec0) \\ \zeta_b(v) =& d_W(A\:v, b) \end{align}$$ The point of the Tikhonov problem is to make a tradeoff between these two cost functions, i.e. you minimise $$ \zeta_\lambda := \zeta_b(v) + \lambda\cdot\zeta_V(v). $$ But why would you want that? Basically, $\zeta_b$ is what we really care for, because it tells us by how much we're missing our target point. The problem is when $A$ fails to be injective, because in that case $\zeta_b$ will not be strictly convex and you have a whole set of solutions $\Xi_b\subset V$ on which $\zeta_b$ is minimised – some of which are very bad solutions, in the sense of, unbounded as visible by huge $d_V$ values.Those solutions can be eliminated by even an arbitrarily small $\lambda$, because $\zeta_b$ is constant on $\Xi_b$. So, in the limit $\lambda\to 0$, you're only really minimising $\zeta_b$, but still preventing solutions that have a needlessly big norm in $V$. In actual applications though, you're already in trouble even if $A$ is injective but badly conditioned, i.e. when there are $v\in V$ for which $d_W(A\:v,\vec0)$ happens to be very small. Because then, just a small bit of measurement noise on $b$ could cause the minimum of $\zeta_b$ to be thrown off by a big amount, even though the actual cost is barely changed. That can still be prevented by the $\zeta_V$ contribution, but in this case you can't make $\lambda$ arbitrarily small anymore but have to select an application-appropriate finite value. †Really, there's no reason for $W$ to actually be a vector space, it could as well be any metric space – but only for affine spaces and affine mappings can the problem be solved so easily.<|endoftext|> TITLE: Statement of classical Ramanujan-Petersson conjecture QUESTION [6 upvotes]: I'm preparing for an expository talk on some topics in the representation theory of reductive p-adic groups, including tempered representations and Whittaker models, and as motivation I wanted to mention the classical Ramanujan-Petersson conjecture. From my point of view what's really interesting is its generalization to automorphic forms, which, as I understand it, says that the local components of a globally generic cuspidal automorphic representation are tempered. But for a talk it's nice to connect something abstract to something classical. My problem is that I haven't been able to find a clear statement of the Ramanujan-Petersson conjectures anywhere. Here are some of the places I've looked. The Wikipedia article clearly states Ramanujan's conjecture for the modular discriminant function, but doesn't state the generalizations clearly. The 1930 paper of Petersson states his generalization of the conjecture, but I don't know German well enough to muddle my way through it. The 1965 Boulder paper of Satake, where he explains how to reformulate the conjecture using automorphic forms for GL$_2$, probably has the statement somewhere, but I'm not comfortable enough with the dictionary between modular forms and automorphic forms to extract it. A paper of Blomer and Brumley states the conjecture for Maass forms in Section 3.1. This makes me wonder if perhaps there are separate statements for Maass forms and holomorphic forms. So I was hoping that the conjecture was written down clearly and simply somewhere, and that perhaps someone on this site could point me to a reference. But barring that, it would be great if someone could just state the conjecture -- I'm hoping that it's ultimately a simple estimate on the Hecke eigenvalues, which is what I've been led to believe. I understand that the conjecture is known for holomorphic forms thanks to the work of Deligne, and that the conjecture for Maass forms is still open. Are these the only essential cases? (Part of my problem is that I haven't thought as much as I should about classical modular forms, and it is not clear to me what general functions, if any, encompass both holomorphic modular forms and Maass forms. I would assume the conjecture says something about such functions.) REPLY [4 votes]: For classical modular forms and Maass forms, you can find the definition of the associated $L$-function in Section 5.11 of the book Analytic number theory by Iwaniec-Kowalski. The statement of the Ramanujan-Petersson conjecture is given at the bottom of page 95 (a definition of general $L$-functions is also given on pages 93-94).<|endoftext|> TITLE: Solidification of free abelian group on compact Hausdorff space QUESTION [7 upvotes]: In the lecture notes on condensed mathematics the solidification of the free condensed abelian group $\mathbb{Z}[S]$ on a profinite set $S$ is defined as the inverse limit $\lim_{\leftarrow} \mathbb{Z}[S_i]$, but it can also alternatively be described as $$ \mathbb{Z}[S]^\blacksquare \cong \underline{\mathrm{Hom}}(C(S,\mathbb{Z}), \mathbb{Z}). $$ My question is whether this also holds for general compact Hausdorff spaces $S$. In the case that $X$ is a CW complex this seems to follow from example 6.5 where $\mathbb{Z}[X]^{\blacksquare} \cong H_\bullet(X)$ is shown. This also seems to follow from it's derived analogue, which was stated in part 2 of the answer here, if we can show that $\mathbb{Z}[X]$ is pseudo-coherent for any compact Hausdorff space $X$. I think this holds because we can resolve $X$ by extremally disconnected spaces, which should yield a resolution of $\mathbb{Z}[X]$ by compact projectives. By analogy I also have the same question for the $\mathcal{M}$-completeness of condensed $\mathbb{R}$-vector spaces. Is it true that the condensed vector space of signed Radon measures on a compact Hausdorff space $X$ can be given as: $$ \mathcal{M}(X) \cong \underline{\mathrm{Hom}}_{\mathbb{R}}( C( X, \mathbb{R}) , \mathbb{R}) $$ Again it seems like all the hints are in the lecture notes, but I'm not comfortable enough with the subject to pin down the details. A short note on why I'm asking this: the main reason is that I want to better understand the analogy with measure theory where one says that a condensed $\mathbb{R}$-vector space $V$ is $\mathcal{M}$-complete if for any continuous $f: K \to A$ and measure $\mu$ on $K$ one can form the "integral" $\int f\ d\mu$. I would find this much more convincing if I can let $K$ be a finite CW-complex rather than just a profinite set. On this note I also have a third, optional question: if the above two descriptions work, is there something similar one can say in the case of $\mathcal{M}_p(S)$ and for $p$-liquid vector spaces? REPLY [10 votes]: Good question! You essentially already give the answer, but let me spell it out. First, in the solid case, as you say one can compute the derived solidification $\mathbb Z[S]^\blacksquare$ for any compact Hausdorff $S$ as $$ \mathbb Z[S]^\blacksquare = R\underline{\mathrm{Hom}}(R\Gamma(S,\mathbb Z),\mathbb Z), $$ where $R\Gamma(S,\mathbb Z)$ is the Cech cohomology of $S$. The answer is nicer for profinite $S$ as then this is concentrated in degree $0$. If you are only interested in the solidification on the abelian level, then you can actually deduce from this that the map $S\to \pi_0 S$ (a profinite set) induces an isomorphism $H_0(\mathbb Z[S]^\blacksquare)\to \mathbb Z[\pi_0 S]^\blacksquare$. In the case of $\mathcal M$-complete $\mathbb R$-vector spaces, it is actually true that the $\mathcal M$-completion of $\mathbb R[S]$ is $\mathcal M(S,\mathbb R)$, the space of signed Radon measures on $S$ (with its Smith space topology), for any compact Hausdorff $S$. One can actually show that for any simplicial resolution $S_\bullet\to S$ by profinite sets $S_i$, the complex $$ \ldots \to \mathcal M(S_1,\mathbb R)\to \mathcal M(S_0,\mathbb R)\to \mathcal M(S,\mathbb R)\to 0 $$ is exact. This follows from the anti-equivalence of Smith spaces with Banach spaces, and the exact sequence $$ 0\to C(S,\mathbb R)\to C(S_0,\mathbb R)\to C(S_1,\mathbb R)\to \ldots, $$ the latter being exact by the computation $H^i_{\mathrm{cond}}(S,\mathbb R)=0$ for $i>0$ (and $C(S,\mathbb R)$ for $i=0$). [I actually find the definition of signed Radon measures on general $S$ a bit hard to process (while for profinite sets, it is completely transparent, one only has to give measures on open and closed subsets). So I actually prefer to think of the above as the definition of $\mathcal M(S,\mathbb R)$ for general compact Hausdorff $S$. But I guess it's nice that one can describe this space explicitly, intrinsically on $S$.] Finally, in the $p$-liquid case, one can also show that for any compact Hausdorff $S$, there is a $p$-Smith space $\mathcal M_p(S,\mathbb R)$ such that for any resolution $S_\bullet\to S$ as above, the complex $$ \ldots \to \mathcal M_p(S_1,\mathbb R)\to \mathcal M_p(S_0,\mathbb R)\to \mathcal M_p(S,\mathbb R)\to 0 $$ is exact. This follows from Proposition 10.1 (i) in Analytic Geometry by passing to the quotient $\mathbb Z((T))_r\to \mathbb R$ (sending $T$ to $r^{1/p}$) as usual (using also Proposition 7.2). Unfortunately, I don't really know how to describe $\mathcal M_p(S,\mathbb R)$ intrinsically on $S$, without choosing this surjection $\mathbb Z((T))_r\to \mathbb R$; I'd be happy to see a description!<|endoftext|> TITLE: Listing all posets on 9 points? QUESTION [10 upvotes]: I'm looking for a list of all (non-isomorphic) posets on 9 points. I know there are 183231 of them (OEIS A000112), but in order to progress with a problem I'm working on, I'd need the posets themselves, not only their number. Background: A poset is a set $S$ together with a relation $\le$ on $S$ which is reflexive, transitive, and antisymmetric. Elements of $S$ are called points here. Two posets are isomorphic if there is a relation-respecting bijection between them. Deciding whether two posets are isomorphic is known to be a difficult problem, and the numbers of isomorphism classes of finite posets are known only for posets with up to 16 points (see the OEIS entry cited above). State of the art is a 2002 paper by Brinkmann and McKay: "Posets on up to 16 Points", Order 19 (2) (2002) 147-179. REPLY [2 votes]: The following file lists adjacency matrices of all 183231 inequivalent partial orders (i.e., $T_0$ topologies) on 9 points, with spaces separating rows: https://www.mathtransit.com/finite_topological_spaces/9-spaces_T_0_binary.txt.gz I created this file from one that was at the now-defunct Chapel Hill Poset Atlas, the front page of which can still be viewed here: http://web.archive.org/web/20190905161049/https://lists-of-posets.math.unc.edu/ In late 2019 I posted files containing all nonhomeomorphic topologies on spaces up to and including 9 points, here: https://www.mathtransit.com/finite_topological_spaces.php I am currently in the process of expanding this page to include files for cardinalities 10 and 11, the latter of which will probably take the form of a simple cookbook recipe due to its large size.<|endoftext|> TITLE: Image of $L^2M$ inside $L^1M$, for $M$ a von Neumann algebra QUESTION [6 upvotes]: Let $M$ be a factor (von Neumann algebra with trivial center), and let $L^1M:=M_*$ be its predual. Let $\omega:M\to\mathbb C$ be a faithful normal state. The Hilbert space $L^2M:=L^2(M,\omega)$ admits a distinguished vector $\omega^{1/2}$, defined as the image of $1\in M$ under the inclusion $M\hookrightarrow L^2M$. There is also a ``multiplication'' map $L^2M \times L^2M \to L^1M$ given by $\xi\cdot \eta := (x\mapsto \langle x \xi,J\eta\rangle)$, where $J$ denotes the modular conjugation. For $\xi\in L^2M$, we write $\lambda_\xi:L^2M \to L^1M$ for left multiplication by $\xi$. For every $t\in\mathbb R$, consider the composite $F_t:L^2M \to L^1M$ of the modular flow $\Delta^{it}:L^2M\to L^2M$ with the map $\lambda_{\omega^{1/2}}$. Equivalently, $F_t$ is the composite of $\lambda_{\omega^{1/2}}$ with the the modular flow $\delta^{it}:M_*\to M_*$ on the predual. Fact: the $1$-parameter family of maps $t\mapsto F_t$ admits an analytic continuation to the complex strip $0 \le Im(t) \le 1/2$, given by the formula $$F_t:L^2M \to L^1M : \xi \mapsto \omega^{it+1/2} \xi \omega^{-it}.$$ Let me now assume that the state $\omega$ has trivial centralizer. Let $\Omega:=\{t\in\mathbb C\mid0 \le Im(t) \le 1/2\}$, and let $$f:\Omega\to L^1M$$ be any holomorphic map (continuous on the closed strip and holomorphic in the interior) such that $f(s+t)=\delta^{it}(f(s))$. Does there exist $\xi\in L^2M$ such that $f(t)=F_t(\xi)$? REPLY [6 votes]: No, such a $\xi \in L^2(M)$ need not exist. Denote by $L$ the space of maps from $\Omega$ to $L^1(M)$ that are continuous on the closed strip $\Omega$, holomorphic on the interior of $\Omega$ and satisfy $f(s+t) = \delta^{it}(f(s))$ for all $s \in \Omega$ and $t \in \mathbb{R}$. One can identify $L$ with the subspace of $L^1(M)$ consisting of all $\mu \in L^1(M)$ that belong to the domain of the analytic continuation $\delta^{-1/2}$. One can then turn $L$ into a Banach space with norm $$\|\mu\|_L = \|\mu\|_1 + \|\delta^{-1/2}(\mu)\|_1 \; .$$ As explained in the question, the map sending $\eta \in L^2(M)$ to $\theta_\eta \in L^1(M)$ given by $\theta_\eta(x) = \langle x \omega^{1/2},J \eta \rangle$ is a bounded, injective linear map from $L^2(M)$ to $L$. Assume that this map is surjective. By the open mapping theorem, we find $\kappa > 0$ such that $\|\eta\|_2 \leq \kappa \, \|\theta_\eta\|_L$ for all $\eta \in L^2(M)$. Fix $a,b \in M$ and put $\eta = a J b^* \omega^{1/2}$. Then, $\theta_\eta(x) = \langle x J a^* \omega^{1/2},b^* \omega^{1/2} \rangle$ and $$(\delta^{-1/2}\theta_\eta)(x) = \langle x a \omega^{1/2} , J b \omega^{1/2} \rangle$$ for all $x \in M$. Therefore, $$\|\theta_\eta\|_L \leq \|a\|_2 \, \|b\|_2 + \|a^*\|_2 \, \|b^*\|_2 \; .$$ So if we assume moreover that $b$ belongs to domain $D(\sigma_{-i/2})$ of the analytic continuation $\sigma_{-i/2}$ of the modular automorphism group on $M$, we conclude that $$ \|a \sigma_{-i/2}(b)\|_2 \leq \kappa \, \bigl( \|a\|_2 \, \|b\|_2 + \|a^*\|_2 \, \|b^*\|_2 \bigr) \; . \hspace{2cm} (\ast) $$ Already at this stage, we should not expect that such an estimate can hold for all $a \in M$ and $b \in D(\sigma_{-i/2})$. Let's produce an explicit counterexample. So we need a factor $M$ with a normal faithful state $\omega$ with trivial centralizer such that we can easily make computations inside $(M,\omega)$. We use the free Araki-Woods factor associated with the regular representation of $\mathbb{R}$. So, denote $H = L^2(\mathbb{R})$ and define the anti-unitary operator $J : H \to H$ and unitary representation $U_t$ on $H$ by $$(J \xi)(x) = \overline{\xi(-x)} \quad\text{and}\quad (U_t \xi)(x) = \exp(itx) \xi(x) \; .$$ Denote by $S$ the closed densely defined antilinear involution on $H$ given by $S = J U_{-i/2}$, so that $(S \xi)(x) = \exp(-x/2) \overline{\xi(-x)}$ and with the domain of $S$ consisting of those $\xi \in L^2(\mathbb{R})$ with $S(\xi) \in L^2(\mathbb{R})$. Denote by $K \subset H$ the closed real subspace of all $\xi \in D(S)$ with $S(\xi) = \xi$. Consider the full Fock space $\mathcal{F}(H)$ and denote for every $\xi \in H$ by $\ell(\xi)$ the left creation operator. The free Araki-Woods factor $M$ is the von Neumann algebra acting on the full Fock space $\mathcal{F}(H)$ generated by the operators $\ell(\xi) + \ell(\xi)^*$ with $\xi \in K$. This representation of $M$ on $\mathcal{F}(H)$ is standard, with the vacuum vector serving as the canonical implementation of the free quasi-free state $\omega$. Choose a sequence of unit vectors $\xi_k \in K$ with support contained in $[-1/k,1/k]$. Define $$a_k = (\ell(\xi_k) + \ell(\xi_k)^*)/2 \; .$$ Then, $a_k$ is a sequence of self-adjoint elements of $M$. We have $a_k \in D(\sigma_{-i/2})$ and, by our choice of support, $\|a_k - \sigma_{-i/2}(a_k)\|_2 \to 0$. Also, each $a_k$ is distributed, w.r.t.\ the vacuum vector, as a semicircular random variable $S_1$ with radius $1$. Applying $(\ast)$ to $a = a_k^n$ and $b = a_k^n$ and taking $k \to +\infty$, it follows that $$\|S_1^{2n}\|_2 \leq 2 \kappa \, \|S_1^n\|_2^2 \; .$$ This means that $$ E(S_1^{4n}) \leq (2\kappa)^2 \, E(S_1^{2n})^2 \quad\text{for all $n \in \mathbb{N}$.}\hspace{2cm} (\ast\ast) $$ One knows that $E(S_1^{2n}) = 2^{-2n} C_n$, where $C_n$ is the $n$'th Catalan number. It follows that $$E(S_1^{2n}) \sim \pi^{-1/2} n^{-3/2} \; .$$ This is incompatible with $(\ast\ast)$. Concluding remark. The inequality $(\ast)$ shouldn't hold in any infinite-dimensional factor. The above concrete counterexample relies on sequences $b_k$ that are almost invariant under $\sigma_{-i/2}$. Such sequences always exist and it should then be possible to deduce from $(\ast)$ a contradiction in general.<|endoftext|> TITLE: Prime generating arithmetical dynamical system QUESTION [5 upvotes]: Is there a prime generating arithmetical dynamical system, by which I mean, is there a rational function $f$ and a prime $p$ such that the set of values of iterates of $f$ starting at $p$, $I(f) = \{f^{(n)}(p): n \in \mathbb{N}\}$, consists precisely of all primes? Or is there an argument showing that there can not be such an $f$? If there is no $f$ as above, is there a rational function which, starting at finitely many different primes, altogether produces all the primes when iterated? REPLY [8 votes]: First, except in some relatively easy to characterize cases, for most rational functions $f(x)\in\mathbb Q(x)$ and most starting rational points $\alpha\in\mathbb Q$, the set of iterates (which is called the orbit) $$ O_f(\alpha) := \bigl\{ f^{(n)}(\alpha) : n\in\mathbb N \bigr\} $$ will contain only finitely many integers, so it contains only finitely many primes. So you pretty much need to start with a polynomial $f(x)\in\mathbb Q[x]$. Then at least the orbit of an integer may contain infinitely many integers, e.g., this will be the case if the coefficients are integers. But unless the orbit is finite, the size of these integers will grow enormously fast. Again more generally, if you take a rational function $f(x)\in\mathbb Q(x)$ of degree $d\ge2$ and a point $\alpha\in\mathbb Q$ having infinite orbit, and if you write $$ f^{(n)}(\alpha) = \frac{A_n}{B_n} $$ as a fraction in lowest terms, then $$ \lim_{n\to\infty} \frac{\log \max\bigl\{ |A_n|,|B_n|\bigr\}}{n^2} $$ will converge to a positive number, so either the numerator or the denominator will grow at least as fast as $C^{d^n}$ for some $C>1$. (And except for those cases mentioned earlier, the numerator $|A_n|$ and denominator $|B_n|$ will grow at about the same rate; but that's harder to prove.)<|endoftext|> TITLE: An elementary question in bond percolation QUESTION [9 upvotes]: Consider a locally finite, connected graph and "bond (edge) percolation" on this graph. Each edge is open with probability $p.$ There is a parameter $\alpha$, $0<\alpha<1.$ The question: Is it possible to divide the set of nodes into "sources" and "receivers" so that the following two conditions are fulfilled? (i) For any "receiver", the probability that it ends up in a connected component with a "source" is larger or equal than $\alpha$; (ii) For any "source", the probability that it ends up in a connected component with another "source" is smaller than $\alpha$. I can prove, by elementary methods, that this is true for linear graphs, finite or infinite, complete graphs, and star graphs (graphs, in which $n-1$ nodes have degree $1$ and one node has degree $n-1$). Also, there are many other situations in which existence is trivial. In a finite graph, when $p$ is close to zero, the solution is that every node is a source, and the set of receivers is empty. If the graph is $\mathbb{Z}^2,$ then, if percolation function $\theta(p)>\sqrt{\alpha},$ then making one node a source and all other nodes receivers is a solution. (The case of an infinite graph might be very different. Interestingly, it is possible to have $p>\frac{1}{2}$ and a solution with an infinite number of sources on $\mathbb{Z}^2$.) I wonder if this is true for any finite (and locally finite) graph? REPLY [2 votes]: To indicate the non-triviality of this problem, consider a symmetric star graph with 3 rays (X,Y, and Z), each having 2 vertices (thus the graph has in total 7 vertices, including the center $C$, let us denote them $X_1,X_2,Y_1,Y_2,Z_1,Z_2$,and $C$). For this graph, the answer is "yes" for all possible values of $\alpha$ and $p$. Star graph with 3 rays One of the following SEVEN possible configurations will work (I indicate which vertices are sources): (i) $\{C\}$ (ii) $\{X_2,Y_2,Z_2\}$ (iii) $\{X_2,Y_2,Z_2,C\}$ (iv) $\{X_1,X_2,Y_2,Z_2\}$ (v) $\{X_1,X_2,Y_1,Y_2,Z_2\}$ (vi) $\{X_1,X_2,Y_1,Y_2,Z_1,Z_2\}$ (vii) $\{X_1,X_2,Y_1,Y_2,Z_1,Z_2,C\}$ It turns out that this is the minimal number of configurations needed to cover all possible values of the parameters. Especially difficult cases, requiring non-symmetric (!) configurations are $p=0.8, \alpha=0.95$ and $p=0.74, \alpha=0.9$. The plot below indicates areas of possible pairs $(p,\alpha)$ covered by the seven configurations above. Phase diagram for the pair $(p,\alpha)$<|endoftext|> TITLE: Examples of creative experiments by mathematicians in modern days QUESTION [41 upvotes]: I'm reading Random Circles on a Sphere and the authors did the following to empirically check their results: To make a partial test of the accuracy of the above approximations an experiment was carried out using table tennis balls. These had a mean diameter of 37.2 mm. with a standard deviation around this mean of 0.02mm. One hundred holes of diameter 29.9mm. were punched in an aluminium sheet forming one side of a flat box. The balls were held firmly against the holes by a foam rubber pad, and sprayed with a duco paint. After drying they were removed and replaced at random by hand. Forty sprayings were done in each of three sets of 100 balls. The number of balls not completely covered after N sprayings are shown in Table 2, and fit the theoretical curve rather more closely than the roughness of the approximations used would lead one to expect. The angle α was about 53.43° as used in the calculations. What are other examples of mathematicians turning into carpenters to test theories in modern days (post 60s)? REPLY [32 votes]: I was amused by the work of Scott Aaronson on soap bubbles and Steiner trees. Given $n$ points $p_1$, $p_2$, ..., $p_n$ in $\mathbb{R}^2$, the Steiner tree through these points is the connected planar graph of shortest length containing these points. Computing the topology of the Steiner tree, given the coordinates of the points, is NP-hard. On the other hand, if two parallel glass plates are separated by rods in positions $p_1$, $p_2$, ..., $p_n$ and dipped into a bucket of soapy water, the shape of the resulting soap film is a local minimum for this problem. Which lead to the question: How close is the local minimum, formed by whatever complicated PDE governs soap film, to the NP-hard global minimum? Aaronson decided to try the experiment, and reported on his results in Section 3 of "NP-complete problems and physical reality". The answer was not very close: The stable shape of the soap film was often not even a tree! REPLY [19 votes]: Psychologist Frank Rosenblatt built the first neural networks in 1957/8. Today it is trivial to build a neural network using software, but Rosenblatt built a neural network using analogue hardware. The perceptron could tell the difference between triangles, circles and and squares, or between different letters of the alphabet. Most critically, it could get better at doing so as it was given feedback on whether it was correct or incorrect about previous predictions. In this sense, it could 'learn'. A hardware implementation of a neural network The input to the perceptron was an array of 20 x 20 grid of light sensitive resistors And the weights/bias values of the primitive neural network were stored in racks of cylindrical objects each consisting of an electrical motor and potentiometer (rotary resistor).<|endoftext|> TITLE: What is known about exotic spheres up to stable diffeomorphism? QUESTION [13 upvotes]: In even dimensions $n=2k$ we can define two smooth manifolds $M$ and $N$ to be stably diffeomorphic if they become diffeomorphic after the connect sum with $r$ many copies of $S^k \times S^k$ for some natural number $r$. Question: What is known about the stable diffeomorphism classification of exotic $2k$-spheres? Are they all stably diffeomorphic to the standard sphere? If not does this happen sometimes? Are there examples which are stably diffeomorphic to the standard sphere? Or are they all distinct up to stable diffeomorphism? Here is some background about what I know so far. Given a manifold $M$, its stable normal bundle is classified by a map $\nu:M \to BO$. This factors through a space $B_M$ called the normal $(k-1)$-type of $M$. $$ M \to B_M \to BO$$ The map $M \to B_M$ is an isomorphism on homotopy groups $\pi_i$ with $i < k$ and is surjective on $pi_k$. The map $B_M \to BO$ is injective on $\pi_k$ and an isomorphism on $\pi_i$ with $i>k$ . The fiber is a $(k-1)$-type. Fixing $B \to BO$, and a manifold $M$, we can consider all maps $M \to B$ which lift the stable normal bundle $M \to BO$ and realize $B$ as the normal $(k-1)$-type of $M$. These are called normal $(k-1)$-smoothings. We can think of this as equipping the normal bundle of $M$ with a $B$-structure. As explained here the notion of $B$-structured stable diffeomorphism also makes sense. We have the following Theorem of Kreck. Theorem: Let $M$ and $N$ be $2k$-dimensional closed smooth manifolds with the same normal $(k-1)$-type $B$. Then two normal $(k-1)$-smoothings $(M, \theta_M)$ and $(N, \theta_N)$ are stably diffeomorphic if and only if the bordism classes of $(M, \theta_M)$ and $(N, \theta_N)$ agree in the $B$-bordism group $\Omega_{2k}^B$ and the Euler characteristics agree. The choice of the normal smoothing can be dealt with by considering the action of the automorphisms of $B \to BO$ on the bordism group. If $M$ is a sphere (or exotic sphere) then the normal $(k-1)$-type is $B = BO\langle k\rangle$. Now here it describes a way to get invariants of exotic spheres using $B$-bordism (general $B$). Any framed manifold admits a $B$-structure. If $B$ is such that $[S^n, \theta] = 0 \in \Omega^B_n$ for any framing $\theta$, then there is a well-defined homomorphism $$\eta^B: \Theta_n \to \Omega_n^B$$ which sends an exotic sphere $\Sigma$ to $[\Sigma, \theta]$ where $\theta$ is any framing. The manifold atlas website I linked to above gives some examples. One is where $B = BO\langle \ell\rangle$ with $k+1 < \ell < 2k + 2 = n + 2$. In that case $\Omega^B_n \cong \pi_n(G/O)$, this map is very interesting and related to the famous work of Kervaire-Milnor. The case relevant to the stable diffeomorphism classification is just outside this range, however. We can re-phrase the main question as follows. Question: When $B = BO\langle k\rangle$, what is known about the map $\eta^B: \Theta_{2k} \to \Omega^B_{2k}?$ Specifically what is the kernel? The same website I linked above also notes that $\eta^\textrm{Spin}$ is non-trivial in dimension $8m + 2$. So I expect that this map won't be zero in general, but I am not sure. REPLY [18 votes]: The inertia group $I_M$ of a closed oriented $d$-manifold $M$ is the subgroup of $\theta_d$ of h-cobordism classes of homotopy spheres $\Sigma$ such that $\Sigma \# M$ is diffeomorphic to $M$. Wall and Kosinski proved that $I_{W_g}$ is trivial in all dimensions, where $W_g := \#^g S^n \times S^n$. This resolves your main question: two exotic spheres that are stably diffeomorphic are already diffeomorphic. I learned this from conversations with Manuel Krannich. As you explain, it then follows from Kreck's result that the kernel of the map $\eta^B$ agrees with $\theta_{2k}$ itself.<|endoftext|> TITLE: Representable cohomology theories in motivic homotopy theory QUESTION [6 upvotes]: I am reading Mazza's, Voevodsky's and Weibel's book Lecture Notes on Motivic Cohomology and have grown curious about the following question: Which cohomology theories on $Sm/k$ are representable, i.e. appear as Hom-groups, in Voevodsky's construction of $DM_{\text{Nis}}^{eff,-}(k, R)$? If we let $X$ be a scheme of finite type over $k$ and $R$ a commutative unital ring, then we have (by construction) motivic homology with coefficients in $R$ $$H_{n,i}(X, R) = \text{Hom}_{DM_{\text{Nis}}^{eff,-}}(R(i)[n], R_{\text{tr}}(X)),$$ and motivic cohomology $$H^{n,i}(X, R) = \text{Hom}_{DM_{\text{Nis}}^{eff,-}}(R_{\text{tr}}(X), R(i)[n]).$$ From this, we obtain the algebraic singular homology (and hence Suslin's singular homology) as $$H^{\text{sing}}_n(X,R) \cong H_{n,0}(X,R)$$ and, if $X$ is a smooth separated scheme of finite type over some perfect field $k$, the higher Chow groups $$CH^q(X, 2q-p) = H^{p,q}(X,\mathbb{Z}).$$ It is clear to me that such cohomology theories must satisfy certain properties: Mayer-Vietories $\mathbb{A}^1$-invarience Künneth-theorem ... But do we have any other examples of cohomology theories that appear as Hom-groups in $DM_{\text{Nis}}^{eff,-}(k, R)$? How about Betti cohomology? $l$-adic cohomology? Crystalline cohomology? Algebraic de Rham cohomology? Another way of framing the question is: How far off is Voevodsky's-Morel's $DM_{\text{Nis}}^{eff,-}(k, R)$ construction from being the category of pure/mixed motives in the sense of Grothendieck. (I realize this might be a very hard question to which not much is known.) REPLY [3 votes]: Recall that $\mathrm{DM}(k)$ can be described as the subcategory of $\mathrm{SH}(k)$ made of modules over the motivic cohomology. This implies that cohomologies which are representable in $\mathrm{DM}(k)$ receive a cycle class map, so they must admit additive Chern classes ($c_1(L\otimes L')=c_1(L)+c_1(L')$). In fact, I think this is an if and only if, in the sense that motivic cohomology is the universal among cohomologies with additive Chern classes, but I do not know right now if this is written anywhere in the motivic setting with integral coefficients. In any case, if you put $R=\mathbb{Q}$ then you can have any type of Chern classes, for example $K$-theory. This is all written at Cisinki-Déglise's Triangulated categories..., chapter 14, which is in general a great reference for many motivic results up until 2009. Your question is more natural for $\mathrm{SH}(k)$, as it happens already in topology, and then you know your cohomology would belong to $\mathrm{DM}(k)$ depending on if it is a module over motivic cohomology or not. An answer to your question in the setting of $\mathrm{SH}$ is in the paper of Cisinski and Déglise, Mixed Weil cohomologies, at 2.1.5. Loosely speaking, for any "sheaf cohomology" they construct an object of $\mathrm{SH}(k)$ representing such cohomology. Their hypothesis must imply having additive Chern classes, since they are willing to define a mixed variant of Weil cohomologies (which have additive Chern classes), so their cohomologies are in $\mathrm{DM}(k)$. At the end of the paper you have their main examples. I recall that Cisinski-Déglise's method was later rewritten and developped by Holmsotrom-Scholbach in Arakelov motivic cohomology, section 3, to represent the Deligne cohomology, and by Déglise-Mazzari in The rigid syntomic spectrum at 1.4.10 where they proved that the rygid syntomic cohomology is representable.<|endoftext|> TITLE: Betti numbers and lower central series quotients of finite-index subgroups of nilpotent groups QUESTION [6 upvotes]: Let $G$ be a finitely generated nilpotent group and let $A\le G$ be a finite-index subgroup. I have two questions about $A$: Is it true that the inclusion map $A \rightarrow G$ induces isomorphisms $H_k(A;\mathbb{Q}) \rightarrow H_k(G;\mathbb{Q})$ for all $k$, and thus that the Betti numbers of $A$ and $G$ are the same? This is true if $G$ is abelian and I suspect that it is true in general, but I can't seem to prove it or find an appropriate reference. Letting $\gamma_k$ denote the kth term of the lower central series, is it true that the abelian groups $\gamma_k(A)/\gamma_{k+1}(A)$ and $\gamma_k(G)/\gamma_{k+1}(G)$ have the same rank (i.e. become the same after tensoring with $\mathbb{Q}$) for all $k$? REPLY [5 votes]: Both statements are true. To prove them, we will need the following lemma: Lemma: Let $G$ be a finitely generated nilpotent group of class $k$ and let $H$ be a subgroup of $G$. Then $\gamma_k(H)$ is a finite-index subgroup of $\gamma_k(G)$. Proof: Iterated $k$-fold commutators induce a surjective homomorphism $\phi\colon \wedge^k G^{\text{ab}} \rightarrow \gamma_k(G)$ of finitely generated abelian groups. Our hypotheses imply that the image of $\wedge^k H^{\text{ab}}$ in $\wedge^k G^{\text{ab}}$ is finite-index, so the image of the composition $$\wedge^k H^{\text{ab}} \rightarrow \wedge^k G^{\text{ab}} \stackrel{\phi}{\rightarrow} \gamma_k(G)$$ has finite-index. But this implies that $\gamma_k(H)$ is finite-index in $\gamma_k(G)$, as desired. $\square$ Here's the proof of 1: Theorem: Let $G$ be a finitely generated nilpotent group and let $H < G$ be a finite-index subgroup. Then $H_n(H;\mathbb{Q}) \cong H_n(G;\mathbb{Q})$ for all $n$. Proof: The proof will be by the degree $k$ of nilpotency of $G$. The base case $k=1$ means that $G$ is abelian, and the theorem is easy. Assume now that the theorem is true for some $(k-1)$, and we will prove it for $k$. Observe that $\gamma_k(H) < \gamma_k(G)$, and the lemma above says that the abelian group $\gamma_k(H)$ is a finite-index subgroup of the finitely generated abelian group $\gamma_k(G)$. Setting $\overline{H} = H / \gamma_k(H)$ and $\overline{G} = G/\gamma_k(G)$, we have a morphism between the central extensions $$1 \rightarrow \gamma_k(H) \rightarrow H \rightarrow \overline{H} \rightarrow 1$$ and $$1 \rightarrow \gamma_k(G) \rightarrow G \rightarrow \overline{G} \rightarrow 1.$$ We therefore get a morphism between the associated Hochschild-Serre spectral sequences. The map on the $E^2$ page is of the form $$H_p(\overline{H};H_q(\gamma_k(H);\mathbb{Q})) \rightarrow H_p(\overline{G};H_q(\gamma_k(G);\mathbb{Q})).$$ Since these are central extensions, this simplifies to a map $$H_p(\overline{H};\mathbb{Q}) \otimes H_q(\gamma_k(H);\mathbb{Q}) \rightarrow H_p(\overline{G};\mathbb{Q}) \otimes H_q(\gamma_k(G);\mathbb{Q}).$$ By our inductive hypothesis, this is a tensor product of isomorphisms, so we conclude that our two spectral sequences are the same and thus $H_n(H;\mathbb{Q}) \cong H_n(G;\mathbb{Q})$ for all $n$. $\square$ Here's the proof of 2: Theorem: Let $G$ be a finitely generated nilpotent group and let $H < G$ be a finite-index subgroup. Then the abelian groups $\gamma_n(H) / \gamma_{n+1}(H)$ and $\gamma_n(G) / \gamma_{n+1}(G)$ have the same ranks for all $n$. Proof: The proof will be by the degree $k$ of nilpotency of $G$. The base case $k=1$ means that $G$ is abelian, and the theorem is easy. Assume now that the theorem is true for some $(k-1)$, and we will prove it for $k$. Using our inductive hypothesis, it is enough to just deal with the bottom of the lower central series and prove that the ranks of $\gamma_k(H)$ and $\gamma_k(G)$ are the same. But the lemma above says that the abelian group $\gamma_k(H)$ is a finite-index subgroup of $\gamma_k(G)$, so their ranks are definitely the same. $\square$ By the way, I can't help but point out a connection between your two questions. In his beautiful paper Stallings, John, Homology and central series of groups. J. Algebra 2 (1965), 170-181. Stallings proves the following theorem. Let $\gamma_k^{tf}$ be the torsion-free lower central series you define in your other question. Theorem (Stallings): Let $f\colon H \rightarrow G$ be a group homomorphism inducing an isomorphism on $H_1(-;\mathbb{Q})$ and a surjection on $H_2(-;\mathbb{Q})$. Then $f$ induces isomorphisms between $(\gamma^{tf}_k(H)/\gamma^{tf}_{k+1}(H)) \otimes \mathbb{Q}$ and $(\gamma^{tf}_k(G)/\gamma^{tf}_{k+1}(G)) \otimes \mathbb{Q}$ for all $k$. $\square$ This can be applied to the inclusion of a finite-index subgroup by the first theorem above, and together with the positive answer to your other question you can deduce a positive answer to your second one (but one that is obviously way over-powered!).<|endoftext|> TITLE: Lower central series vs torsion-free lower central series QUESTION [6 upvotes]: $\newcommand\tf{\text{tf}}\newcommand\tor{\text{tor}}$Let $G$ be a finitely generated group. Let $\gamma_k(G)$ denote the $k$th term in the lower central series for $G$, so $\gamma_1(G) = G$ and $\gamma_{k+1}(G) = [\gamma_k(G),G]$ for all $k \geq 1$. There is also a natural central series that it makes sense to call the "torsion-free lower central series", and is defined inductively as follows. First, define $\gamma_1^{\tf}(G) = G$. Now assume that $\gamma_k^{\tf}(G)$ has been defined for some $k \geq 1$. We then have a finitely generated abelian group $$V_k = \gamma_k^{\tf}(G) / [G,\gamma_k^{\tf}(G)].$$ Let $V_k^{\tor}$ be the torsion subgroup of $V_k$, and define $\gamma_{k+1}^{\tf}(G)$ to be the pullback of $V_k^{\tor}$ under the projection $\gamma_k^{\tf}(G) \rightarrow V_k$. It follows from the definitions that each $\gamma_k^{\tf}(G)/\gamma_{k+1}^{\tf}(G)$ is a finitely generated free abelian group. Moreover, $\gamma_k(G) < \gamma_k^{\tf}(G)$ for all $k$. Question: Is it true that $\gamma_k^{\tf}(G)/\gamma_{k+1}^{\tf}(G)$ and $\gamma_k(G)/\gamma_{k+1}(G)$ have the same rank (i.e. become isomorphic after tensoring with $\mathbb{Q}$) for all $k$? REPLY [5 votes]: You can prove this by repeatedly applying the following lemma: Lemma: Let $G$ be a group and let $A,B \lhd G$ be finitely generated normal subgroups. Assume that $A$ and $B$ are commensurable. Then $[G,A]$ and $[G,B]$ are commensurable and $$\left(A / [G,A]\right) \otimes \mathbb{Q} \cong \left(B / [G,B]\right) \otimes \mathbb{Q}.$$ At the kth stage of the argument, you've proven that $\gamma_k(G)$ is a finite-index subgroup of $\gamma^{tf}_k(G)$. Applying the lemma (with $G$ replaced with $G/\gamma_{k+1}(G)$ —- this doesn’t change the relevant subquotients, and since it makes the group finitely generated nilpotent it make all its subgroups finitely generated) we deduce that the ranks of $\gamma_k(G)/\gamma_{k+1}(G)$ and $\gamma^{tf}(G)/[G,\gamma^{tf}_{k}(G)]$ are the same, and also that $\gamma_{k+1}(G)$ is a finite-index subgroup of $[G,\gamma^{tf}_{k+1}(G)]$. By construction, the ranks of $\gamma^{tf}(G)/[G,\gamma^{tf}_{k}(G)]$ and $\gamma^{tf}(G)/\gamma^{tf}_{k+1}(G)$ are the same (giving one conclusion!), and also that $\gamma^{tf}_{k+1}(G)$ is commensurable with $[G,\gamma^{tf}_{k}(G)]$ (letting us continue the induction!). Proof of lemma: We can assume without loss of generality that $A$ is a finite-index subgroup of $B$. The purported isomorphism can be rewritten as $$H_1(A;\mathbb{Q})_G \cong H_1(B;\mathbb{Q})_G,$$ where the subscripts indicate that we are taking the $G$-coinvariants. Since $B/A$ is a finite group, the Hochschild-Serre spectral sequence of the extension $$1 \rightarrow A \rightarrow B \rightarrow B/A \rightarrow 1$$ degenerates to show that $$H_k(B;\mathbb{Q}) \cong H_0(B/A;H_k(A;\mathbb{Q})) = H_k(A;\mathbb{Q})_B \quad \text{for all $k \geq 0$},$$ and taking the $G$-convariants of this gives that $$H_k(A;\mathbb{Q})_G = \left(H_k(A;\mathbb{Q})_B\right)_G \cong H_k(B;\mathbb{Q})_G \quad \text{for all $k \geq 0$}.$$ The special case $k=1$ of this is what we were supposed to prove. To see that $[G,A]$ and $[G,B]$ are commensurable, we will need the Hirsch index of a general group $\Gamma$, which is the supremum of the $n$ such that there exists a chain $$\Gamma_0 < \Gamma_1 < \cdots < \Gamma_n$$ of subgroups of $\Gamma$ such that $\Gamma_i$ is an infinite-index subgroup of $\Gamma_{i+1}$ for all $0 \leq i < n$ (this is usually only defined for polycyclic groups, but it is not hard to see that this reduces to the usual definition in that case; see this answer for more details). Let $r$ be the common dimension of $\left(A / [G,A]\right) \otimes \mathbb{Q}$ and $\left(B / [G,B]\right) \otimes \mathbb{Q}$. Since $A$ is a finite-index normal subgroup of $B$, we can apply the additivity result proved here to the short exact sequence $$1 \longrightarrow A/[G,A] \longrightarrow B/[G,A] \longrightarrow B/A \longrightarrow 1$$ to deduce that $$h(B/[G,A]) = h(A/[G,A]) + h(B/A) = r + 0 = r.$$ Applying this additivity now to the short exact sequence $$1 \longrightarrow [G,B]/[G,A] \longrightarrow B/[G,A] \longrightarrow B/[G,B] \longrightarrow 1,$$ we see that $$r = h(B/[G,A]) = h([G,B]/[G,A]) + h(B/[G,B]) = h([G,B]/[G,A]) + r.$$ It follows that $h([G,B]/[G,A])=0$, so $[G,A]$ is a finite-index subgroup of $[G,B]$, as desired.<|endoftext|> TITLE: When do flat holomorphic connections exist? QUESTION [10 upvotes]: Let $X$ be a smooth projective variety over $\mathbb{C}$. I know that a vector bundle $\mathcal{E}$ on $X$ admits a holomorphic/algebraic connection iff its Atiyah class vanishes, $A(\mathcal{E}) = 0$. If we choose a Hermitian structure on $\mathcal{E}$ giving a Chern connection $\nabla$ then $A(\mathcal{E}) = [\omega_\nabla]$ where $\omega_\nabla$ is the curvature. Therefore, if $\mathcal{E}$ admits a flat Hermitian structure then it admits a holomorphic connection. I am wondering to what extent this has a converse. Precisely, there are four properties I am interested in: (1) $\mathcal{E}$ admits a flat connection, (2) $\mathcal{E}$ admits a flat Hermitian structure, (3) $\mathcal{E}$ admits a holomorphic connection, (4) $\mathcal{E}$ admits a flat holomorphic connection. What are the implications between these properties? We know (2) $\implies (3)$ and obviously (4) $\implies$ (3) and (2) $\implies$ (1) and (4) $\implies$ (1). What about (1) $\implies$ (2) and (3) $\implies$ (4)? If $\mathcal{E}$ admits a holomorphic connection then we know that $[\omega_\nabla] = 0$ for any Chern connection but I cannot see how to conclude that there exists a flat Chern connection. I know from How many flat connections has a line bundle in algebraic geometry? that if $\mathcal{E}$ is a line bundle then any holomorphic connection is automatically flat, but it is clear that this is false for rank at least two. Explicit counterexamples would be helpful. REPLY [9 votes]: The comment of HYL should be an answer. Since the OP has asked for explicit counterexamples, I will give an example that 1) does not imply 2): Consider a compact Riemann surface $\Sigma$ of genus $g\geq 2.$ A complex projective structure on $\Sigma$ is given by an atlas of holomorphic coordinates which are related to each other by Moebius transformations (as given by $PSL(2,\mathbb C)$). The developing map $\widetilde\Sigma\to\mathbb CP^1$ is well-defined on the universal covering and induces a $PSL(2,\mathbb C)$ monodromy. It is well-known that (for compact Riemann surfaces) there always exists a lift to a $SL(2,\mathbb C)$ monodromy. The corresponding flat $SL(2,\mathbb C)$-bundle $(V\to\Sigma,\nabla)$ induced from the representation is unstable. In fact, the projective structure can be recovered from $\nabla$ as follows. There is a holomorphic subbundle $S\to V$ such that $$\nabla \colon S\to K_\Sigma V/S$$ is an isomorphism. As $V/S=S$ this implies that $S^2=K_\Sigma.$ Thus, the holomorphic bundle $V$ is unstable. On the other hand, every flat Hermitian bundle must be semi-stable (stable or totally reducible).<|endoftext|> TITLE: Explicit abelianization functor for groups QUESTION [13 upvotes]: Assume that I have a short exact sequence of finitely presented groups $$1 \longrightarrow K \longrightarrow H \longrightarrow G \longrightarrow 1,$$ where $G$ is finite (but I do not know whether this is relevant for what follows). Applying abelianization, we get an exact sequence $$K_{\mathrm{ab}} \longrightarrow H_{\mathrm{ab}} \to G_{\mathrm{ab}} \longrightarrow 0.$$ I would like to have a "quantitative" measure of the lack of left-exactness for the above sequence. For instance, I would like to know if it is possible to find an explicit sequence $\{L_i\}$ of (finitely presented) groups sitting in a long exact sequence of the form $$\ldots \longrightarrow L_3 \longrightarrow L_2 \longrightarrow L_1 \longrightarrow K_{\mathrm{ab}} \longrightarrow H_{\mathrm{ab}} \to G_{\mathrm{ab}} \longrightarrow 0.$$ By "explicit" I mean (for instance) that it is, at least in principle, possible to find presentations for the $L_i$ once one has presentations for $K$, $H$, $G$. Since the category of groups is not abelian (or even additive) we cannot perform the usual construction of the derived functors for the abelianization functor. I am aware that some more refined constructions have been presented (cotangent complex, André-Quillen homology, etc, see for instance the comments to this MSE question) but they look very technical and perhaps overkill in the simple case I have in mind. I am not an expert, but it seems to me that for the case of groups there should be some more down-to-earth construction, possibly related to the usual group (co)homology, but I looked in some standard textbooks and I did not find any. So, let me ask the following Question. Is it possible to construct groups $L_i$ as above, in some (at least in principle) computable way, providing a sort of "explicit derived functor" for the abelianization functor? If so, what are some references? REPLY [4 votes]: Here is a comment about the other approach proposed in the original question, using nonabelian derived functors. It is too long to be posted as a comment, so unfortunately I have had to post it as an answer. The category of groups is not an abelian category, but the category of modules over a fixed group G is an abelian category, and derived functors in that abelian category yield group (co)homology. We could instead directly consider the abelianization functor $ab$ from groups to abelian groups, and we could consider nonabelian derived functors of $ab$. This would involve choosing a simplicial resolution of a group $G$, applying $ab$ to that simplicial group to get a simplicial abelian group, taking the normalized Moore chain complex of that simplicial group, and then taking the homology groups of that chain complex. As far as I know, the standard textbook reference for these kinds of nonabelian derived functors is chapter 2 of H. Inassaridze's book "Non-abelian homological algebra and its applications." It is not a trivial matter to check that a short exact sequence of groups induces a long exact sequence of nonabelian derived functors: see Theorem 2.63 in Inassaridze's book for some sufficient conditions. In particular, I do not know if a short exact sequence of groups $1 \rightarrow K \rightarrow H \rightarrow G \rightarrow 1$ induces a long exact sequence $$\dots \rightarrow L_{n+1}ab(G) \rightarrow L_n ab(K) \rightarrow L_nab(H) \rightarrow L_nab(G)\rightarrow \dots $$ This would require checking the conditions of Inassaridze's Theorem 2.63, or something similar. In the absence of such a long exact sequence, I do not think the approach by nonabelian derived functors yields the long exact sequence $ \dots \rightarrow K_{ab} \rightarrow H_{ab}\rightarrow G_{ab} \rightarrow 1$ which you ask for. The only positive result on nonabelian derived abelianization of groups which I was able to locate in the literature is mentioned in the proof of Theorem 3 in the paper "N-fold Cech derived functors and generalized Hopf type formulas" by G. Donadze, N. Inassaridze, and T. Porter: the first nonabelian derived functor of abelianization, $L_1ab(G)$, is apparently isomorphic to the second group homology group $H_2(G,Z)$ with integer coefficients, that is, the classical Schur multiplier of $G$. Perhaps someone with more knowledge of nonabelian derived functors can contribute further and give a better answer to your question. It appears to me that going this route is not only more technically demanding than using group homology and the LHSSS, but there seem to be fewer tools available for making calculations (compared to the wealth of tools for computing group homology), and fewer connections to other areas of mathematics; and, possibly most troubling, it is not clear that you get the desired long exact sequence, in nonabelian derived abelianization, that you're looking for.<|endoftext|> TITLE: Vinogradov-Korobov for Dirichlet L-functions? QUESTION [6 upvotes]: Where can one find a Vinogradov-Korobov zero-free region for Dirichlet L-functions? It has to be in a standard reference, but I'm having a non-trivial time finding it. REPLY [2 votes]: In Theorem 2 of Mark Coleman's paper below, a full proof of the Vinogradov-Korobov zero-free region for the $L$-function of a Grossencharacter twisted by a Hecke character over an arbitrary number field is provided. Restricting to $K=\mathbb{Q}$, one recovers the result for Dirichlet $L$-functions. This is the only place where I have seen a full proof of such a zero-free region published. Coleman, M. D., A zero-free region for the Hecke L-function, Mathematika 37, No. 2, 287-304 (1990). ZBL0721.11050.<|endoftext|> TITLE: Condensed / pyknotic sets in terms of forcing over Boolean-valued models of set theory / multiverse concepts? QUESTION [14 upvotes]: Here is one way of saying what a pyknotic set is. Fix an inaccessible cardinal $\kappa$, and let $Proj_\kappa$ be the category of $\kappa$-small, extremally disconnected compact Hausdorff spaces. Recall that Stone duality restricts to an equivalence between the opposite category $Proj_\kappa^{op}$ and the category $CBool_\kappa$ of $\kappa$-small, complete Boolean algebras, and all Boolean algebra homomorphisms. Barwick and Haine define a pyknotic set $X$ to be a sheaf on the category $Proj_\kappa$ with respect to the canonical Grothendieck topology. It turns out to be very easy to say what this means explicitly. That is, we have: Definition: A pyknotic set comprises A functor $X: CBool \to Set$; (i.e. for each complete boolean algebra $B$ we have a set $X(B)$, and for each boolean algebra homomorphism $B \to B'$ we have a function $X(B) \to X(B')$ respecting identities and composition) such that $X$ respects finite products. (i.e. $X(1)$ is a singleton, where $1$ is your favorite 1-element Boolean algebra, and the canonical map $X(B \times B') \to X(B) \times X(B')$ is a bijection for any pair of complete Boolean algebras $B,B'$.) I've chosen to state the definition of a pyknotic set this way; the condensed sets of Clausen and Scholze are similar, but avoid the requirement of choosing an inaccessible cardinal. This is done by defining $Proj_\kappa$ as above where $\kappa$ is merely a strong limit cardinal, and then taking a direct limit over all $\kappa$'s to arrive at the final notion. Now, the thing about complete Boolean algebras is that it seems they're more commonly studied by set theorists than by anybody else. My understanding is that in the "Boolean-valued models" approach to forcing, the forcing extension is more-or-less identified with sheaves on the forcing poset, which is a complete Boolean algebra. From this perspective, it sounds like a pyknotic set is somehow a set which "lives in all forcing extensions at once". Question: Do set theorists have a notion of a "set which lives in all forcing extensions at once"? If so, how similar is such a thing to the data of a pyknotic / condensed set? In any event, do there exist theoretical frameworks in set theory for talking about objects like pyknotic / condensed sets? My initial guess is that in multiverse-type frameworks, one probably considers morphisms of forcing posets which preserve a little more structure than just an arbitrary boolean algebra homomorphism, so that perhaps the connection is not so tight. But I really have no idea! REPLY [10 votes]: Great question — for some reason this tight relation between extremally disconnected profinite sets and forcing had elapsed me! I've just been trying to read a bit about it. From what I understand, the sheaf-theoretic approach to forcing, as in MacLane-Moerdijk "Sheaves in geometry and logic" Chapter VI, consists of three steps. Start with any extremally disconnected profinite set $S$. Consider the category of open and closed subsets of $S$, with the following notion of cover: $\{U_i\subset U\}_i$ is a cover of $U$ if $\bigcup_i U_i\subset U$ is dense in $U$. (This is what, I believe, the "double negation topology" amounts to.) In this topology, the subsheaves of $\ast$ are exactly given by the open and closed subsets $U\subset S$, so one has a boolean topos. The first step is thus completed: The construction of the boolean topos of sheaves $\mathrm{Sh}(S)$ on $S$. The second step is to pick any point $s\in S$, and take the colimit $\varinjlim_{U\ni s} \mathrm{Sh}(U)$. Here, the subsheaves of $\ast$ are just $\emptyset$ and $\ast$, so it's a boolean topos with only two truth values. The third step is to start with the topos $\varinjlim_{U\ni s} \mathrm{Sh}(U)$ and somehow make it into a model of ZFC. When I've previously tried myself to contemplate forcing from the sheaf-theoretic point of view, this is the point that got me very confused: In a model of ZFC, elements of sets should be sets, but there's no meaningful way to talk about elements of objects in this topos, and certainly they won't be objects of this topos. I have to read more about this step; MacLane-Moerdijk cite work of Fourman. Apparently the idea is to redo the iterative construction of $V_\alpha$'s by iteratively taking the powerset, but now internally in this topos. [Edit: This third step deals with the problem of turning a structural set theory into a material set theory. A very nice discussion of this is in the paper Comparing material and structural set theories of Shulman. In particular, he explains how to very cleanly go back and forth between models of ECTS + a structural form of replacement, which $\varinjlim_{U\ni s} \mathrm{Sh}(U)$ satisfies, and models of ZFC, see Corollary 9.5.] (I might at this point be sold on structural set theory — forcing seems to have an extremely clean formulation in terms of structural set theory, namely just $\varinjlim_{U\ni s} \mathrm{Sh}(U)$. Please correct me if I'm misunderstanding something!) In any case, the third step seems to be orthogonal to the question at hand. More salient is that the category of sheaves on $S$ is actually incompactible with the category of sheaves on $S$ that we would consider, where covers are just open covers. This is critical! If $S$ is the Stone-Cech compactification of a discrete set $S_0$, then sheaves in our sense are equivalent to functors on subsets of $S_0$, taking finite disjoint unions to products. But in the forcing-sense, they are equivalent to such functors taking all disjoint unions to products; equivalently, they are just sheaves on the discrete $S_0$. Most condensed sets of interest (like $\mathbb Z$ or $\mathbb R$) do not have this property; actually, the condition singles out the compact Hausdorff condensed sets if I'm not mistaken. So even before analyzing the question of how to put this together for varying $S$, I think there are slightly different things happening even for individual $S$. But I agree that it's definitely worth finding out if there's something more to this! Addendum in response to Mike Shulman's question in the comments below: I think $\varinjlim_{U\ni s} \mathrm{Sh}(U)$ is well-pointed. The key seems to be the following: If $f: B\to A$ is a map in $\mathrm{Sh}(S)$ that is surjective in the stalk at $s$, then $f|_U$ is surjective for some $U$ containing $s$. (This property does seem surprising to me. Is it a formal consequence of being a boolean topos?) To prove this, we prove that if $f|_U$ is not surjective for all such $U$, then also $f$ is not surjective in the stalk at $s$. Look at pairs $(V\subset S,a\in A(V))$ of an open and closed subset $V\subset S$ and a section $a$ of $A$ over $V$, such that $a\times_{A|_V} B|_V=\emptyset$. There is an obvious partial order on such; by Zorn's lemma and as any section over a union $\bigcup_i V_i$ extends to the closure by the notion of covering, there is some maximal such $(V,a)$. If $s\in V$, then in particular $f$ is not surjective on the stalk at $s$, as desired. Otherwise, let $U=S\setminus V$, which contains $s$. By assumption, $f|_U$ is not surjective, so there is some $U'\subset U$ and $a'\in A(U')$ such that $a'\times_{A|_{U'}} B|_{U'}$ (a subsheaf of $\ast|_{U'}$) is not all of $\ast|_{U'}$, and thus given by $\ast_{U''}$ for some $U''\subsetneq U'$; replacing $U'$ by $U'\setminus U''$, we can assume that $a'\times_{A|_{U'}} B|_{U'}=\emptyset$. But then $(V\sqcup U',a\sqcup a')$ extends $(V,a)$, contradicting maximality of $(V,a)$.<|endoftext|> TITLE: Wiener Corollary in "An introduction to harmonic analysis" by Yitzhak Katznelson QUESTION [6 upvotes]: I can't understand a lemma in "An introduction to harmonic analysis" by Yitzhak Katznelson which is stated as follows: Corollary. Let $\mu\in M(\mathbb T)$. Then $$\sum\limits_{\tau\in\mathbb T}|\mu(\{\tau\})|^2=\lim\limits_{N\rightarrow\infty }\frac{1}{2N+1}\sum\limits_{-N}^N|\hat\mu(n)|^2.$$ It looks bizzare to me since the left hand sum up all variable $\tau$, while the other side not. What I only know is that $\mu(\{\tau\})=\frac{1}{2N+1}\sum\limits_{-N}^N|\hat\mu(n)|,$ but it seems impossible to derive the corollary from this relation. (note added by YC: the result is stated without proof at the end of Chapter I Section 7.11 in the 1976 Dover edition) REPLY [7 votes]: As far as I understand, $\mu$ is a Borel probability measure on the unit circle $\mathbb{T}$. Then $$ \frac{1}{2N+1}\sum\limits_{n=-N}^N|\hat\mu(n)|^2=\int\int\frac1{2N+1}\sum_{n=-N}^N (x/y)^nd\mu(x)d\mu(y). $$ The integrand $$ \frac1{2N+1}\sum_{n=-N}^N (x/y)^n $$ has absolute value at most 1 and converges pointwise to 1 for $x=y$ and to $0$ for $x\ne y$ (the sum is a geometric progression; you may sum it up to see that it is bounded when $x\ne y$). Thus, by the Dominated Convergence Theorem, the limit is the $\mu\times \mu$-measure of the diagonal $\{x=y\}$, which is exactly the LHS.<|endoftext|> TITLE: Are at most $1/3$ vertices "kings"? QUESTION [12 upvotes]: If $G=(V,E)$ is a finite, simple, undirected graph, and $v\in V$, we set $N(v) = \{w\in V:\{v,w\}\in E\}$, and $\text{deg}(v)= |N(v)|$. We say a vertex $v\in V$ is a king if $\text{deg}(v) > \text{deg}(w)$ for all $w\in N(v)$. In the graph $G=(\{0,1,2\}, \big\{\{0,1\}, \{1,2\}\big\})$, one of the $3$ vertices is a king. Let $\text{King}(G)$ be the set of king vertices. Question. Is it true that for any finite connected graph $G=(V,E)$ with $|V|>1$ we have $|\text{King}(G)|/|V|\leq 1/3$? If not, how large can this value get? REPLY [5 votes]: I wanted to find a proof that uses Hall's marriage theorem [1] instead of double-counting. Given a graph $G=(V,E)$, let $K$ be the set of kings in $V$, and $R:=V \setminus K$ the rest. Claim: $|K| < |R|$. Proof Let $G=(V,E)$ be a counterexample that minimizes the sum $|V|+|E|$, so $|K| \ge |R|$. Then $G$ is bipartite, since any edge between nodes in $R$ can be removed. If $|K|>|R|$ then removing one king would yield a smaller counterexample, so $|K|=|R|$. If there was a subset $S$ of $K$ where its neighborhood satisfies $|N(S)|< |S|$, then the induced graph on $S \cup N(S)$ would be a smaller counterexample. Thus the Hall condition is met in $G$. Removing from $G$ a perfect matching of $K$ to $R$ yields a smaller counterexample. [1] https://en.wikipedia.org/wiki/Hall%27s_marriage_theorem<|endoftext|> TITLE: Deriving an asymptotic for $\pi(x)$ directly from $\log \zeta(s)$? QUESTION [9 upvotes]: Denote by $\pi(x)$ the number of primes $p\leq x$. We generally give approximations for $\pi(x)$ by first approximating $\psi(x) = \sum_{n\leq x} \Lambda(n)$. Part of the reason is presumably that, if we apply Perron's formula directly, without going through $\psi(x)$, we end up with an expression for $\pi(x)$ in terms of an integral involving $\log \zeta(s)$. Here we see what the issue is: $\log \zeta(s)$ has a branch point at $s=1$, and that is unpleasant. At the same time, it is not that unpleasant; you can do a little loop ("truncated Hankel contour") around $s=1$. Is there a standard reference where $\pi(x)$ is estimated in this way? REPLY [2 votes]: When $T\le\sqrt x$ and $x=\frac12+\mathbb Z^+$, it is possible to show using Perron's formula that $$ \pi(x)={1\over2\pi i}\int_{k-iT}^{k+iT}{x^s\over s}\log\zeta(s)\mathrm ds+\mathcal O\left(x\log x\over T\right) $$ After applying certain analytic properties of $\zeta(s)$ in the critical strip, the task turns into evaluating the following residue integral: $$ {1\over2\pi i}\oint_{(1+)}{x^s\over s}\log{1\over s-1}\mathrm ds $$ where $(1+)$ represents any counterclockwise path containing $s=1$ but not $s=0$. A possible approach to handle the integral, inspired by Riemann (see chapter 1 of Edwards's 1974 classic Riemann's zeta function), is to introduce a new parameter $r$ to the problem: $$ f(r)={1\over2\pi i}\oint_{(r+)}{x^s\over s}\log{1\over s/r-1}\mathrm ds $$ Taking derivatives on both side gives $f'(r)=x^r/r$, and integrating back we get $$ f(r)=\operatorname{li}(x^r)+\mathcal O(1) $$ where $\operatorname{li}(x)$ is the logarithmic integral function. The reason why there is an O-term is because the sign of $\Im(r)$ will introduce certain $i\pi$ deviation to the final result. A full derivation of the PNT using this idea can be found here.<|endoftext|> TITLE: Is the solution to this trig function known to be algebraic or transcendental? QUESTION [5 upvotes]: This largest solution to this gorgeous equation is the first local extremum on a function related to the Fibonacci sequence: $$x^2 \cdot \sin \left(\frac{2\pi}{x+1} \right) \cdot \left(3+2 \cos \left(\frac{2\pi}{x} \right) \right) = (x+1)^2 \cdot \sin \left(\frac{2\pi}{x} \right) \cdot \left(3+2 \cos \left(\frac{2\pi}{x+1} \right) \right)$$ This is as simplified as I could get it. The largest solution to this equation is around $x = 2.1392.$ It appears there is no closed-form solution for this; is there any way to prove if the solution is algebraic or transcendental? P.S. Can anyone approximate this constant to more decimal places? ANSWERED REPLY [4 votes]: It should be possible to show that $x$ is irrational using Theorem 7 of Trigonometric diophantine equations (On vanishing sums of roots of unity) by J. H. Conway and A. J. Jones, Acta Arithmetica 30 (1976), 229–240, although I have not carried out the full calculation. By letting $\alpha = 2\pi/(x+1)$ and $\beta = 2\pi/x$ and using standard trig identities, we can rewrite the given equation as $$3x^2\sin \alpha -3(x+1)^2\sin\beta - (2x+1)\sin(\alpha+\beta) + (2x^2+2x+1)\sin(\alpha-\beta) = 0.$$ We can convert from sines to cosines via $\sin \gamma \equiv \cos(\pi/2 - \gamma)$. Then Theorem 7 of Conway and Jones tells us that there are only a few "primitive" ways of getting a rational linear combination of four cosines of rational multiples of $\pi$ to vanish: $$\eqalign{{1\over2} &= \cos{\pi\over 3}\cr 0 &= -\cos\phi + \cos\biggl({\pi\over 3}-\phi\biggr) + \cos\biggl({\pi\over 3}+\phi\biggr)\cr {1\over2} &= \cos{\pi\over 5} - \cos{2\pi\over 5}\cr {1\over2} &= \cos{\pi\over 7} - \cos{2\pi\over 7} + \cos{3\pi \over 7}\cr {1\over2} &= \cos{\pi\over 5} - \cos{\pi\over 15} + \cos{4\pi\over 15}\cr {1\over2} &= -\cos{2\pi\over 5} + \cos{2\pi\over 15}-\cos{7\pi\over 15}\cr {1\over2} &= \cos{\pi\over 7} + \cos{3\pi\over 7} - \cos{\pi\over 21} + \cos{8\pi\over21}\cr {1\over2} &= \cos{\pi\over 7} -\cos{2\pi\over7}+\cos{2\pi\over21}-\cos{5\pi\over 21}\cr {1\over2} &= -\cos{2\pi\over 7} + \cos{3\pi\over 7}+ \cos{4\pi\over21} + \cos{10\pi\over 21}\cr {1\over2} &= -\cos{\pi\over 15}+\cos{2\pi\over15}+\cos{4\pi\over 15}-\cos{7\pi\over 15}\cr }$$ One should be able to go through this list and check case by case that no rational value of $x$ can yield the desired equation. I do not know how to show that $x$ must be transcendental.<|endoftext|> TITLE: A tricky integral to evaluate QUESTION [9 upvotes]: I came across this integral in some work. So, I would like to ask: QUESTION. Can you evaluate this integral with proofs? $$\int_0^1\frac{\log x\cdot\log(x+2)}{x+1}\,dx.$$ REPLY [11 votes]: For $t\in(0,1]$, let \begin{equation} I(t):=\int_0^1\frac{\log(x)\,\log(1+t(x+1))}{x+1}\,dx \end{equation} (so that the integral in question is $I(1)$), \begin{equation} \begin{aligned} J(t)&:=-\text{Li}_3(2+1/t)+\text{Li}_3(-2 t-1)+\text{Li}_3(t+1) \\ &-\text{Li}_3(2t+1) +\text{Li}_2(2+1/t) (\log (2 t+1)+i \pi ) \\ &+\text{Li}_2(-2 t-1)(-\log (2 t+1)-i \pi ) \\ &+\text{Li}_2(t+1) (-\log (t+1)-i \pi )+\text{Li}_2(2 t+1) (\log (2 t+1)+i \pi ) \\ &+\frac{1}{6} (\log ^3(t)+(-3 \log ^2(t+1)-6 i \pi \log (t+1)+4 \pi ^2) \log (t) \\ &+3 \pi (-i \log ^2(t+1)+2 i \log ^2(2 t+1)+2 \pi \log (t+1)-4 \pi \log (2 t+1))) \\ &+\frac{3 \zeta (3)}{4}-\frac{5 i \pi ^3}{12}, \end{aligned} \end{equation} \begin{equation} \begin{aligned} I_1(t)&:=6tI'(t)=6t\int_0^1\frac{\log(x)}{1+t(x+1)}\,dx \\ & =6 \text{Li}_2\left(\frac{t+1}{2 t+1}\right)-3 \log ^2(t+1)+3 \log ^2(2 t+1) \\ &+6 \log (t) \log \left(\frac{t+1}{2 t+1}\right)-\pi ^2, \end{aligned} \end{equation} \begin{equation} \begin{aligned} J_1(t)&:=6tJ'(t)=6 \text{Li}_2\left(2+1/t\right)+3 \log ^2(t)-3 \log ^2(t+1) \\ &-6 (\log (2 t+1)+i \pi ) \log (t)+6 \log (t+1) \log (2 t+1) \\ &+6 i \pi \log (2 t+1)-2 \pi ^2. \end{aligned} \end{equation} Then $I'_1=J'_1$ and $I_1(0+)=J_1(0+)$, so that $I_1=J_1$, and hence $I'=J'$. Also, $I(0)=I(0+)=0=J(0+)$, so that $I=J$, and the integral in question is \begin{equation} \begin{aligned} I(1)&=J(1)=\text{Li}_3(-3)-2 \text{Li}_3(3)+i \pi \left(-\text{Li}_2(-3)+2 \text{Li}_2(3)+\log ^2(3)\right) \\ &-\text{Li}_2(-3) \log (3)+\text{Li}_2(3) \log (9)+\frac{13 \zeta (3)}{8}-\frac{2 i \pi ^3}{3}-\pi ^2 \log (9) \\ &=-0.651114\dots. \end{aligned} \end{equation}<|endoftext|> TITLE: On modified Euler product QUESTION [6 upvotes]: Consider the modified Euler product as follows: $$F(s) = \prod_{p} \left( 1 - \frac{c}{p^s} \right)^{-\ln(p)}$$ Here $c$ is a constant My questions are Is there a compact representation for this product? What are some non-trivial properties of this product? Value of the regularized sum: $$-\sum_{p}\ln\left( 1 - \frac{1}{(ep)^{1/2}} \right){\ln(p)}$$ Help me understand the analytic continuation of this function. Or consider $$f(s) = \sum_{p}\left( 1 - \frac{1}{(ep)^{1/2}} \right)^{s\ln(p)}$$ $p$ belongs to set of primes We can find f'(x)'s analytic continuation for some region containing s=0. (This is the main motivation behind the question) Edit: Now by using the answer by R.Furman how can we calculate the product's value 'uniquely' at 1/2. So, I need more insight into this. Any comments regarding this are welcome. Any ideas about attacking this are welcome. References: [1] Germund Dahlquist ; "On the analytic continuation of Eulerian products" [2] https://mathworld.wolfram.com/PrimeProducts.html ( and all the references therein) [3] https://en.m.wikipedia.org/wiki/Euler_product [4] Kimoto , Wakayama "Remarks on Zeta Regularized Products" Progress post: On infinite sum containing logarithmic derivative of Zeta function and Möbius function: Recently I found a similar post while surfing through the site: Does this product have analytic continuation? If anyone could give answer in the context of the above post it will be very nice. REPLY [2 votes]: I'll expand on Ralph's answer to describe how to evaluate $F(1/2)$. Ralph's main point is that $\log F(s)$ is well-approximated by a sum of logarithmic derivatives of $\zeta(s)$. Writing it out explicitly, we have that $$ \begin{align} \log(F(s)) &= -c \frac{\zeta'(s)}{\zeta(s)} - \big(\tfrac{c^2}{2} - c\big) \frac{\zeta'(2s)}{\zeta(2s)} \\ &\qquad + \sum_p \log p \Big( \frac{c^3}{3} p^{-3s} + \frac{c^4}{4} p^{-4s} + \cdots\Big) \\ &\qquad + c\sum_p \log p \big(p^{-3s} + p^{-4s} + \cdots) \\ &\qquad + (\tfrac{c^2}{2} - c)\sum_p \log p \big( p^{-4s} + p^{-6s} + p^{-8s} + \cdots \big) \end{align}.$$ The first line comes from matching the first two terms in the expansion of $\log F(s)$, as in Ralph's answer. It would be possible to carry this on for mor terms if desired. The second line is the remaining terms of the original expansion of $\log F(s)$. The third line are the remaining terms from $c \zeta'(s)/\zeta(s)$. And the fourth line are the remaining terms from $(c^2/2 - c) \zeta'(2s)/\zeta(2s)$. For evaluation, one can use typical continuations for $\zeta$ and $\zeta'$ to evaluate the first line, and the remaining three lines are all absolutely convergent as long as $\lvert c p^{-s} \rvert < 1$. For evaluation, it's easier to recollect the various infinite expansions and to have something instead like $$ \begin{align} \log(F(s)) &= -c \frac{\zeta'(s)}{\zeta(s)} - \big(\tfrac{c^2}{2} - c\big) \frac{\zeta'(2s)}{\zeta(2s)} \\ &\qquad + \sum_p \log p \Big( - \log(1 - cp^{-s}) - \big( cp^{-s} + c^2 p^{-2s}/2 \big) \Big) \\ &\qquad + c\sum_p \log p \Big(\frac{p^{-3s}}{1 - p^{-s}}\Big)\\ &\qquad + (\tfrac{c^2}{2} - c)\sum_p \log p \Big(\frac{p^{-4s}}{1 - p^{-2s}}\Big) \end{align}.$$ With this, using the first primes up to $10^5$ in sage, I estimate $\log(F(0.5)) \approx -0.3312$ when $c = e^{-1/2}$. The (very simple, hastily written) code for this is below. c = e**-0.5 tot1 = 0 for p in primes(100000): tot1 += RR(log(p) * (- log(1 - c * p**-0.5) - (c * p**-0.5 + c*c * p**-1 / 2))) tot2 = 0 for p in primes(100000): tot2 += RR(log(p) * p**-1.5/(1 - p**-0.5)) tot3 = 0 for p in primes(100000): tot3 += RR(log(p) * p**-2 / (1 - p**-1)) actual_total = tot1 + c * tot2 + (c*c/2 - c) * tot3 - c * zetaderiv(1, 0.5)/zeta(0.5) # Note the other zeta term divides by zeta(1), and thus vanishes print(actual_total) ```<|endoftext|> TITLE: Cohomology of finite $p$-groups over integers in local fields QUESTION [6 upvotes]: Let $p$ be a prime, $G$ be a finite group of order $p^a$. Let $M$ be a $\mathbb{Z}[G]$-module. Then $H^n(G, M)$ is annihilated by $p^a$ for all $n \geq 1$ (see e.g. Brown, Corollary III.10.2). In particular this is true for $\mathbb{Z}_p[G]$-modules, and I was wondering if this version of the result can be generalized to local fields other than $\mathbb{Q}_p$. More precisely, let $\mathbb{K}$ be a non-Archimedean local field with ring of integers $\mathfrak{o}$, uniformizer $\omega$, maximal ideal $\mathfrak{p}$, residue field $\mathfrak{k}$ of characteristic $p$. Let $G$ be a finite $p$-group and $M$ an $\mathfrak{o}[G]$-module. What is the least $a$ such that $H^n(G, M)$ is annihilated by $\omega^a$? It should be easy to use the statement at the beginning of this post to generalize this to any local field of characteristic $0$, and $a$ should still be uniformly bounded, with the bound depending only on the order of $\mathfrak{k}$ and the order of $G$. But what about characteristic $p$? This case seems to be substantially different. I think this question is interesting in general, but there is probably no chance to get a general answer without assuming anything extra. So let me point out the specific setting that I am interested in: $n = 1$ and especially $n = 2$: this comes up in an extension problem so I only really care about these degrees. $M$ is a free $\mathfrak{o}/\mathfrak{p}^k$-module. So I already know that $H^n(G, M)$ is annihilated by $\omega^k$, that is $a \leq k$. But I would like to get better estimates, ideally I would like to find a bound of the form $a \leq a(k)$ such that $(k - a(k)) \to \infty$ as $k \to \infty$. In characteristic $0$ I can get $a(k)$ to be a constant so this is a significant weakening, but still somewhat strong. Even though I am interested in the general case, I have trouble understanding even the first example: $G = \mathbb{Z}/2\mathbb{Z}$ and $\mathbb{K} = \mathbb{F}_2((X))$, so $\omega = X$ and $\mathfrak{o} = \mathbb{F}_2[[X]]$. Explicitly: if $M$ is a free $\mathbb{F}_2[X]/X^k$-module with an action of $G$, what is the least $a$ such that $X^a H^n(G, M) = 0$ for $n = 1, 2$? After trying a bit I think that one cannot do better than $a(k) = k/2$ (see the previous point), but I am not able to prove that this works. REPLY [2 votes]: If you take $G=\mathbb Z/p^a\mathbb Z$, then the cohomology of any $M$ are computed by the Tate complex $M\xrightarrow{\sigma-\mathrm{id}} M \xrightarrow{1+\sigma+\ldots+ \sigma^{p^a-1} }M\xrightarrow{\sigma-\mathrm{id}} M \xrightarrow{1+\sigma+\ldots+ \sigma^{p^a-1} }M\xrightarrow{\sigma-\mathrm{id}}\ldots$. In particular if the action on $M$ is trivial then the norm map $1+\sigma+\ldots+ \sigma^{p^a-1}$ is given by multiplication by $p^a$ and $\sigma-\mathrm{id}=0$. If $M$ is of characteristic $p$ then $H^{i}(G,M)\simeq M$ for all $i$. Note that if $M$ had a structure of an $\mathfrak o$-module this is an isomorphism of $\mathfrak o$-modules (I assume here that $G$ acts $\mathfrak o$-linearly): this shows that in general $a(k)=k$. This is also true for any $G$: you can argue inductively by considering lower central series and the Hochschield-Serre spectral sequence to reduce to the case of products of cyclic groups (where you can use Kunneth formula). In more detail the $E^{0,1}$ term there is never hit by a differential and thus by induction $H^1(G, M)$ naturally contains $M$. Allowing $M$ (and $k$) vary for a fixed $G$ you can always make $a(k)$=0. Indeed in the case of a cyclic group $\mathbb Z/p^a\mathbb Z$ we can take $M=\mathfrak o/\omega^{i\cdot p^a}$ and the action of $\sigma$ by $1+\omega^i$. Then $\sigma^{p^a}=1$, so we have an action and $1+\sigma+\ldots \sigma^{p^a-1}=\frac{(1+\omega^i)^{p^a}-1}{\omega^i}=\omega^{i\cdot (p^a-1)}$, while $1-\sigma=\omega^i$. Looking at the Tate complex we get that $H^{>0}(G,M)=0$. Then for any $G$ you can take the a cyclic subgroup $H$ in the center $Z(G)\subset G$ and induce the above representation to $G$. In fact you could also induce from the trivial subgroup right away (in other words just take the regular module $M=\mathfrak o[G]$). The above construction might be slightly better because there the annulating power of $\omega$ for $H^0$ also decreases (so if you also include $H^0$ in the definition of $a(k)$ you still get $a(k)\sim\frac{p^a-1}{p^a}k$)<|endoftext|> TITLE: The tensor product of two topological complexes with closed range QUESTION [5 upvotes]: A Künneth formula by Grothendieck/Schwartz states the following: Let $A, B$ be chain complexes of nuclear Fréchet spaces. If the differentials $d_A, d_B$ are topological homomorphisms (meaning in this setting: if they have closed ranges), then we have the Künneth formula $$H(A \, \hat \otimes \, B ) \cong H(A) \, \hat \otimes \, H(B). \quad \quad\quad (*)$$ I would like to apply this statement to more than two product factors, and to do so, it would suffice that the range of the differential on the product space be closed again. Is there an easy way to see why/if this is true? I've tried to deduce it from the proof given in their paper, but somehow I don't see it, because the range of the product differential does not seem to come up explicitly. On the one hand, the closed range property seems quite fickle and is in general not closed under linear combinations, and I fear that an expression like $d_A \otimes \text{id} + \text{id} \otimes d_B$ might be too general to hope for the closed range property again. On the other hand, the RHS of $(*)$ is canonically a Fréchet space again, the LHS only if the range of the product differential is closed. If this isomorphism held, but the product differential range was not closed, I feel that would be quite strange... REPLY [2 votes]: The discussion in the comments catapulted me onto the right track! It seems the solution is exactly to note that the isomorphism $H(A \, \hat \otimes \, B ) \cong H(A) \, \hat \otimes \, H(B)$ is not only an isomorphism of abstract vector spaces, but indeed an isomorphism of topological vector spaces, where all homology spaces $\ker d/\text{im }d$ are equipped with the respective quotient topologies. If this is done, then the right-hand-side is a Fréchet space, so the left-hand-side is, too, and the quotient topology of a Fréchet space by a subspace if Fréchet if and only if the subspace was closed. Proving this takes a bit more effort than what is done in the Schwartz/Grothendieck paper, but the paper "A Künneth formula in topological homology and its applications to the simplicial cohomology of $ \ell^1 (\mathbb{Z}_+^{k}) $." by Gourdeau, Lykova and White, mentioned by Yemon Choi in the comments, deals with this. They shows the precise statement in Corollary 5.3, under the assumption that $A$ and $B$ are bounded from below as complexes.<|endoftext|> TITLE: Set theoretic equation for Veronese varieties QUESTION [9 upvotes]: Consider the embedding $f:\mathbb{P}^n\rightarrow\mathbb{P}^N$ induced by the complete linear system of degree $d$ hypersurfaces of $\mathbb{P}^n$. Its image $V_{n,\,d}$ is degree $d$ Veronese variety of dimension $n$. The polynomials generating the ideal of $V_{n,\, d}$ are well-known: they are minors of a suitable matrix. Question. Is it possible to cut-out $V_{n, \, d}$ set-theoretically with fewer equations? For instance, we need three quadrics to generate the ideal of the twisted cubic $V_{1,3}\subset\mathbb{P}^3$, but $V_{1,3}$ is the set-theoretic intersection of a quadric and a cubic. I am particularly interested in the case $V_{2,\,3}\subset\mathbb{P}^9$. REPLY [4 votes]: This is an attempt to address the special case of $V_{2, 3} \subseteq \mathbb{P}^9$. Denote the homogeneous coordinates on $\mathbb{P}^9$ by $z_{ij}$, $i + j \leq 3$, so that the embedding $\mathbb{A}^2 \hookrightarrow V_{2, 3} \subseteq \mathbb{P}^9$ is given by $z_{ij} = x^iy^j$. Then I believe the following $7$ equations cut $V_{2, 3}$ as a set theoretic complete intersection on $\mathbb{P}^9$: $$z_{20}z_{00} = z_{10}^2 \qquad z_{30}^2z_{00} = z_{20}^3 \qquad z_{02}z_{00} = z_{01}^2 \qquad z_{03}^2z_{00} = z_{02}^3$$ $$z_{12}z_{30} = z_{21}^2 \qquad z_{03}^2z_{30} = z_{12}^3 \qquad z_{11}^3 = z_{30}z_{03}z_{00} $$ (The first six equations are precisely the cubics and quadrics giving as set theoretic intersections the three twisted cubics coming from the "coordinate lines" on $\mathbb{P}^2$, and the last equation takes care of the remaining coordinate. An inductive procedure like this might work in general.)<|endoftext|> TITLE: The "square root" of a graph? QUESTION [51 upvotes]: The number $f(n)$ of graphs on the vertex set $\{1,\dots,n\}$, allowing loops but not multiple edges, is $2^{{n+1\choose 2}}$, with exponential generating function $F(x)=\sum_{n\geq 0} 2^{{n+1\choose 2}}\frac{x^n}{n!}$. Consider $$ \sqrt{F(x)} = 1+x+3\frac{x^2}{2!}+23\frac{x^3}{3!} +393\frac{x^4}{4!}+13729\frac{x^5}{5!}+\cdots. $$ It's not hard to see that the coefficients 1,1,3,23,393,13729,$\dots$ are positive integers. This is A178315 in OEIS. Do they have a combinatorial interpretation? More generally, we can replace $2^{{n+1\choose 2}}$ with $\sum_G t_1^{c_1(G)} t_2^{c_2(G)}\cdots$, where $G$ ranges over the same graphs on $\{1,\dots,n\}$, and where $c_i(G)$ is the number of connected components of $G$ with $i$ vertices. Now we will get polynomials in the $t_i$'s with positive integer coefficients, the first four being $$ t_1 $$ $$ t_1^2+2t_2 $$ $$ t_1^3+6t_1t_2+16t_3 $$ $$ t_1^4+12t_1^2t_2+64t_1t_3+12t_2^2+304t_4. $$ Again we can ask for a combinatorial interpretation of the coefficients. Note. What happens if we don't allow loops, so we are looking at $\sqrt{\sum_{n\geq 0}2^{{n\choose 2}}\frac{x^n}{n!}}$? Now the coefficient of $\frac{x^n}{n!}$ is equal to the coefficient of $\frac{x^n}{n!}$ in $\sqrt{F(x)}$, divided by $2^n$, which in general is not an integer. Hence it makes more sense combinatorially to allow loops. REPLY [40 votes]: These numbers count balanced signed graphs (without loops). A signed graph is a graph in which every edge has a sign, either positive or negative. It is balanced if every cycle has an even number of negative edges. We will use the following lemma of Harary [Frank Harary, On the notion of balance of a signed graph, Michigan Math. J. 2 (1953/54), 143–146 (1955), Theorem 3]. I'll omit the proof, which is not difficult. A signed graph is balanced if and only if it is possible to color the vertices in black and white so that every positive edge joins two vertices of the same color and every negative edge joins two vertices of opposite colors. I'll call these colored graphs bicolored balanced signed graphs. It is easy to see that every connected bicolored balanced signed graph corresponds to exactly two balanced signed graphs The number of bicolored balanced signed graphs on $[n]:=\{1,2,\dots,n\}$ is $2^{\binom{n+1}2}=2^n\cdot 2^{\binom n2}$. To see this, we construct all bicolored balanced signed graphs on $[n]$ in the following way: We start with an arbitrary graph on $[n]$. Then we color the vertices arbitrarily in black and white. Finally we make edges between vertices of the same color positive edges and we make edges between vertices of opposite colors negative edges. There are $2^{\binom n2}$ graphs on $[n]$ and $2^n$ ways of coloring the vertices of each graph in black and white, so there are $2^n\cdot 2^{\binom n2}$ bicolored balanced signed graphs. The exponential generating function for bicolored balanced signed graphs is thus $F(x)$. So by the exponential formula, the exponential generating function for connected bicolored balanced signed graphs is $\log F(x)$. Since each connected balanced signed graph has two colorings, the exponential generating function for connected balanced signed graphs is $\tfrac12\log F(x)$, and by the exponential formula again, the exponential generating function for balanced sign graphs is $\sqrt{F(x)}$. The same approach shows that the polynomials in the $t_i$ count balanced signed graphs where each component with $i$ vertices is weighted $t_i$. It is curious that the enumeration of unlabeled balanced signed graphs was accomplished by Harary and Palmer in 1967 [F. Harary and E. M. Palmer, On the number of balanced signed graphs, Bulletin of Mathematical Biophysics 29 (1967), 759–765] but the considerably easier enumeration of labeled balanced signed graphs did not appear, as far as I know, until much later [F. Ardila, F. Castillo, Federico, and M. Henley, The arithmetic Tutte polynomials of the classical root systems, Int. Math. Res. Not. IMRN 2015, no. 12, 3830–3877]. There is another interpretation for the coefficients of $\sqrt{F(x)}$ which is less natural but easier to see: they count graphs on $[n]$ with loops allowed in which the least vertex in each component has a loop.<|endoftext|> TITLE: $a^{th}$-root of exponential generating functions QUESTION [6 upvotes]: This is a quick follow up on R. Stanley's interesting post on MO in a different direction, which might be easier. For positive integers $a$, define the family of functions (infinite series) given by $$F_a(x)=\sum_{n\geq0}a^{\binom{n+1}2}\frac{x^n}{n!}.$$ Then, $F_2(x)$ becomes Stanley's $F(x)$. QUESTION. Is it true that the coefficients of $\sqrt[a]{F_a(x)}$ are all positive integers, as an exponential generating function? REPLY [7 votes]: In fact, we can let $a$ be an indeterminate, and the coefficient of $x^n/n!$ in $\sqrt[a]{F_a(x)}$ will be a polynomial with integer coefficients. This is because $a^{{n+1\choose 2}}$ counts graphs with edges colored with $a$ colors (including the color 0, which denotes no edge). Thus the number $c_a(n)$ of connected such graphs on $n≥1$ vertices is a polynomial in $a$ with integer coefficients and is divisible by $a$. Hence the coefficient of $\frac{x^n}{n!}$ in $\exp \frac 1a\sum_{n\geq 1}c_a(n)\frac{x^n}{n!}$ is a polynomial in $a$ with integer coefficients. Addendum. Let me explain why $c_a(n)$ is divisible by $a$. Let $G$ be a connected graph on $n$ vertices with no loops. If $G$ has $q$ edges, then there are $(a-1)^q$ ways to color them. (The nonedges, corresponding to the color 0, are already determined by $G$.) At each vertex we are free to add a loop or not without affecting connectedness. If we add a loop, we can color it in $a-1$ ways, so $a$ possibilities in all for each vertex. Hence the total number of coloring of all graphs obtained from $G$ by adding loops is $(a-1)^q a^n$. This is divisible by $a$ (since $n\geq 1$), and summing over all $G$ without loops will therefore give a polynomial divisible by $a$, as claimed. Now note that if $n\geq 2$, then $q\geq 1$ so $(a-1)^q a^n$ is divisible by $a^2(a-1)$. Hence $$ \exp \frac 1a\sum_{n\geq 1}c_a(n)\frac{x^n}{n!} = \exp\left( x+\sum_{n\geq 2}a(a-1)d_a(n)\frac{x^n}{n!}\right), $$ for some $d_a(n)\in\mathbb{Z}[a]$. Taking this modulo $a(a-1)$ gives $e^x=\sum_{m\geq 0}\frac{x^m}{m!}$. It follows that the conjecture in OEIS A178319 mentioned in the comment by Max Alekseyev is true.<|endoftext|> TITLE: Can a nonstandard model of $\mathsf{PA}$ be "$\Delta^1_1$-well-ordered?" QUESTION [5 upvotes]: This was asked and bountied at MSE with no response: My question is the following: Is there a nonstandard model $\mathcal{M}\models\mathsf{PA}$ such that $\mathcal{M}$ has no $\Delta^1_1$-with-parameters-definable nonempty proper successor-closed initial segments? Here "$\Delta^1_1$" is meant in the sense of the standard semantics of second-order logic - so a $\Delta^1_1$ subset of $\mathcal{M}$ doesn't need to be "internal" to $\mathcal{M}$ in any nice sense. If we replace $\Delta^1_1$ with $\Pi^1_1$ the answer is trivially negative since the cut of standard naturals is $\Pi^1_1$; no parameters are needed here. If we replace $\Delta^1_1$ with $\Sigma^1_1$ this answer again becomes negative since for each nonstandard $a\in\mathcal{M}$ the set of elements infinitely below $a$ is $\Sigma^1_1(a)$ over $\mathcal{M}$. (See here.) However, I don't see a way to get a $\Delta^1_1$ cut in a nonstandard model of $\mathsf{PA}$. On the other hand, I don't see how to build a nonstandard model without a $\Delta^1_1$ cut. In particular, a natural hope might be to look at a nontrivial ultrapower of $\mathbb{N}$, but while $\Sigma^1_1$ formulas are preserved by taking ultrapowers, $\Delta^1_1$-ness (= an equivalence between a $\Sigma^1_1$ formula and a $\Pi^1_1$ formula) doesn't obviously need to be. REPLY [5 votes]: I believe you don't need this, but assume that there is a strongly inaccessible cardinal $\kappa$. Fix a first-order completion $T$ of $\mathsf{PA}$ and let $\mathcal{M}$ be a saturated model of $T$ of cardinality $\kappa$. I will show that $\mathcal{M}$ has no $\Delta^1_1$-with-parameters-definable cuts. Claim. For any $\Sigma^1_1(a)$ formula $\varphi(x,a)$, there is a closed set $F_\varphi \subseteq S_{x}(a)$ of types such that $\mathcal{M} \models \varphi(b,a)$ if and only if $\mathrm{tp}(b/a) \in F_\varphi$ (where $\mathrm{tp}(x/y)$ is the first-order type of $x$ over $y$). Proof. Given $\varphi(x,a)$, by compactness, there is a set of formulas $\Lambda(x,a)$ such that for any $b \in \mathcal{M}$, there exists an elementary extension $\mathcal{N} \succeq \mathcal{M}$ for which $\mathcal{N} \models \varphi(b,a)$ if and only if $\mathcal{M} \models \Lambda(b,a)$. Since $\mathcal{M}$ is saturated, it is resplendent, and if there is such an elementary extension for a given $b$, then we actually have that $\mathcal{M} \models \varphi(b,a)$ (since some expansion of $\mathcal{M}$ by a predicate satisfies the part of $\varphi(b,a)$ after the set quantifier). Clearly the other direction holds, so we have that $F_\varphi$ is the set of types corresponding to the partial type $\Lambda(x,a)$. $\square_{\text{claim}}$ Assume that there is a $\Delta^1_1$-with-parameters-definable cut, so in other words, assume that we have two $\Sigma^1_1$ formulas $\varphi(x,a)$ and $\psi(x,a)$ such that $\varphi(\mathcal{M},a)$ is the cut and $\psi(\mathcal{M},a)$ is the complement of the cut. Let $F_\varphi$ and $F_\psi$ be as in the claim. Since $\mathcal{M}$ is saturated, it is $\omega$-saturated. This implies that $F_\varphi$ and $F_\psi$ are disjoint (otherwise a type in their intersection would be realized) and cover $S_x(a)$ (otherwise a type in the complement of their union would be realized). Therefore, they are actually clopen, and correspond to some first-order formula $\chi(x,a)$ and its negation, but then $\chi(x,a)$ is a definable cut, which cannot happen, as $T$ is an extension of $\mathsf{PA}$. $\square$ As for getting rid of the inaccessible, I believe a special model will be sufficient, since they are resplendent for finite expansions. I actually think a computably saturated model might be sufficient too, since the first-order theory we're trying to expand to is c.e. (finitely axiomatizable, even).<|endoftext|> TITLE: Latest "A Term of Commutative Algebra" by Altman and Kleiman? QUESTION [9 upvotes]: Where can I find the latest revision of A term of Commutative Algebra by Allen B. ALTMAN and Steven L. KLEIMAN? Is my 2013 version ok? It is hard to locate the latest one; many old revisions and pointers to them are randomly scattered across the web. (Details: The first page of a web search showed me all 4 versions below. Furthermore, none of the seemingly official sites shows the revision date, only saying "2013 issue". You can't check the version until you actually donwnload one and open it.) This free textbook is intended to be an update of, and an improvement to "A & M", i.e. Introduction to Commutative Algebra by Atiyah and MacDonald. REPLY [15 votes]: You can get the latest from these sites: ResearchGate DSpace @ MIT (As of Jul 2021, this version is not the latest, but of the year 2018.) Worldwide Center of Mathematics Notice they are surely the latest, although all these three sites only show the initial release year, 2013. Download is open to everyone, including ResearchGate. Make sure you have at least the 2017 version or later, which was a great expansion. Compare these: Ver 2021-04-11: 441 pages, 612 exercises. Ver 2018-03-11: 426 pages, 594 exercises. Ver 2017-08-06: 423 pages, 585 exercises. Ver 2013-09-01: 258 pages, 324 exercises. Ver 2012-09-03: 208 pages, $\gg 200$ exercises. (Let us define the page number by the last page number printed in Arabic numerals. ResearchGate adds an extra front cover, so the pdf page number is not necessarily well-defined.) Acknowledgement: Emeritus Professor Steven Kleiman kindly answered my question concerning this point.<|endoftext|> TITLE: Hom between Brody hyperbolic varieties QUESTION [5 upvotes]: If $X$ and $Y$ are smooth projective Brody hyperbolic varieties is $\mathrm{Hom}(X, Y)$ also Brody hyperbolic? REPLY [7 votes]: I assume that for $\operatorname{Hom}(X,Y)$ you mean $\operatorname{Hol}(X,Y)$, that is the family of all holomorphic maps from $X$ to $Y$, endowed with its universal complex structure (which exists since your $X$ is compact). As you probably know, for a compact complex space being Kobayashi hyperbolic is equivalent to being Brody hyperbolic. Now, Theorem (6.4.1) in S. Kobayashi "Hyperbolic complex spaces" gives you the answer. Among other things it is stated there what follows. Even if you assume $X,Y$ to be merely compact complex spaces (no need of any projectivity assumption), and only $Y$ to be hyperbolic, then $\operatorname{Hol}(X,Y)$ is compact and any of its connected component is compact hyperbolic (and hence Brody hyperbolic). Aside comment (related to Piotr Achinger): if you look at the space of surjective holomorphic maps (or, more generally, dominant meromorphic maps), then under the same assumption as above, it is finite (Theorem (6.6.2) op. cit.).<|endoftext|> TITLE: Can every set of points with rational distance squares be isometrically embedded in $\Bbb Q^d$? QUESTION [6 upvotes]: Suppose we are given a finite family of points $p_1,...,p_n\in \Bbb R^d$, so that any two points have a rational distance square, that is, $$\|p_i-p_j\|^2\in\Bbb Q,\quad\text{for all $i,j\in\{1,...,n\}$}.$$ Is it know whether these points can be isometrically embedded into $\Bbb Q^{D}$ for some sufficiently large $D\ge d$? That is, are there $q_1,...,q_n\in\smash{\Bbb Q^D}$ so that $$\|p_i-p_j\|=\|q_i-q_j\|,\quad\text{for all $i,j\in\{1,...,n\}$}?$$ If yes, what are reasonable lower/upper bounds for $D$? REPLY [6 votes]: The following argument gives a bound of $D =4d$, based on a suggestion of LSpice in the comments. Set $p_1=0$. Using the distances and the fact that $p_1=0$, we can find the dot products $p_i \cdot p_j$, which are all rational. Assume we've embedded $p_1,\dots,p_k$ in $\mathbb Q^m$ and let's embed $p_{k+1}$. The result will then follow by induction on $k$. Among $p_2,\dots, p_k$ we can find $r$ of them that form a basis for the space $V_k$ generated by $p_2,\dots, p_k$. Then every vector in $V_k$ whose dot products with these $r$ vectors are rational is a rational linear combination of these vectors (invert the matrix) and thus lies in $\mathbb Q^m$. Hence the projection of $p_{k+1}$ onto $V_k$ lies in $\mathbb Q^m$. If $p_{k+1}$ lies in $V_k$, we are done. Otherwise, $p_{k+1}$ is equal to its projection onto $V_k$ plus a vector orthogonal to $V_k$. The length-squared of this orthogonal vector is rational. Choose a vector of that length-squared in $\mathbb Q^4$, and let $p_{k+1}$ equal its projection plus this vector in $\mathbb Q^{m+4}$. The total value of $D$ produced this way is $4d$ since $D$ goes up by $4$ every time $d$ goes up by $1$. For a better value of $D=d+3$, we can use the following trick to reduce the dimension: Let $W$ be the orthogonal complement in $\mathbb Q^{4d}$ of the vector space generated by $p_1,\dots, p_n$. If we find a vector in $W$ with length $1$, its orthogonal complement is isomorphic, as a vector space over $\mathbb Q$ with quadratic form, to $\mathbb Q^{4d-1}$. Iterating this, we can embed $p_1,\dots, p_n$ into $\mathbb Q^N$ for some $N$ in such a way that $p_1=0$ and the orthogonal complement to the vector space generated by $p_1,\dots, p_n$ contains no vectors of length $1$. By the Hasse-Minkowski theorem, the only way a positive definite quadratic form over the rationals can fail to represent $1$ is if there is a $p$-adic obstruction for some $p$, which can only happen if the dimension is at most $3$, so the dimension of the orthogonal complement is at most $3$ and $N \leq d+3$, as desired.<|endoftext|> TITLE: Characterization of functors whose right adjoint is monadic? QUESTION [10 upvotes]: Let $F: \mathcal A^\to_\leftarrow \mathcal B: U$ be an adjunction, and suppose we want to know whether the comparision functor $\mathcal B \to Alg^{UF}$ is an equivalence, where $Alg^{UF}$ is the category of algebras for the monad $UF$. The Beck monadicity theorem gives a necessary and sufficient condition for this to be the case, which can be verified by looking just at $U$. That is, all we need to know about $F$ in order to apply Beck's theorem is that $F$ exists; the condition "$U$ creates coequalizers of $U$-split pairs" refers only to $U$ and not to $F$. I wonder if there is dually some necessary and sufficient criterion for monadicity which can be checked by looking at $F$ only (so that all we need to know about $U$ is that it exists)? Question: Given a functor $F$ which is known to have a right adjoint $U$, is there some way to check whether $U$ is monadic by looking just at $F$ (and $\mathcal A, \mathcal B$), so that all we need to know about $U$ is that it exists? For my purposes, I'm not at all averse to making strong assumptions about $\mathcal A, \mathcal B$, like (co)completeness assumptions, exactess assumptions, etc. Just so long as I don't have to explicitly consider $U$. To put a finer point on it, if the hypotheses of the adjoint functor theorem hold, then Beck's theorem can be used to show that $U$ is monadic without referring to $F$ at all -- $F$ can be verified to exist by verifying that $U$ preserves limits and satisfies the solution set condition, and then the other condition for the monadicity theorem likewise refers only to $U$. So dually, I'm looking for a criterion which would give monadicity of the right adjoint of $F$ which, in the presence of the adjoint functor theorem, might never require me to explicitly refer to that right adjoint at all. REPLY [11 votes]: Let $F: C \to D$ be a left adjoint functor. I hope I'm not saying anything stupid, but I think you can just rephrase the two conditions of Beck Monadicity theorem in terms of the left adjoint: The condition that $U$ is conservative translate as: The $Hom(F(x),\_)$ are jointly conservatives. It can also be replaced by the apparently stronger condition but more familiar: The essential image of $F$ is dense, As this condition is known to holds for monadic functors and implies the previous one. The other condition is a bit harder. But I think we can manage if we assume that the domain of $F$ is Cauchy complete using the following lemma. Lemma: Let $C$ be a Cauchy complete category. Then a pair $X \rightrightarrows Y$ admit a split coequalizer if and and only if $Hom(A,X) \rightrightarrows Hom(A,Y)$ admits a split coequalizer for each object $A \in C$ functorially in $A$. I suspect one can remove the "functorially in $A$", but that was anoying to check, if someone has the motivation to do it, let me know ! In any case, the functoriality makes the proof very simple: this provides a split hence absolute coequalizer in the category of presheaves on $C$, but as $C$ is Cauchy complete any absolute colimits of representables is representable, hence the split coequalizer is already in $C$. So assuming $C$ is Cauchy complete the second condition can be rephrased as: Every pair $X \rightrightarrows Y$ in $D$ such that $Hom(F(A),X) \rightrightarrows Hom(F(A),Y)$ admit (functorially in $A$) a split coequalizer for each $A \in C$, admits a coequalizer in $D$ that is preserved by $Hom(F(A),\_)$ for each $A \in C$. Of course if you can remove the functoriality in the lemma, then you can remove it here as well.<|endoftext|> TITLE: Construction of the universal covering space via compact-open topology QUESTION [10 upvotes]: This is a re-post of a question I asked a month ago on MSE, but unfortunately didn't receive any answers. I'm hoping someone could help me with it. Here it goes: Recently I've been self-studying the theory of covering spaces from "Introduction to Topological Manifolds", by John M. Lee. At the end of Chapter 11, there is an explicit construction of the universal covering space for a connected, locally path connected and semilocally simply connected topological space $X$. Very briefly, the idea is as follows: take $x_{0}\in X$, define $P(X,x_{0})$ to be the set of all paths in $X$ which start at $x_{0}$, and let $\tilde{X}$ be the quotient set of $P(X,x_{0})$ by the relation: $\alpha \sim \beta$ if and only if $\alpha(1)=\beta(1)$ and $\alpha\simeq \beta$ are path-homotopic. We define $q\colon \tilde{X}\to X$ to be the map $q([\alpha]):=\alpha(1)$. We later define a topology on $\tilde{X}$ by means of a basis, which turns $\tilde{X}$ into a simply connected, locally path connected space and $q$ into a covering map. However, some authors suggest that an alternative way to construct $\tilde{X}$ is to give $P(X,x_{0})$ the compact-open topology (that is, the subspace topology it inherits as a subset of $\mathcal{C}([0,1],X)$), and $\tilde{X}$ the quotient topology induced by the canonical map $\pi \colon P(X,x_{0})\to \tilde{X}$. Nevertheless, I haven't been able to find a clear proof that $q\colon \tilde{X}\to X$ is the universal cover of $X$. What I'm trying to do is imitate the proof in Lee's book, but using this alternative topology: most of the details of the proof are rather set-theoretic, so I've managed to reduce the problem of rewriting the proof to just having to prove these 4 facts: $\tilde{X}$ is a path connected topological space. $q\colon \tilde{X}\to X$ is continuous. $q\colon \tilde{X}\to X$ is open. For every $\alpha \in P(X,x_{0})$ and every open set $U\subseteq X$ such that $\alpha(1)\in U$, the subset $$ [\alpha \cdot U]=\{ [\alpha \cdot \beta]\colon \beta \text{ is a path in } U \text{ such that } \beta(0)=\alpha(1) \} $$ is open in $\tilde{X}$. Here are my ideas so far: $P(X,x_{0})$ is path connected (and hence $\tilde{X}$ is too): let $\tilde{x}_{0}=[c_{x_{0}}]$ be the class of the constant path at $x_{0}$ and $\alpha\in P(X,x_{0})$ be arbitrary. Define $F\colon [0,1]\times [0,1]\to X$ by $F(t,s)=\alpha(ts)$. This is a continuous map, so by properties of the CO-topology, the map $f\colon [0,1]\to \mathcal{C}([0,1],X)$ given by $f(t)(s)=F(t,s)$ is continuous. Since $f(0)(s)=x_{0}$, $f(1)(s)=\alpha(s)$ and $f(t)(0)=x_{0}$, $f$ is a path in $P(X,x_{0})$ from $c_{x_{0}}$ to $\alpha$. Therefore, $P(X,x_{0})$ is path-connected. Let $\operatorname{ev}\colon P(X,x_{0})\to X$ be the map defined by $\operatorname{ev}(\alpha)=\alpha(1)$. This map is continuous (again, this is a general property of the CO-topology) and passes continuously to the quotient, inducing precisely the map $q$, so $q$ is continuous. As of openness of $q$, I'm not sure how to proceed: what I'm sure of is that, if $\operatorname{ev}$ is an open map (which I think it is), then $q$ is too. I believe that it should be possible to prove that basic open subsets of $P(X,x_{0})$ are mapped into open sets of $X$. Let $[\alpha \cdot U]$ defined as above, we need to see that $\pi^{-1}([\alpha \cdot U])$ is open in $P(X,x_{0})$. Take any $\gamma \in \pi^{-1}([\alpha \cdot U])$, so that $\gamma \simeq \alpha \cdot \beta$, where $\beta$ is a path in $U$ starting at $\alpha(1)$. Let $V$ be the path component of $U$ which contains $\gamma(1)$ (and therefore, it contains $\alpha(1)$). Then $\gamma\in [\{1\},V]$, where $[\{1\},V]$ is the basic open subset of $P(X,x_{0})$ given by all paths which end at some point of $V$. From here I don't know how to continue: I think this open set is contained in $\pi^{-1}([\alpha \cdot U])$, but I haven't got an idea to prove it (certainly, since we know "a posteriori" that $\tilde{X}$ is simply connected, every element of $[\{1\},V]$ is homotopic to $\alpha\cdot \eta$, where $\eta$ is a path in $V$ from $\alpha(1)$ to the endpoint of such element). So far, are my ideas on the right track? Can someone help me especially with the last two statements? Thank you in advance! SOME INSIGHTS I MADE AFTER ASKING THE QUESTION (these appear as edits on the MSE post) First, what I thought about the third statement: I think this could be a valid proof for openness of $\operatorname{ev}$: let $A=[K_{1},U_{1}]\cap \dots \cap[K_{n},U_{n}]$ be a basic open set of $\tilde{X}$, we need to prove that $\operatorname{ev}(A)$ is an open subset of $X$. Take $x\in \operatorname{ev}(A)$, so that there exists some $\alpha \in A$ such that $x=\alpha(1)$. Let $B\subseteq X$ be the intersection of all $U_{i}$'s such that $1\in K_{i}$ (if none of the $K_{i}$ contains $1$, then $B:=X$), so $x\in B$, and let $V$ be the path component of $B$ that contains $x$ (so $V$ is open in $X$). I claim that $V\subseteq \operatorname{ev}(A)$. To see this, let $y\in V$. For every $j$ such that $1\notin K_{j}$, let $a_{j}=\max \{t\colon t\in K_{j}\}<1$, and let $a=\max a_{j}$. By continuity, one can choose some $a TITLE: How do $\infty$-categories allow us to do descent on the derived level? QUESTION [9 upvotes]: I have heard that one application of $\infty$-categories is that they allow us to formulate a meaningful theory of descent for derived categories (say of sheaves on a scheme). While I'm sure the details are somewhere in Lurie's exposition of stable $\infty$-categories, I was hoping that someone familiar with the process could explain in broad strokes why we can't do this in the classical setting, and what $\infty$-categories add to the picture that changes the situation. REPLY [22 votes]: Let $X$ be a topological space covered by open sets $U$ and $V$. Let $\mathscr{F}$ and $\mathscr{G}$ be complexes of sheaves defined on $U$ and $V$, respectively. Suppose you are given an isomorphism $\alpha: \mathscr{F}|_{ U \cap V} \rightarrow \mathscr{G}|_{ U \cap V}$ in the derived category of the intersection $U \cap V$. You would like to use these to glue $\mathscr{F}$ and $\mathscr{G}$ together to obtain a complex of sheaves on $X$. Let $j: U \hookrightarrow X$, $j': V \hookrightarrow X$, and $j'': U \cap V \hookrightarrow X$ denote the inclusion maps. Then the "glued" complex should be the fiber of the map $$j_{\ast} \mathscr{F} \oplus j'_{\ast} \mathscr{G} \rightarrow j''_{\ast} \mathscr{G}|_{U \cap V},$$ which is given on the first factor by $\alpha$. Working at the level of triangulated categories, this characterizes the glued complex up to non-canonical isomorphism. But for many purposes, producing a complex which is only well-defined up to non-canonical isomorphism is probably not good enough: you would like to define something that depends functorially on the input. The formalism of triangulated categories is poorly suited to this, because taking the fiber (or cocone) of a morphism is not a functorial operation. This is the sort of thing that is "corrected" by working with $\infty$-categories.<|endoftext|> TITLE: a Vandermonde-type of determinants summed over permutations QUESTION [15 upvotes]: Let $S_n$ be the symmetric group. Consider $$D:=\sum_{\sigma\in S_n} \text{sgn}(\sigma)\cdot \det\begin{pmatrix}1 & a_{\sigma(1)}-0 & (a_{\sigma(1)}-0)^2 & \cdots & (a_{\sigma(1)}-0)^{n-1} \\1 & a_{\sigma(2)}-a_{\sigma(1)} & (a_{\sigma(2)}-a_{\sigma(1)})^2 & \cdots & (a_{\sigma(2)}-a_{\sigma(1)})^{n-1} \\ \vdots & & \vdots & & \vdots \\1 & a_{\sigma(n)}-a_{\sigma(n-1)} & (a_{\sigma(n)}-a_{\sigma(n-1)})^2 & \cdots & (a_{\sigma(n)}-a_{\sigma(n-1)})^{n-1} \end{pmatrix} .$$ If $a_i=a_j$ then it is easy to see $D=0$. Thus, the Vandermonde determinant $V:=\prod_{1\le i TITLE: Is there a known classification of regular multiplicity-free permutation groups? QUESTION [6 upvotes]: The question is in the title, but let me clarify the terminology. I consider a permutation group $\Sigma\subseteq\mathrm{Sym}(\Omega)$ on a finite set $\Omega$. $\Sigma$ is regular if it acts transitively and freely on $\Omega$, i.e., for any two $i,j\in \Omega$ there is a unique $\sigma\in\Sigma$ with $\sigma(i)=j$. $\Sigma$ is multiplicity-free if its permutation character (the character of the linear representation of $\Sigma$ by permutation matrices) is the sum of distinct irreducible characters. Examples are the permutation groups generated by a single cyclic permutation. REPLY [9 votes]: These are the abelian regular permutation groups. The permutation character in this case is the character of the regular representation and in the regular representation a character appears with multiplicity equal to the dimension of the irreducible representation. So it can be multiplicity free iff all irreducibles are one dimensional which is equivalent to abelian.<|endoftext|> TITLE: How special is first-order $\mathsf{PA}$? QUESTION [14 upvotes]: This is a modified version of a question which was asked and bountied at MSE without success. Below, "$\mathsf{PA}$" refers to first-order Peano arithmetic. There are various "schematic" theories out there, like $\mathsf{PA}$ and $\mathsf{ZFC}$, which basically consist of three components: a "base" set of axioms which are more-or-less taken for granted (e.g. the discrete nonnegative ordered semiring axioms for $\mathsf{PA}$, and Pairing-Union-Extensionality-Foundation-Powerset-Infinity-Choice for $\mathsf{ZFC}$), an informal idea(s) for a further set of rules indexed by formulas (e.g. induction for $\mathsf{PA}$, and separation/replacement for $\mathsf{ZFC}$), and a choice of logic for implementing the latter (first-order logic for $\mathsf{PA}$ and $\mathsf{ZFC}$, and indeed in general). For any such theory we can hope to produce interesting variants by fixing the first two parts and varying the third - see e.g. here for a question about the case of $\mathsf{ZFC}$. However, $\mathsf{PA}$ seems remarkably stubborn here: every natural logic (= regular logic possibly without negation) I can think of lands in one of two extremes. To be precise, given a logic $\mathcal{L}$ let $\mathfrak{PA}(\mathcal{L})$ be the class of models of $\mathsf{PA}$ with no $\mathcal{L}$-definable nontrivial proper cuts, and say that $\mathcal{L}$ is: strong for induction if $\mathfrak{PA}(\mathcal{L})$ consists of just $\mathbb{N}$ up to isomorphism; weak for induction if every complete first-order extension of $\mathsf{PA}$ is satisfied by some element of $\mathfrak{PA}(\mathcal{L})$; and intermediate for induction if neither of the above cases holds. My question is: Is there any natural logic which is intermediate for induction? (Here, by "natural" I mean "has appeared in at least two different papers whose respective authorsets are $\subseteq$-incomparable.") In the MSE version I asked for an even stronger property, namely not pinning down $\mathbb{N}$ even up to elementary equivalence, but that seems overly optimistic in retrospect. Here are some quick negative observations: Uniformly across models of $\mathsf{PA}$, standardness is definable by a $\Pi^1_1$ formula or by a very simple infinitary formula. So neither $\Pi^1_1$ nor any reasonable infinitary logic will be intermediate for induction; in fact, they'll pin down $\mathbb{N}$ up to isomorphism, not just up to first-order-elementary-equivalence. $\Sigma^1_1$ also pins down $\mathbb{N}$ up to isomorphism, although parameters now appear necessary; see here. For the same reason, first-order logic + the equicardinality quantifier pins down $\mathbb{N}$ up to isomorphism. On the opposite end of things, $\Delta^1_1$ is weak for induction. (OK, $\Delta^1_1$ may not look like a logic since the $\Delta^1_1$-ness of a $\Sigma^1_1$ formula is structure-dependent, but we can handwave past this: define an abstract logic with a formula $\hat{\varphi}$ for each $\Sigma^1_1$ formula $\varphi$, where $\hat{\varphi}$ is interpreted in $\mathcal{M}$ as $\varphi$ if $\neg\varphi^\mathcal{M}$ is $\Sigma^1_1$ over $\mathcal{M}$ and as $\top$ otherwise.) Similarly, $\mathsf{FOL}$ + the quantifier "At least as many $x$ satisfy $\varphi$ as satisfy $\neg\varphi$" is weak for induction, although its model class contains no countable nonstandard models; to see weakness, consider the $\omega_1$-like models of $\mathsf{PA}$. The most promising approach to me at the moment is to look for fragments of second-order logic between $\Delta^1_1$ and $\Sigma^1_1$, since we see a genuinely interesting and nontrivial change in behavior there. However, I can't at the moment think of a good candidate fragment. (I've added the set theory tag since, while set theory isn't built into the question a priori, it seems relevant to all the ideas I've had so far.) REPLY [13 votes]: There are several notable papers, starting with a key paper of Angus Macintyre (Ramsey quantifiers in arithmetic, Model theory of algebra and arithmetic, Lecture Notes in Math., 834, Springer, 1980), which explore natural extensions of Peano Arithmetic formulated in "well-behaved" logics extending first order logic. The most well-known of these extensions are obtained by adding the so-called Ramsey quantifiers $Q^{n}$ (where $n\in \omega,~n>0$) to first order logic; these quantifiers are also known as Magidor-Malitz quantifiers. The extension of PA formulated using Ramsey quantifiers is usually referred to as $PA(Q^{<\omega})$. Given a model $\mathcal{M}$ of PA, the semantics of $Q^{n}$ is guided by: $\mathcal{M}\models Q^{n}x_1,...,x_n~ \varphi(x_1,...,x_n)$ iff there is an unbounded subset $H$ of $\mathcal{M}$ such that $\mathcal{M}\models \varphi(h_1,...,h_n)$ for every increasing sequence $h_1,...,h_n$ from $H$. Macintyre proved various interesting results in his paper, including a completeness theorem (with the help of the combinatorial principle $\diamond_{\omega_1}$). This allowed him to show that $PA(Q^{<\omega})$ has an $\omega_1$-like model (an uncountable model every proper initial segment of which is countable). As also noted by Macintyre, the Paris-Harrington extension of the finite form of Ramsey's theorem is provable in $PA(Q^{<\omega})$, thus the 1-consistency of PA is provable in $PA(Q^{<\omega})$ [indeed proving the Paris-Harrington principle in a natural extension of PA is what led Macintyre to introduce $PA(Q^{<\omega})$]. Schmerl and Simpson refined Macintyre's work by eliminating the need for $\diamond_{\omega_1}$ in the results of Macintyre in the aforementioned paper (On the role of Ramsey quantifiers in first order arithmetic, Journal of Symbolic Logic, Vol.47, 1982). Their refinement allowed them to calibrate the arithmetical strength of $PA(Q^{<\omega})$ by proving the following theorem: Theorem (Schmerl and Simpson). The following statements are equivalent for an arithmetical statement $\varphi$: (a) $PA(Q^{<\omega}) \vdash \varphi$. (b) $PA(Q^{2}) \vdash \varphi$. (c) $\Pi^1_1-\mathrm{CA_0} \vdash \varphi$. In a later paper by Schmerl, entitled PA(aa), in Notre Dame Journal of Formal Logic, Vol. 36, 1982, Schmerl studied the extension of PA with stationary quantifiers (originally introduced and studied in a landmark 1978 paper of Barwise, Kaufmann, and Mekler) and established that the purely arithmetical consequences of PA(aa) + Det [where Det is the so-called scheme of finite determinateness of stationary logic] coincides with the purely arithmetical consequences of second order arithmetic $Z_2$ (the first order theory whose subsystems are studied in reverse mathematics, as opposed to PA formulated in full second logic, which is a much stronger theory which only has one model up to isomorphism). Another important paper of Schmerl (Peano arithmetic and hyper-Ramsey logic, Trans. Amer. Math. Soc., 1986) connects PA formulated in so-called hyper-Ramsey logic with subsystems of second order arithmetic.<|endoftext|> TITLE: Group with non-trivial center containing trivially-intersecting copies of itself QUESTION [16 upvotes]: I'm trying to think of an example of a group $G$ with non-trivial center such that there exist subgroups $H_1,H_2\le G$ both isomorphic to $G$ and satisfying $H_1\cap H_2=\{1\}$. Does such a group exist? Preferably an easy-to-state example that I'm just not thinking of? (Ideally finitely generated, but I won't insist on it.) If the center is trivial then this is easy (free groups work, I think any centerless RAAG will work, various Thompson-like groups work), but requiring non-trivial center seems to complicate things. I would also be happy with an example that requires intersecting more than two copies of $G$ inside $G$ to ensure a trivial intersection (as long as it's finitely many copies). REPLY [5 votes]: An immediate variant of $V$ works. View Thompson's $V$ with its natural action on $[0,1[$. Let $G$ be the centralizer of the involution $s:x\mapsto x+c(x)$ with $c(x)=1/4$ if $x\in [0,1/4\mathclose[\cup [1/2,3/4\mathclose[$ and $c(x)=-1/4$ otherwise. Let $H_1$ (resp. $H_2$) be the intersection of $G$ and the pointwise stabilizer of $[1/2,1\mathclose[$ (resp. of $[0,1/2\mathclose[$). Clearly $H_1,H_2\simeq G$ and $H_1\cap H_2=\{\mathrm{id}\}$. [Edit: I initially defined things in a clumsy way, fixed after Matt Zaremsky's comment.]<|endoftext|> TITLE: Drinfeld Sokolov and the semiinfinite flag variety QUESTION [5 upvotes]: For a long time I've been confused about Drinfeld Sokolov/BRST reduction/semiinfinite cohomology for affine Lie algebras. Most treatments define it in what to me feels like a fairly ad-hoc way, by choosing a nilpotent element then applying an elaborate construction. (Of course it's not unmotivated: it generalises the BRST construction for finite dimensional Lie algebras, and in a precise sense it is a quantisation of Hamiltonian reduction of $LN$ acting on a certain subspace of $\widehat{\mathfrak{g}}^*$). However, until I see a geometric interpretation of what's going on I think I will continue to be confused. Question: is there a "geometric" interpretation of the Drinfeld-Sokolov functor, e.g. one living over the semiinfinite flag variety? If so, what is the relation to the above remarks in parentheses? REPLY [4 votes]: Maybe let me try to synthesize my comments into an answer. All of this is contained in Raskin's beautiful paper arxiv.org/abs/1611.04937 on Whittaker categories. Convention: We work here in the derived world, i.e., all our categories are assumed pretriangulated dg (equivalently one can take stable $\infty-$categories). Let $LG$ be the loop group of $G$, considered as a group ind-scheme. Convolution endows the category of D-modules $D(LG)$ with the structure of a monoidal category, and we can consider module categories $C$ for $D(LG)$. Two important examples: The category of D-modules $D(X)$ on an ind-scheme $X$ with a $LG$ action. The category $\hat{g}\operatorname{-mod}$ of representations of the affine Lie algebra. The $D(LG)$ action on the first example is relatively straightforward to construct, but the second example merits some explanation. The easiest way to see that there should be such an action is to interpret objects $\hat{g}\operatorname{-mod}$ as D-modules on $LG$ weakly equivariant for the left $LG$ action. Such objects are preserved by the right action of $LG$, and this induces the desired action. (Making this precise is very technical, and I will avoid saying more about it.) More generally, you can include a level here; this is very useful but doesn't alter anything I will say below so I suppress it. Now take the subgroup $LN$ of $LG$, and choose a nondegenerate character $\chi$ of $LN$, just as you would for Drinfeld-Sokolov. For any category $C$ as above, we define the Whittaker category $\operatorname{Whit}(C)$ to be the category of $(LN,\chi)$-equivariant objects in $C$. More precisely, it is $\operatorname{Hom}_{D(LN)}(\operatorname{Vect},C)$, where the action on $\operatorname{Vect}$ is twisted by $\chi.$ For $C\cong D(X)$, this recovers exactly D-modules on $X$ which are $(LN,\chi)$-equivariant. There is also a dual construction, which I denote by $\operatorname{Whit}_{co}(C),$ given by Whittaker coinvariants, i.e., $C\otimes_{D(LN)}\operatorname{Vect}.$ Raskin proves in the aforementioned paper that $$\operatorname{Whit}(C)\cong\operatorname{Whit}_{co}(C).$$ As $\operatorname{Whit}_{co}(C)$ is a category of coinvariants, it comes with a natural functor from $C$. So the equivalence between invariants and coinvariants gives us a natural functor $C\rightarrow\operatorname{Whit}(C)$ as well, which we call !-averaging and denote by $\operatorname{Av}_!$. (For $C=D(X)$ it can be defined explicitly as a $!$-pushforward of D-modules, but it is not a priori obvious that this $!$-pushforward is well defined.) Now what happens for $C\cong\hat{g}\operatorname{-mod}$? In this case, there is an equivalence $\operatorname{Whit}(\hat{g}\operatorname{-mod})\cong\mathcal{W}\operatorname{-mod},$ for $\mathcal{W}$ the W-algebra. Furthermore, the functor $\operatorname{Av}_!:\hat{g}\operatorname{-mod}\rightarrow\mathcal{W}\operatorname{-mod}$ is exactly the operation of Drinfeld-Sokolov reduction. So this provides a geometric interpretation of Drinfeld-Sokolov reduction. Let me give one quick (slightly silly) application of this framework. Denote the affine Grassmannian of $G$ by $\operatorname{Gr}$. Then we have a global sections functor $D(\operatorname{Gr})\rightarrow\hat{g}\operatorname{-mod}.$ Because this functor is $LG$-equivariant, it gives a functor $$\operatorname{Whit}(D(\operatorname{Gr}))\rightarrow\operatorname{Whit}(\hat{g}\operatorname{-mod})$$ such that the two compositions $$D(\operatorname{Gr})\rightarrow\operatorname{Whit}(D(\operatorname{Gr}))\rightarrow\operatorname{Whit}(\hat{g}\operatorname{-mod})$$ and $$D(\operatorname{Gr})\rightarrow\hat{g}\operatorname{-mod}\rightarrow\operatorname{Whit}(\hat{g}\operatorname{-mod})$$ are equivalent. So for any D-module $M$ on $\operatorname{Gr}$, the Drinfeld Sokolov-reduction $\operatorname{DS}(\Gamma(M))$ only depends on the image of $M$ inside $\operatorname{Whit}(\operatorname{Gr})$ (which is a small and pretty well understood category.)<|endoftext|> TITLE: Quiver representations of type $D_n$ mutation class QUESTION [13 upvotes]: I was wondering if there is a classification of the indecomposable quiver representations of (not necessarily acyclic) quivers that are mutation equivalent to the $D_n$ Dynkin diagram. Such quivers are classified by Vatne, see https://arxiv.org/abs/0810.4789. For a small example, what are the indecomposable representations of the "kite" quiver given by: REPLY [5 votes]: Up to possibly learning some new technology, you can get a good answer to this question using the cluster category of type $D_n$ (or indeed of any Dynkin type). Given a Dynkin quiver $Q$, its associated cluster category $\mathcal{C}_Q$ which has finitely many indecomposable objects up to isomorphism, and it is possible to draw its Auslander–Reiten quiver, the vertices of which represent isoclasses of indecomposables and the arrows represent irreducible maps, via a completely combinatorial procedure. The original reference for this category is Tilting theory and cluster combinatorics by Buan, Marsh, Reineke, Reiten and Todorov. (Roughly, you take the translation quiver $\mathbb{Z} Q$, which has one copy of $Q$ for each integer joined together by additional arrows $(i,n)\to (j,n+1)$ for each arrow $j\to i$ in $Q$, where $(i,n)$ is vertex $i$ in the $n$-th copy of $Q$, and $(j,n+1)$ is vertex $j$ in the $(n+1)$-st copy. Then you have to quotient out by a symmetry which depends on the Dynkin type of $Q$. This is explained in Section 1 of BMRRT and Figure 1 shows the result in the case that $Q$ is of type $A_3$.) This category contains special objects, called cluster-tilting, and if $T$ is such an object, the endomorphism algebra $A_T=\operatorname{End}_{\mathcal{C}_Q}(T)$ (or its opposite, depending on your conventions) is presented by a quiver $Q'$ mutation equivalent to $Q$ (with relations, as Hugh indicated in the comments), and all quivers mutation equivalent to $Q$ appear in this way. There is then an equivalence of categories $$\mathcal{C}_Q/(T)\simeq\operatorname{mod}{A_T}$$ where the left hand-side means $\mathcal{C}_Q$ modulo the ideal of morphisms which factor over a direct sum of copies of $T$, and the right-hand side is the category of finite-dimensional $A_T$-modules (i.e. representations of the quiver $Q'$ satisfying the appropriate relations). In this context I think this is due to Buan, Marsh and Reiten in Cluster-tilted algebras, but it has been generalised many times. (Sometimes, as in this reference, an automorphism of $\mathcal{C}_Q$ appears in front of $T$ on the left-hand side, giving a different but equivalent statement.) Incidentally, cluster-tilted algebras are by definition the algebras of the form $A_T$, for $T\in\mathcal{C}_Q$ cluster-tilting, and $Q$ Dynkin. The payoff then is that if you have drawn the Auslander–Reiten quiver of $\mathcal{C}_Q$, and you can find in it the vertices corresponding to indecomposable summands of a cluster-tilting object $T$, then you can just delete these vertices (and all incident arrows) to get the Auslander–Reiten quiver of $A_T$. So not only do you 'classify' the indecomposable $A_T$-modules (by vertices of this quiver), you can also see all of the morphisms between them. It is also not so hard to work out from the Auslander–Reiten quiver of $\mathcal{C}_Q$ what each vertex really is as a quiver representation.<|endoftext|> TITLE: Derivation on $SO(3)$ QUESTION [5 upvotes]: Let $$u:\mathbb{R}\ni t \mapsto u(t)\in\mathcal{S}, \quad v:\mathbb{R}\ni t \mapsto v(t)\in\mathcal{S}$$ where $\mathcal{S}$ is the unit sphere of $\mathbb{R}^3$. Consider \begin{align} R:\ \mathcal{S}^2 &\to SO(3) \\\ (u,v) & \mapsto (u\cdot v)I_3 + \operatorname{hat}(u\times v) + \dfrac{(u\times v) \otimes (u\times v)}{1+u\cdot v} \end{align} with $\operatorname{hat}$ the hat operator. $R\in SO(3)$ so $R^\top \dot R$ is an antisymmetric matrix and there exists a $\omega$ such that $R^\top \dot R = \hat{\omega}$ (note: $\dot R$ is $\frac{dR}{dt}$). After much effort and many steps I eventually found out with the help of symbolic computation program that $\omega$ simply reduces to: $$\omega = \dfrac{-1}{1+u\cdot v} \Big((\operatorname{hat}(u+v))\dot u + (-\operatorname{hat}(u+v)+2uw^\top)\dot v\Big)$$ with $w=u\times v$. Question Is there a simple way of finding such a simple formula? REPLY [4 votes]: Let me work out the case where $\dot{v} = 0$. The case $\dot{u} = 0$ is similar and you can add using linearity of the derivative mapping. The matrix $R(u,v)$ is the rotation of the vector $u$ to the vector $v$, in the plane spanned by $u$ and $v$. For our purposes it is convenient to define $$ e_1 = \frac{u - (u\cdot v) v}{\| u - (u\cdot v) v\|}, \quad e_2 = v, \quad e_3 = e_1 \times e_2 = \frac{u \times v}{\|u \times v\|} $$ For small $t$, we can write $u(t) = \exp(\phi(t)e_1\wedge e_3)\exp( \theta(t) e_1 \wedge e_2) u$, with $\phi(0) = \theta(0) = 0$. (Imagine $v$ as the north pole, we first move on latitude and then change longitude.) Notice that $$\dot{u}(0) = \dot{\phi}(0) (e_1 \wedge e_3) u + \dot{\theta}(0)(e_1\wedge e_2) u $$ Using the definitions of the frame we note that $e_3 \cdot u = 0$ and so $$ \dot{\phi}(0) = - \frac{\dot{u}(0) \cdot e_3}{e_1 \cdot u} $$ Notice that $R(u,v)$ is of the form $\exp(\tilde{\theta} e_1\wedge e_2)$ for some $\tilde{\theta}$. Thus $$ R(u(t),v) = \exp(\phi(t) e_1 \wedge e_3)\exp((\tilde{\theta} - \theta(t))e_1\wedge e_2)\exp(-\phi(t) e_1 \wedge e_3) $$ And so $$ R^T \dot{R}|_{t = 0} = \dot{\phi}(0) \cdot \left[ R^T (e_1 \wedge e_3) R - e_1 \wedge e_3\right] - \dot{\theta}(0) e_1 \wedge e_2$$ Since $e_3$ is fixed by $R$, we have that $$ R^T \dot{R} = \dot{\phi}(0) (R^T e_1)\wedge e_3 - \dot{\phi}(0)(e_1 \wedge e_3) - \dot{\theta}(0) e_1 \wedge e_2 $$ Now, $R^T v = u$ and $R^T u = - v + 2(v\cdot u) u$ is the reflection of $v$ to the other side of $u$, we find $R^T e_1 = \frac{(v\cdot u) u - v}{\| (v\cdot u) u - v\|}$. Now, we see that $u, R^T e_1, e_3$ form another orthonormal frame. Noting that $e_1 \wedge e_2 = R^T e_1 \wedge u$, and $$ e_1 = \sqrt{1 - (u\cdot v)^2} u - (u\cdot v) R^T e_1 $$ we find $$R^T \dot{R} = u \wedge \dot{u} - (1 - (u\cdot v))\frac{\dot{u}\cdot e_3}{e_1 \cdot u} (R^Te_1) \wedge e_3 $$ Using that $R^T e_1 \times e_3 = - u$, we conclude finally $$ \widehat{R^T \dot{R}} = u \times \dot{u} + \frac{\dot{u} \cdot(u\times v)}{1 + (u\cdot v)} u $$ To compare this result to what you found, notice that $$ v = (u\cdot v) u + \sqrt{1 - (u\cdot v)^2} R^T e_1 $$ So $v \times \dot{u} = (u\cdot v) u \times \dot{u} + \sqrt{1 - (u\cdot v)^2} R^T e_1 \times \dot{u}$. For the second term, since $\dot{u}\perp u$, we have that the cross product is actually proportional to $(\dot{u} \cdot e_3) u$. (Our formulas do not 100% agree, but I do not rule out making a small mistake here and there with numerical factors.)<|endoftext|> TITLE: Graphs with linear Ramsey number for two colors, but super-linear Ramsey number for three colors? QUESTION [5 upvotes]: Given a graph $H$, let $R_k(H)$ be the smallest integer $N$ such that in every $k$-coloring of the edges $K_N$ there is a monochromatic copy of $H$ (in other words, $R_k(H)$ is the ordinary $k$-color Ramsey number of $H$). Does there exist a sequence of graphs $(H_n)_{n\in \mathbb{N}}$ (where $H_n$ has $n$ vertices) such that $R_2(H_n)=O(n)$, but $R_3(H_n)=\omega(n)$? REPLY [6 votes]: There is an almost example, 'almost' in three senses: it's for $3$-uniform hypergraphs rather than graphs; for $4$ colours rather than $3$; and the $2$-colour case isn't quite known to be linear (though it is conjectured to be). To be more specific, the example is the $3$-uniform hypergraph known as the hedgehog. This is the $3$-uniform hypergraph $H_t$ with vertex set $[t + \binom{t}{2}]$ such that for every pair $(i, j)$ with $1 \leq i < j \leq t$ there is a unique vertex $k > t$ for which $ijk$ is an edge. It is known that $R_2(H_t) = O(t^2 \log t)$, which is almost linear in the number of vertices, while $R_4(H_t) \geq 2^{\Omega(t)}$. For more information, see https://arxiv.org/pdf/1511.00563.pdf and https://arxiv.org/pdf/1902.10221.pdf.<|endoftext|> TITLE: Does Grothendieck's algebraization imply existence of colimits of schemes? QUESTION [6 upvotes]: I am a little bit confused about two lemmas regarding Grothendieck's algebraization. Assume all algebras are defined over some field. Here is the short version of my question: Does Tag 09ZT ("Lemma 1") plus Tag 0A42 ("Lemma 2") imply that colimit of $Y_i$'s in Lemma 1 exists in the category of schemes under mild assumptions? (this would mean the construction of $Y$ in the proof of Lemma 1 does not depend on $X$ and is the colimit) If so what is the minimal requirement for that to happen. The longer version: Let $A$ be complete with respect to $I$. Let $S=\text{Spec}(A)$ and $S_n=\text{Spec}(A/I^n)$ Let $Y_1\rightarrow Y_2 \rightarrow Y_3 \rightarrow \ldots$ be an infinite sequence of morphism of schemes. Assume $Y_i$ has a structure of a scheme over $S_i$. Let $X$ be a scheme which receives morphisms from $Y_i$'s in a compatible manner. Because of the condition on $Y_i$'s we can deduce that $X\times S$ also receives same type of compatible morphisms from $Y_i$'s. This specially induces a diagram of commutative morphisms in the following form: Assuming $X_i:=X\times S_i$ let's denote the maps from $Y_i$ to $X_i$ by $f_i$. Furthermore all $f_i$ are finite morphism and $Y_1\rightarrow S_1$ is proper then by Tag 09ZT, there is some scheme $Y$ such all the maps from $Y_i$ to $X\times S$ factors through $Y$ and $Y_i=Y\times S_i$. Now I want to deduce that the scheme $Y$ does not depend on $X$ and is the colimit of the diagram of $Y_i$'s. Now assume the same situation, we are going to replace $X$ with $X'$ with exactly the same role. Assume $X'$ is receiving compatible morphisms from $Y_i$'s. Then we can form a compatible commutative diagram between $Y_i$ and $X'_i:=X'\times S_i$. Applying Tag 0A42 (for some reason $X$ and $Y$ is switched in this lemma), implies that the map from $Y_i$'s to $X'\times S$ factors uniquely through $Y$. Then we can project $X'\times S$ on the first factor to get the same conclusion for $X'$. This means $Y$ is the colimit of the diagram of $Y_i$'s. REPLY [10 votes]: Here's one way to see what's going on. I will use the Tannakian duality theorem of Hall and Rydh (see Theorem 1.1 here). It is stated for algebraic stacks, but if you replace the word "algebraic stack" with "scheme'' and forget about stabilizers, you're golden. We will use your notation above. Let $T$ be an $S$-scheme (we will worry about finiteness conditions on $T$ later). Then $$\begin{eqnarray*}\text{Hom}(Y,T) &=& \text{Hom}(\text{Coh}(T),\text{Coh}(Y)) \\ &=& \text{Hom}(\text{Coh}(T),\varprojlim\text{Coh}(Y_n))\\ &=&\varprojlim \text{Hom}(\text{Coh}(T), \text{Coh}(Y_n)) \\ &=&\varprojlim \text{Hom}(Y_n, T).\end{eqnarray*}$$ The first line is Hall-Rydh. Since $Y$ is proper over $S$ (the spectrum of a complete Noetherian local ring), the second line is Grothendieck's existence theorem. The third line is from the definition of the inverse limit, and the fourth from Hall-Rydh again. This shows that we may define the direct limit $\varinjlim Y_n$ (in whatever category for the moment) to be $Y$. Let us now worry about the relevant category of schemes/finiteness assumptions on $T$ that we must impose. First we check that the assumptions of Hall-Rydh are satisfied. The theorem assumes: $Y$ is locally excellent. $T$ is Noetherian. Assumption (1) is automatically satisfied since $Y$ is finite type over a complete Noetherian local ring and thus is excellent. (2) is satisfied if we work in the category of finite type $S$-schemes (the category of Noetherian $S$-schemes is bad since it does not have fiber products). Conclusion: The direct limit of the $Y_n$'s in the category of finite type $S$-schemes "is" $Y$. P.S. The reader may be confused and think: "Wait a minute, the Hom scheme is only representable when the target scheme is separated. Doesn't this mean that the direct limit only exists in the category of finite type separated $S$-schemes?" Wrong! It turns out that the Hom scheme is representable even when the target scheme is not separated. This is Theorem 1.2 of the the same Hall-Rydh paper. The usual representability theorems for Hom schemes assume separatedness of the target scheme for the reason that it has simply been tradition to check that formal deformations are algebraizable by a "graph trick." More precisely, we may realize a morphism of formal $\widehat{S}$-schemes $\widehat{T} \to \widehat{Y}$ as a closed formal subscheme of $\widehat{Y \times_S T}$ (crucially using separatedness!). Algebraizing the coherent ideal cutting out this closed formal subscheme gives the required morphism of $S$-schemes $Y \to T$. My argument above shows that we can avoid all this ballyhoo with separatedness and instead invoke the deus ex machina of Tannakian duality. For more on why we may want to avoid separatedness, the reader may refer to the introduction here (specifically Sections 1.4 and 1.6). Short answer: Stacks with infinite stabilizers are almost never separated.<|endoftext|> TITLE: Does every group arise as the fundamental group of a complete Kähler manifold? QUESTION [20 upvotes]: The fundamental group of a manifold is countable, and every countable group $G$ arises as the fundamental group of a (smooth) manifold; see this comment or this answer for a construction of an open subset $U \subset \mathbb{R}^5$ with $\pi_1(U) \cong G$. Note that every smooth manifold admits a complete Riemannian metric. In fact, every conformal class contains a complete metric, see The Existence of Complete Riemannian metrics by Nomizu and Ozeki. Therefore, every countable group arises as the fundamental group of a complete Riemannian manifold. As the hermitian property is preserved under conformal change, every conformal class of a hermitian metric on a complex manifold contains a complete hermitian metric. Replacing $U$ with $V := U\times\mathbb{R} \subset \mathbb{R}^6 = \mathbb{C}^3$, we see that every countable group arises as the fundamental group of a complete hermitian manifold. Note that $V$ also admits a Kähler metric. However, unlike the hermitian case, the Kähler property is not preserved under non-constant conformal change, see this question. In fact, not every Kähler manifold admits a complete Kähler metric, see this question. Despite this, do we still have the Kähler analogue of the two bold statements above? Does every countable group arise as the fundamental group of a complete Kähler manifold? It's worth pointing out that the question of which groups arise as the fundamental groups of compact Kähler manifolds (which are necessarily complete) is an active area of research. Such groups are known as Kähler groups and much is known about them, see this question. REPLY [26 votes]: Any Stein manifold admits a complete Kähler metric. Start with a connected real analytic manifold with the given fundamental group. A suitable tubular neighbourhood of the complexification will be Stein. This Stein manifold will have the same fundamental group as the original manifold. For details see page 383 of Narasimhan, R., On the homology groups of Stein spaces, Invent. Math. 2, 377-385 (1967). ZBL0148.32202.<|endoftext|> TITLE: Spectral sequences in algebraic topology QUESTION [8 upvotes]: What books/articles do you recommend for learning spectral sequences? I am interested in their applications to algebraic topology, particularly to understand the homology of fibre bundles. I have a good backgroud on differential geometry and a reasonable background on modules and algebras. REPLY [4 votes]: I have learnt it from Mosher and Tangora. But, perhaps John McCleary’s book is a reference to be mentioned.<|endoftext|> TITLE: Almost commuting matrices, one a projection, is there a nearby projection that commutes? QUESTION [5 upvotes]: Suppose that $P, A, Q \in \mathbb{M}^{n \times n}(\mathbb{R})$ (I'm still interested if it must be done over $\mathbb{C}$), (EDIT:) suppose that $A$ is given, $P$ is an orthogonal projection, and $\lVert PA-AP \rVert < \delta$, is there some $Q$ an orthogonal projection such that $\Vert P-Q \rVert < \epsilon(\delta)$ (such that as $\delta \to 0$, so does $\epsilon$) and $QA = AQ$? Obviously if there's a broader result than just the finite-dimensional case, then that's good too. I've seen results vaguely like this, e.g. Lin's Theorem is a result about almost-commuting normal matrices. However I haven't seen a result about projections of this kind. It would be nice if this were true, I hope to learn more about this type of result, but if the good people here could point me to this result or a similar result from which I could jump off, I'd very much appreciate it. EDIT: A simple counterexample to my suggestion: $$ A = \begin{bmatrix}1 & \delta \\ 0 & 1\end{bmatrix} $$ $$ P = \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix} $$ The only non-trivial invariant subspaces are $\mathbb{R}^2$ and the span of $\begin{bmatrix}1 & 0\end{bmatrix}$, but P is not near to the projection for either. But $\lVert PA-AP \rVert = \delta$. Another counterexample is given in the comments. REPLY [3 votes]: EDIT: I realized that this does not work, because $PA$ has one nonzero too much, sorry. $A$ and $P$ can be simultaneously almost-triangularized, but they don't almost-commute. I'm leaving it up as an attempt, but it doesn't deserve acceptance / upvotes. Suppose such $\delta<1$ and $\epsilon(\delta)$ exist. Take $A = J + \delta e_1 e_n^T$, $P = e_1 e_1^T$ where $J$ is a nilpotent Jordan block of size $n$, and $e_1,e_n$ are the first and last column of the identity matrix $I$. Since $\|P-Q\|<1$, and orthogonal projection matrices have integer trace equal to their rank, $Q$ must have rank 1. Then $\operatorname{Im} Q$ must be a real eigenvector of $A$, and the only one is $[1\, \delta^{1/n}\, \delta^{2/n}\, \dots \, \delta^{1-1/n}]^T$ (and its multiples). So $Q$ is at a distance $O(\delta^{1/n})$ from $P$. Hence $\epsilon(\delta) > C\delta^{1/n}$ for some $\delta$. This holds for all $n$, and thus $\epsilon(\delta) \geq 1$, contradicting $\epsilon(\delta) \to 0$.<|endoftext|> TITLE: What is the Turing degree of the monadic theory of the real line? QUESTION [10 upvotes]: The monadic theory of the real line is the set of all sentences in the monadic second-order language of order which are true in $\mathbb{R}$. In this 1982 paper, Gurevich and Shelah show that true first-order arithmetic is Turing-reducible to the monadic theory of the real line. (Shelah first showed it in the 1970's, but his original result assumed the continuum hypothesis.) But my question is, what is the Turing degree of the monadic theory of the real line? Clearly the monadic theory of the real line is Turing reducible to true third-order arithmetic, but how does it compare to true second-order arithmetic, for instance? REPLY [6 votes]: Gurevich and Shelah showed in The monadic theory and the “next world” that the monadic theory of the real line (or even just the Cantor Discontinuum) can compute - the $V^B$ theory of second order arithmetic, - the $V^B$ theory of third order arithmetic if CH holds (or if the union of $<\!c$ meager sets is meager), where $B$ is the Boolean algebra of regular open sets, equivalently the algebra for adding a Cohen real. Note the sharp contrast with the decidability of S2S. Also, despite its power, the monadic theory of the real line does not interpret (allowing parameters) even Robinson arithmetic $\mathrm{Q}$ (On the strength of the interpretation method by Gurevich and Shelah; the theory there is mutually interpretable with $\mathrm{Q}$, $\mathrm{I}Σ_0$, and other typical weak arithmetical theories without exponentiation). See also Peano Arithmetic may not be interpretable in the monadic theory of orders by Lifsches and Shelah. The reason is that we have certain symmetries that break any purported interpretation. However, the Cohen algebra $B$ is interpretable in the monadic theory of the reals, and with some effort, the theory can be used to reason about $B$-valued models, hence the unusual form of the complexity result. I suspect the qualification $V^B$ can be removed in the 'true' $V$, but I only know the trivial (given known absoluteness results) consequences: - under $V=L$, the theory of third order arithmetic is computable from the theory, - under projective determinacy, the theory of second order arithmetic is computable from the theory, - under a measurable Woodin cardinal + CH, $Σ^2_1$ truth is computable from the theory, - there is a generic extension of $V$ (such as $\mathrm{Add}(ω_1,1) \times \mathrm{Add}(ω,1)$) in which the theory of third order arithmetic is computable from the theory (both as defined in the extension).<|endoftext|> TITLE: Homotopy groups $\pi_{4n-1}(SO(4n))$ QUESTION [9 upvotes]: There is a very natural way to define generators of $\pi_{4n-1}(SO(4n))\cong \mathbb{Z}\oplus \mathbb{Z}$ in terms of quaternions when $n=1$ and octonions when $n=2$ (see for example Tamura, On Pontrjagin classes and homotopy types of manifolds, 1957). Since there are no normed division algebras in higher dimensions, it is not possible to do the same for $n>2$. I was wondering whether there still is a natural identification between $\pi_{4n-1}(SO(4n))$ and $\mathbb{Z}\oplus\mathbb{Z}$ using explicit generators when $n>2$? REPLY [14 votes]: One way to think about this would be to consider the (injective!) map $\pi_{4n-1} SO(4n) = \pi_{4n} BSO(4n) \to \mathbf Z^2$ that sends the classifying map of a $4n$-dimensional oriented real vector bundle $\xi\colon E \to S^{4n}$ to the pair $(e,p_n)$ consisting of its Euler and $n$th Pontryagin class evaluted against the fundamental class of the base sphere. This map is injective and for $n \neq 1,2,4$ its image is given by $2\mathbf{ Z} \times a_n(4n-1)!\mathbf{Z}$ where $a_n = 2$ if $n$ is odd, and $1$ otherwise: the Euler number of any bundle is even by the Hopf invariant $1$ problem, and the possible values of $p_n$ of (stable) vector bundles over$ S^{4n}$ is known from homotopy theory, see Theorem 3.8. in these notes by Levine, for instance. The generators you are looking for are then simply given by two bundles that are sent to $(2,0)$ and $(0,a_n(2n-1)!)$. The first one is clearly the tangent bundle of $S^{4n}$. I am not sure if the second one, which has vanishing Euler class and minimal $p_n$, has a nice geometric description.<|endoftext|> TITLE: Sheaf-theoretic approach to forcing QUESTION [47 upvotes]: Inspired by the question here, I have been trying to understand the sheaf-theoretic approach to forcing, as in MacLane–Moerdijk's book "Sheaves in geometry and logic", Chapter VI. A general comment is that sheaf-theoretic methods do not a priori produce "material set theories". Here "material set theory" refers to set theory axiomatized on the element-of relation $\in$, as usually done, in ZFC. Rather, they produce "structural set theories", where "structural set theory" refers to set theory axiomatized on sets and morphisms between them, as in the elementary theory of the category of sets ETCS. I will always add a collection (equivalently, replacement) axiom to ETCS; let's denote it ETCSR for brevity. Then Shulman in Comparing material and structural set theories shows that the theories ZFC and ETCSR are "equivalent" (see Corollary 9.5) in the sense that one can go back and forth between models of these theories. From ZFC to ETCSR, one simply takes the category of sets; in the converse direction, one builds the sets of ZFC in terms of well-founded extensional trees (modeling the "element-of" relation) labeled by (structural) sets. So for this question, I will work in the setting of structural set theory throughout. There are different ways to formulate the data required to build a forcing extension. One economic way is to start with an extremally disconnected profinite set $S$, and a point $s\in S$. (The partially ordered set is then given by the open and closed subsets of $S$, ordered by inclusion.) One can endow the category of open and closed subsets $U\subset S$ with the "double-negation topology", where a cover is given by a family $\{U_i\subset U\}_i$ such that $\bigcup_i U_i\subset U$ is dense. Let $\mathrm{Sh}_{\neg\neg}(S)$ denote the category of sheaves on the poset of open and closed subsets of $S$ with respect to this topology. Then $\mathrm{Sh}_{\neg\neg}(S)$ is a boolean (Grothendieck) topos satisfying the axiom of choice, but it is not yet a model of ETCSR. But with our choice of $s\in S$, we can form the ($2$-categorical) colimit $$\varinjlim_{U\ni s} \mathrm{Sh}_{\neg\neg}(U)$$ called the filter-quotient construction by MacLane–Moerdijk. I'm highly tempted to believe that this is a model of ETCSR — something like this seems to be suggested by the discussions of forcing in terms of sheaf theory — but have not checked it. (See my answer here for a sketch that it is well-pointed. Edit: I see that well-pointedness is also Exercise 7 of Chapter VI in MacLane–Moerdijk.) Questions: Is it true that $\varinjlim_{U\ni s} \mathrm{Sh}_{\neg\neg}(U)$ is a model of ETCSR? If the answer to 1) is Yes, how does this relate to forcing? Note that in usual presentations of forcing, if one wants to actually build a new model of ZFC, one has to first choose a countable base model $M$. This does not seem to be necessary here, but maybe this is just a sign that all of this does not really work this way. Here is another confusion, again on the premise that the answer to 1) is Yes (so probably premature). An example of an extremally disconnected profinite set $S$ is the Stone-Cech compactification of a discrete set $S_0$. In that case, forcing is not supposed to produce new models. On the other hand, $\mathrm{Sh}_{\neg\neg}(S)=\mathrm{Sh}(S_0)=\prod_{S_0} \mathrm{Set}$, and if $s$ is a non-principal ultrafilter on $S_0$, then $\varinjlim_{U\ni s} \mathrm{Sh}_{\neg\neg}(U)$ is exactly an ultraproduct of $\mathrm{Set}$ – which may have very similar properties to $\mathrm{Set}$, but is not $\mathrm{Set}$ itself. What is going on? REPLY [17 votes]: Thanks for all the enlightening answers! Let me summarize my understanding now. (Please correct me if I'm saying something stupid!) First, as explained by Mike Shulman in his answer, the answer to Question 1) is Yes: $\varinjlim_{U\ni s} \mathrm{Sh}_{\neg\neg}(U)$ is a model of ETCSR. So if you, like me, always wanted to have a hands-on understanding of what forcing does, I think you can go with this explicit model (and do away with all meta-mathematical baggage surrounding either boolean models, or countable base models, and generic filters). It remains to answer Question 2): How is this related to forcing? The answer seems to be that it is a generic extension $\mathrm{Set}_s[G]$ of $\mathrm{Set}_s$ where $\mathrm{Set}_s$ is an elementary extension of $\mathrm{Set}$. More precisely, recall the concept of "boolean ultrapowers", which is "ultrapowers with Stone-Cech compactifications replaced by a general extremally disconnected profinite set $S$". Namely, given $S\ni s$ as above, and any model $M$ of some first-order theory, we can look at the constant sheaf with value $M$ on $S$ with respect to the double-negation topology, and take its stalk at $s\in S$. (Warning: This is not really a stalk! I.e., $s$ does not define a point of the topos of double-negation sheaves. By "stalk" I just mean the colimit $\varinjlim_{U\ni s}$; equivalently, push forward to usual sheaves on the topological space $S$, and take the stalk there.) This defines a new model $M_s$ of the first-order theory. (If $S$ is the Stone-Cech compactification of $S_0$ and $s\in S$ is an ultrafilter on $S_0$, then the sheafification is taking any open and closed subset $U\subset S$, corresponding to some subset $U_0\subset S_0$, to $\prod_{U_0} M$, and then the stalk is $\varinjlim_{s\in \beta U_0} \prod_{U_0} M$, i.e. an ultraproduct.) This procedure can be applied in particular to a model of ZFC, or a model of ETCSR, with the procedures being equivalent. This construction features prominently in this beautiful article of Hamkins–Seabold. The Boolean ultrapower $\mathrm{Set}_s$ of $\mathrm{Set}$ (at $s\in S$) is thus given by the stalk at $s\in S$ of the constant sheaf of categories with value $\mathrm{Set}$. The constant sheaf of categories with value $\mathrm{Set}$ is different from set-valued sheaves! However, there is a natural functor, in particular giving a functor $$\mathrm{Set}_s\to \varinjlim_{U\ni s}\mathrm{Sh}_{\neg\neg}(U).$$ As in Zhen Lin's answer, the topos $\mathrm{Sh}_{\neg\neg}(S)$ classifies generic filters $G$. In particular, it contains such a generic filter itself, given by $$ G=\bigsqcup_{U\subset S} U.$$ (Here, $U$ is identified with the sheaf it represents.) Now I believe that the discussion around Theorem 22 in Hamkins-Seabold should translate into the following statement. $\mathrm{Set}_s$ is an elementary extension of $\mathrm{Set}$, $G$ is a generic filter over $\mathrm{Set}_s$, and $\varinjlim_{U\ni s} \mathrm{Sh}_{\neg\neg}(U)$ identifies with the generic extension $\mathrm{Set}_s[G]$ of $\mathrm{Set}_s$. (What is $\mathrm{Set}_s[G]$ here? It corresponds to the (material) forcing extension by $G$ under translating back-and-forth between structural and material set theory.) Finally, how is all of this related to usual forcing? In usual forcing, one first makes an auxiliary enlargement $V\subset \tilde{V}$ (i.e. $\mathrm{Set}\to \tilde{\mathrm{Set}}$) so that after this enlargement, "there are new elements of $S$" in the sense that the corresponding complete Boolean algebra $B$ admits new homomorphisms $B\to \{0,1\}$. One can then redo the previous construction, but take the stalk at a "new point" $s\in \tilde{S}$ (where $\tilde{S}$ is the spectrum of $B$ in $\tilde{V}$). All of the above should still work in this setting, giving us $\mathrm{Set}\to \mathrm{Set}_s\to \mathrm{Set}_s[G]$. However, if $s$ corresponds to a $V$-generic filter, then it turns out that $\mathrm{Set}_s=\mathrm{Set}$. So in this case the forcing extension is really one of $\mathrm{Set}$ itself, not of an elementary extension. This also answers my final confusion in the OP: If $S$ is a Stone-Cech compactification, then $\mathrm{Sh}_{\neg\neg}(S)$ is in fact the constant sheaf of categories with value $\mathrm{Set}$, so in particular $G$ already lies in $\mathrm{Set}_s$ for all $s\in S$, and $\mathrm{Set}_s[G]=\mathrm{Set}_s$. Thus, the forcing extension is trivial, but the elementary extension $\mathrm{Set}\to \mathrm{Set}_s$ to the ultrapower remains. Why does one usually want generic filters? I think the point is that the extension $\mathrm{Set}_s\subset \mathrm{Set}_s[G]$ always preserves ordinals (in its material incarnation). By contrast, $\mathrm{Set}\subset \mathrm{Set}_s$ very much does not, and in fact $\mathrm{Set}_s$ usually has nonstandard integers; in the material incarnation, this leads to non-wellfounded models. So one uses generic filters to make $\mathrm{Set}=\mathrm{Set}_s$. From the structural point of view, wellfoundedness is not visible, but of course it may still be disconcerting that the extension has nonstandard integers.<|endoftext|> TITLE: Question about definition of Hopf algebra in Hatcher QUESTION [13 upvotes]: I have two questions about the definition of a Hopf algebra in Hatcher's book on algebraic topology. He defines it as follows (see Section 3.C, page 283): Definition: A Hopf algebra is a graded algebra $A = \oplus_{n \geq 0} A^n$ over a commutative base ring $R$ satisfying the following two conditions: There is an identity element $1 \in A^0$ such that the map $R \rightarrow A^0$, $r \mapsto r \cdot 1$, is an isomorphism; one says that $A$ is connected. There is a diagonal or coproduct $\Delta\colon A \rightarrow A \otimes A$, a homomorphism of graded algebras satisfying $\Delta(\alpha) = \alpha \otimes 1 + 1 \otimes \alpha + \sum_{00$, and $\alpha'_j,\alpha''_j \in A^j$. Of course, in other settings Hopf algebras don't need to be graded, but I guess graded ones are the ones he cares about. Here are my two questions: In condition 2, one could instead ask for the a priori weaker condition that $\Delta(\alpha) = \alpha \otimes 1 + 1 \otimes \alpha + x$ with $x \in \oplus_{0 TITLE: What are some compact Hessian manifolds? QUESTION [7 upvotes]: In case this is too general, here is a more specific question. Is there a hyperbolic threefold which admits a Hessian metric (hyperbolic or otherwise)? Background A Hessian manifold is a Riemannian manifold which admits an atlas of coordinate charts whose transition maps are affine (i.e. $x \mapsto Ax+b$) and whose metric satisfies $$ g_{ij} = \frac{\partial^2}{\partial x^i \partial x^j} \phi $$ in each coordinate chart for some potential function $\phi$ (which may depend on the chart). These spaces are also known as affine Kahler manifolds. If one prefers a coordinate-free definition, this is equivalent to a Riemannian manifold admitting a curvature- and torsion-free connection $D$ which satisfies $$D_X g(Y,Z) = D_Y g(X,Z) $$ for all vector fields $X,Y,Z$. Three preliminary examples I'm aware of three examples of Hessian manifolds. If one considers a convex domain $\Omega \subset \mathbb{R}^n$ and a strongly convex potential $\phi: \Omega \to \mathbb{R}$, one can construct a Hessian manifold by setting $g = \frac{\partial^2}{\partial x^i \partial x^j} \phi$. In this case $\Omega$ serves as a global coordinate chart. The torus with its standard affine structure (i.e. $\mathbb{R}^n$ modulo a discrete lattice) can be made into a Hessian manifold. To do so, you can take a convex potential on $\mathbb{R}^n$ whose Hessian is invariant under the action of the discrete lattice. Less obviously, it is possible to another Hessian structure on the circle. For instance, if one considers the circle as the space $\mathbb{R}^+$ modulo the ''multiplicative lattice" $\{ 2^k | k \in \mathbb{Z} \}$, the resulting space is a Hessian manifold with the potential $-log (x)$ (defined on the affine universal cover). There have been a considerable number of papers (and even a book) written on the geometry of Hessian manifolds. However, apart from these three examples (and their products), I have not been able to find any other examples in the literature. As such, I'm curious if there are other examples which are known, especially those which are compact. Some relevant results Hessian manifolds manifold are necessarily affine, which greatly restricts their geometry. For instance, the fundamental group of a Hessian manifold must be infinite. However, many (most?) affine manifolds do not admit Hessian structures at all. Shima that the universal affine cover of a compact Hessian manifold is a convex domain [1]. As a result, the Hopf affine manifolds $\mathbb{S}^{n-1}\times \mathbb{S}^1$ do not admit Hessian metrics for $n>1$. In dimension 2, it is possible to classify compact Hessian manifolds completely. The only compact Riemann surface which admits an affine structure (i.e. a flat connection) is the torus and there are 6 inequivalent affine structures. Yagi showed the only affine structures which admit Hessian metrics are products of the standard affine structure on the circle and the affine structure in Example 3 [2]. One can deform the convex potential (in fact, you can find a potential which induces an arbitrary analytic metric on some small neighborhood). However, the global structure of these spaces is fairly simple. [1 ]Shima, Hirohiko, Hessian manifolds and convexity, Manifolds and Lie groups, Pap. in Honor of Y. Matsushima, Prog. Math. 14, 385-392 (1981). ZBL0481.53038. [2] Yagi, Katsumi, On hessian structures on an affine manifold, Manifolds and Lie groups, Pap. in Honor of Y. Matsushima, Prog. Math. 14, 449-459 (1981). ZBL0495.53011. REPLY [7 votes]: Here is a collection of examples: Let $Q:\mathbb{R}^{n+1}\to\mathbb{R}$ be a (Lorentzian) quadratic form of type $(n,1)$, and let $L^+\subset\mathbb{R}^{n+1}$ be one component of the cone of time-like vectors with respect to $Q$ (i.e., where $Q$ is negative). Set $\phi = -\log(-Q)$, which is a well defined function on $L^+$, and let $g$ be the Hessian metric of $\phi$, i.e, for (any set of) linear coordinates $x^0,\ldots, x^n$, we have $$ g = \frac{\partial^2\phi}{\partial x^i\partial x^j}\,\mathrm{d}x^i\mathrm{d}x^j. $$ This $g$ is positive definite on $L^+$. Moreover, since $\phi$ is invariant under the group $\mathrm{O}(Q)$ and satisfies $\phi(rv) = \phi(v)-2\log r$ for any constant $r>0$, it follows that $g$ is invariant under the action of $\mathbb{R}^+{\cdot}\mathrm{O}(Q)$. Now let $\Gamma\subset\mathrm{SO}(Q)$ be a discrete subgroup that acts freely and co-compactly on the $n$-dimensional level set $Q=-1$ in $L^+$ (which is a copy of hyperbolic space), so that the quotient of this level set by $\Gamma$ is a compact hyperbolic manifold. Also, let $\mathbb{Z}$ act on $L^+$ by $n\cdot v = r^n v$ for some real number $r>1$. The action generated by the commuting groups $\Gamma$ and $\mathbb{Z}$ is free and discrete and cocompact on $L^+$ and the metric $g$, which is obviously Hessian, is invariant under this action. Thus, $g$ descends to a Hessian metric on the compact quotient $M$ of $L^+$ by the action of $\Gamma\times\mathbb{Z}$. Note that $M$ is diffeomorphic to the product of a circle and a compact hyperbolic $n$-manifold. I imagine that there are other examples of this nature constructed on other invariant cones in vector spaces that have cocompact actions.<|endoftext|> TITLE: Hybrid online/ in-person workshops, conferences, and summer schools QUESTION [7 upvotes]: I am writing to ask if people in the community can post any noteworthy experiences they have had with hybrid online/in-person workshops. Does anyone have experience with these, as either organizer or participant? The hybrid aspect would of course be before the pandemic, but my institution is thinking towards a future scenario where in-person workshops can return, and how to capture some of the gains made in online workshops in the past year, with a twin goal of widening participation and reducing unnecessary travel. Of course if you have insights to share which have been gleaned from fully online workshops in the past year, please do share them as well. I suppose the actual filming and streaming of the lectures is fairly self-explanatory, given enough resources. Has anyone had good experience moderating Q&A in a hybrid situation? How can one include remote participants on a similar footing to local participants? Thinking also about bringing people together to actually meet and discuss rather than just listen to talks, has anyone ever had an experience with discussing mathematics that was superior to simply discussing on Zoom? I have heard about a PI/Bonn conference where there were some screens and cameras set up for spontaneous conversations, but I am not sure how well it worked. That raises the possibility of clustered conferences, hosted in multiple locations in conjunction. Besides the PI/Bonn workshop have there been others, and did anyone attend one they thought was particularly successful? If one imagined a budget to buy kit at the conference host -- but presumably requiring remote participants to interact on their own kit -- is there anything non-obvious one can do to make this more of a success? Any ideas/suggestions to ensure quality delivery of remote lectures? REPLY [3 votes]: I haven't participated in any hybrid workshops, but I have had some positive experiences with discussing math in fully remote workshops. Forgive me if this is all old news to you! I ran a program last summer that used Discord as its main platform; this had the advantage that written messages (e.g. scheduling announcements, questions) were organized, saved, and in the same platform as the video "in person" sessions. Another advantage was that everyone could see who was in the voice channels and choose to join particular conversations themselves. Zoom now allows participants to choose their own breakout rooms, but this wasn't implemented yet over the summer when my program ran. To discuss math together, we used shared online whiteboards--one per voice channel to simulate each room having a blackboard. I recently participated in a workshop where we worked together on mathematics in small groups in Zoom breakout rooms, and it went pretty well. People were mostly working on paper, but since we were all educators, many of us had document cameras and could switch our cameras or share our screens. This didn't have the same collaborative feel of working at the same board, but the discussions were still pretty good. This semester, I am teaching a hybrid class, and I'm encouraging students to work together in groups. From day 1, the students who are attending in person insisted that they did not want to work with the people who were online. Understandably, they much preferred to talk to the other people in the room with them. The fully remote groups are working together on shared online whiteboards, and they're doing well (though still not as engaged or animated as the in person groups). My impression so far is that the remote students are having a good (or good enough) experience and the in person students are having a good experience, but the interaction between the two groups is minimal. Unfortunately, I don't have great ideas about how to make a hybrid workshop better, with both populations on equal footing. One tool that people have suggested to me is gather.town. I've only used it for social events, but you can create a layout that simulates your favorite place to do mathematics, including having boards that people (or their avatars) can stand at together. You walk your avatar through the map and can only hear/see people who are within a certain radius of you. I've heard good things, but not used it extensively myself. Good luck! I'll be interested to see other people's experiences.<|endoftext|> TITLE: Randomized version of Turán's theorem QUESTION [5 upvotes]: Turán's theorem says the following. Take any natural $n$ and $r$. Suppose that \begin{equation*} |G|>\Big(1-\frac1r\Big)\frac{n^2}2, \tag{0} \end{equation*} where $|G|$ is the number of edges of an (undirected) graph $G$ with $n$ vertices. Then $G$ contains an $(r+1)$-vertex clique. The bound on $|G|$ in (0) is the best possible one. The extreme graphs not containing an $(r+1)$-vertex clique, called Turán's graphs, are complete $r$-partite graphs with maximally balanced numbers of vertices in the $r$ parts. Taking here, e.g., $r=2$, we see that Turán's theorem requires that more than half of all $\binom n2$ possible edges between the $n$ vertices be included into $G$ in order to guarantee, even in a worst possible case, the existence of a triangle in $G$. Suppose, though, that we are interested, not in a worst possible case, but in a case typical to a however degree. To be more specific, suppose $G$ is a random graph with the set $[n]:=\{1,\dots,n\}$ of vertices. For any $i$ and $j$ in $[n]$, let $G_{\{i,j\}}$ denote the indicator of the inclusion of the edge $\{i,j\}$ into $G$, with $G_{\{i,i\}}=0$ for all $i\in[n]$, so that $|G|=\sum_{\{i,j\}\subseteq[n]}G_{\{i,j\}}$. Further, assume the following conditions: For each pair of distinct $i$ and $j$ in $[n]$, the r.v. $G_{\{i,j\}}$ has the same Bernoulli distribution with parameter $1-p$. For each $J\subseteq[n]$, the families $(G_{\{i,j\}}\colon\{i,j\}\subseteq J)$ and $(G_{\{k,l\}}\colon\{k,l\}\subseteq[n]\setminus J)$ are independent of each other. For each $i\in[n]$, the random variables (r.v.'s) $G_{\{i,1\}},\dots,G_{\{i,i-1\}},G_{\{i,i+1\}},\dots,G_{\{i,n\}}$ are independent. The distribution of $G$ is invariant with respect to all permutations of $[n]$. (No conditions other than 0--3 are assumed here. In particular, it is not necessary that all the indicators $G_{\{i,j\}}$ be jointly independent.) Then, by condition 0, \begin{equation*} E|G|=\binom n2(1-p). \end{equation*} So, with a nonzero probability \begin{equation*} |G|\ge\binom n2(1-p)\ge\Big(1-\frac1{cn}\Big)\frac{n^2}2 \end{equation*} if $00$ (and still have, with a nonzero probability, a clique of a size $\asymp n$ in $G$)? To put this more formally: Is it true that there exist universal real constants $b>0$ and $c\in(0,1)$ such that -- for each natural $n$, each $p\in[0,b\frac{\ln n}n]$, and each random graph $G$ with the set $[n]$ of vertices satisfying conditions 0--3 -- the probability that $G$ has a clique of a size $\ge cn$ is nonzero? REPLY [3 votes]: No. Consider the following distribution: Let $M$ be an integer, say $M= \frac{n}{\log n}$. Then for each vertex $v \in [n]$ assign an integer $m(v)$ where $m(v)$ is chosen according to the uniform distribution on $\{0,1,\ldots, M-1\}$, and where the $m(v)$s; $v \in [n]$; are mutually independent. Then for each pair $u$,$v \in [n]$ of vertices, $uv$ is an edge if and only if $m(v)\not = m(u)$. Then this distribution satisfies (0)--(3) of your conditions, and furthermore the probability that 2 vertices form an edge is $\frac{M-1}{M} = 1 -\theta(\frac{\log n}{n})$, but there is necessarily no clique with more than $M$ vertices.<|endoftext|> TITLE: If $M\otimes_S T$ is an $A$-module, is $M$ an $A$-module? QUESTION [6 upvotes]: Let $\mathbb{C}$ be the field of the complex numbers. Let $R=\mathbb{C}[x]$, $T=\mathbb{C}\langle x\rangle$ be the ring of entire series with convergence radius at least $1$, and let $S=\mathbb{C}\langle\langle x\rangle\rangle$ be the ring of entire series with infinite convergence radius. We have $R\subset S \subset T$. Let $A$ be an $R$-algebra which is flat and finitely generated as an $R$-module. Let $M$ be a finitely generated $S$ module such that the $R$-module $M\otimes_S T$ admits an $A$-module structure extending the one of $R$-module. Can I conclude that the $A$-module structure descends to $M$, extending the $R$-module structure? EDIT: I should state what I have in mind. Let $X$ be the affine algebraic curve over $\mathbb{C}$ given by the spectrum of $A$. The choice of a coordinate $x$ corresponds to the choice of a non constant morphism $f:X\to \mathbb{A}^1_{\mathbb{C}}$ of algebraic curves. By GAGA, we associate a morphism of Riemann surfaces $X^{an}\to (\mathbb{A}^1_{\mathbb{C}})^{an}$. Note that $\mathcal{O}_{\mathbb{A}^{1,an}}(\mathbb{A}^{1,an})\cong S$ and $\mathcal{O}_{X^{an}}(X^{an})\cong A\otimes_R S$. Let $U$ be open subset of $X$ given by the inverse image of the open unit ball via $f$. Again, note that $\mathcal{O}_{X^{an}}(U)\cong A\otimes_R T$. Now, suppose I have a coherent sheaf $\mathcal{F}$ on $U\subset X^{an}$. Assume further that $f_*\mathcal{F}$ can be analytically continuated to $(\mathbb{A}^1_{\mathbb{C}})^{an}$. Can $\mathcal{F}$ be analytically continuated to $X^{an}$? Many thanks! REPLY [2 votes]: I may have a counter-example. Suppose $A$ is $\mathbb{C}[x,t]/(t^2=x+1)$, and let $M$ be $\mathbb{C}\langle \langle x\rangle\rangle $ as an $S$-module. Then, $M\otimes_S T=\mathbb{C}\langle x\rangle $ is given an $A$-module structure by defining the action of $t$ as the multiplication by $\sqrt{x+1}$ where $$\sqrt{x+1}=1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3-\frac{5}{128}x^4+...$$ But $\sqrt{1+x}$ does not defines an entire function and hence the $A$-module structure does not desend to $M$.<|endoftext|> TITLE: Where should I learn about the p-adic L-functions of elliptic curves? QUESTION [11 upvotes]: Where is the best place to learn about the p-adic L-functions of Elliptic Curves? Doing a bit of research I have found books like "An Introduction to Cyclotomic Fields" by Washington, but those only seem to discuss the p-adic analogues of Dirichlet L-functions. As background, I am a student whose main relationship with algebraic Number Theory is Joe Silverman's "The Arithmetic of Elliptic Curves". I have also taken a course on Analytic Number Theory. REPLY [17 votes]: We know how to attach $p$-adic $L$-function to elliptic curves over $\mathbb Q$ only because we know how to attach them to cuspidal modular eigenforms, and we know by Breuil-Conrad-Diamond-Taylor that every elliptic curves over $\mathbb Q$ are themselves attached to a cuspidal modular eigenform of weight $2$. Historically, the construction of $p$-adic $L$-functions attached to modular forms is the work of Mazur-Swinnerton-Dier, Manin, Amice-Velu, Shokurov. The article of Mazur-Tate-Teitelbaum quoted in comment contains a synthesis of all that. But I think the best way to learn the theory now is to learn the version of the construction (not fundamentally different than the previously mentioned, but much simpler and clearer) given by Pollack and Stevens. For this, see the two papers by Pollack-Stevens: Overconvergent modular symbols and p-adic L-functions, Ann. Sci. Ec. Norm. Sup. (4) 44 (2011), no. 1, 1–42, and Critical slope $p$-adic $L$-functions, J. Lond. Math. Soc. (2) 87 (2013), no. 2, 428--452. I have also written a book, an expanded version of a course I gave at Brandeis, on the subject of p-adic L-functions of modular forms (and families thereof). It exposes the Pollack-Stevens method, as well as their generalization and mine for the so-called "critical" case, that was missing at the time of Mazur-Tate-Teitelbaum, and many other things. The book is called The Eigenbook. Eigenvarieties, families of Galois representations, p-adic L-functions and it will be published soon by Springer-Birkhauser, but you can also find a non-final version on my webpage.<|endoftext|> TITLE: Linear independence of algebraic integers of equal norm QUESTION [5 upvotes]: Let $K$ be a number field with $[K:\mathbb{Q}]=n$ with $n \geq 2$ and let $\mathcal{O}_K$ be its ring of integers. Suppose that $\alpha_1, \cdots, \alpha_n \in \mathcal{O}_K$ are distinct algebraic integers such that $N_{K/\mathbb{Q}}(\alpha_j) = a$ for some fixed rational integer $|a| > 1$ and the principal ideals $(\alpha_j)$ are distinct for $j =1, \cdots, n$. Does it follow that $\alpha_1, \cdots, \alpha_n$ are $\mathbb{Q}$-linearly independent? When $n = 2$ this is obvious. Indeed, a quadratic algebraic integer is of the form $u = u_1 + \omega u_2$ with $u_1,u_2 \in \mathbb{Z}$ and where $\omega = \sqrt{d}$ for some integer $d$ or $\omega = \frac{1 + \sqrt{d}}{2}$ for some rational integer $d$. Then $u,v$ are $\mathbb{Q}$-linearly dependent in $\mathcal{O}_K$ if and only if there exist rational integers $\ell,m$ such that $\ell u = mv$. Since $u,v$ have the same norm, it follows that $N_{K/\mathbb{Q}}(\ell) = N_{K/\mathbb{Q}}(m)$, and from here we see that $u,v$ are necessarily associates. But then we must have $u = v$. Is the result above true when $n \geq 3$? If not, what is a simple counterexample to the claim? REPLY [5 votes]: We adapt an idea from a now-deleted answer by Kenny Lau to construct examples for any $n>2$ with the $\alpha_j$ all contained in 2-dimensional space. Let $a$ be prime, and choose distinct integers $x_1,\ldots,x_n$ that remain different mod $a$ for which $$ P(x) := \left[\prod_{j=1}^n (x-x_j)\right] - a $$ is irreducible. (By Hilbert's irreducibility theorem, most $x_j$ will satisfy the last condition.) Then take $\alpha_j = x - x_j$ in the field ${\bf Q}[x] / (P(x))$. Each of these has norm $\pm a$, and they generate distinct ideals because the ideals above $a$ correspond to factors of $P \bmod a$. For example, when $n=7$ we may take $a=11$ and $\alpha_j = j-4$ ($j=1,\ldots,7$) to make $P(x) = x^7 - 14 x^5 + 49 x^3 - 36 x - 11$ with $\alpha_j = x+3, \, x+2, \, x+1, \, x, \, x-1, \, x-2, \, x-3$.<|endoftext|> TITLE: Translation lengths in CAT(0) spaces QUESTION [10 upvotes]: Let $a,b$ be two loxodromic isometries of a CAT(0) space. Assume that, for every $n \geq 1$, $a^nb$ is also loxodromic. Is it possible for the translation length of $a^nb$ to be bounded independently of $n$? First, I thought as obvious that the translation length of $a^nb$ has to tend to $+ \infty$ as $n \to + \infty$, but I may have been misled by the CAT(-1) case (where this is clearly true). Now, I go back and forth between a possible counterexample and an easy argument I am missing... REPLY [14 votes]: Consider the following two transformations of $\mathbb{R}^3$: $a:(x,y,z)\mapsto (x+1,y,z)$ and $b:(x,y,z)\mapsto (-x,-y,z+1)$. The translation axis for $a^nb$ is the line $x=n/2$, $y=0$ and $a^nb$ translates this line by a distance of 1, independent of $n$.<|endoftext|> TITLE: Origin and context of adjunctions inducing equivalences between full subcategories QUESTION [9 upvotes]: The following is well-known. Theorem. Let $F\dashv U$ be a pair of adjoint functors $$F\colon \mathcal C\to \mathcal D, \qquad U\colon \mathcal D\to\mathcal C$$ with unit $(\eta_A\colon A\to U(F(A)))_{A\in\mathcal C}$ und counit $(\varepsilon_B\colon F(U(B)) \to B)_{B\in\mathcal D}$. Then $F$ and $G$ restrict to equivalences $\mathcal C'\to\mathcal D'$ and $\mathcal D'\to\mathcal C'$ between the full subcategories $\mathcal C'$ and $\mathcal D'$ given by $$\mathcal C':=\{A\in\mathcal C\mid \eta_A\text{ is an isomorphism}\}$$ and $$\mathcal D':=\{B\in\mathcal D\mid \varepsilon_B\text{ is an isomorphism}\}.$$ Posets can be considered as categories. This yields: Corollary. Let $P$ and $Q$ be posets, and $f\colon P\to Q$ and $u\colon Q\to P$ monotone maps with $$\forall a\in P.\, \forall b\in Q.\,a\leq u(b)\iff f(a)\leq b.$$ Then $f$ and $u$ restrict to isomorphisms between the induced sub-posets $P'$ and $Q'$ given by $$P':=\{a \in P\mid a = u(f(a))\}$$ and $$Q':=\{b\in Q\mid f(u(b))=b\}.$$ Questions. Can someone give me the original references for the theorem and the corollary above? Of course, in some sense, the corollary must be hidden in the works of Galois (or in the works of people who cleaned up his theory), since taking $P$ to be $(\mathcal P(F), \subseteq)$ for any field $F$, $Q$ to be $(\mathcal P(G), \supseteq)$ for any group $G$ of automorphisms of $F$, and considering $$f\colon \mathcal P(F) \to \mathcal P(G),\, a\mapsto\{g\in G\mid \forall x\in a. \, g(x)=x\}$$ and $$u\colon \mathcal P(G) \to \mathcal P(F), \, b\mapsto \{x\in F\mid \forall g\in b. \, g(x)=x\}$$ yields the fundamental theorem of Galois theory up to a concrete characterization of $P'$ and $Q'$. But I'm interested in who extracted the "abstract content" of this proof in the sense of the corollary and who, after that, formulated the generalization to categories in the sense of the theorem. Not just the relation "$g(x)=x$" between field elements $x$ in $F$ and automorphisms $g$ in $G$ yields a pair of functions $f\colon \mathcal P(F) \to \mathcal P(G)$ and $u\colon \mathcal P(G) \to \mathcal P(F)$ satisfying the conditions of the corollary, but in fact any relation $R\subseteq A\times B$ yields such a pair of functions $f\colon \mathcal P(A) \to \mathcal P(B)$ and $u\colon \mathcal P(B) \to \mathcal P(A)$. (Such a pair functions constituting an adjunction between posets is called a Galois connection, by the way.) Thus relations are a powerful tool to generate Galois connections. Is there an analogue of relations that allows one to generate pairs of adjoint functors between categories? Not just the fundamental theorem of Galois theory can be formulated as an instance of the above theorem, but also the Gelfand–Naimark theorem, the Stone duality, the duality between affine schemes and commutative rings, and the Pontrjagin duality are corollaries of the above theorem up to concrete characterization of $\mathcal C'$ and $\mathcal D'$. I guess most of them are recast as applications of the above theorem only with hindsight. But can the above theorem be also used as a guide that helps and inspires proving completely new equivalences? REPLY [2 votes]: Question 1 I don't know if they are the first publications, but two very early references are Garrett Birkhoff (1940). Lattice Theory. Øystein Ore (1944). "Galois Connexions". (see Theorem 2 for a formulation of the "Corollary") Remarks in these texts indicate that Galois connections between power set lattices $\mathcal P(X)$ and $\mathcal P(Y)$ induced by relations $R\subseteq X\times Y$ are due to Birkhoff, while Ore introduced the theory of Galois connections between arbitrary posets. From 2: It has already been pointed out by Garrett Birkhoff that any binary relation defines a correspondence of the type of a Galois connexion between the subsets of two sets In the 1948 revised edition of 1, Birkhoff remarks in a footnote that the "Corollary" in the special case of Galois connections induced by relations is due to himself, while he refers to 2 as the origin of a particular axiomatization of Galois connections between arbitrary posets that he gives in the text. As remarked by Marc Olschok in the comments, the "Theorem" occurs implicitly in Saunders Mac Lane. Categories for the Working Mathematician. in the form of an exercise (Exercise 2, §IV-5). I don't have the first edition of that book to check whether this exercise already appears in the first edition. Given the similarity between adjunctions (due to Kan) and Galois connections, the generalization from the "Corollary" to the "Theorem" may well be a "folklore result", if not due to Mac Lane.<|endoftext|> TITLE: Tameable hypergraphs QUESTION [6 upvotes]: Let $H=(V,E)$ be a hypergraph. We say that $I\subseteq V$ is an independent set if $e\not\subseteq I$ for all $e\in E$. We say that $H$ is tameable if every independent set is contained in a maximal independent set. Every graph is tameable and, more generally, so is every hypergraph with finite edges. There are easy examples of non-tameable hypergraphs, and I use this one given by user @bof in the comment section of this answer: Let $H=(\omega,[\omega]^\omega)$, where $[\omega]^\omega$ denotes the collection of infinite subsets of $\omega$. (The only independent subsets of this graph are the finite sets, and there is no maximal finite set.) If $(P,\leq)$ is a poset, then with $\text{Max}(P)$ we denote the collection of maximal elements of $P$. (Note that $\text{Max}(\omega) = \varnothing$, for instance.) Given a hypergraph $H=(V,E)$ we let $$\text{Tame}(H) = \{E'\subseteq E: (V, E') \text{ is tameable}\}.$$ Question. Given a hypergraph $H=(V,E)$ with $V\neq\varnothing\neq E$ and $\varnothing\notin E$, do we necessarily have $\text{Max}(\text{Tame}(H)) \neq \varnothing$? REPLY [3 votes]: No. For a counterexample let $H=(\omega,E)$ where $E=\{e_n:n\in\omega\}$ and $e_n=[n,\omega)=\{x\in\omega:x\ge n\}$. If a subset $E'\subseteq E$ is finite then $(\omega,E')$ is tameable; every vertex cover contains a finite vertex cover which contains a minimal vertex cover, and (equivalently) every independent set is contained in a cofinite independent set which is contained in a maximal independent set. If $E'\subseteq E$ is infinite then $(\omega,E')$ is not tameable; in fact, it has no minimal vertex cover and (equivalently) no maximal independent set, since the vertex covers are just the infinite subsets of $\omega$ and the independent sets are just the coinfinite subsets of $\omega$. Thus $\operatorname{Tame}(H)=\{E'\subseteq E:E'\text{ is finite}\}$ and $\operatorname{Max}(\operatorname{Tame}(H))=\varnothing$. More generally, if $H=(V,E)$ is a hypergraph ($\varnothing\notin E$), then $\operatorname{Max}(\operatorname{Tame}(H))=\varnothing$ except in the trivial case when $H$ is tameable and $\operatorname{Max}(\operatorname{Tame}(H))=\{E\}$. Theorem. If $H=(V,E)$ is an untameable hypergraph (with nonempty edges), then $\operatorname{Max}(\operatorname{Tame}(H))=\varnothing$. Since the vertex set $V$ is fixed throughout the discussion, to save typing I will identify a hypergraph with its edge set; so I am going to show that the untameable set $E$ has no maximal tameable subset. For some reason I find it easier to work with minimal vertex covers than maximal independent sets. Proof. Assume for a contradiction that $E_0$ is a maximal tameable subset of $E$, and choose an edge $f\in E\setminus E_0$. Then $E_0\cup\{f\}$ is untameable, so there is a vertex cover $S$ of $E_0\cup\{f\}$ which contains no minimal vertex cover of $E_0\cup\{f\}$. Since $S$ is a vertex cover of the tameable set $E_0$, $S$ contains a minimal vertex cover $S_0$ of $E_0$. Now $S_0$ can't be a vertex cover of $E_0\cup\{f\}$, as it would then be a minimal vertex cover of $E_0\cup\{f\}$, which is impossible since $S_0\subseteq S$. Choose a vertex $y\in S\cap f$. Then $S_0\cup\{y\}$ is a vertex cover of $E_0\cup\{f\}$, but it can't be a minimal vertex cover of $E_0\cup\{f\}$ since $S_0\cup\{y\}\subseteq S$. Since $S_0$ is not a vertex cover of $E_0\cup\{f\}$, there is a vertex $x\in S_0$ such that $(S_0\setminus\{x\})\cup\{y\}$ is a vertex cover of $E_0\cup\{f\}$ and therefore of $E_0$. Since $E_0$ is tameable, there is a set $S_1\subseteq(S_0\setminus\{x\})\cup\{y\}\subseteq S$ such that $S_1$ is a minimal vertex cover of $E_0$. Morever $y\in S_1$, as otherwise we would have $S_1\subseteq S_0\setminus\{x\}$, contradicting the fact that $S_0$ is a minimal vertex cover of $E_0$. Thus $S_1$ is a vertex cover of $E_0\cup\{f\}$, and being a minimal vertex cover of $E_0$ it is a minimal vertex cover of $E_0\cup\{f\}$. Since $S_1\subseteq S$, this contradicts the fact that $S$ contains no minimal vertex cover of $E_0\cup\{f\}$.<|endoftext|> TITLE: First time appearance of Lie crossed module (crossed module of Lie groups) in literature QUESTION [5 upvotes]: Can someone point me to a reference where the notion of "Lie crossed module" appeared for the first time? I see many papers "recall" the definition of the Lie crossed module but, I do not see any mention of a "first-time" reference. The definition of Lie crossed module I am referring to is mentioned as Definition 1.3 in João Faria Martins and Roger Picken, On Two-Dimensional Holonomy, Trans. Amer. Math. Soc. 362 (2010) pp. 5657-5695, doi:10.1090/S0002-9947-2010-04857-3, arXiv:0710.4310. REPLY [4 votes]: For some more context, Kassel and Loday's 1982 paper defining crossed modules of Lie algebras cites an early paper by Loday that discusses crossed modules of (ordinary) groups, so it would appear that this citation trail won't give you crossed modules of Lie groups. Elsewhere I've seen crossed modules mentioned in conjunction with "abstract kernels", which date back to Eilenberg and Mac Lane's work on classifying nonabelian extensions of groups (which is secretly controlled by non-abelian cohomology, and the explicit group cohomology $H^3$—with values in an abelian group—that they use also helps classify 2-groups): Cohomology theory in abstract groups. II. Group extensions with a non-Abelian kernel. Mackenzie in his 1989 paper defining crossed modules of Lie groups cites a 1960 paper on analytic abstract kernels: R. A. Macauley, Analytic group kernels and Lie algebra kernels, Trans. Amer. Math. Soc. 95 (1960) pp. 530-553, doi:10.1090/S0002-9947-1960-0122908-6 Analytic groups here can be thought of as at least a special case of Lie groups (at the very least, classical groups are analytic), and the Macauley paper deals with constructions that turn up in the study of crossed modules of Lie groups, like induced maps to outer automorphism groups. So one could say that this latter paper almost invented crossed modules of Lie groups. But it's not too surprising that it doesn't, since even Eilenberg and Mac Lane's Annals paper involving abstract kernels (from 1947) don't cite Whitehead, even though they are even closer in subject matter (Mac Lane and Whitehead collaborated later in 1950, producing On the 3-type of a complex, and there crossed modules turn up). I think it unlikely someone working with Lie groups would have been familiar with work in algebraic homotopy theory or algebraic topology so as to even know of Whitehead's and/or Mac Lane–Whitehead's work. If Mackenzie, who was an early proponent of Lie groupoids, could only cite someone who is a near-miss for crossed modules of Lie groups, then it's safe to say his paper is probably the earliest. The only others I could have imagined independently developing the notion would be Jean Pradines (another Lie groupoidist, from the Ehresmann school) or perhaps someone in Ronnie Brown's orbit arriving at the idea of a topological crossed module via an analogue of the Brown–Spencer theorem relating groupoids internal to $\mathbf{Grp}$ and crossed modules), but I briefly checked for the latter and couldn't find anything.<|endoftext|> TITLE: Relation between Schanuel's theorem and class number equation QUESTION [26 upvotes]: (Crossposted on math stack exchange: https://math.stackexchange.com/questions/4040249/relation-between-schanuels-theorem-and-class-number-equation) It was recently brought to my attention that there is a striking similarity between the Class Number Formula and Schanuel's Theorem. See for yourself: Notation: \begin{align*} &K, \text{ number field}\\ &d, \text{ degree of $K$ over $\mathbb{Q}$}\\ &h_K, \text{ class number of $K$}\\ &R_K, \text{ regulator of $K$}\\ &D_K, \text{ discriminant of $K$}\\ &\mu_K, \text{ number of roots of unity contained in $K$}\\ &r_1, \text{ number of real places of $K$}\\ &r_2, \text{ number of complex places of $K$}\\ &\zeta_K(s), \text{ Dedekind zeta function of $K$}\\ &H_K(P), \text{ height relative to $K$ on $\mathbb{P}^n$} \end{align*} Theorem (Class Number Formula) $$\lim_{s\to 1}(s-1)\zeta_K(s)=\frac{h_K R_K 2^{r_1}(2\pi)^{r_2}}{\mu_K \sqrt{|D_K|}}$$ Theorem (Schanuel's Theorem) \begin{align*} \#\{P\in \mathbb{P}^n(K): H_K(P)\leq T\}=\frac{h_K R_K}{\mu_K\ \zeta_K(n+1)}\left(\frac{2^{r_1} (2\pi)^{r_2}}{\sqrt{|D_K|}}\right)^{n+1} (n+1)^{r_1+r_2-1}T^{n+1}+O(T^{n+1-1/d}) \end{align*} Both of these results can be proven using a geometry of numbers argument. So one reason for the similarity might simply be that they end up being answers to similar counting problems. But this is not very satisfying, and it would be nice to have a more conceptual reason that they are so similar. Main Question: Is there an intuitive (or deep) reason that the analytic class number formula is so similar to the coefficient of the leading term in Schanuel's Theorem? Secondary Question: What other examples are there of values of $L$-functions showing up in main terms of asymptotics? REPLY [19 votes]: A spirit. A general approach that may interest you is the following: counting laws can be obtained by studying suitable generating functions, via Tauberian arguments: the rightmost pole (resp. residue) of the generating function gives the growth order (resp. leading constant) in the counting law. In the case of Schanuel's theorem, the generating function is in fact linked to the Dedekind zeta function $\zeta_K$, showing how the residue of $\zeta_K$ naturally appears as part of the main term. With this point of view, it is not a surprise. Let me give some details and references below. Counting laws from zeta functions. In many counting problems, we introduce a generating function (Fourier series, Dirichlet series, zeta functions, etc.) as an analytic tool to study the underlying combinatorics. In the realm of counting rational points on algebraic varieties $V$ over a number field $K$, a suitable such generating function is the height zeta function defined by $$Z_V(s) := \sum_{x \in V(K)} h(x)^{-s}.$$ For $s$ large enough, this series is convergent (as a consequence for instance of Northcott's theorem, stating that the number of rational points of bounded height grows polynomially). The spirit now is the same as for the prime number theorem: obtain enough analytic properties of $Z_V(s)$. Suppose you are able to prove that $Z_V(s)$ has meromorphic continuation to a certain right half-plane, say with a single pole at $s=a$ and then remains holomorphic up to $\Re(s)>a-\delta$ for a certain $\delta>0$. Then, the (Wiener-Ikehara) Tauberian theorem gives you straight away the analogue of Schanuel's theorem for the variety $V$: $$N_V(X) := \# \{x \in V(K) \ : \ h(x) \leq X\} = \underset{s=a}{\rm Res} \ Z_V(s) \cdot X^a + O(X^{a-\delta}).$$ You can of course obtain finer asymptotic information on $N_V(X)$ with finer analytic information on $Z_V(s)$ (and may pick extra $\log(X)$ factors in case there is a vertical growth). The case of Schanuel's theorem. Now, how do we study $Z_V(s)$? This is done in a great paper of Franke-Manin-Tschinkel (1989): they relate $Z_V(s)$ to an explicit Eisenstein series (see their Proposition 3), hence deducing its nice analytic properties, and then they evaluate precisely the residue to obtain the leading coefficient (see their equation (2.12)). Without much surprise if you are acquainted with Eisenstein series: it is expressed in terms of the Dedekind zeta function. Hence, the leading coefficient in Schanuel's law is essentially the residue at $s=1$ of $\zeta_K$. A nice introduction to these geometric methods can be found for instance in the short and recent lecture notes by Chambert-Loir. Another expression for the constant. Note that you may also be willing to forget about the expressions in terms of class numbers and Dedekind zeta functions (which bears a lot of arithmetics) and look for a more geometric expression. This is provided for instance in Chambert-Loir-Tschinkel where the leading constant is rephrased essentially as a height zeta integral (see their Theorem 1.3.1, I try to be consistent with my notations above) $$\int_{V(\mathbb{A}_K)} h(x)^{-a} dx$$ for a certain suitably normalized Tamagawa measure $dx$. Recall that $a$ is the growth order in the counting law. A glimpse to the second question. What I said above already gives you a whole bunch of analogous phenomena: zeta functions are generating functions, hence their residues naturally appear in counting laws for the underlying objects. An example with a pretty different flavor is the following: you can consider the analogous zeta functions (of the conductor) for automorphic forms, use to trace formulas to study it and in particular see that the identity term (which is the most significant one) can be expressed by the Dedekind zeta function of the underlying field. Hence, you get counting (Weyl) laws displaying a leading constant expressed with the same residue $\zeta_K^\star(1)$. Hope it helps!<|endoftext|> TITLE: Units in the group ring over fours group after Gardam QUESTION [21 upvotes]: Giles Gardam recently found (arXiv link) that Kaplansky's unit conjecture fails on a virtually abelian torsion-free group, over the field $\mathbb{F}_2$. This conjecture asserted that if $\Gamma$ is a torsion-free group and $K$ is a field, then every invertible element in the group ring $K\Gamma$ is a nonzero scalar multiple of a group element. Consider the fours group $$ P = \langle a, b, A, B \;|\; a = A^{-1}, b = B^{-1}, (a^2)^b = a^{-2}, (b^2)^a = b^{-2} \rangle $$ (Gardam calls it the Promislow group). This group is a join of two Klein bottle groups and fits in a non-split exact sequence $1 \rightarrow \mathbb{Z}^3 \rightarrow P \rightarrow C_2^2 \rightarrow 1$. Now in the group ring $\mathbb{F}_2[G]$ we have $pq = 1$ where $$ p = 1 + aa + aaa + aab + AABA + aaBA + aaBABA + aabABA + aabb + ab + aba + abaBa + ababa + aBAbb + Abb + b + bA + BABA + bABA + BAbb + bb $$ and $$ q = 1 + a + AA + aaab + aaababb + AAAbb + aaB + AABB + Ab + aB + AbaB + Abab + abaB + abab + AbaBa + Ababa + abaBB + abbb + B + BA + BB, $$ so Kaplansky's unit conjecture is false. So an obvious question is: What can we now say about the group of units in $\mathbb{F}_2[G]$? I claim no originality in observing that this question can be asked (I saw Avinoam Mann discuss this question on Twitter). I gave this five minutes of thought myself and I see that the group is residually finite and contains a copy of $P$ (:P). Two random questions that are particularly interesting to me (but any information is welcome). Is this group finitely-generated? Is it amenable? Such questions do not seem obviously impossible on a quick look (for the first, some kind of Gaussian elimination? for the latter, use the fact the group has very slow growth somehow?), but they don't seem obvious either. I assume there is no information in the literature explicitly about the units in group rings of torsion-free groups, because we didn't know they can be nontrivial. But people have studied group rings a lot and I don't have much background on them, so maybe more is known than I was able to see. One fun thing to look at is the following: What is the isomorphism type of the subgroup $\langle p, P \rangle$ of the group of units? I checked that p has order at least $10$, and I suppose it must have infinite order. I checked also that the $3$-sphere generated by $p$ and $a$ has no identities, but $p$ and $b$ satisfy some identities. REPLY [4 votes]: This is an extended comment, follow-up to Giles Gardam's answer and the comments therein. Recall that the group of nonzero scalars $K^\times$ is a central subgroup of $(KG)^\times$ (for each group $G$ and field $K$). Proposition. Let $G$ be a group and $K$ a field, such that $\bar{K}G$ has no zero divisor. Then for every field $K$, the group $(KG)^\times/K^\times$ is torsion-free. In particular, every torsion element in $(KG)^\times$ is a scalar. Here $K\subset \bar{K}$ is an algebraic closure. For arbitrary fields, the assumption on $G$ is satisfied by torsion-free virtually solvable groups (and hence residually torsion-free solvable groups, such as free groups and many more). Lemma 1. Let $K$ be an algebraically closed field. Let $A$ be a $K$-algebra (=unital associative $K$-algebra) with no zero divisors. If $P\in K[t]\smallsetminus\{0\}$ and $P(x)=0$ for $x\in A$ then $x$ is a scalar. Proof: just consider the unital $K$-subalgebra generated by $x$. It is commutative, finite-dimensional over $K$, and has no nonzero divisor, hence is a field, and hence is 1-dimensional since $K$ is algebraically closed. In turn this implies: Lemma 2. Let $G$ be a group and $K$ a field for which $\bar{K}G$ has no zero divisor. Then for every nonzero polynomial $P\in K[t]$, every element $x$ in $KG$ such that $P(x)=0$ for some nonzero polynomial $P$, is a scalar. Proof: by Lemma 1, $x$ is a scalar in $\bar{K}G$, and hence is a scalar in $KG$. Lemma 2 in turn, applied to $P(t)=t^n-\lambda$ for any fixed $\lambda\in K$, immediately implies the proposition. Edit: the second sentence of the proposition can be restated as: if $K$ is a domain and $\overline{\mathrm{Frac}(K)}G$ has no zero divisors, then every torsion element in $(KG)^\times$ is a scalar. This is obtained in ARG's answer just assuming that $KG$ itself has no zero divisors.<|endoftext|> TITLE: Techniques from number theory used in algorithmic compositions? QUESTION [8 upvotes]: I have been experimenting with number theoretic techniques to get some algorithmic compositions. I use python (mingus) to create the midi, import it on musescore 2 and have the corresponing score. The method is based on a few ingredients: I use C-major or C-minor and use a symmetric function $f(a,b,c,\cdots)$ and plug in numbers $a,b,c,$ from $1 \le x \le 31$ or some other intervall and compute $f(a,b,c,..) \mod 8$ or $\mod 5$ to generate the pitches. To generate the rhythms / bars, I use a technique to map natural numbers to binary trees (divisorTree, sumTree). This way I have some control over the rhythm fast / slow by plugging in different numbers. In a mixed ensemble I use binary digits to determine at which bar the instrument plays. This creates attention to the listener, so that he/she asks himself/herself, what comes next. Which other algortihmic techniques from number theory do you know which are used for algorithmic compositions? Thanks for your help. The scores can be found at: https://musescore.com/user/37663311 Edit: I use e new technique to create a piece of 48 min with 8 instruments and 650 pages scores: (I hope you enjoy:) https://musescore.com/user/37663311/scores/6651858 Another edit: Voronoi diagram / Delaunay triangulation of pitch consonance similarity, as measured with the kernel $k(a,b) = \frac{\gcd(a,b)^2}{ab}$: https://www.reddit.com/r/musictheory/comments/tcamot/an_image_of_pitches_for_pitch_consonance/ REPLY [3 votes]: You may add this recently published paper to your reading list. P. Špaček and P. Sobota, Musipher: Hiding information in music composition, Rad HAZU, Matematičke znanosti, Vol. 25 (2021), p. 161-179, full text online. Abstract. In this paper, we present a new way of hiding information. We store the information directly in the process of composing music, based on musical theory. We created an algorithm to produce music based on binary string, where each bit is transformed into a music composition decision. We follow simple rules to make music, which sounds good. We conducted survey to find whether our solution works, and found promising results of our approach.<|endoftext|> TITLE: Proofs of theorems that proved more or deeper results than what was first supposed or stated as the corresponding theorem QUESTION [54 upvotes]: Recently, I figured out that a colleague of mine has had published during recent years a proof of a theorem in which he was actually proving a deeper result which we both thought to be still open. After a closer look at his proof I found that, taking a bit more care and putting some additional emphasis in certain parts of his previous proof, he was actually proving the other still-thought-to-be-open problem: the construction was absolutely the same and therefore the proof of the previously published theorem was certainly a better argument than we first thought. I am curious now about this phenomenon happening more often. Do you know some other recent (let's say from 1700 to the current day) examples of this phenomenon of proofs being stronger than initially stated or proving more than thought at first? REPLY [2 votes]: In the paper P. Erdős and A. Hajnal, On the structure of set-mappings, Acta Math. Acad. Sci. Hungar. 9 (1958), 111-131, the authors came close to solving Ulam's measure problem by proving that the first uncountable inaccessible cardinal is not measurable, as they pointed out in their subsequent paper P. Erdős and A. Hajnal, Some remarks concerning our paper "On the structure of set-mappings" — non-existence of a two-valued $\sigma$-measure for the first uncountable inaccessible cardinal, Acta Math. Hungar. 13 (1962), 223-226: In accordance with the notations of [4] we say that a cardinal $m$ possesses property $P_3$ if every two-valued measure $\mu(X)$ defined on all subsets of a set $S$ of power $m$ vanishes identically, provided $\mu(\{x\})=0$ for every $x\in S$ and $\mu(X)$ is $m$-additive. It was well known that $\aleph_0$ fails to possess property $P_3$ and that every cardinal $m\lt t_1$ possesses property $P_3$ where $t_1$ denotes the first uncountable inaccessible cardinal. Recently A. Tarski has proved, using a result of P. Hanf, that a certain wide class of strongly inaccessible cardinals possesses property $P_3$ (called strongly incompact cardinals). H. J. Keisler gave a purely set-theoretical proof of this result. After having seen these papers we observed that the special case of this result that $t_1$ possesses property $P_3$ follows almost trivially from some of our theorems proved in [1]. We are going to give this simple proof in §2. Our method for the proof is of purely combinatorial character, and although it is certainly weaker than that of A. Tarski and H. J. Keisler, we think that it is of interest to formulate how far one can go with these methods at present.<|endoftext|> TITLE: Seeking for references on some PDEs QUESTION [5 upvotes]: This is not a technical mathematical question. I came across some PDEs with no references nor their names. $$-\Delta u + \int_\Omega udx = f\qquad \hbox{in $\Omega$} \label{1}\tag{Eq1}$$ The above equation can be augmented either with Dirichlet boundary condition $u=g$ on $\partial\Omega$ or with Neumann boundary condition $\partial_n u=g$ on $\partial\Omega$. I would like to know the name and the application of this type of problem. Or any good reference to this. A slightly more general setting would be considering $$-\Delta u + \varphi\Big(\int_\Omega udx\Big) = f\qquad \hbox{in $\Omega$} \label{2}\tag{Eq2}$$ For an appropriate function $\varphi$. I have the feeling this must have a good application that because assuming, for instance, $\varphi=0$ and \eqref{2} is augmented with the Neumann boundary condition $\partial_n u=g$ on $\partial\Omega$ then it becomes the classical Neumann problem whose solutions exist if and only if we have the compatibility condition $$ \int_\Omega fdx+ \int_{\partial\Omega} gdx =0.\label{c}\tag{C}$$ In practice, $g$ is the flux term, $f$ is the source term. I forgot the meaning of the compatibility condition \eqref{c}. Question 1: What is the name and application of the problem \eqref{1} or \eqref{2}. Or what are references to this problem? Another problem is the following Dirichlet problem with $f\in L^p(\Omega)$ $1 TITLE: Is this operator bounded? QUESTION [7 upvotes]: Let $T$ be an invertible positive operator and $S$ be another positive operator on a complex Hilbert space. We then study $$ \Vert (T+S)^{-1/2}T(T+S)^{-1/2}\Vert$$ I would assume that this norm is bounded by one. But I fail to see how one could actually show this? Cause the definition of the square root using the functional calculus is rather abstract. REPLY [13 votes]: Denote $Q=(T+S)^{-1/2}T(T+S)^{-1/2}$. The inequality $\|Q\|\leqslant 1$ is equivalent to $\langle Qx,x\rangle\leqslant \langle x,x\rangle$ for all vectors $x$. Denote $(T+S)^{-1/2}x=y$, we get $$\langle Qx,x\rangle=\langle (T+S)^{-1/2}Ty,x\rangle=\langle Ty,(T+S)^{-1/2}x\rangle=\langle Ty,y\rangle\leqslant \langle (T+S)y,y\rangle\\= \langle (T+S)^{1/2}y,(T+S)^{1/2}y\rangle=\langle x,x\rangle.$$<|endoftext|> TITLE: A weak form of countable choice QUESTION [8 upvotes]: Let $\Omega$ be the set/type of truth values. We're using constructive logic. Define $AC_{0, 0} = \forall P : \mathbb{N}^2 \to \Omega, (\forall n \in \mathbb{N}, \exists m \in \mathbb{N}, P(n, m)) \to \exists f : \mathbb{N} \to \mathbb{N}, \forall n \in \mathbb{N}, P(n, f(n))$. It is well-known that $AC_{0, 0}$ is sufficient to prove that the Cauchy and Dedekind reals coincide. I have determined that a weaker form of this axiom also suffices. $AC_{weak} = \forall P : \mathbb{N} \times 2 \to \Omega, (\forall n \in \mathbb{N}, \exists m \in 2, P(n, m)) \to \exists f : \mathbb{N} \to 2, \forall n \in \mathbb{N}, P(n, f(n))$ Or to put it another way, $\forall P : \mathbb{N} \to \Omega, \forall Q : \mathbb{N} \to \Omega, (\forall n \in \mathbb{N}, P(n) \lor Q(n)) \to \exists f : \mathbb{N} \to 2, \forall n \in \mathbb{N}, (f(n) = 0 \to P(n)) \land (f(n) = 1 \to Q(n))$ Why does $AC_{weak}$ suffice? Consider the locatedness axiom for a Dedekind cut $(L, U)$, which states $\forall a, b \in \mathbb{Q}, a < b \to (a \in L \lor b \in U)$. Clearly, $S = \{(a, b) \in \mathbb{Q}^2 : a < b\}$ is a decidable infinite subset of $\mathbb{Q}$; hence, it can be put into bijection with $\mathbb{N}$. Then by $AC_{weak}$, we will have a choice function $f : S \to 2$ such that $f(a, b) = 0$ implies $a \in L$, and $f(a, b) = 1$ implies $b \in U$. This allows us to use the classic "trisect the interval" trick to get a Cauchy sequence. My primary question is this. Does $AC_{weak}$ imply $AC_{0, 0}$? My intuition says no. It's clear that $AC_{weak}$ easily allows us to prove something similar about predicates $P : \mathbb{N} \times k \to \Omega$, where $k$ is any finite set. This can be shown by induction. But extending to the case where $k$ is infinite doesn't seem possible. If $AC_{weak}$ does not imply $AC_{0, 0}$, does anyone know of a topos in which $AC_{weak}$ holds but not $AC_{0, 0}$? REPLY [6 votes]: It turns out, I believe, that there's actually a fairly simple counterexample. The example is sheaves on the topological space given by the product of countably many copies of $\mathbb{N}$ with downwards closed set topology. Explicitly, the underlying set of the space is $\mathbb{N}^\mathbb{N}$ and a set $U \subset \mathbb{N}^\mathbb{N}$ is open when it is downwards closed (according to the pointwise ordering) and for every $f \in U$ there exists $n \in \mathbb{N}$ such that $g \in U$ whenever $g|_n = f|_n$. For each open set $U \subseteq \mathbb{N}^\mathbb{N}$ and each $f \in U$, write $U_f$ for the open neighbourhood defined as below, where $n$ is least ensuring that $U_f \subseteq U$. $$U_f := \{ g \in \mathbb{N}^\mathbb{N} \;|\; g(i) \leq f(i) \text{ for } i \leq n \}$$ Observe that the space is locally connected in the very strong sense that any inhabited open set contains the function $\lambda n.0$, so any two inhabited open sets have an inhabited intersection. We also have the following lemma. Lemma Whenever $U_f$, as above, is the union of two open subsets $U_f = V \cup W$, either $V = U_f$, or $W = U_f$. Proof We show the contrapositive. Suppose $g \in U_f \setminus V$ and $h \in U_f \setminus W$. Note $U_f$ is closed under binary joint (wrt the pointwise order), and so $g \vee h \in U_f \setminus (V \cup W)$. We first check that $\mathbf{AC}_{\mathbb{N}, 2}$ holds in sheaves. Suppose $R : \mathbb{N} \times 2 \to \Omega$ is a total relation over an open set $U$. Let $f \in U$. Then restricting $R$ to $U_f$, for each $n \in \mathbb{N}$ we have $U_f = [[ R(n, 0) ]] \cup [[ R(n, 1) ]]$. Hence by the lemma we have either $[[ R(n, 0) ]] = U_f$ or $[[ R(n, 1) ]] = U_f$. Applying countable choice externally gives us a function $c : \mathbb{N} \to 2$ such that $[[ R(n, c(n)) ]] = U_f$, which we can then use to give an internal function over $U_f$. However, $U$ is covered by open sets of the form $U_f$, so we have choice functions everywhere in $U$, as required. We now show that $\mathbf{AC}_{\mathbb{N}, \mathbb{N}}$ does not hold in the topos. Define $R : \mathbb{N} \times \mathbb{N} \to \Omega$ so that $[[R(n, m)]] = \{ f : \mathbb{N} \to \mathbb{N} \;|\; f(i) \leq m \text{ for } i \leq n \}$. Observe that we do have $\bigcup_{m \in \mathbb{N}} [[R(n, m)]] = \mathbb{N}^\mathbb{N}$ for each $n$, and so this does give a total relation in the topos. Write $d$ for the identity function $\mathbb{N} \to \mathbb{N}$. We will show that there is no open neighbourhood $V$ of $d$ with a choice function for $R$ defined everywhere on $V$. By restricting to $V_d$, we may assume without loss of generality that $V$ is of the form $\{ g \in \mathbb{N}^\mathbb{N} \;|\; g(i) \leq i \text{ for } i \leq n \}$ for some $n$. Suppose that $C$ is a choice function defined on $V$ internally in the topos. Since $V$ is connected, we in fact have an external underlying function $c : \mathbb{N} \to \mathbb{N}$ such that for all $i$ we have $[[ R(i, c(i)) ]] = V$. In particular we have $[[ R(n + 1, c(n + 1)) ]] = V$ for the $n$ above. However, this gives a contradiction, since $[[ R(n + 1, c(n + 1)) ]]$ does not contain the element $g$ of $V$ defined below: $$g(i) := \begin{cases} i & i \leq n \\ c(n + 1) + 1 & \text{otherwise} \end{cases} $$ Hence we have confirmed the topos of sheaves on $\mathbb{N}^\mathbb{N}$ satisfies $\mathbf{AC}_{\mathbb{N}, 2}$ but not $\mathbf{AC}_{\mathbb{N}, \mathbb{N}}$. Finally, regarding the implication $\mathbf{AC}_{\mathbb{N}, 2} \Rightarrow \mathbb{R}_c = \mathbb{R}_d$, I'll say the same as Andrej Bauer: I've seen the result before, and I think it's fairly well known, but I can't point to anywhere specific in the literature where it is mentioned.<|endoftext|> TITLE: Ordering preference for two zero mean Gaussian outcomes QUESTION [6 upvotes]: Let $X\sim \mathcal{N}(0,1)$ be a standard Gaussian random variable. If we let $f_a(x)\triangleq\mathbb{E}[\max\{aX,x\}]$ for $a,x >0$, how to prove that $$f_a(f_b(1))2\,. $$ Now $e^7>1; 1+0.7=1.7; 1+0.7+\frac 12 0.49=1.7+0.245=1.945$, so we lack just $0.055$ for the next term. It is $0.49\cdot 0.7/6=0.049+0.049/6>0.049+0.048/6=0.049+0.008=0.057$, which is good for us. The whole computation can be verified in one's head. Note that I switched several times from what has been already gained to what is left to gain once I computed the "easy" pieces and simplified afterwards. That is a normal feature of the human bookkeeping: get what you can easily get and when comparing the remaining parts, make the inequality cruder but simpler. Dividing $49$ by $6$ in decimals would take forever, so we have to resort to truncation again and if so, why not to start with the natural truncation that produces the whole quotient? If it is insufficient look at what is left, add one digit a time, never rush to the entire series unless you can compute it and try to simplify the inequality if it doesn't look obvious yet. It may all sound idiotic now, but when we come to estimating some non-elementary integrals, the technique will be exactly the same, so I'm dwelling on it now on an example that can be explained to a 5-year old. The next one is a bit harder but I'll go over it faster. Since you know what to look for now, it will be the same technique. Inequality 2 $$ 3.14<\pi<3.2 $$ We shall use Machin's representation $\pi=16\arctan \frac 15-4\arctan \frac1{239}$. One can have an interesting discussion on how a CAS could come up with it and choose it among other possible representations, but let's skip it for now. The upper bound is now immediate for human verification: $16\arctan \frac 15<\frac{16}5=3.2$. The point is to get the lower bound. We'll truncate the power series for $\arctan\frac 15$ to $\frac 15-\frac 1{3\cdot 5^3}$ and bound $\arctan\frac 1{239}$ by $\frac{4}{239}$. We thus need to check that $$ \frac{16}{3\cdot 5^3}+\frac 4{239}<0.06 $$ or $$ \frac{8}{3\cdot 5^3}+\frac 2{239}<0.03 $$ *First attempt at a human verifiable proof: (the model of a human I'm considering cannot multiply 2-digit numbers but knows some algebra (of the type $\frac 1n=\frac 1{n+1}+\frac 1{n(n+1)}$) Note that $\frac{8}{375}=\frac 8{376}+\frac 8{375\cdot 376}= \frac 1{47}+\frac 8{375\cdot 376}<\frac 1{47}+0.0001$ ($300\times 300=90000$ already). Similarly, $\frac 2{239}=\frac{2}{240}+\frac 2{239\cdot 240}< \frac{2}{240}+\frac 2{239\cdot 240}$. Thus, we need to show that $$ \frac 1{47}+\frac 2{240}+0.0001+\frac 2{239\cdot 240}<0.03\,. $$ But $\frac 1{47}=\frac{1}{48}+\frac 1{47\cdot 48}<\frac{1}{48}+0.001$, $\frac 1{48}+\frac 2{240}=\frac 7{240}$, $\frac 7{240}+\frac 2{239\cdot 240}\le \frac 7{240}+\frac {7\cdot 2}{238\cdot 240}=\frac 7{238}=\frac 1{34}$, so we are left with $$ \frac 1{34}+0.0001<0.03\,, $$ or $\frac{3}{100}-\frac{1}{34}=\frac 2{3400}>\frac 2{20000}=0.0001$ Did you like that? Probably not, and neither did I. It is human verifiable, but it needs more than 4 short term memory registers. Not good. However this is typical of what first comes to your head when you try playing such games. Let's try again: Second attempt at a human verifiable proof: $$ \frac{8}{3\cdot 5^3}+\frac 2{239}<0.03 $$ $$ \frac{8}{5^3}+\frac 6{239}<0.09 $$ $$ \frac{8\cdot 2^3}{10^3}+\frac 6{239}<0.09 $$ $$ 0.064+\frac 6{239}<0.09 $$ $$ \frac 6{239}<0.09-0.064=0.026 $$ $$ \frac 6{240}+\frac 6{240\cdot 239}=0.0025+\frac 6{240\cdot 239}<0.026 $$ $$ \frac 6{240\cdot 239}<\frac 6{40000}<\frac 6{6000}=0.001 $$ Much better, isn't it? One short term memory register used and all multiplications/divisions are trivial. That is the whole difference between a "cumbersome" approach and an "elegant" one and the formal score function is pretty clear. Can one do even better? Perhaps. Computers can consider thousands of possible derivation chains per second and I tried just 3 or 4 in half an hour before presenting these two here. So, it would be nice to teach machines (or, at least, students) some arithmetic like that before they start handling integral inequalities. "Arithmetic for analysis" might be quite a useful course, making much more sense IMHO than the widespread "Algebra for calculus". I thanks everybody who had patience to read all this. I couldn't explain what I meant on anything simpler than inequalities for rational numbers with seemingly ugly denominators. Now we will play exactly the same game with (occasionally double) integrals and functional inequalities. I'll go faster from now on and spare you from seeing imperfect attempts (but they were there and related to the final argument as Attempt 1 to Attempt 2). Convenient renormalization Let $g(x)=E(\max X,x)$, $x>0$ where $X$ is a standard normal. The inequality we want can be rewritten as $$ ag(\tfrac{g(x)}{a})0, 0x$ plays some prominent role here, so let us rewrite it as $$ \varphi(x)+a\varphi(y+\tfrac{\varphi(x)}a)y$ such that $\varphi(z)=a\varphi(y)$. Then the inequality will become $$ \varphi(x)+a\varphi(y+\tfrac{\varphi(x)}a)<\varphi(z)+\varphi(x+\varphi(z))\, 00$ now becomes free parameter, so let us call it $T$ (small but tangible gain in the simplicity score a machine can easily detect). Also let us bring $z$'s to one side and $y$'s to the other as much as we can (another purely mechanical move). We'll get $$ \frac{\varphi(y+\tfrac{\varphi(y)}{\varphi(z)}T)}{\varphi(y)}< \frac{\varphi(z+T)}{\varphi(z)}\,. $$ Now let us use one more standard technique: elimination of trivial cases for the gain of extra assumptions. Since $\varphi(y)>\varphi(z)$, if $y+\tfrac{\varphi(y)}{\varphi(z)}T>z+T$, there is nothing to prove. Thus we may assume the contrary. Then $y+\tfrac{\varphi(y)}{\varphi(z)}t0$ (not because one can take the derivative using the quotient rule, but because $\frac {x}{x+1}=\frac 1{1+\frac 1x}$ and $x\mapsto \frac 1x$ is obviously decreasing, of course). Elementary lemma (see the post mentioned in the beginning for the proof). If $\mu$ is a positive measure on $(0,+\infty)$ and $f$ is a decreasing (increasing) function, then $$ x\mapsto \frac{\int_0^\infty f(t)e^{-xt}\,d\mu(t)}{\int_0^\infty e^{-xt}\,d\mu(t)} $$ is increasing(decreasing) in $x>0$. Is it possible to teach a machine to recognize symbolic instances of such principle (and a few others)? I see no reason why not. The necessary and sufficient condition on the family $e^{-xt}$ for this principle to hold is easy to formulate and the recognition of common factor depending on $x$ and carrying it out is a child game for the modern CAS. For a human it is the already discussed on MO principle that "every mathematician has just a few tricks" (but can learn new ones!) so one can start with a CAS that has a few tricks too (even to implement human verifiable rational arithmetic would be a step forward from what outputs we typically get from our machine partners now). We will use this lemma to show that $\frac{|\varphi'|}{\varphi}=\frac{\psi_1}{\psi}$ is increasing (immediate) and convex (will be done in the next section). Also we trivially have that $\frac{\varphi''}{|\varphi'|}=\frac 1{\psi_1}$ is increasing. Convexity of $\frac{\psi_1}{\psi}$ We have $$ \left(\frac{\psi_1}{\psi}\right)'=\frac{\psi_1'\psi-\psi_1\psi'}{\psi^2} \\ = \frac 12\frac{\iint(t-s)^2e^{-x(t+s)}e^{-(s^2+t^2)/2}\,dt\,ds} {\iint tse^{-x(t+s)}e^{-(s^2+t^2)/2}\,dt\,ds} $$ (I symmetrized the numerator integral, of course; all double integrals are over $(0,+\infty)^2$ unless specified otherwise). Switching to variables $\sigma=s+t$ and $\gamma=s-t$ and using symmetry, we can rewrite this ratio (up to a positive constant factor) as $$ \frac{\iint_{0<\gamma<\sigma}\gamma^2e^{-x\sigma}e^{-(\sigma^2+\gamma^2)/4}\,d\gamma\,d\sigma} {\iint_{0<\gamma<\sigma}(\sigma^2-\gamma^2)e^{-x\sigma}e^{-(\sigma^2+\gamma^2)/4}\,d\gamma\,d\sigma}\,. $$ For fixed $\sigma$ it makes sense to integrate over $\gamma$ using the change of variable $\gamma=\tau\sigma$, $0<\tau<1$, to get the ratio as $$ \frac{\int_0^\infty F(\sigma)e^{-x\sigma}e^{-\sigma^2/4}\,d\sigma} {\int_0^\infty G(\sigma)e^{-x\sigma}e^{-\sigma^2/4}}\,. $$ where $$ F(\sigma)=\sigma^3\int_0^1 \tau^2e^{-\tau^2\sigma^2/4}\,d\tau $$ and $$ G(\sigma)=\sigma^3\int_0^1(1-\tau^2)e^{-\tau^2\sigma^2/4}\,d\tau $$,. By our Lemma, to show that the initial ratio of integrals is increasing in $x$, it is enough to show that $\frac{F}{G}$ is decreasing in $\sigma$, which, by the lemma again (just use $\tau^2$ as a new variable and $\sigma^2$ as a new parameter) follows from the fact that $\frac{\tau^2}{1-\tau^2}$ is increasing in $\tau$. Is there a reason beyond this? For a CAS it is enough to implement the principle "check all standard properties of the function and simple combinations of its derivatives you can". The calculus students are taught the lists of those standard properties and their verification techniques all the time with some partial success. Computers, which have perfect memory and no need to spend most of their time making a living, can learn them much faster if somebody bothers to teach them, of course. For a human a reason for choosing just these properties will be clear from what goes next. We also have $\psi(0)=1, \psi_1(0)=\sqrt{\pi/2}$. The first one is an elementary integral; the second one is well-known to humans and machines alike. Now again, let us eliminate the trivial case. Note that $$ \varphi(x)-\varphi(x+\varphi(z))\le |\varphi'(x)|\varphi(z) $$ and $$ \varphi(z)-\varphi(z+\varphi(x))\ge \varphi(z)(1-e^{-Z\varphi(x)}), Z=\frac{|\varphi'(z)|}{\varphi(z)}\,. $$ Thus, if $Z\varphi(x)>\log\frac{1}{1-|\varphi'(x)|}$, the inequality holds. Since $|\varphi'(x)|$ is $\frac 12$ at $0$ and decreasing, we are fine every time $Z\varphi(x)>\frac 75|\varphi'(x)|$ (Inequality 1). Now it is time for Taylor expansions. We'll use the integral form of the remainder. It is something we teach our analysis students to do every time they see the difference with presumably small increment. $$ \varphi(x)-\varphi(x+\varphi(z))=-\varphi'(x)\varphi(z)-\varphi(z)^2\int_0^1(1-t)\varphi''(x+t\varphi(z))\,dt \\ = -\varphi'(x)\varphi(z)-\varphi(z)^2\varphi''(x)\int_0^1(1-t)e^{-x\varphi(z)t-[t\varphi(z)]^2/2}\,dt $$ Doing the same representation for $\varphi(z)-\varphi(z+\varphi(x))$ and dividing both parts by $\varphi(x)\varphi(z)$, we get the inequality $$ \frac{|\varphi'(z)|}{\varphi(z)}- \frac{|\varphi'(x)|}{\varphi(x)}\ge \frac{\varphi''(z)}{\varphi(z)}\varphi(x) \int_0^1(1-t)e^{-z\varphi(x)t-[t\varphi(x)]^2/2}\,dt- \frac{\varphi''(x)}{\varphi(x)}\varphi(z) \int_0^1(1-t)e^{-x\varphi(z)t-[t\varphi(z)]^2/2}\,dt\,. $$ Note now that $\varphi(x)>\varphi(z)$ and $z\varphi(x)>x\varphi(z)$, so we can replace the integral in the subtrahend by the (smaller) one in the minuend and thus strengthen the inequality simultaneously simplifying it. After we make the integrals common, we raise them to $$ I=\int_0^1(1-t)e^{-z\varphi(x)t}\,dt $$ just dropping the $[t\varphi(x)]^2/2$ part. Of course, the natural move would be to drop the $e^{-z\varphi(x)t}$ as well and come up with a clean neat $\frac 12$. This is exactly what I did initially, but that turned out to be going too far in the simplification game, so I had to step back and keep the integral as written for the moment. It is not that I drop and keep the terms for no apparent reason: I drop all I can drop at the first attempt and then reintroduce the dropped terms one by one starting with most "heavy" ones until I make the ends meet. Let's think of that symbol $I$ as "improvable $\frac 12$" for now. Denoting $X=\frac{|\varphi'(x)|}{\varphi(x)}$, $Z=\frac{|\varphi'(z)|}{\varphi(z)}$ and recalling the identity $\varphi''(x)=\varphi(x)+x|\varphi'(x)|$, so $\frac{\varphi''(x)}{\varphi(x)}=1+xX$, we can write the resulting inequality as $$ Z-X\ge I[(zZ+1)\varphi(x)-(xX+1)\varphi(z)]=I[(zZ-xX)\varphi(x)]+(xX+1)(\varphi(x)-\varphi(z))] \\ =I z\varphi(x)(Z-x)+I[(z-x)X\varphi(x)+(xX+1))(\varphi(x)-\varphi(z))] \,. $$ Now recall that $X\varphi(x)=|\varphi'(x)|$ and $xX+1=\frac{\varphi''(x)}{\varphi(x)}$. Also move the first term to the left, replace $I$ on the LHS by \frac 12 (but not on the RHS) and estimate $\varphi(x)-\varphi(z)\le|\varphi'(x)|(z-x)$. You'll get $$ (Z-X)(1-\tfrac 12z\varphi(x))\ge I(z-x)|\varphi'(x)|(1+\tfrac{\varphi''(x)}{\varphi'(x)}) $$ to prove. Why this form and not another? You have finite choice here with the general guidance that factorization is good. Of course, I initially kept it in just some equivalent form with the options to replace some expressions by identically equal ones. It took a couple of hours to finalize on this particular one but it was just routine trial and error, nothing ingenious. Recall now that $x\mapsto \frac{|\varphi'(x)|}{\varphi(x)}=\frac{\psi_1(x)}{\psi(x)}$ is increasing convex. At $x=0$ its derivative is $\frac{\pi}2-1$. At $+\infty$ you can easily check by, say, Laplace asymptotic formula (though more elementary ways are also available) that $\psi_1(x)\sim 1/x$ and $\psi(x)\sim 1/x^2$ at $+\infty$, so the ratio is asymptotic to $x$, which for a convex increasing function is possible only if its derivative is asymptotic to $1$. Thus $$ \frac\pi 2-1\le \left(\frac{|\varphi'(x)}{\varphi(x)|}\right)'\le 1. $$ Why? Again the same principle: if you don't know what to do next, figure out a few things that look useful. Why is that one useful? For a human it is obvious: the difference ratio $\frac{Z-X}{z-x}$ cries for the MVT so loudly that you can hear that cry near Jupiter, if not Betelgeuse. And that requires the bounds for the derivative we have just done. Now, plugging the lower bound for the derivative on the LHS (and using the MVT, of course), we reduce the task to proving that $$ (\tfrac \pi 2-1)(1-\tfrac 12 z\varphi(x))\ge I|\varphi'(x)|(1+\tfrac{\varphi''(x)}{\varphi(x)}) $$ We want now to investigate the product of the last two factors on the RHS. It is a function of $x$ that rather obviously quickly decays at $\infty$. So, faithful to our principle "Investigate all functions you meet for standard properties" let's see if it is decreasing all the way through. Taking its derivative, we get $$ -\varphi''(x)+\varphi'''(x)\tfrac{|\varphi'(x)|}{\varphi(x)}+\varphi''(x)(\tfrac{|\varphi'(x)|}{\varphi(x)})'. $$ But $\varphi'''(x)<0$ and $(\tfrac{|\varphi'(x)|}{\varphi(x)})'\le 1$ as we figured out short time ago. So yeah, it is decreasing. The value at $0$ is $\frac 12(1+1)=1$, so we can just drop this factor altogether. We are now left with $$ (\tfrac \pi 2-1)(1-\tfrac 12 z\varphi(x))\ge I $$ Clearly, at $z=0$ we have the LHS equal to $\tfrac \pi 2-1>0.57$ (Inequality 2). Thus if the inequality fails, $z\varphi(x)$ should be not too small, which means that $Z\varphi(x)$ should be even larger and we can hope that it will then beat $\frac 75|\varphi'(x)|$ throwing us into the domain when the inequality is trivial. Let's try to implement that strategy (it is not guaranteed to succeed and it actually won't if we just replace $I$ by $\frac 12$ on the RHS but, fortunately, keeping $z\varphi(x)$ in $I$ helps a lot and, even more fortunately, we have exactly the same $z\varphi(x)$ on the LHS). Let's denote it by $\rho$. In the bad case we have $$ 0.57(1-\tfrac\rho 2)<\int_0^1(1-t)e^{-\rho t}\,dt\,. $$ Tayloring the RHS to the quadratic terms, which is a clear overestimate (it should be obvious at least for $\rho<1$, which is enough for our purposes and is machine detectable), we get $$ RHS\le \frac 12-\frac\rho 6+\frac{\rho^2}{24}. $$ Now the rate at which the RHS decays is at most $\frac 16$ while the rate at the LHS is $>\frac 14$, so the curves cross just once. We will show that the crossing point is beyond $\rho=\frac 12$. Why? Here the AI shines: it can easily find the root of the quadratic equation (or even the original transceendental one) with high precision. However, IMHO, it should keep that root to itself and present a human verifiable fraction instead. Finding that fraction is not hard either. I don't think anybody will argue that it is difficult to program. However, it has never been programmed and we still get human unverifiable statements like $\rho>0.12345678910$ from our computers when we ask them to prove our numeric inequalities. They can just well output "(out 15) true" and some people start believing that it is all we can possibly hope for. I don't think so. Thus we need to check that $0.57(1-\frac 14)>\frac 12-\frac{1}{6\cdot 2}+\frac 1{24\cdot 2^2}$. Multiplying by $24\cdot 2^2$, we get $$ 0.57\cdot 3\cdot 24> 48-8+1\,, $$ i.e. $$ 1.14\cdot 36>41 $$ $$ 0.14\cdot 36=3.6+1.44=5.04>5 $$ finishing the proof. Thus $\rho>\frac 12$. Since $\rho=z\varphi(x)$, we have $$ Z\varphi(x)\ge \sqrt{\tfrac \pi 2}\varphi(x)+0.57z\varphi(x)\ge \sqrt{\tfrac \pi 2}\varphi(x)+0.285 $$ (I estimated $Z$ by its value at zero plus the minimum of the derivative times $z$) This has to beat $\frac 75|\varphi'(x)|$. Thus we need to show that $$ \max_{x>0}(\tfrac 75|\varphi'(x)|-\sqrt{\tfrac \pi 2}\varphi(x))=\frac 75\max_{x>0}(|\varphi'(x)|-\tfrac 57\sqrt{\tfrac \pi 2}\varphi(x))\le 0.285\,. $$ Now, $\frac 57\sqrt{\tfrac \pi 2}>\sqrt{\frac 2\pi}$ (this is equivalent to $\frac\pi 2>\frac 75$, i.e., $\pi>\frac{14}{5}=2.8$), so it is enough to show that $$ \frac 75\max_{x>0}(|\varphi'(x)|-\sqrt{\tfrac 2 \pi}\varphi(x))\le 0.285\,. $$ However the function under the maximum sign now has $0$ derivative at $0$ and, since $\frac{\varphi''(x)}{|\varphi'(x)|}$ is increasing, negative derivative for all $x>0$. Thus this maximum is at $x=0$ where the value of the LHS is $\frac 75(\frac 12-\frac 1\pi)<\frac 75(\frac 12-\frac 1{3.2})$. Thus we are left with $$ \frac 75\left(\frac 12-\frac{10}{32}\right)=\frac 75\frac{6}{32}=\frac{42}{160} \\ =\frac{40}{160}+\frac 2{160}<0.25+0.02=0.27<0.285. $$ The end. So now I showed the complete game. It is akin to chess: at each step you have finitely many moves and just try to simplify, simplify, and simplify with the score function that I tried to present and even occasionally formalize. Like in the chess I do not see the final winning position until I nearly arrive there; I just try to improve the current one a little bit. If you want to understand why I made each move the way I did, just ask yourself what moves you would consider instead. It is not a rhetorical question: each position I passed through had several legitimate moves. I didn't show them except on the simple rational arithmetic example because this post is already too long to the extent of possibly being "inappropriate for MO", but if you want to really understand the strategy, you should find those alternative moves yourself in this particular game, see what positions they lead to, and compare the resulting positions to my resulting positions and the positions before the move. Then, if you feel like the move is disadvantageous even compared to the original position (or advantageous even compared to mine), try to discern the criteria you are using for such evaluation and formalize them. Mine are embarrasingly simple: the total formula length, the number of different symbols, the number of occurrences of each symbol, the possibility to factor, etc. Most of the time I sacrifice precision, i.e., make the inequality stronger or more general, but if I can gain simplification with an identity, I certainly play it first. This particular game took 7 evenings (mainly due to considering the alternative moves that led nowhere). An intelligent AI would spend a few milliseconds on it. How much you spend on its full analysis depends on your basic skills. But I want to emphasize that there is nothing "amazing" in it and if you find the argument reasonably elegant, I showed where this elegance comes from by comparing Attempt 1 and Attempt 2 for rational arithmetic problem. The remarks of the kind "I have absolutely no idea how you came up with this!" please my ego of course, but also give me deep frustration because they mean that I failed to communicate the underlying ideas entirely and that instead of enabling the reader to prove his next elementary inequality by himself, I just pulled out one more rabbit from the hat. This game is really simple enough to learn and even to "communicate to computers". The price of this communication is as usual: time and effort. The ultimate gain is machines that can produce easily verifiable arguments instead of thousands of megabytes of incomprehensible gibberish that are currently called "computer assisted proofs". Not every such game will be winnable, of course. But not every true statement can be proved either. Whether the gain is worth the price is for everyone to decide by yourself. The life is short and everybody has his or her only priorities. I just tried to convince you that this game is close enough to chess to use the same general techniques in programming. Of course, if you understand and can play it better than I, I will be happy to hear your opinions about it. Your techniques may easily be superior to mine. Just show them on an example of a real proof construction for an unknown problem like I did and we'll compare both the results and the techniques. At last one may ask whether one should learn to play such games at the expense of the time he could do "serious research" instead. I don't think there is anything "serious" enough in this life to justify the total abandonment of the pleasure of playing but if you prefer to be a "mathematical samurai" I see no problem with that either. I'll leave it to other people to judge how much (or whether) I'm capable of the so called "serious research". I just want to say that for me the only two tangible distinctions between different mathematical problems are whether I can understand their statements or not and whether I can solve them or not and that I employ in "serious research" pretty much the same techniques like father Brown from Chesterton novels used exactly the same method to catch thieves and to talk to angels. I just cannot explain them on examples of that level. I honestly tried. The message did not go through. Let's see if it will at the level of elementary arithmetic and functional inequalities for explicit functions. Thanks to all who had patience to read up to this point!<|endoftext|> TITLE: Are locally compact, Hausdorff, locally path-connected topological groups locally Euclidean? QUESTION [9 upvotes]: Is every locally compact, Hausdorff, locally path-connected topological group $G$ locally Euclidean? (That would imply of course also being a Lie group.) Is it true when countable basis is assumed? I wasn't able to find a discussion of this question in the literature on topological groups and the Hilbert 5th problem. EDIT: As YCor rightly pointed out, let's assume that $G$ is of finite topological dimension. The general theme of my question is to identify a minimal set of conditions on a topological group making it into a Lie group. REPLY [2 votes]: This is an addendum to Pedro's answer: Theorem 1. Suppose that $G$ is a locally compact finite dimensional (Hausdorff) topological group. Then $G$ is locally homeomorphic to the product of a totally disconnected space and ${\mathbb R}^n$. A reference for this result (that should be better known) is Montgomery, Deane; Zippin, Leo, Topological transformation groups, Mineola, NY: Dover Publications (ISBN 978-0-486-82449-9). xi, 289 p. (2018). ZBL1418.57024. section 4.9.3. (The actual theorem says a bit more, namely that locally, at $e$, $G$ is a product of a local Lie group and a totally disconnected group. In the same book you will also find Theorem 4.10.1 answering your original question: Theorem 2. If $G$ is a locally compact, locally connected and finite dimensional topological group, then $G$ is a Lie group (possibly non-2nd countable). Of course, this is a consequence of Theorem 1.<|endoftext|> TITLE: Putting sheaves to work for algebraic topology? QUESTION [6 upvotes]: This is cross-posted from math.se after receiving points and no answers. I apologise if this question is too basic for MathOverflow. I'm refreshing my memory of covering space theory, and this time around, I know some sheaf theory. It feels like arguments are used to prove results about covering spaces, such as uniqueness of lifts, having something "sheafy" about them. For example, to prove uniqueness of lifts, we argue by trying to extend "equality at a point" to "equality over a neighbourhood" to "equality over the entire domain". It seems like the language of sheaves may make this clearer? Similarly, when it comes to covering spaces, there is something "etale-like" about them. Is there a reference that expands on this perspective? REPLY [11 votes]: For sufficiently nice topological spaces $X$ (e.g., locally connected for the last two categories to be equivalent, and semilocally simply connected and locally path-connected for all three to be equivalent), the following three categories are equivalent: Functors from the fundamental groupoid of $X$ to the category of sets; Covering spaces over $X$; Locally constant sheaves of sets on $X$. This is an extremely primitive baby version of the Riemann–Hilbert correspondence. References specifically for this elementary case are sparse, but there is an extensive discussion on locally constant sheaves at the nCafé.<|endoftext|> TITLE: About locally compact groups without compact subgroups QUESTION [8 upvotes]: Is every Hausdorff, locally compact group that does not contain any non-trivial compact group, finitely dimensional? REPLY [12 votes]: Yes, it's even a Lie group whose unit component is a semidirect product $R\rtimes S^n$, where $R$ is a simply connected solvable Lie group and $S$ is the universal covering of $\mathrm{SL}_2(\mathbf{R})$. Indeed, by van Dantzig, every locally compact group $G$ has an open subgroup $U$ such that $U/U^\circ$ is compact. By the solution to Hilbert's fifth problem, $U$ has a compact normal subgroup $W$ such that $U/W$ is Lie (necessarily with finitely many components). Hence, in the current setting, $W=1$, so $U$ is Lie and $U^\circ=G^\circ$ is open in $G$. Every connected Lie group with maximal compact subgroup $K$ is homeomorphic to $\mathbf{R}^d\times K$ for some $d$ (Iwasawa). Hence, $G^0$ is homeomorphic to $\mathbf{R}^d$ for some $d$. Being simply connected, it is semidirect product $R\rtimes T$ with $R$ its radical, necessarily simply connected, and $T$ a semisimple Levi factor, simply connected and hence $T$ is a direct product of simple simply connected Lie group. The only possibility for such simple factor to be contractible is indeed the 3-dimensional $S$. Added: here's a characterization. The contractible Lie groups are characterized above. Proposition. A locally compact group $G$ has no nontrivial compact subgroup iff $G$ is Lie (i.e., $G^0$ is Lie and open) and the discrete quotient $G/G^0$ is torsion-free. Proof: clearly, these are necessary conditions. Conversely, suppose that $G$ has no nontrivial compact subgroup. By the above, $G^0$ is open and is a contractible Lie group. We claim that $G/G^0$ is torsion-free (note that this is not so immediate since the surjection $G\to G/G^0$ may not be split). Otherwise, let $F$ be a nontrivial finite subgroup of $G/G^0$, and let $H$ be its inverse image in $G$: this is an open subgroup of $G$, which is Lie, virtually connected but not connected. A result of Mostow (valid in any virtually connected Lie group $H$) is that $H^0K=H$ for some compact subgroup $K$ of $H$. So $K=\{1\}$. Hence $H^0=H$, thus $H$ is connected, contradiction.<|endoftext|> TITLE: Superharmonicity at infinity QUESTION [5 upvotes]: Some authors define superharmonicity at infinity in the following way. A function $u$ is superharmonic on an open set $V\subset\mathbb{R}^m\cup\{\infty\}$ (one point compactification), containing infinity, if it is superharmonic on $V\setminus\{\infty\}$ in the regular way, and at infinity, $u$ is lower semicontinuous and $u(\infty)$ is bounded below by the average of $u$ over any sphere $S(y,r)$, if the complement of the ball $B(y,r)$ lies in $V$ (see Introduction to potential theory, by Helms, pg. 198). Some other authors give a different definition. In this new definition, a function $u$ is superharmonic on $V$ if it is superharmonic on $V\setminus\{\infty\}$ and if the Kelvin transform $u^*$ of $u$, that is already superharmonic on $V^*\setminus\{\infty\}$, can be defined at $0$ so that it is superharmonic on $V$ (see Classical Potential Theory, by Papadimitrakis, available freely on the web, pg. 154). Here, $$u^*(x)=\left(\frac{1}{|x|}\right)^{m-2}u(x^*)$$ and $x^*=\frac{1}{|x|^2}x.$ It seems that these two defintions are not equivalent for $m\geq3$, because constant functions are superharmonic everywher according to the first definition but according to the second definition the only constant function that is superharmonic is $\equiv0$. My questions are: suppose $m\geq3$. Are there some conditions under which these two definitions are equivalent? Does the second definition implies the first? REPLY [2 votes]: None of the definitions implies the other: The function $|x|^{2-m}$ is superharmonic (in fact: harmonic) in $\mathbb R^m \cup \{\infty\} \setminus \{0\}$ according to the second definition, but it is not according to the first one. Conversely, the function $-1$ is superharmonic (in fact: harmonic) in $\mathbb R^m \cup \{\infty\} \setminus \{0\}$ in the sense of the first definition and it is not according to the second one. If we assume that $u \geqslant 0$, then every $u$ superharmonic in $\mathbb R^m \setminus K$ for a compact $K$ is superharmonic in $\mathbb R^m \cup \{\infty\} \setminus K$ according to the second definition, so for positive superharmonic function, the first definition clearly implies the second one (but not vice versa). I believe things become clearer if one writes both definitions in terms of the Kelvin transform $u^*(x) = |x|^{2 - m} u(|x|^{-2} x)$. The second definition requires that $u^*$ is superharmonic in a neighbourhood of $0$, so in a sufficiently small ball $B_r$ we have $$ u^*(x) = \int_{B_r \setminus \{0\}} (|x - y|^{2 - m} - |y|^{2 - m}) \mu(dy) + (h(x) - h(0)) + a |x|^{2 - m} + b$$ for some $\mu \geqslant 0$, some $h$ harmonic in $B_r$, some $a \geqslant 0$ and some $b \in \mathbb R$. On the other hand, the first definition requires that $u^*$ is superharmonic in a punctured neighbourhood of $0$, and $|x|^{m - 2} u^*(x)$ satisfies the "super-mean-value property at $0$", so in a sufficiently small ball $B_r$ we should have $$ u^*(x) = \int_{B_r \setminus \{0\}} (|x - y|^{2 - m} - |y|^{2 - m}) \mu(dy) + (h(x) - h(0)) + a |x|^{2 - m} + b$$ for some $\mu \geqslant 0$, some $h$ harmonic in $B_r$, some $a \in \mathbb R$ and some $b \leqslant 0$. Note: I might be wrong here, I did not have time to think about this carefully. The difference is of course in the admissible range of $a$ and $b$. Edit: Some additional comments to the above item 3. (A) If $u^*$ is superharmonic in a ball $B_R$, then in a smaller ball $B_r$ we have $$ u^*(x) = \int_{B_r} |x - y|^{2 - m} + h(x) $$ for some harmonic $h$; this is the Riesz decomposition theorem. If we take out the atom at $0$, then we can write $$ u^*(x) = \int_{B_r \setminus \{0\}} |x - y|^{2 - m} \mu(dy) + h(x) + a |x|^{2 - m} $$ for some $a \geqslant 0$, and this is equivalent to what I wrote above. (B) Suppose now that $u^*$ is the Kelvin transform of a function superharmonic at infinity in the sense of the first definition. Then $u^*$ is superharmonic in $B_R \setminus \{0\}$, and since $u$ is bounded from below by a constant $-M$ in a neighbourhood of $\infty$, we have $$ u^*(x) + M |x|^{2 - m} \ge 0 $$ in some ball $B_R$. This implies that $u^*(x) + M |x|^{2 - m}$ is superharmonic in $B_R$, not just $B_R \setminus \{0\}$. By the same argument as in (A) we find that $$ u^*(x) = \int_{B_r \setminus \{0\}} |x - y|^{2 - m} \mu(dy) + h(x) + a |x|^{2 - m} $$ for some harmonic $h$, but this time $a$ need not be nonnegative (we only know that $a \ge -M$, but $M$ can be arbitrarily large). This leads us to the expression given in the original answer, with an arbitrary real $b$. Now why in fact we need $b \leqslant 0$? I did not check this carefully, but the average of $|x|^{m - 2} u^*(x)$ over a small ball $B_s$ seems to be equal to $a + c_m b s^{m - 2} + o(s^{m - 2})$ for an appropriate constant $c_m > 0$, and this is no greater than the limit $a$ only if $b \leqslant 0$. The above shows that $b \leqslant 0$ is necessary. The next question is whether is is also sufficient. I guess it is, and this should be fairly easy to verify, but I have to stop here for the time being.<|endoftext|> TITLE: 'Fattest' polygons based on diameter and 'least width' QUESTION [8 upvotes]: Definitions: The diameter of a convex region is the greatest distance between any pair of points in the region. The least width of a $2$D convex region can be defined as the least distance between any pair of parallel lines that touch the region. Let us refer to a polygon as 'fat' if the ratio between its diameter and least width (call this the 'thinness ratio') is low. General Question: Given a general $n$, which $n$-gon is the fattest? For which values of $n$ (if at all) is the fattest n-gon a regular $n$-gon? 2 Special cases: -For $n =3$, the equilateral triangle is the fattest. The thinness ratio is $2/\sqrt(3) \approx 1.15$. -For $n = 4$, the square is not the fattest - its thinness ratio is $\sqrt{2}$. As pointed out by Yakov Baruch below, if we form a quad by adding a vertex arbitrarily close to one of the vertices of an equilateral triangle, its thinness ratio is only about $1.15$. Some Further Queries: Thanks very much to Prof. O'Rourke for the pointer to Bezdek and Fodor's work (answer below) that answers just the above question. One could ask a further question: Given values for area and perimeter, which convex shape (not necessarily polygonal) is the most/least fat? For fixed $A$ and $P$ within a suitable range (from perimeter equal to that of circle of area $A$ to perimeter of a Reuleux triangle with area $A$), it appears that fatness is maximized by curves of constant width (https://nandacumar.blogspot.com/2012/11/maximizing-and-minimizing-diameter-ii.html). For $P$ values greater than this range, I don't know. For specified $A$ and $P$, diameter is maximized by a 'convex lens' shape as given in above linked page. But this shape does not seem to minimize fatness. REPLY [7 votes]: The paper Audet, Charles, Pierre Hansen, and Frédéric Messine. "Extremal problems for convex polygons-an update." J Global Optimization 55 (2009): 1-16. Springer link. discusses this problem, which I believe is equivalent to your question: Find for all $n$ which convex polygon with unit diameter has the largest width. They cite Bezdek and Fodor for these results: (a) The maximum width satisfies $W_n \le \cos \frac{\pi}{2n}$ for $n \ge 3$ and equality holds if $n$ has an odd divisor greater than $1$. (b) If $n$ has an odd prime divisor then a polygon $V_n$ is extremal if and only if it is equilateral and is inscribed in a Reuleaux polygon of constant width $1$, so that the vertices of the Reuleaux polygon are also vertices of $V_n$. (c) Results are also obtained for quadrilaterals: $W_4 = \sqrt{3}/2 \; (=W_3)$. Moreover, all extreme quadrilaterals have the property that three of their vertices form a regular triangle and the fourth vertex is contained in the Reuleaux triangle determined by the three vertices. A. Bezdek and F. Fodor, On convex polygons of maximal width, Archiv der Mathematik, Vol. 74, No. 1, pp. 75–80, 2000.<|endoftext|> TITLE: Constructing the inverse of a braiding in a braided pivotal category QUESTION [5 upvotes]: Assume we have a braided pivotal monoidal category. This means we assume the braiding $c$ to be a natural isomorphism. But looking at the corresponding string diagram, it seems to me as if we could obtain $c^{-1}$ as the composition of the following morphisms (treating the monoidal structure as strict): $$ A \otimes B \xrightarrow{\eta_{B^{*}} \otimes A \otimes B} B^{**} \otimes B^{*} \otimes A \otimes B \xrightarrow{B^{**} \otimes c_{A, B^{*}} \otimes B} B^{**} \otimes A \otimes B^{*} \otimes B \xrightarrow{B^{**} \otimes A \otimes \varepsilon_{B}} B^{**} \otimes A = B \otimes A $$ As a string diagram, this construction would look like this: This begs the questions: Is it enough to assume the braiding as a morphism instead of assuming it to be an iso? When working 2-categorically, we now have a non-trivial 2-cell at $c^{-1}$, should this be filled to make the notion well-behaved? REPLY [6 votes]: As Tim says this should perhaps be an answer: yes this is true, and doesn't require pivotal. As long as the category is right rigid, the braiding is automatically invertible. See e.g. Prop 1.3 in Day, Panchadcharam, and Street's "Lax Braidings and the Lax Centre" (http://science.mq.edu.au/~street/laxcentre.pdf).<|endoftext|> TITLE: Parabolics and simple roots for a special unitary group: reference request QUESTION [8 upvotes]: I am looking for a reference where the relative root system, the relative system of simple roots, and parabolic $\Bbb R$-subgroups for the real algebraic group ${\rm SU}(p,q)$ are explicitly computed. More generally, let $F$ be a field of characteristic 0, $L/F$ a quadratic extension, and $H$ be an $L/F$-Hermitian form on $L^n$. Write $G={\rm SU}(L^n,H)$, which is a semisimple $F$-group. I am looking for a reference where the relative system of simple roots and the conjugacy classes of $F$-parabolics in $G$ are explicitly computed. Here "relative" means with respect to a maximal split $F$-torus. The case of a special orthogonal group is treated in Borel's 1966 paper "Linear algebraic groups" in the Boulder proceedings "Algebraic Groups and Discontinuous Subgroups", Proc. Sympos Pure Math. vol. 9. Borel writes that for an orthogonal group, the $F$-paraboics are the stabilizers of the $F$-rational isotropic flags. Question. Is it true for $G={\rm SU}(L^n,H)$ that the $F$-parabolics in $G$ are the stabilizers of the $F$-rational isotropic flags in $(L^n, H)$ ? REPLY [2 votes]: A slightly off-target answer, but possibly of some use. First, indeed, there is a large population who deliberately take the viewpoint, attributed to Harish-Chandra, that everything should be done for arbitrary reductive/semi-simple groups, not just "examples". Thus, as @LSpice's comment, "there's just one reductive group, $G$". :) But, with some substance for the theory of integral representations of automorphic $L$-functions, the obvious/natural relations of classical groups prove non-trivial theorems... that do not have "intrinsic" analogues, in any useful sense, apparently. So! "The classical groups". Over $\mathbb C$, there are really just $3$: general or special linear, orthogonal (nevermind the parity of dimension), and symplectic. Over $\mathbb R$ we have the subdivision... Over $\mathbb R$, we can have Sylvester's Inertia theorems for the real forms that are described by "signatures", the simplest ones being $O(p,q)$, $U(p,q)$, $Sp^*(p,q)$. Yes, modeled by reals, complex, and quaternions. In light of Weil's "Classical groups/algebras with involutions", this is not a coincidence... So, operationally, there are two types of classical groups: general-linear, and isometry (or similitude). In general-linear, parabolics are stabilizers of flags. In isometry... groups, parabolics are stabilizers of totally isotropic flags. Choice of Levi component is equivalent to choice of complementary isotropic flag. One can continue, as desired, to be able to re-specify "roots", etc. Witt's theorem(s) show that there is only one isomorphism class of choices... The "physicality" of Inertia Theorems as a part of "classification over $\mathbb R$" is (to my mind) use of inequalities, the intermediate value theorem, and such. Yes, this is all subsumed by the less-"physical" Witt-theorem business, but some of it is (to my mind) much easier to believe. Yes, for example, Witt's Thm assures us that the leftover, after removing $W\oplus W'$ for maximal totally isotropic $W$ and $W'$ complementing each other, is independent of choices. The latter bit is perhaps the easiest thing to talk about from a classical-groups viewpoint, but a bit clumsy (so far as I can understand) instrinsically. Anisotropic groups in the Levi component(s) of minimal ($\mathbb R$-) parabolics. (I must confess that the times I've tried to give courses on "Lie theory" or "algebraic groups" emphasizing "classical groups", I've met considerable resistance from a considerable fraction of the audience, who could not find corroboration for the viewpoint "in the wild".) After quite a few years of looking at reductive groups from both ends... I do finally think that even the basic facts are fairly well determined by the smallish number of data points we have from the split and quasi-split classical groups. Perhaps more usefully, the phenomena beyond that are easy to exemplify among classical groups, but somewhat clumsy to get-under-the-umbrella of intrinsic description, it seems to me. (A funny business...) (To be clearer: A. Weil's "Algebras with involutions and the classical groups"...)<|endoftext|> TITLE: Definition of a profinite category QUESTION [7 upvotes]: When studying objects like profinite groups, profinite spaces and profinite rings, I have noticed that some properties just remain the same. For example they will always be inductive limits of some discrete finite spaces, or equivalently totally disconnected compact Hausdorff spaces. I was therefore wondering, if there is some kind of a categorical generalization of these notions. Something I would call a profinite category. I was thinking defining it either in terms of a subcategory of profinite spaces, where the objects would be profinite spaces and morphisms some subsets of continuous maps between two profinite spaces. (For profinite groups we'd require them to be morphisms of groups, for rings morphisms of rings and for modules, morphisms of modules etc...) Other possible definition, I had in mind would be taking a subcategory of the category of finite sets with some properties that would make it possible to do projective limits and then define the objects in the profinite category as these projective limits. Are there any references for these kinds of notions? REPLY [5 votes]: There are two natural definitions of a profinite category. You can look at inverse limits of finite categories or you can look at topological categories whose underlying spaces are profinite (call these Stone categories). Any profinite category is Stone and any Stone category with finitely many objects is profinite. But there are Stone categories that are not profinite. Many years ago I asked George Bergman for an example, and in typical Bergman fashion he immediately provided me one. It's been many years, so I might be slightly off, but there is a well known example of a Stone lattice that is not a profinite lattice. Any poset can be viewed as a category and I believe just used that poset, viewed as a category, as his example or something very close to that. Stone categories come up very frequently in monoid theory via taking the free Stone category on a profinite graph. This comes up when trying to understand semidirect product decompositions of monoids. It is typical given a homomorphism of profinite monoids $f\colon M\to N$ to build the category of elements of $N$ viewed as a right $M$-set with $M$ acting on $N$ via $f$. This is a profinite category and so can be generated by a profinite graph. People then like two write it as a quotient of a free Stone category on that profinite graph to get profinite identity theory involved. But because that free Stone category might not be profinite, troubles arise. There is lots of work on this by Jorge Almeida and his former students. The original goal was to find profinite identities defining semidirect products of pseudovarieties of monoids, but the approach has not been 100% successful and the profinite versus Stone issue is part of the difficulties. This paper might give some flavor of what people look at. There are older papers referred to within.<|endoftext|> TITLE: Maximal independent sets in MAD families QUESTION [6 upvotes]: We call ${\cal A}\subseteq {\cal P}(\omega)$ almost disjoint if ${\cal A}\neq \varnothing$, every member of ${\cal A}$ is infinite, and for $A_1\neq A_2\in {\cal A}$ we have that $A_1\cap A_2$ is finite. Zorn's Lemma implies that every almost disjoint family is contained in a maximal almost disjoint (MAD) family. If ${\cal A}$ is a maximal almost disjoint family, we say $I\subseteq \omega$ is independent if $I\not \supseteq A$ for every $A\in{\cal A}$. Question. Given a MAD family ${\cal A}$, is there necessarily a maximal independent subset $I$ for ${\cal A}$, with respect to $\subseteq$? REPLY [5 votes]: No, a given MAD family does not necessarily admit a maximal independent subset. The key notion here is that of a completely separable MAD family. This means a MAD family $\mathcal A$ having the property that for every $X \subseteq \omega$, either $X \subseteq \bigcup \mathcal A_0$ for some finite $\mathcal A_0 \subseteq \mathcal A$, or else there is some $A \in \mathcal A$ with $A \subseteq X$. It is consistent that such families exist. It is still (I think) an open question whether $\mathsf{ZFC}$ proves the existence of a completely separable MAD family. The question of whether this is true was raised by Erdős and Shelah in the early 70's. Balcar and Simon proved that it is consistent for completely separable MAD families to exist under a wide variety of hypotheses ($\mathfrak{a} = \mathfrak{c}$, or $\mathfrak{b} = \mathfrak{d}$, or $\mathfrak{d} \leq \mathfrak{a}$, or $\mathfrak{s} = \aleph_1$). Years later, Shelah proved that the existence of completely separable MAD families follows from either $\mathfrak{s} \leq \mathfrak{a}$, or else $\mathfrak{a} < \mathfrak{s}$ plus a certain PCF hypothesis that might be (but is not known to be) a theorem of $\mathsf{ZFC}$. The failure of this hypothesis implies $\mathfrak{c} > \aleph_\omega$ and (I think) the consistency of large cardinals. More information can be found in this paper of Osvaldo Guzman (see especially the bottom of page 2 and top of page 3). This is relevant to your question because: Observation: If $\mathcal A$ is a completely separable MAD family (and $\mathcal A$ is infinite), then there is no maximal independent set for $\mathcal A$. The reason: If $I \subseteq \omega$ and $I \not\supseteq A$ for every $A \in \mathcal A$, then because $\mathcal A$ is completely separable, there is some finite $\mathcal A_0 \subseteq \mathcal A$ with $I \subseteq \bigcup \mathcal A_0$. But (as $\mathcal A$ is infinite) $\omega \setminus \bigcup \mathcal A_0$ is infinite. But if $n \in \omega \setminus \bigcup \mathcal A_0$, then $I \cup \{n\}$ is still "independent" in the sense that $I \cup \{n\} \not\supseteq A$ for every $A \in \mathcal A$. Thus $I$ is not maximal with respect to this property.<|endoftext|> TITLE: Is $\mathsf{ZFC+V=L}$ consistently $\omega$-complete? QUESTION [7 upvotes]: This was previously asked and bountied on MSE: For brevity, let $T$ be $\mathsf{ZFC+V=L}$. Say that an extension of $\mathsf{ZFC}$ is $\omega$-complete iff it has exactly one $\omega$-model up to elementary equivalence. While the $\omega$-incompleteness of $T$ is easily provable in theories only slightly stronger than $T$ itself, I don't immediately see how to do it in $T$ alone. My question is: Is the theory $S:=T+$ "$T$ is $\omega$-complete" consistent? Here are a couple observations: If we replace "$\omega$-model" by "well-founded model," the answer is obviously yes under standard assumptions. Let $\alpha$ be the second-smallest ordinal such that $L_\alpha\models\mathsf{ZFC}$. Then $L_\alpha$ also satisfies "$\mathsf{ZFC+V=L}$ has exactly one well-founded model." Unfortunately, we have no analogous hierarchy of $\omega$-models, so this is a non-starter here. As to the specific choice of theory in question, the point is that (something like) $\mathsf{V=L}$ is needed to block an easy proof of a negative answer via forcing. For example, reasoning in $\mathsf{ZFC}$, if $\mathsf{ZFC}$ had an $\omega$-model $\mathcal{M}$ it would have a countable one $\hat{\mathcal{M}}$, and we could force over $\hat{\mathcal{M}}$ to get a non-elementarily-equivalent $\omega$-model $\hat{\mathcal{N}}$. (Forcing over ill-founded countable models is no harder really than forcing over well-founded ones.) The key point here is that forcing preserves $\mathsf{ZFC}$. This breaks down of course for $\mathsf{V=L}$ and so this argument is irrelevant here. Given the paucity of techniques we currently have for building models of $\mathsf{ZFC}$ in the first place, this seems to be a real issue. Ultimately I suspect that the answer is negative, but the above two points between them rule out all the lines of attack I've been able to think of so far. EDIT: In light of Farmer S's answer below, let me explicitly mention a rule of thumb which I forgot: when thinking about properties which are not too far from first-order definable, always consider the hyperarithmetic hierarchy! For example, for every $\mathcal{L}_{\omega_1,\omega}$-sentence $\varphi$, if $\varphi$ has a model then it has a model $M$ which is countable in $L$, and moreover the $L$-least (real coding a) model of $\varphi$ is hyperarithmetic relative to (any real coding) $\varphi$. The property "Is an $\omega$-model of $T$" is expressible as a computable $\mathcal{L}_{\omega_1,\omega}$-sentence, and this drives Farmer S's point that $T^+\in L_{\omega_1^{CK}}$. REPLY [12 votes]: Claim: $T+$"$T$ is $\omega$-complete" is inconsistent. For suppose it's consistent and now work in a model $V$ of this theory. Let $T^+$ be the resulting completion of $T$ (i.e. the unique theory of the $\omega$-models of $T$ in the sense of $V$). Then note that $T^+$ is a $\Delta^1_1$ real, so $T^+\in L_{\omega_1^{\mathrm{ck}}}$. But $L_{\omega_1^{\mathrm{ck}}}\subseteq\mathrm{wfp}(M)$ whenever $M\models T$ is an $\omega$-model, and therefore every real $x\in L_{\omega_1^{\mathrm{ck}}}$ is such that $x\leq_{\mathrm{T}} T^+$ (the sub-$\mathrm{T}$ there being "Turing", as opposed to the theory $T$). (Given $x$, fix a wellorder $W$ of $\omega$ in ordertype $\alpha$ with $x\in L_\alpha$. Then (roughly) $T^+$ models "$W$ is a wellorder", and can recover $x$ from $W$. (Edited in:) Formally, fix an integer $e$ which indexes a recursive wellorder of $\omega$ in ordertype $\alpha$ with $x\in L_\alpha$. Recall $L_\alpha$ projects to $\omega$, and $x\leq_{\mathrm{T}} t^{L_\alpha}$, the first-order theory of $L_\alpha$. Fix a Turing reduction $n$ of $x$ from $t^{L_\alpha}$. Then for $m<\omega$, we have $m\in x$ iff $T^+$ contains the statement "Let $\beta$ be the ordertype of the wellorder coded by $e$, and let $y\leq_{\mathrm{T}} t^{L_\beta}$ via the $n$th Turing program; then $m\in y$".) But with $T^+\in L_{\omega_1^{\mathrm{ck}}}$, this gives a contradiction.<|endoftext|> TITLE: Analogues of worldly cardinals for (an unusual version of) second-order $\mathsf{ZFC}$ QUESTION [6 upvotes]: Let $\mathfrak{ZFC}(\mathsf{SOL})$ be the theory in second-order logic (with the standard semantics) gotten from $\mathsf{ZFC}$ by modifying the Separation and Replacement schemes to apply to arbitrary second-order formulas. For example, for each second-order formula $\varphi$ with only first-order free variables $x_1,...,x_n,x_{n+1}$ we have the Separation instance $$\forall x_1,...,x_n,y\exists z\forall w(w\in z\leftrightarrow w\in y\wedge \varphi(x_1,...,x_n,w)).$$ Note that $\mathfrak{ZFC}(\mathsf{SOL})$ is not the same as the system called "second-order $\mathsf{ZFC}$" - see e.g. here. Building off of this MSE question of mine, I'm interested in understanding the models of this theory. I'm hoping that by restricting to levels of the cumulative hierarchy we can get a clean answer: Question 1: For which cardinals $\alpha$ do we have $V_\alpha\models\mathfrak{ZFC}(\mathsf{SOL})$? It's easy to show that every (strongly) inaccessible cardinal has this property; the converse, however, is not immediately clear to me. My current suspicion in fact is that (very) large cardinals imply the existence of non-inaccessible $\alpha$s with the above property, but I don't immediately see how to prove this. As a secondary question, I'm also interested in a "Henkin version" of this question: Question 2: For which $\alpha$ is there an $X\subseteq \mathcal{P}(V_\alpha)$ such that $(V_\alpha,X)$ forms a Henkin model of $\mathfrak{ZFC}(\mathsf{SOL})$? At a glance I think that the least such $\alpha$ has countable cofinality, by a quick modification of the analogous argument for worldly cardinals, but I haven't had time yet to check the details. REPLY [8 votes]: Complementing @JasonChen's answer: Assume ZFC+$I_1$ and let $j:V_{\lambda+1}\to V_{\lambda+1}$ be elementary, so $\lambda$ is the sup of the critical sequence of $j$. Then $V_{\lambda}$ models $\mathfrak{ZFC}(\mathsf{SOL})$, but $\mathrm{cof}(\lambda)=\omega$. For suppose $f:V_\alpha\to\lambda$ is cofinal and definable over $V_{\lambda+1}$ from the parameter $p\in V_{\lambda}$. Let $n<\omega$ be such that $\alpha,p\in V_{\mathrm{crit}(j_n)}$ (where $j_n=$ the $n$th iterate of $j$). Note that $j_n\circ f\neq f$, because taking $x\in V_\alpha$ with $f(x)>\mathrm{crit}(j_n)$, we get $j_n(f(x))>f(x)$. But $j_n\circ f=f$ because $j_n:V_{\lambda+1}\to V_{\lambda+1}$ is elementary and $j_n(p,\alpha)=(p,\alpha)$. Edit, considering @AsafKaragila's comment on consistency strength: Consistency-wise, the assumption above was overkill; a measurable suffices. Assume ZFC + $\kappa$ is measurable. Let $G$ be Prikry generic at $\kappa$. So $\kappa$ has cofinality $\omega$ in $V[G]$. Claim: In $V[G]$, $V_\kappa$ models $\mathfrak{ZFC}(\mathsf{SOL})$. In fact, if $f:\omega\to\kappa$ is cofinal and $f$ is definable over $V_{\kappa+1}^{V[G]}$ from parameters in $V_\kappa$, then $f\in V$, so $f$ is bounded. Since $V_\kappa^{V[G]}=V_\kappa^V$, this is a consequence of the fact that $\mathrm{HOD}^{V[G]}_V=V$, i.e. if $X\in V[G]$ and $X\subseteq V$ and $X$ is definable over $V[G]$ from parameters in $V$, then $X\in V$. (This follows from the fact that if $p,q$ are Prikry conditions then there are generics $G_p,G_q$ with $p\in G_p$ and $q\in G_q$ and $V[G_p]=V[G_q]$.) Edit 2: On the other hand, the kind of argument used in the paper "Inner models from extended logics: Part 1" referred to in @JasonChen's answer to show that in $L$, $V_\alpha$ models $\mathfrak{ZFC}(\mathsf{SOL})$ iff $\alpha$ is inaccessible, also works for the standard fine structural $L[\mathbb{E}]$ models $M$ for short extenders, for instance if $M$ has no largest cardinal, and assuming $M$ has Mitchell-Steel indexing, though I expect it would also work with Jensen indexing. So if those models are indeed consistent through ZFC + superstrongs, then one would need more than ZFC + ``There is a superstrong extender'' to prove there is a non-inaccessible $\alpha$ with $V_\alpha$ modelling $\mathfrak{ZFC}(\mathsf{SOL})$. (The paper "The definability of the extender sequence $\mathbb{E}$ from $\mathbb{E}\upharpoonright\aleph_1$ in $L[\mathbb{E}]$" contains enough to generalize the argument of Kennedy, Magidor, Väänänen for $L$. The definability there is all done over $\mathcal{H}_\kappa$s, as it's more convenient, but that can be translated into the cumulative hierarchy with the usual coding; in the present case that's only actually needed at the very top, since we can assume $V_\alpha\models\mathrm{ZFC}$ to start with.) Edit 3: Following @AsafKaragila's suggestions in the comments, we have: Claim: Suppose $V_\lambda$ models $\mathfrak{ZFC}(\mathsf{SOL})$ but $\lambda$ is singular. Then for every $X\in V_\lambda$, $X^\#$ exists. Moreover, there is a proper class inner model $M$ with a measurable cardinal. Proof: For simplicity take $X=\emptyset$. Suppose first that $0^\#$ does not exist. Note first that since $V_\lambda$ models ZFC, $\lambda$ is a (singular) strong limit cardinal. By Jensen's covering lemma, $\lambda$ is singular in $L$. Let $B$ be the constructibly least singularization. Then $B$ can be defined over $V_{\lambda+1}$ (without parameters), which contradicts $\mathfrak{ZFC}(\mathsf{SOL})$. The argument for an inner model $M$ with a measurable is likewise, but using the Dodd-Jensen core model: We also have the appropriate version of covering for that core model $K=K^{\mathrm{DJ}}$, and $K|(\lambda^+)^K$ can also be defined in the codes over $V_{\lambda+1}$, and hence the least singularization of $\lambda$ in the $K$-order is definable. So Edits 1 and 3 together give that ZFC + "There is a singular $\lambda$ such that $V_\lambda\models\mathfrak{ZFC}(\mathsf{SOL})$" is equiconsistent with ZFC + "There is a measurable cardinal".<|endoftext|> TITLE: Generalization of Hopf invariant QUESTION [7 upvotes]: This may be a dumb question, but I ask it here. In ordinary cohomology, we can construct a Hopf invariant for $f \colon S^{2n-1} \to S^{n}$ by applying $H^{*}(- \colon \mathbb{F}_p)$ to the cofibre sequence, so that the Hopf invariant measures the non-triviality of $$ 0 \to H^{*}(S^{2n}) \to H^*(C_f) \to H^{*}(S^n) \to 0 \in \mathrm{Ext}_A^1(H^{*}(S^n), H^{*}(S^{2n})) $$ where $A$ is the Steenrod algebra. Similarly, Adams constructed the $e$-invariant by doing an analogy of Hopf invariant in complex K-theory and Adams operation and he computed the image of the $J$-homomorphism. My question is: Is there a generalization of this method to another cohomology theory (e.g. $MU$, Morava K-theory, $tmf$, etc.)? If there was, what kind of elements in $\pi_i^{S}$ could be detected in this invariant? Can we view this invariant in another way? I am not sure there must be a modern view of this or not, but I would like to know it if there was a more way to view Hopf invariant than just apply a cohomology and measure how non trivial the cohomology operation is. REPLY [6 votes]: The $E$-based Adams spectral sequence is the homotopy spectral sequence associated to the tower of spectra $\dots \to Y_2 \to Y_1 \to Y_0 = S$ with $Y_{s+1} \to Y_s \to E \wedge Y_s$ a homotopy fiber sequence for each $s\ge0$. The edge homomorphism to filtration $s=0$ detects the Hurewicz image of $\pi_*(S)$ in $\pi_*(E) = E_*$. On its kernel there is defined a homomorphism to filtration $s=1$, which specializes to the Hopf-Steenrod invariant when $E$ is mod $p$ homology. I believe Chapter 19 of Switzer's book "Algebraic Topology -- Homotopy and Homology" gives a classical introduction. In particular, for $E = KU$ and $E = KO$ he connects this generalization to Adams' $e$-invariant (in the complex and real cases). For $E = ko$ there is work by Mahowald at $p=2$. His Bulletin of the AMS announcement (1970) discusses a splitting of $ko \wedge ko$ (closely related to $E \wedge Y_1$ in this case), which was improved by Milgram and Carlsson, and Mahowald's Pacific J. Math. (1981) paper titled "$bo$-resolutions" uses this approach to determine the Bott periodic homotopy of the mod $2$ Moore spectrum. For $E = MU$ (or $E = BP$) the $E$-based Adams spectral sequence is the Adams-Novikov spectral sequence. The $s=1$-line was calculated by Miller, Ravenel and Wilson (Annals of Math., 1977), and detects the image-of-J, much like topological $K$-theory does. For $E = \ell$ (the Adams summand of $p$-local connective $K$-theory, with $p$ odd), the cooperations $\ell \wedge \ell$ were studied by Kane (1981) and Lellmann (1984), while Miller (JPAA, 1981) determined the periodic homotopy of the mod $p$ Moore spectrum. For $E = Ell$ a form of elliptic cohomology, an early reference is Clarke-Johnson (Adams memorial symposium, 1992). For $E = tmf$ (topological modular forms), the Behrens-Ormsby-Stapleton-Stojanoska paper (J. of Topology 2019) is a good place to look. The $tmf$-Hurewicz image and the $tmf$-based $e$-invariant detect significantly more of $\pi_*(S)$ than the image-of-J. The Behrens-Mahowald-Quigley preprint (and my forthcoming book with Bruner) also contain more about this.<|endoftext|> TITLE: Local topology of Whitney stratified spaces QUESTION [11 upvotes]: Let $M$ be a smooth manifold, let $\mathcal{P}$ be a Whitney stratification of $M$ and let $S\subset M$ be a stratum with closure $\overline{S}$. Question: Does there exist an open neighborhood $U\subset M$ of $\overline{S}$ such that $U$ deformation retracts onto $\overline{S}$? In the case when $\overline{S}\subset M$ is a smooth submanifold, this follows from the tubular neighborhood theorem. For another example, let $M=\mathbb{R}^{2n}$ and consider the stratification with three strata given by $S_1=0\in \mathbb{R}^{2n}$, $S_2=(\mathbb{R}^n\times\{0\}\cup\{0\}\times\mathbb{R}^n)-S_1$ and $S_3=\mathbb{R}^{2n}-\overline{S_2}$. Then $M$ deformation retracts onto $\overline{S_2}$ (e.g. along the connected components of the quadics $||x||^2-||y||^2=t$ for $(x,y)\in \mathbb{R}^{2n}$ and $t\in\mathbb{R}$). I hope that an affirmative answer in general should follow from the Thom/Mather theory of tubular neighborhoods/control data, but I keep getting turned around. A proof or counter-example would be greatly appreciated! Thanks! Update: For conically stratified manifolds in the sense of Ayala-Francis-Tanaka, this follows from Proposition 8.2.5 of https://arxiv.org/abs/1409.0501. Under the expectation that Whitney stratified spaces should be conically stratified, there should be an analogous result for Whitney stratified manifolds. Is this the case? If so, a reference would be greatly appreciated! REPLY [3 votes]: In a paper written with a collaborator that we have recently uploaded on the arxiv, we show that indeed the conical charts of a Whitney stratified space provided by Thom and Mather induce a conically smooth structure: thus, you may freely apply Proposition 8.2.5 in the paper by Ayala-Francis-Tanaka to get what you want.<|endoftext|> TITLE: Correlations of $\phi(n)/n$ QUESTION [15 upvotes]: We know that the average of $\phi(n)/n$ is approximated by a constant. Here $\phi $ is the Euler quotient function. One can furthermore show asymptotics with a secondary main term, at least for the smooth sum $$ \sum_{n \in \mathbb{N} } \frac{\phi(n) } {n} w(n/x)=c_0(w) x+c_1(w) (\log x ) +o(\log x ) ,$$ where $w$ is a smooth weight and $c_0,c_1$ are constants depending on $w$. Can we prove asymptotics for the secondary term regarding the shifted sum $$ \sum_{n \in \mathbb{N} } \frac{\phi(n) } {n}\frac{\phi(n+1) } {n+1} w(n/x) $$ for some $w$? It is not clear to me whether the secondary term here should be oscillating or like $\sim c \log^2 x$ or something else. The standard approach to prove the previous asymptotic relies on the fact that $\frac{\phi(n) } {n} $ is multiplicative, whereas $\frac{\phi(n) } {n}\frac{\phi(n+1) } {n+1}$ is clearly not. REPLY [6 votes]: If $w$ is smooth and compactly supported in $[1/2,2]$, say, then $$\sum_{n}\frac{\varphi(n)}{n}\frac{\varphi(n+1)}{n+1}w(n/x) = Cx + O_A((\log x)^{-A}).$$ Write $$\frac{\varphi(n)}{n} = \sum_{d\mid n} \frac{\mu(d)}{d}$$ and $$\frac{\varphi(n+1)}{n+1} = \sum_{e \mid n+1} \frac{\mu(e)}{e},$$ then interchange the orders of summation. Perform Poisson summation on the $n$ variable to turn the sum into something like $$\sum_d \frac{\mu(d)}{d} \sum_{(e,d)=1}\frac{\mu(e)}{e} \frac{x}{d^2e^2}\sum_{|k|\leq de/x} e\left(-k\frac{\overline{d}}{e}\right)\widehat{w}(kx/de),$$ where $\overline{d}$ denotes the inverse of $d$ modulo $e$. The $k=0$ term gives the main term, which is seen to be $Cx + O_A((\log x)^{-A})$ by using the cancellation in the mobius function. The nonzero frequencies contribute an error of size $O_A((\log x)^{-A})$. For these terms, break $d$ and $e$ into dyadic ranges $d \asymp D, e \asymp E$. Clearly we may assume $DE \gg x$. Also, we may assume without loss of generality that $E \ll D$, otherwise use the reciprocity relation $$\frac{\overline{d}}{e} + \frac{\overline{e}}{d}\equiv \frac{1}{de}\pmod{1}$$ to switch $d$ and $e$ in the exponential. After some work, Siegel-Walfisz and the large sieve cover the case when $E\leq D (\log x)^{-B}$, so we may assume $D \approx E$. The double sum over $d$ and $e$ may then be suitably bounded using results of Duke-Friedlander-Iwaniec on bilinear forms in Kloosterman fractions (see also work of Bettin and Chandee). [Edit: March 15] I'm adding a bit more detail, as requested. I'm assuming the smoothing $w$ is a smooth bump function, so the Fourier transform satisfies $|\widehat{w}(y)| \ll \exp(-|y|^{1/2})$, say. The arguments below might have to be adapted or substituted entirely for less smooth weights. With $w$ as above, one finds that $$\sum_n \varphi(n)/n w(n/x) = Cx + O_A((\log x)^{-A})$$ by Mellin inversion, contour shifting, and the zero-free region for zeta. Now, as to the correlation sum. Begin as above, and perform Poisson summation. We separate the $k=0$ term which gives the main term, and the contribution of the nonzero frequencies is $$\sum_{d\ll x}\frac{\mu(d)}{d}\sum_{\substack{e \ll x \\ (e,d)=1}}\frac{\mu(e)}{e}\frac{x}{de}\sum_{k \in \mathbb{Z}\backslash \{0\}} e\left(-k\frac{\overline{d}}{e}\right) \widehat{w}\left(k\frac{x}{de} \right).$$ Dyadically decompose $d\asymp D$ and $e \asymp E$. As mentioned above, up to changing the sign of $k$ in the exponential one may assume that $E \ll D$ by using reciprocity. Now use the rapid decay of $\widehat{w}$ to truncate the sum on $k$. Up to a minor change in the coefficients and an acceptable error term, we have $$\frac{x}{D^2E^2}\sum_{e\asymp E} \mu(e) \sum_{\substack{d \asymp D \\ (d,e)=1}}\mu(d) \sum_{0<|k|\leq (\log x)^{10}DE/x} e\left(k\frac{\overline{d}}{e}\right)e\left(\frac{k}{de} \right) \widehat{w}\left(k\frac{x}{de} \right).$$ Clearly, we may assume $DE > x(\log x)^{-11}$, say. We are going to use the large sieve inequality, but in order to do so we need to separate the $d$ and $e$ variables from each other inside $\widehat{w}$. Change variables in the definition of $\widehat{w}$ and interchange to obtain $$\int_{u \asymp 1} \frac{x}{D^2E^2}\sum_{0<|k|\leq (\log x)^{10}DE/x} \sum_{e \asymp E}\mu(e)w(\tfrac{eu}{E})\sum_{\substack{d \asymp D \\ (d,e)=1}}\mu(d) e\left(-k\frac{x}{dE}u \right) e\left(-k\frac{\overline{d}}{e}\right) du,$$ where this holds up to harmless changes in the $d$ and $e$ coefficients. We work uniformly in $u\asymp 1$, so we can drop the integral. We give two arguments. The first works when $E$ is a bit smaller than $D$, and the other works when they are about the same size. The first argument uses the large sieve. Split the sum over $d$ into arithmetic progressions modulo $e$ to separate the additive character, then apply multiplicative characters to detect the congruence condition modulo $e$. This turns the sum over $d$ into $$\frac{1}{\varphi(e)}\sum_{\chi (e)}\sum_{(a,e)=1}e\left(-k\frac{\overline{a}}{e}\right) \overline{\chi}(a)\sum_{\substack{d \asymp D}}\mu(d)\chi(d) e\left(-k\frac{x}{dE}u \right).$$ We take out the contribution of the principal character $\chi_0 \pmod{e}$. We get cancellation in the sum over $d$ using partial summation. The sum over $a\pmod{e}$ becomes a Ramanujan sum, which has size $\leq (|k|,e)$, the GCD of $|k|$ and $e$. It is easy to see that the total contribution from $\chi_0$ after summing over all the variables is $\ll_A (\log x)^{-A}$. We have to bound the contribution of the non-principal characters: $$\frac{x}{D^2E^2}\sum_{0<|k|\leq (\log x)^{10}DE/x} \sum_{e \asymp E}\mu(e)w(\tfrac{eu}{E})\frac{1}{\varphi(e)}\sum_{\substack{\chi (e) \\ \chi \neq \chi_0}}\sum_{(a,e)=1}e\left(-k\frac{\overline{a}}{e}\right) \overline{\chi}(a)\sum_{\substack{d \asymp D}}\mu(d)\chi(d) e\left(-k\frac{x}{dE}u \right).$$ We fixed $k$, since we will not need to sum over this variable, and then apply Cauchy-Schwarz to the sum over $e$. This gives $$\sum_e \ll (\Sigma_1 \Sigma_2)^{1/2},$$ where $$\Sigma_1 = \sum_{e \asymp E} \frac{1}{\varphi(e)}\sum_{\substack{\chi (e) \\ \chi \neq \chi_0}} \left|\sum_{(a,e)=1}e\left(-k\frac{\overline{a}}{e}\right) \overline{\chi}(a) \right|^2$$ and $$\Sigma_2 = \sum_{e \asymp E} \frac{1}{\varphi(e)}\sum_{\substack{\chi (e) \\ \chi \neq \chi_0}} \left|\sum_{\substack{d \in I_D}}\mu(d)\chi(d) e\left(-k\frac{x}{dE}u \right) \right|^2.$$ We easily bound $\Sigma_1$ by using positivity to include $\chi_0$, then opening the square and using orthogonality of characters. It follows that $$\Sigma_1 \ll E^2.$$ The argument for $\Sigma_2$ is slightly more involved. First, we reduce to summing over primitive Dirichlet characters $\psi$ (cf. any standard proof of the Bombieri-Vinogradov theorem). We apply a dyadic decomposition to the conductor of the primitive characters, so we derive something like $$\Sigma_2 \ll \sum_{1 \ll R = 2^j\ll E}\sum_{f \ll E} \frac{1}{\varphi(f)}\sum_{r \asymp R} \frac{1}{\varphi(r)} \sum_{\substack{\psi( r) \\ \psi \text{ prim}}}\Big|\sum_{\substack{d \in I_D \\ (d,f)=1}}\mu(d)\psi(d) e\left(-k\frac{x}{dE}u \right) \Big|^2.$$ If $R \leq (\log x)^C$ then we use partial summation and the Siegel-Walfisz theorem to bound the sum over $d$. If $R > (\log x)^C$ we use the multiplicative large sieve inequality $$\sum_{r \asymp R} \frac{1}{\varphi(r)} \sum_{\substack{\psi( r) \\ \psi \text{ prim}}}\Big|\sum_{\substack{d \in I_D \\ (d,f)=1}}\mu(d)\psi(d) c(d)e\left(-k\frac{x}{dE}u \right) \Big|^2 \ll \frac{1}{R}\left(R^2 + D \right)D.$$ We deduce that $$\Sigma_2 \ll D^2 (\log x)^{-2A} + DE.$$ Recalling the application of Cauchy-Schwarz and the bounds for $\Sigma_1$ and $\Sigma_2$, we sum over $k$ to see that the total contribution is $$\ll (\log x)^{-A + O(1)} + (\log x)^{O(1)} \left(\frac{E}{D} \right)^{1/2}.$$ This is acceptably small if $E \leq D(\log x)^{-B}$. Therefore, we may assume that $D x^{-o(1)} \ll E \ll D$. Since $DE \gg x^{1-o(1)}$ this implies $D \gg x^{1/2-o(1)}$. For this second argument we do not need any properties of the coefficients attached to the $d$ and $e$ variables other than that they are 1-bounded, so we wish to get a bound on $$\frac{x}{D^2E^2}\sum_{1\leq k\leq (\log x)^{10}DE/x}\Big| \mathop{\sum\sum}_{\substack{d \asymp D \\ e \asymp E \\ (d,e)=1}}\alpha_d \beta_e e \left(k \frac{\overline{d}}{e} \right)\Big|$$ for 1-bounded coefficients $\alpha_d,\beta_e$. Theorem 2 of the above-mentioned Duke-Friedlander-Iwaniec paper gives $$\Big| \mathop{\sum\sum}_{\substack{d \asymp D \\ e \asymp E \\ (d,e)=1}}\alpha_d \beta_e e \left(k \frac{\overline{d}}{e} \right)\Big| \ll (DE)^{1/2} (k+DE)^{3/8} (D+E)^{11/48}\ll D^{2-1/48+o(1)},$$ which saves $D^{1/48}$ over the trivial bound. Therefore $$\frac{x}{D^2E^2}\sum_{1\leq k\leq (\log x)^{10}DE/x}\Big| \mathop{\sum\sum}_{\substack{d \asymp D \\ e \asymp E \\ (d,e)=1}}\alpha_d \beta_e e \left(k \frac{\overline{d}}{e} \right)\Big| \ll \frac{D^{2 - \frac{1}{48} + o(1)}}{DE} \ll D^{-1/48+o(1)} \ll x^{-1/96+o(1)}.$$<|endoftext|> TITLE: Is higher-order excision related to higher-order cohomology operations? QUESTION [6 upvotes]: Here are some things which are almost, but not quite, [representable by] (co)homology theories: $n$-excisive functors (in the sense of Goodwillie calculus), for $n \geq 2$. The domains and codomains of secondary, tertiary, etc. cohomology operations. The $E^r$ pages of various spectral sequences, for fixed $r$. Now, (3) gives examples of (2), since the differentials of a spectral sequence are typically higher-order cohomology operations. I'm not sure to what extent (2) always arises from (3). But what I'm most interested in is the relationship between (1) and (2/3). Questions: To what extent do higher-order cohomology operations necessarily arise as the differentials for some spectral sequence? Are the homotopy groups of the values of a 2-exisive functor, say, canonically defined using primary cohomology operations? "Conversely," is the $E^r$ page, of the Atiyah-Hirzebruch spectral sequence, say, for a fixed spectrum $F$, represented as the homotopy groups of the output of some $r+1$-excisive functor? Or is there some other relationship between (1,2,3) that I'm missing? I suspect that a positive answer to (1) might come from the Goodwillie spectral sequence. (My apologies for having two numbered lists!) REPLY [5 votes]: This probably falls short of an answer, more like a rambling sequence of comments. Regarding question 2, notice that $E^r$ is an additive functor. I.e., $E^r(X\vee Y)\cong E^r(X)\oplus E^r(Y)$. This implies that if $E^r(X)$ is realized as the homotopy groups of an $n$-excisive functor for some $n$, then it would be a linear functor, i.e, $n$ would be 1. But in this case $E^r(X)$ would be a generalized homology theory, which (unless I am very confused) it is not for $r>2$. So the answer to question 2 is No. The moral that I would draw from this is that while there may be interesting connections between Goodwillie towers of specific functors and higher cohomology operations, a very general connection between $r$-excisive functors and higher cohomology operations or higher pages in a spectral sequence is perhaps too much to expect. In particular, the Atiyah-Hirzebruch spectral sequence does not appear to have much to do with higher excision. Added later: A particular example of a Goodwillie tower that obviously has a lot to do with homology operations is the tower of the functor from spectra to spectra $X\mapsto H \wedge \Omega^\infty X$, where $H$ is the spectrum representing your favorite homology theory. The tower of this functor calculates the homology of $\Omega^\infty X$ in terms of the homology of $X$. The differentials have a lot to do with homology operations, specifically Dyer-Lashof operations. All this was studied in considerable detail (for $H=H\mathbb Z/2$) in the following paper of N. Kuhn and J. McCarty The mod 2 homology of infinite loopspaces, Algebr. Geom. Topol. 13 (2013) Regarding question 1, you could say that the answer is yes, for trivial reasons, if you interpret broadly enough what it means to be "defined using primary cohomology operations". A $2$-excisive functor is the homotopy fiber of a map between two spectra, so if any map between spectra is a cohomology operation, then a functor is determined by a cohomology operation. To be more specific and a little less trivial, let me recall a general description of $2$-excisive functors. We restrict ourselves to pointed functors between the categories of pointed spaces or spectra. Then any quadratic functor is the homotopy fiber of a natural transformation of the following form, where $C$ is a spectrum and $D$ is a spectrum with an action of $\Sigma_2$: $$ C\wedge X\to (D\wedge X^{\wedge 2})_{h\Sigma_2} $$ (in the case of space-valued functors you should put $\Omega^\infty$ in front of these functors, and there may be non-deloopable natural transformation. These correspond to the unstable homotopy operations that were mentioned in a comment). Natural transformations of this form are well-understood, and can be described in terms of maps from $C$ to (some spectrum constructed out of) $D$. If you call these maps cohomology operations, then you can say that $2$-excisive functors with prescribed first and second layers are classified by cohomology operations of certain type from the first layer to the second. In case of functors from/to the category of spaces, the derivatives have the structure of a module over the spectral Lie operad. There is a relationship between this module structure and the differentials in the Goodwillie spectral sequence. So you can say that to some extent the cohomology operations that determine the differentials in the Goodwillie spectral sequence come from this Lie module structure. But this is not the whole story.<|endoftext|> TITLE: When is a nilpotent Lie algebra isomorphic to the associated graded of its lower central series? QUESTION [7 upvotes]: All Lie algebras in this question are finite-dimensional and defined over a field $k$ of characteristic $0$ which I'm happy to take to be $\mathbb{R}$ or $\mathbb{C}$. $\DeclareMathOperator\gr{gr}$Let $L$ be a nilpotent Lie algebra. It is then filtered by its lower central series, and we have an associated graded nilpotent Lie algebra $\gr L$. It is definitely not the case that $L$ and $\gr L$ have to be isomorphic; see Malcev Lie algebra and associated graded Lie algebra for some examples. Question: what kinds of conditions can I put on $L$ that ensure that it is isomorphic to $\gr L$? E.g. if the field is $\mathbb{R}$ are the there geometric/topological/algebraic conditions on the associated simply-connected nilpotent Lie group that ensure this? REPLY [2 votes]: $\DeclareMathOperator\gr{gr}$Too tired to think clearly, but it looks like a standard Deformation Theory thingy. We have natural linear maps $\gamma_n (L)\rightarrow \gr_n L$. Split them as linear maps. We get a bijective linear map $L\rightarrow \gr L$. Use it to equip $\gr L$ with a second Lie algebra structure $[,]^{\prime}$, coming from $L$. Now consider the difference $$\mu:L\otimes L \rightarrow L, \ \ \mu (x\otimes y) = [x,y]-[x,y]^{\prime}.$$ I claim that $\mu$ is a 2-cocycle on $\gr L$ and that finding an isomorphism $\gamma_n (L)\cong \gr_n L$ is equivalent to finding a 1-cochain $\theta$ such that $\mu=d\theta$.<|endoftext|> TITLE: A theorem about the symplectic geometry of projective bundles QUESTION [10 upvotes]: I am trying to understand the following theorem about symplectomorphisms of projective bundles. Theorem 1.5 of "Characteristic Classes in Symplectic Topology" A.G. Reznikov. Selecta Mathematica, volume 3, pages 601–642(1997). Theorem: Let $E_i \rightarrow M_i$ be Hermitian vector bundles, $i = 1, 2$. Let $ \mathbb{P}(E_i)$ be the projectivization of $E_i$. Let $f : \mathbb{P}( E_1) → \mathbb{P} (E_2)$ be a fiber-like symplectomorphism, covering a map $\varphi : M_1 \rightarrow M_2$. Then $\varphi^{*}(c_k(E_2)) = c_k(E_1)$ for $k \geq 2$. Here is where I get confused: taking a simple case where $M_1=M_2=\,\mathbb{CP}^2$. Suppose that $E_{1} = \mathcal{O} \oplus \mathcal{O}$. $E_{2} = \mathcal{O}(1) \oplus \mathcal{O}(1)$. Then, if I am not mistaken, there should be a fibre-wise symplectomorphism $\mathbb{P}(E_1) \rightarrow \mathbb{P}(E_2)$ since they are both isomorphic to the $3$-fold $\mathbb{CP}^2 \times \mathbb{CP}^1$. But there is no diffeomorphism of $\mathbb{CP}^2$ mapping $c_{2}(\mathcal{O} \oplus \mathcal{O}) = 0$ to $c_{2}(\mathcal{O}(1) \oplus \mathcal{O}(1)) = H^2$, where $H$ is the hyperplane class (it is not hard to see that a fibre preserving symplectomorphism must induce a diffeomorphism of the base). I am thinking possibly that this theorem is for a specific choice of symplectic form? I would be grateful if somebody could clear up my confusion. I am by no means questioning this theorem, I just want to understand the statement correctly. thanks. REPLY [9 votes]: I think you spotted an imprecision in Reznikov's paper. Clearly, the statement has a problem because it is not robust under tensoring with a line bundle. It seems that by "$c_k(E_i)$ for $k\geq 2$" Reznikov really means the characteristic classes of the $PU(n)$ principal bundle obtained by quotienting the $U(n)$ bundle underlying $E_i$ by its center $U(1)$. This is the algebra generated by classes $\tilde{c}_k, 2\leq k \leq n$ of degree $2k$ which are sort of ``projections of $c_k$ orthogonally to $\mathbb Z[c_1]$''. More concretely, I think the theorem works if you replace $c_k$ by $\tilde c_k$, where $$1 + \sum_{k=2}^n \tilde c_k(E) t^k = \left [\mathrm{det}\left( \mathrm I_n + \frac{i t}{2\pi} \left( \Omega - \frac{1}{n} \mathrm{Trace}(\Omega) \mathrm{I_n}\right) \right)\right]~,$$ where $\Omega$ is the curvature form of some Hermitian connection.<|endoftext|> TITLE: Is there a straightforward generalization of min(x,y) to positive-semidefinite Hermitian matrices? QUESTION [12 upvotes]: This is an open-ended question I have. Is there a function of two positive-semidefinite hermitian operators $\min(A,B)$ returning another positive-semidefinite Hermitian operator such that: If A and B commute, the eigenvectors of $\min(A,B)$ should be the shared eigenvectors of A and B, with eigenvalues $\min (\lambda_A , \lambda_B)$ where $\lambda_A$ , $\lambda_B$ are the corresponding eigenvalues for A and B. When A and B do not necessarily commute, $A - \min(A,B)$ and $B - \min(A,B)$ should always be positive semidefinite, so that both A and B are greater than $\min(A,B)$ in the Löwner order. If yes, is there a straightforward proof or construction of such a function? If not, is there any theorem limiting what properties such a function can have? Now, these two axioms by themselves are fairly weak. You can easily find a trivial but not so powerful example by defining $\min(A,B)$ as being defined by axiom 1) if the two operators exactly commute and $\min (\lambda^{\min}_A , \lambda^{\min}_B) I$ otherwise, where $\lambda^{\min}_A$, $ \lambda^{\min}_B$ are the smallest eigenvalues of A and B. So I'll reformulate the question: is there a reasonable such function with more useful properties? Possible other useful properties that it could have (optional, should satisfy as many of them as possible): Min should be associative and commutative. It should distribute over addition if possible, like in the tropical semiring for real numbers (this is most likely way too strong of a condition but I would be happy to be proven wrong). Translation invariance: $\min(A + C,B + C) = \min(A,B) + C$ at least when C commutes with A and B (in particular when C is a multiple of the identity matrix). This is somewhat weaker than 4) $\min(A,B)$ should ideally be continuous in A and B. the smallest eigenvalue of $\min(A,B)$ should not be "too much smaller" than the smallest eigenvalues of A and B (a strong version of this with equality would follow from 5 and positive definiteness). REPLY [8 votes]: I don't know if this is the kind of answer that you were looking for, but (hat tip this math.SE answer) the preprint Nikolas Stott, Maximal lower bounds in the Löwner order [arXiv:1612.05664] seems to have a detailed discussion of maximal lower bounds of two symmetric matrices with respect to the Löwner order, which is precisely the partial order generated by the cone of positive semi-definite matrices. The conclusion seems to be that maximal lower bounds are generally non-unique (in a partial order a strict maximum may not exist due to incomparable elements), but can be usefully parametrized. Whether there could be a useful way to pick a preferred maximal lower bound by a formula, I can't say.<|endoftext|> TITLE: How can we make precise the notion that a finite-dimensional vector space is not canonically isomorphic to its dual via category theory? QUESTION [9 upvotes]: There are quite a few questions both on this site and math.SE related to this topic as well as what we mean when we say "natural" or "canonical". For the purposes of this question, I'm going to consider canonical and natural to be synonyms, and use wikipedia's definition of an unnatural isomorphism: A particular map between particular objects may be called an unnatural isomorphism (or "this isomorphism is not natural") if the map cannot be extended to a natural transformation on the entire category. From here: https://en.wikipedia.org/wiki/Natural_transformation There's an excellent question here: https://math.stackexchange.com/q/622589/816960 which raises the issue that it seems to be an artefact of the definition of a natural transformation that there is no canonical isomorphism between $V$ and $V^*$, since the dual functor is contravariant. One of the answers suggests that the only way to resolve this is by showing that every dinatural transformation from the identity functor to the dual functor must be zero. Personally I feel this doesn't get around the issue, because it seems to require us to redefine a canonical isomorphism between objects in a category as one which can be extended to a nontrivial dinatural transformation, and abandon the definition given above. Otherwise we're still left with a "definition not applicable"-based proof. Another candidate is given here: https://mathoverflow.net/a/345148/175537 in which we change the definition of the dual functor. One way to get around this is by working instead with the core groupoid $\mathbf{Vect}_{core}$, consisting of vector spaces and invertible linear transformations, and defining $*:\mathbf{Vect}_{core} \to \mathbf{Vect}_{core}$ to be the functor taking $f:V \to W$ to $(f^{-1})^{\ast}: V^\ast \to W^\ast$, the linear adjoint of its inverse. Then one can ask whether the identity is naturally isomorphic to the covariant dual functor $\ast$. It is not. But it seems like in both cases, we need a different definition of something to prove the nonexistence of a canonical isomorphism. My questions are: Can we prove that any isomorphism between a finite-dimensional vector space and its dual is unnatural using the definition above, without appealing to the fact that the definition of a natural transformation doesn't allow comparisons of covariant functors with contravariant ones? Since I suspect the answer to the above is "no", does this mean the definition of "unnatural isomorphism" isn't quite right? what is the right definition of "canonical isomorphism" to use in order to do this properly with category-theoretic machinery? REPLY [15 votes]: Some people (including me) think that "canonical" should be synonymous with "natural on isomorphisms" or "Functorial on isomorphisms" (depending on if you are talking of a "canonical object" or a "canonical arrow"). Doing so solves the problem of variance in the definition. To be clear, by "Functorial on isomorphism" I justs means that we have a functor $F:Core(C) \to D$ where $Core(C)$ is the subcategory of $C$ containing all objects but only isomorphisms as arrows. And by "Natural on isomorphisms", I'm asking for things that are natural transformation between such functors, i.e. that satisfies the naturality condition only with respect to isomorphisms. If you look at the category of finite dimensional vector spaces and linear isomorphisms between them, here $V \mapsto V^*$ can be made into an actual (covariant) endofunctor of this category as you can fix the contravariance by inverting the morphisms, so that an invertible arrow $f:V \to W$ induces $(f^*)^{-1} : V^* \to W^*$. And there you can concretely show that there is no natural isomorphism between this functor and the identity. Indeed, chosing an isomorphism $V \simeq V^*$ gives you a non-degenerate bilinear form on $V$ and you can always find an automorphism of $V$ that does not preserves this bilinear form, which is exactly what the naturality on isomorphisms would mean! So at the end of they day, one recovers exactly the argument that Paul Taylor or Chris Schommer-Pries made in the comments, but starting with a concrete categorical definition of "canonical".<|endoftext|> TITLE: Cohomological dimension bounds on the fundamental group of a manifold QUESTION [8 upvotes]: Suppose $M$ is a (closed, connected, oriented, smooth) manifold. If $M$ is aspherical, i.e., if the inversal covering $\tilde{M}$ is contractible, $M$ is a $B\pi_1(M)$. This is often enforced by geometry, for instance it holds if $M$ admits a metric of non-positive sectional curvature (Cartan--Hadamard). We deduce that the inequality $$\text{vcd}(\pi_1(M)) \leq \text{dim}(M)$$ holds (it is actually an equality in this case). I learned recently that for a simply-connected Lie group $G$ and a lattice $\Gamma < G$, we can pick a maximal compact $K < G$, then $K \backslash G$ is contractible, hence $K \backslash G / \Gamma \sim_{\mathbf Q} B\Gamma$. Thus for $M = G/\Gamma$, the above inequality is also satisfied. Furthermore, and going in the same direction, Mostow proved that fundamental groups of arbitrary homogeneous spaces, if they are solvable, have rank at most the dimension of the space. The above inequality thus also holds in this case. However, it is (of course) easy to construct manifolds for which the above inequality fails, as any finitely presented group is the fundamental group of some $d$-manifold for all $d \geq 4$. Is there a common generalization of being aspherical or a homogeneous space of the type described that ensures that the above inequality is satisfied? What are other conditions on $M$ that ensure that it holds? REPLY [7 votes]: Let $M$ be an orientable manifold. The relation between the cohomological dimension of $\pi_1(M)$ and that of $M$ comes from the following: Proposition: $M$ has the homotopy type of a fibration over $B\pi_1(M)$ with fiber $\tilde M$. Proof: Let $B\pi_1(M) = E\pi_1(M)/\pi_1(M) $ be a classifying space for $\Gamma$ and consider the quotient $X = (E\pi_1(M)\times \tilde M )/\pi_1(M)$ where $\pi_1(M)$ acts diagonally on the product. One can view $X$ in two different ways: As a fibration over $M$ with fibers $E\pi_1(M)$. Since $E\pi_1(M)$ is contractible, this is a homotopy equivalence and we deduce that $H^\bullet(M) = H^\bullet(X)$. As a fibration over $B\pi_1(M)$ with fibers $\tilde{M}$. QED Now, the Leray--Serre spectral sequence computes (in theory) the cohomology of $M$ in terms of the cohomology of $\Gamma$ and $\tilde M$. Let us make the following assumptions: (i) $\Gamma$ has finite virtual cohomological dimension $d_1$ (ii) if $d_2$ is the cohomological dimension of $\tilde M$ then the cohomology group with twisted coefficients $$H^{d_1}(\pi_1(M), H^{d_2}(\tilde M))$$ is non-zero. (Here, $H^{d_2}(\tilde M)$ is seen as a $\pi_1(M)$-module). Then the non-zero terms of the second page of the Leray--Serre spectral sequence are all contained in the rectangle $[0,d_1] \times [0,d_2]$ and the top right corner term $E_k^{d_1,d_2}$ stabilizes at page $2$. One deduces that $X$ has cohomological dimension $d_1+d_2$ and $$H^{d_1+d_2}(X) = H^{d_1}(\pi_1(M), H^{d_2}(\tilde M)).$$ On the other side, $X$ is homotopy equivalent to $M$ and thus has cohomological dimension at most $\dim(M)$, with equality if and only if $M$ is closed. We conclude that $$d_1\leq \dim(M) - d_2~,$$ with equality if and only if $M$ is closed. I might be missing some condition to guaranty the convergence of the Leray--Serre spectral sequence, but I know that the argument works in the following setting: let $G/H$ be a homogeneous space an $\Gamma$ a discrete and torsion-free subgroup of $G$ acting properly discontinuously on $G/H$. Then $G$ acts trivially on $H^\bullet(G/H,\mathbb Z)$ by continuity, hence so does $\Gamma$. Applying the argument above to $M= \Gamma \backslash G/H$, one gets that $$\mathrm{coh.dim}(\Gamma)\leq \dim(G/H) - \mathrm{coh.dim}(G/H)~,$$ with equality if and only if the action of $\Gamma$ is cocompact. The cohomological dimension of $G/H$ is easily computed because $G/H$ retracts to $K_G/K_H$ where $K_G$ and $K_H$ are maximal compact subgroups of $G$ and $H$ respectively.<|endoftext|> TITLE: Gradient of solution to heat equation under evolving metric QUESTION [5 upvotes]: The following simple question came to me when I was studying the heat equation on a Riemannian manifold: Suppose $M$ is a closed Riemannian manifold and $g_t$ is a smooth family of Riemannian metrics on $M$. It's a well-known question the study the evolving heat equation on $M$: \begin{equation} \dfrac{\partial}{\partial t}f=\Delta_{g(t)} f \end{equation} Suppose that $V\subset TM$ is a fixed sub-bundle of the tangent space of $M$. Given that at time $t=0$, the gradient of $f_0$ belongs to $V$. Is it true that the gradient of $f$ remains in $V$ the whole time? I have a feeling that this is true because of the equalizing nature of the heat equation, but I don't know how to formally prove or disprove it. Does anyone have any thoughts or helpful references? Please apologize if this is a trivial matter. REPLY [3 votes]: Let me expand a bit on my comment to try to answer your question. In general, there is no relationship between the metrics $g(t)$ and the sub-bundle $V$, so it is possible to find examples where the gradient is contained in $V$ at an initial time but is not contained within it for future times. To get a positive answer, we need to assume that the sub-bundle $V$ interacts with the metric in a nice way. For instance, if $V$ is the tangent bundle of a totally geodesic foliation of $M$, then it might be possible to show that $\nabla f \subset V$ is preserved if you deform the metric by Ricci flow and the function by a heat equation. To be honest though, I don't know if this is the case or not. However, there is one situation where the answer is affirmative. Consider the case where the function $f$ is invariant under the action of some group of isometries $G$. At first, this might seem unrelated to sub-bundles, but the idea is that when the group is infinite, this condition implies that the $\nabla f$ must be perpendicular to the infinitesimal transformations induced by the isometries (picture $O(1)$ spinning a round $\mathbb{S}^2$ around its axis and $f$ only a function of the polar angle). The subset of $TM$ perpendicular to the infinitesimal transformations is not necessarily a bundle (as can be seen with this example), but $f(t)$ will remain invariant under the group action under Ricci flow (so $\nabla f$ remains in a subset of $TM$). To see this, we use the uniqueness of the Ricci flow. For compact manifolds, this was proven by Hamilton and for complete manifolds with bounded curvature, Ricci flow uniqueness is a theorem of Chen-Zhu [1]. These results imply that the isometries at the initial time are preserved in forward time, and when combined with the linearity of the heat equation, it follows that $f(t)$ is preserved under $G$ for positive time as well. Since you are asking this question about Ricci flow, I'm guessing that you might be interested in the case where the heat flow runs in reverse time (since that's how the $\mathcal{W}$-functional is defined). In this case, the flow \begin{equation} \dfrac{\partial}{\partial t}f=\Delta_{g(t)} f \end{equation} no longer preserves mass so you need an extra term to correct for that. After fixing that issue, you can use the same argument as before to show that $f$ remains invariant under the group $G$, except now we use Kotschwar's result showing backwards uniqueness for the Ricci flow [2]. [1] Chen, Binglong; Zhu, Xiping, Uniqueness of the Ricci flow on complete noncompact manifolds, J. Differ. Geom. 74, No. 1, 119-154 (2006). ZBL1104.53032. [2] Kotschwar, Brett L., Backwards uniqueness for the Ricci flow, Int. Math. Res. Not. 2010, No. 21, 4064-4097 (2010). ZBL1211.53086.<|endoftext|> TITLE: Boolean valued models in a general setting QUESTION [5 upvotes]: It is well known that Boolean valued models play significant roles for set-theoretic purposes. But how well-studied are Boolean valued models in a more general setting, as models for random first-order languages? For example, towards the end of chapter 0 of Bell's "Set Theory: Boolean Valued Models and Independence Proofs", Bell gives the definition of a Boolean-algebra-valued $\mathscr{L}$ structure where $\mathscr{L}$ is a first order language whose sole extralogical symbol is a binary predicate $P$: A Boolean-algebra valued $\mathscr{L}$ structure is a quadruple $\mathbf{M} = (M, eq, Q, B)$, where $M$ is a non-empty class, $B$ is a complete Boolean algebra and $eq$ and $Q$ are maps from $M \times M \rightarrow H$ satisfying, for all $m, n, m', n'\in M$, $$ eq(m, m)=1, eq(m, n)=eq(n,m), eq(m,n) \land eq (n, n') \le eq(m, n'), Q(m,n) \land eq(m, m') \le Q (m', n), Q(m,n) \land eq(n, n') \le Q(m, n') $$ For any formula $\phi$ of $\mathscr{L}$ and finite finite sequence $\mathbf{x} = $ of variables of $\mathscr{L}$ containing all the free variables of $\phi$, define for any Boolean-valued $\mathscr{L}$ structure $\mathbf{M}$ a map $$ ‖\phi‖^{M_x}: M^n \rightarrow B $$ recursively as follows: $‖x_p = x_q‖^{M_x} = \mapsto eq(m_p, m_q),$ $‖Px_px_q‖^{M_x} = \mapsto Q(m_p, m_q),$ $‖\phi \land \psi‖^{M_x} = ‖\phi‖^{M_x} \land ‖\psi‖^{M_x}$, and similarly for other connectives, $‖\exists y \phi‖^{M_x} = \mapsto \bigvee_{m \in M} ‖\phi(y/u)‖^{M_{ux}}(m, m_1, ..., m_n)$, $‖\forall y \phi‖^{M_x} = \mapsto \bigwedge_{m \in M} ‖\phi(y/u)‖^{M_{ux}}(m, m_1, ..., m_n)$ That definition, it seems to me, can be easily generalized to a random first order language with other predicate, constant, or function symbols. And Bells mentions that it can be shown that a formula $\phi$ is $\mathbf{M}$-valid for all $\mathbf{M}$ (a forluma $\phi$ is $\mathbf{M}$-valid just in case $‖\phi‖^{M_x}$ is identically 1) iff $\phi$ is provable in classical first-order logic. So I'm wondering if there are other interesting results on Boolean-valued models as models for arbitrary first order languages. Or in general, how well-studied is the theory of Boolean valued models, as models for random first order languages? How much of traditional model theory (the theory of 2-valued models of first order languages) can be generalized to Boolean valued models? Are there any books or articles on this topic? Thanks! REPLY [6 votes]: In 1970th there was more interest on this topic. See for example; Some aspects of Boolean-valued model theory Filter Constructions in Boolean Valued Model Theory On Chang's omitting types theorem in Boolean valued model theory Eastern model theory for Boolean-valued theories Also as you mentioned there are connections with sheaves, see Sheaves and Boolean Valued Model Theory.<|endoftext|> TITLE: Remark 5.4.2.15 in HTT QUESTION [5 upvotes]: In Remark 5.4.2.15 in Higher Topos Theory, Lurie explains in what sense an accessible functor $F:\mathcal{C}\rightarrow \mathcal{D}$ between accessible $\infty$-categories is "determined by small data". In particular he wants to show that $F$ is induced by a functor $F':\mathcal{C}^\kappa\rightarrow \mathcal{D}^\kappa$, where $\mathcal{C}^\kappa$ refers to the $\kappa$-compact objects of $\mathcal{C}$. He argues as follows. We begin by picking a regular cardinal $\tau$ such that $\mathcal{C}$, $\mathcal{D}$ and $F$ are $\tau$-accessible. Now $\mathcal{C}\simeq \mathrm{Ind}_\tau(\mathcal{C}^\tau)$, and so the universal property tells us that $F$ is induced by a functor $F':C^\tau\rightarrow \mathcal{D}$. This is where I lose the plot. Lurie then seems to claim that remark 5.4.2.13 implies that the image of $F'$ is contained in the $\tau'$-compact objects of $\mathcal{D}$ for some other $\tau'$. However all that remark 5.4.2.13 says is that given any regular cardinal $\tau'$, $\mathcal{D}^{\tau'}$ is small. We also know that the image of $F'$ is small, but nevertheless I don't see why this is enough to conclude what Lurie requires. So the question is the following: Why does the image of $F'$ land inside the $\tau'$-compact objects of $\mathcal{D}$ for some regular cardinal $\tau'$? Ps. After concluding that the image of $F'$ lands in $\mathcal{D}^{\tau'}$, Lurie claims that we can pick $\kappa>> \tau'$ and obtain our conclusion. But again I fail to understand how we still know that $F(\mathcal{C}^\kappa)\subset \mathcal{D}^\kappa$. REPLY [10 votes]: He is applying 5.4.2.13 to $C$ and not to $D$: Because $C^\kappa$ is essentially small, and each $F(c)$ is $\lambda_c$-compact for some $\lambda$; there is a $\kappa = \sup_{c \in C^\kappa} \lambda_c$ such that each $F(c)$ is $\kappa'$-compact (well, rather, take some regular cardinal bigger than $\sup_{c \in C^\kappa} \lambda_c$, and by $c \in C^\kappa$, I really, mean for $c$ running in a small set of object representing all isomorphisms class of objects in $C^\kappa$.<|endoftext|> TITLE: Backward heat equation and forward perturbed heat equation well posed? QUESTION [5 upvotes]: I consider the following scenario. Let $I$ be a compact interval in space and $f$ a nice function in the space $C^{\infty}(I)$. In the following we consider a self-adjoint realization of our operators on said interval. We can consider the perturbed heat semigroup $T=e^{(\Delta+f)}$ at fixed time $1$. The heat semigroup, as we all know is smoothing. The unbounded operator $S=e^{-\Delta }$ corresponding to the inverse heat semigroup is also well-defined by the functional calculus for self-adjoint operators. Now one could be tempted to think that the smoothing by the heat semigroup was enough that $ST$ was bounded as an operator on $L^2$ and this is true if there was no $f$. Is it still true with having the $f$ around? REPLY [4 votes]: No, this cannot be true if $f$ is just $C^\infty$. Let $u=e^{(\Delta+f)t}u_0$. At $t=1$, $u=e^{\Delta+f}u_0=e^\Delta v_0$ for some $v_0$. Then, by well known properties of the heat equation, $u$ is spatially analytic. Moreover, $u_t=e^{\Delta+f}(\Delta+f)u_0$. If $u_0$ is sufficiently smooth, then $( \Delta+f)u_0$ is in $L^2$, so $u_t$ would have to equal $e^\Delta w_0$ for some $w_0$. This implies that both $u$ and $u_t$ are analytic, which makes $f$ analytic wherever $u\neq 0$.<|endoftext|> TITLE: How to prove that the solution to $x^{x+1}=(x+1)^{x}$ is transcendental? QUESTION [29 upvotes]: I was asked by an high school student if there is an algebraic way to find the exact value of the solution to the equation \begin{equation}\label{eq} x^{x+1}=(x+1)^x \end{equation} Let us define that with the expression "algebraic way" the student really means "the solution $x$ to the equation is an algebraic number". Now my feeling is that $x$ has to be transcendental but I'm not able to see how to prove it. Note first that there is a unique solution $2 1$. One can then rewrite $x = \frac{a}{b-a}$ and $x+1 = \frac{b}{b-a}$. This gives $$ \frac{a}{b-a} = \left(\frac{b}{b-a}\right)^{\frac{a}{b}}. $$ This leads to $a^{b} (b-a)^{a} = b^{a} (b-a)^{b}$. However, since $\gcd(a,b) = 1$, if $p$ is a prime divisor of $b$, then $p$ cannot divide $a$ and $p$ also cannot divide $b-a$ (since if $p | b-a$ and $p | b$, then $p | b - (b-a) = a$). This makes $a^{b} (b-a)^{a} = b^{a} (b-a)^{b}$ impossible since there is a prime number dividing the right hand side that does not divide the left. This is a contradiction. QED Claim<|endoftext|> TITLE: Upper bounds on the irrationality measure of the arctan of an algebraic number QUESTION [7 upvotes]: Let $x$ be an algebraic number. Must $\arctan(x)/\pi$ have finite irrationality measure? Are there any useful upper bounds? REPLY [5 votes]: Let $\alpha=\frac{1+xi}{\sqrt{1+x^2}}$. There are some cases $(\arctan x) /\pi$ is rational. For example, $x=1, \sqrt3$. In these cases, $(\arctan x)/\pi$ has the irrationality measure $1$. These occur precisely when $\alpha$ is a root of unity. Since $x$ is algebraic, so is $\alpha$. Then $\arctan x = \arg \alpha = (\log \alpha)/i$. Using $\log(-1)=i\pi$, we have $(\arctan x)/\pi= (\log \alpha)/\log (-1)$. Here, we need a fixed determination of logarithms of complex numbers. Suppose that $\alpha$ is not a root of unity. We must have that $(\arctan x)/\pi$ is irrational. That is, there is no nonzero rational solutions to $\beta_1 \log \alpha + \beta_2 \log(-1) = 0$. By Gelfond-Schneider theorem, $(\log \alpha)/\log(-1)$ is transcendental. Thus, the irrationality measure of $(\arctan x)/\pi$ is at least $2$. Note that the transcendence of $(\arctan x)/\pi$ also follows from the argument below. For the upper bound of the irrationality measure, we apply Baker-Wustholz theorem. The logarithmic form $$L=\beta_1 \log\alpha + \beta_2 \log (-1)$$ is nonvanishing for any pair of integers $(\beta_1,\beta_2)\neq (0,0)$. Let $n=2$ and let $d$ be the degree of $\alpha$ over $\mathbb{Q}$. Then the theorem yields $$ \log |L| > -C(2,d)h'(\alpha)h'(-1)h'(L) $$ where $C(2,d)= 905969664 \ d^4\log(4d)$, $h'(-1)=1$, $h'(\alpha)$ is a modified height of $\alpha$ defined by $$ h'(\alpha)=\frac1d \max(h(\alpha), |\log \alpha|, 1) $$ and $h(\alpha)$ is the logarithmic Weil height of $\alpha$. We may use $h'(L)=\log (\max(|\beta_1|/b, |\beta_2|/b))+1$ although the definition is slightly different in the note. Also, $b=\gcd (\beta_1,\beta_2)$. Hence, there are positive numbers $A(d,\alpha)$, $B(d,\alpha)$ such that for any integer $p$ and any positive integer $q$, $$ \left|\frac{\log \alpha}{\log(-1)}- \frac pq \right|> \frac{A(d,\alpha)}{q^{B(d,\alpha)}}. $$ As we see above, the number $B(d,\alpha)$ is quite huge, but it gives an upper bound of the irrationality measure of $(\arctan x)/\pi$.<|endoftext|> TITLE: "Cute" applications of the étale fundamental group QUESTION [43 upvotes]: When I was an undergrad student, the first application that was given to me of the construction of the fundamental group was the non-retraction lemma : there is no continuous map from the disk to the circle that induces the identity on the circle. From this lemma, you easily deduce the Brouwer fixed point Theorem for the circle. This was (for me) one of this "WOOOOW" moments where you realize that abstract constructions and some seemingly innocuous functorial lemmas may yield striking results (especially as I knew a quite long and complicated proof of Brouwer Theorem in dimension 2 before taking this topology class). I was wondering if there exist (+ reference if they do) similarly "cute" applications of the construction of the étale fundamental group in Algebraic Geometry. Of course "cute" is not well-defined and may vary for each one of us, but existence of fixed points for the Frobenius morphism would I find especially cute. Any other relatively elementary result related to algebraic geometry over fields of positive characteristic will be appreciated! Edit : I am obviously curious of any application of the étale fundamental group endowed with the aforementioned "WOOOW feeling". However, I'd be really interested in examples I could explain to smart grad students who are taking a first (but relatively advanced) course in Algebraic Geometry. REPLY [14 votes]: I think a fantastic application of the étale fundamental group is in extensions of the Chabauty-Coleman(-Kim) method. The original idea was that we could bound the rational points on a curve $C$ by embedding it into the Jacobian and under good enough conditions (on the rank of the Jacobian compared to it's dimension), this leads to a proof of finiteness. Well, what happens if these conditions aren't met? The idea is that we can replace the Jacobian (which can be thought of as coming from the abelian quotient of the fundamental group) with more non abelian analogues. This goes under the name of Chabauty-Coleman-Kim and Jennifer Balakrishnan, for instance, has lots of recent work on this topic. Edit (adding in more details): The idea behind the original proof is simple: Suppose $C$ is a curve of genus $\geq 2$, $K$ is a number field and we want to prove that $C(K)$ is finite (Mordell's conjecture). For simplicity of notation, let me assume $K = \mathbb Q$ Under the Abel Jacobi embedding, we can think of $C$ as living inside the Jacobian ( with $g$ it's dimension) $$C(\mathbb Q) \subset J(\mathbb Q_p) \cong \mathbb Z_p^g\times\text{finite group}$$ If we consider the p-adic closure of the Mordell-Weil group (of rank $r$), $\overline{J(\mathbb Q)} \subset J(\mathbb Q_p)$ we get a lattice of rank $\leq r$ inside $\mathbb Z_p^g$ and let us suppose that $r+1 \leq g$. On the other hand $C(\mathbb Q_p)$ is a one dimensional p-adic curve and we have the inclusion: $$C(\mathbb Q) \subset C(\mathbb Q_p) \cap \overline{J(\mathbb Q)}$$ and we expect this intersection to be finite because we have a one dimensional thing intersection something of codimension at least 1. This can be made rigorous and even gives an effective bound under the assumption that $r + 1 \leq g$. That's reasonable straightforward at least in concept but unfortunately, the restriction that $r < g$ is pretty severe and often won't hold. The really cool idea is that we can reinterpret this method in terms of data intrinsic to the curve $C$, in particular in terms of it's étale fundamental group. This paper by David Corwin gives a much better/complete exposition but let me try my best nevertheless. We can reinterpret the Mordel-Weil group $J(\mathbb Q)\otimes\mathbb Z_p$ as living inside the cohomology group $H^1(\mathbb Q, T_p)$ where $T_p$ is the p-adic Tate module (ie, the abelian quotient of the fundamental group) and we identify which subgroup it corresponds to in terms of data intrinsic to $C$.What we need to do next is rewrite the entire Chabauty-Coleman method in terms of data that is intrinsic to $K$ (without reference to an Abelian variety) and then we can replace $T_p = \pi^{et}_C/[\pi^{et}_C,\pi^{et}_C]$ by a larger quotient by going deeper in the central filtration and it gets technically a lot more complicated but I hope that gives a general flavour of the idea.<|endoftext|> TITLE: Tangential harmonic $1$-forms are pullbacks of harmonic functions QUESTION [6 upvotes]: This question has also been posted on MSE, but maybe here is the right place to obtain an answer. Let $(M^3,g)$ be a compact connected oriented Riemannian $3$-manifold with nonempty boundary. The Hodge-de Rham Theorem says that there is an isomorphism between $H^1_{dR}(M)$, the first de Rham cohomology group of $M$, and the space $\mathcal{H}_N^1(M)$ of tangential harmonic $1$-forms on $M$, given by $$\mathcal{H}_N^1(M) = \{ \omega \in \Omega^1(M) : d \omega = 0,\, d^\ast \omega=0 \text{ and } \omega(\nu) = 0 \text{ on }\partial M\},$$ where $\nu$ is a unit normal for $\partial M$. Suppose that $H^1(M; \mathbb{Z}) \neq 0$. Then, since $\mathbb{S}^1$ is a $K(\mathbb{Z},1)$, there is a bijection $$\Phi : [M, \mathbb{S}^1] \to H^1(M; \mathbb{Z})$$ given by $\Phi([u]) = [u^\ast(d\theta)]$, where $d\theta \in \Omega^1(\mathbb{S}^1)$ is the volume element of $\mathbb{S}^1$. Now, de Rham Theorem and the Universal Coefficient Theorem give isomorphisms $$H_{dR}^1(M) \cong H^1(M; \mathbb{R}) \cong H^1(M;\mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{R}$$ How do I conclude from this information that there exists a smooth map $u : M \to \mathbb{S}^1$ such that $u$ is harmonic with Neumann condition and $u^\ast(d\theta)\in \Omega^1(M)$ is tangential and harmonic? REPLY [2 votes]: Any constant map satisfies the requirements of the final question :). More seriously, if you want to find tangential harmonic form of this type, which represents a given $[u]\in [M,S^1]\cong H^1(M,\mathbb{Z})$ in de Rham cohomology, then you can proceed as follows (unless I am missing something): Pick a smooth map $v:M\rightarrow S^1$ representing $[u]\in [M,S^1]$. Then $\sigma:=v^*d\theta$ is in the right de Rham cohomology class, if $\int_{S^1}d\theta=1$. By the relative Hodge-de Rham theorem, there is an exact form $df\in \Omega^1(M)$ such that $\tau:=\sigma+df$ is tangential and harmonic. Since we can write $df$ as $$df=(p\circ f)^*d\theta$$ (where $p:\mathbb{R}\rightarrow S^1$ is a covering map with $p^*d\theta=dt$, with $t$ as the standard coordinate on $\mathbb{R}$), we find that $\tau=(v+p\circ f)^*d\theta$ has the desired form. Moreover $v+p\circ f$ is harmonic since $\tau$ is $d^*$-closed and it satisfies the von Neumann condition since $\tau$ is tangential. Edit: As requested in the comments, here are some more details on how one concludes that $v+p\circ f$ is harmonic: A map $w:M\rightarrow S^1$ is harmonic if and only if for all local isometries $L:S^1\supset U\rightarrow \mathbb{R}$ the function $L\circ w:w^{-1}(U)\rightarrow \mathbb{R}$ is harmonic. A real valued function $g$ is harmonic if and only if $dg=g^*dt $ is $d^*$-closed (as the Laplacian reduces on functions to $d^*d$). If $g=L\circ w$, then $(L\circ w)^*dt= c w^*d\theta$ for a constant $c$. Combining these equivalences with the fact that $w^*d\theta $ is $d^*$-closed for $w=v+p\circ f$ we find that $v+p\circ f:M\rightarrow S^1$ is harmonic as desired.<|endoftext|> TITLE: Modular forms over $\mathbb{Z}$ vs modular forms with integral Fourier coefficients QUESTION [8 upvotes]: It is well known that the ring of modular forms over $\mathbb{C}$ is $$ \mathbb{C}[c_4,c_6] $$ where $$ c_4 = 1+240 q + \cdots,\qquad c_6 = 1-504 q - \cdots $$ are the standard Eisenstein series, and the discriminant $$ \Delta= q - 24 q^2 + \cdots $$ satisfies $$ c_4^3-c_6^2=1728 \Delta. $$ You can use elliptic curves over $\mathbb{Z}$ to define modular forms over $\mathbb{Z}$, and then the resulting ring of integral modular forms is known to be given by $$ \mathbb{Z}[c_4,c_6,\Delta]/(c_4^3-c_6^2-1728 \Delta), $$ cf. Deligne après Tate. My question is as follows: You can instead consider a subring of $\mathbb{C}[c_4,c_6]$ containing modular forms over $\mathbb{C}$ whose Fourier coefficients (in $q$) are all integers. Clearly this subring includes $\mathbb{Z}[c_4,c_6,\Delta]/(c_4^3-c_6^2-1728 \Delta)$. Is it known/shown/easy to see that they are actually equal? REPLY [6 votes]: Apparently it is a classic result, e.g. a paper of Igusa starts by stating this fact saying it's classic without citing any. A big more googling turned up that there's something called Victor Miller's basis which exactly does the job, implemented in Sage, see here and here. Let me present the proof here, which proceeds by induction in the degree. Let $M_k$ be the $\mathbb{C}$-vector space of modular forms of weight $k$, and $S_k$ be the $\mathbb{C}$-vector space of cusp forms of weight $k$ (i.e. those which vanish at $q=0$). It is a standard fact that the multiplication by $\Delta$ gives an isomorphism $M_k \simeq S_{k+12}$. We would like to prove that any element of $M_k$ whose $q$-expansion coefficients are all integral is a $\mathbb{Z}$-linear combination of monomials of $c_4$, $c_6$ and $\Delta$. For $0\le k < 12$ this can be proved by inspection. Now let us assume $k\ge 12$, and say $f\in M_k$ is a modular form with integral $q$-coefficients. When $k$ is odd, $f=0$, and there is nothing to prove. When $k$ is even, there is a pair of nonnegative integers $(a,b)$ such that $c_4^a c_6^b \in M_k$. Writing $f=f_0 + f_1 q + f_2 q^2 + \cdots$, we find $f-f_0 c_4^a c_6^b$ is in $S_k$ and also has integral $q$-expansion coefficients. Dividing by $\Delta$, this gives an element $g\in M_{k-12}$ which has integral $q$-expansion coefficients such that $f=g\Delta + f_0 c_4^a c_6^b$. Applying the inductive hypothesis, we are done.<|endoftext|> TITLE: commutativity of restriction and Gysin morphisms in a cartesian square QUESTION [5 upvotes]: Let $X, Y, Z$ be compact topological manifolds $f: Y \to X, g: Z \to X$ be embeddings of submanifolds meeting transversely and let $W = Y \times_X Z$: $$ \begin{array}{ccc} Y & \to^f & X \\ \uparrow^G & & \uparrow^g\\ W & \to^F & Z \\ \end{array} $$ My question is: How does one show that the two morphisms $F_! \circ G^*$ and $g^* \circ f_!$ from $H^*(Y, \mathbb Q)$ to $H^*(Z, \mathbb Q)$ coincide? Here $F_!, f_!$ are Gysin morphisms; let us not concern ourselves with degree shift for brevity. Using the natural morphisms ${\mathbb Q}_Y \to G_*{\mathbb Q}_W$ and ${\mathbb Q}_W \to RF^!{\mathbb Q}_Z$ and adjunction the first morphism can be represented as the composition of the following morphisms $$ RHom_Y({\mathbb Q}_Y, {\mathbb Q}_Y) \to RHom_Y({\mathbb Q}_Y, G_* {\mathbb Q}_W) \cong RHom_W(G^*{\mathbb Q}_Y,{\mathbb Q}_W) \cong RHom_W({\mathbb Q}_W,{\mathbb Q}_W) \to RHom_W({\mathbb Q}_W, RF^!{\mathbb Q}_Z) \cong RHom_Z(RF_! {\mathbb Q}_W, {\mathbb Q}_Z) \cong RHom_Z({\mathbb Q}_Z, {\mathbb Q}_Z) $$ and using the natural morphisms ${\mathbb Q}_Y \to Rf^!{\mathbb Q}_X$ and ${\mathbb Q}_X \to g_*{\mathbb Q}_Z$ the second morphism can be represented as the composition of $$ RHom_Y({\mathbb Q}_Y, {\mathbb Q}_Y) \to RHom_Y({\mathbb Q}_Y, Rf^!{\mathbb Q}_X) \cong RHom_X(Rf_!{\mathbb Q}_Y, {\mathbb Q}_X) \cong RHom_X({\mathbb Q}_X, {\mathbb Q}_X) \to RHom_X({\mathbb Q}_X, g_*{\mathbb Q}_Z) \cong RHom_Z(g^*{\mathbb Q}_X, {\mathbb Q}_Z) \cong RHom_Z({\mathbb Q}_Z, {\mathbb Q}_Z) $$ The base change theorem (Iversen, "Cohomology of sheaves" VII.2.6) gives an isomorphism of the corresponding functors, i.e. $RF_! \circ G^* \mathcal F \cong g^* \circ Rf_! \mathcal F$ functorially in the object $\mathcal F$ in the derived category. I am not sure how one aplies this statement about functors to deduce the statement about morphisms on cohomology. REPLY [3 votes]: Let me sketch how I would do it in the six functor formalism. Let us consider slightly more generally a cartesian square of spaces $$ \begin{array}{ccc} Y & \to^f & X \\ \uparrow^G & & \uparrow^g\\ W & \to^F & Z \\ \end{array} $$ in which: $f$ is proper we have fixed an isomorphism $f^! \mathbf Z = \mathbf Z[d]$ for some $d$ (i.e. $f$ is oriented) the map $G^\ast f^! \to F^! g^\ast$, obtained as the mate of the inverse of $g^\ast f_!\stackrel \sim\to F_!G^\ast $, is an isomorphism (transversality). In particular $F$ is also oriented. We have a commutative diagram of natural transformations between endofunctors of the derived category $D(X)$ (all functors derived from now on) $$ \begin{array}{ccc} f_!f^! & \to & \mathrm{id} \\ \downarrow & & \downarrow\\ g_\ast g^\ast f_!f^! & \to & g_\ast g^\ast \\ \end{array} $$ and I claim that if we apply this diagram to the unit object $\mathbf Z$ and take global sections then we obtain a commuting diagram $$ \begin{array}{ccc} H(Y)[d] & \to & H(X) \\ \downarrow & & \downarrow\\ H(W)[d] & \to & H(Z) \\ \end{array} $$ where the vertical maps are pullback in cohomology and horizontal are Gysin maps. The point is that using base change ($g^\ast f_! = F_! G^\ast$), transversality ($G^\ast f^! = F^! g^\ast$), and properness+commutativity (in the form $f_!G_\ast = g_\ast F_!$) it follows that $$ f_! G_\ast G^\ast f^! \cong g_\ast g^\ast f_!f^! \cong g_\ast F_!F^! g^\ast $$ where the first expression makes clear that the left vertical map is pullback in cohomology, the second makes clear commutativity of the square above, third makes clear that the lower horizontal map is the Gysin map. What is needed to finish the proof is to check that the isomorphisms above actually make the diagrams commute. So, first of all, both sides of $$f_!G_\ast G^\ast \cong g_\ast g^\ast f_!$$ receive a natural transformation from $f_!$, and one must check that the obvious triangle commutes. This means that we need a stronger statement than just base change in the form of the existence of some isomorphism $F_!G^\ast \cong g^\ast f_!$, we want more specifically that the map $g^\ast f_! \to F_! G^\ast$ obtained as the mate of $ f_! G_\ast\stackrel \sim \to g_\ast F_! $ is an isomorphism. Similarly both sides of $$ g^\ast f_!f^! \cong F_!F^! g^\ast $$ admit a natural transformation to $g^\ast$, and again one should check that the triangle commutes, and again it is because the transversality and base change isomorphisms are obtained as mates.<|endoftext|> TITLE: residue calculation for rational function QUESTION [8 upvotes]: A colleague and I are working on a problem and part of it comes down to evaluating the residue of a rational function. In particular, $$ \mathrm{Res} \left( z^{kn-1} \left( az^{m}+1 \right)^{-k}; r \right), $$ where $a$, $k$, $m$ and $n$ are positive integers satisfying $a \geq 2$ and $0 TITLE: Lemma 5.4.5.11 of HTT QUESTION [5 upvotes]: In Lemma 5.4.5.11 of HTT, the proof given relies on Lemma 5.4.5.10. However it seems that Lurie applies Lemma 5.4.5.10, which requires the given simplicial set to be contractable, to an arbitrary $\kappa$-small simplicial set. This seeming incongruity was pointed out in this question on mathoverflow 2 years ago. However an answer was never given, and therefore I thought I might re-ask this in a new question (Let me know if there is a better way to re-ask an unanswered question). Is there either a) a way to salvage the proof given, or b) a new proof which avoids the issue, or c) is the proof actually correct (and we're all being daft) REPLY [7 votes]: I think there is a typo in Lemma 5.4.5.11: $K$ is supposed to be $\tau$-small and not $\kappa$-small. Note that if $\tau < \kappa$ and $K$ is $\kappa$-small but not $\tau$-small then the statement of the lemma is simply false: e.g., set $\mathcal{I}=\mathcal{J}=K=\mathbb{N}$ to be the poset of natural numbers (with arrows pointing from small numbers to larger numbers) and take both $f$ and $p$ to be the identity. Set $\tau=\omega$ and $\kappa$ an uncountable cardinal. Then $\mathcal{I}_{p/}$ is empty, and so certainly not $\tau$-filtered. Note that the second statement in 5.4.5.11 is independent of this problem, and its proof does not use any size bound on $K$ (there is a small typo though in the first line of the proof of (2), "where $K$ is now $\kappa$-small and weakly contractible" -> $K$ should be $K'$. Note also that 5.4.5.11 is cited two times in HTT, once in 5.4.5.12 and once in 5.4.6.5, but in both cases $K=\Delta^0$ and is in particular $\tau$-small. The issue will hence in principle be resolved if we prove 5.4.5.11 under this modified assumption: Proof of 5.4.5.11(1) when $K$ is $\tau$-small Let $\tilde{q}:K' \to \mathcal{I}_{p/}$ be a $\tau$-small diagram classifying a compatible pair of maps $q: K' \to \mathcal{I}$ and $q':K \star K' \to \mathcal{J}$. Since $\mathcal{I}$ is $\tau$-filtered we can find an extension $\overline{q}:(K')^{\triangleright} \to \mathcal{I}$ of $q$. To facilitate notation later on let us write $L:= (K')^{\triangleright}$ and let $l \in L$ be the vertex corresponding to the cone point of $(K')^{\triangleright}$. Now $q'$ and $f\overline{q}$ combine to give a map $r:[K \star K'] \coprod_{K'}L \to \mathcal{J}$. Since $\mathcal{J}$ is $\tau$-filtered and $K, L$ are $\tau$-small we can find an extension of $r$ to a map $$ \overline{r}:\Big[[K \star K']\coprod_{K'} L \Big]^{\triangleright} \to \mathcal{J} .$$ Let $x := \overline{q}(l)$ and $\alpha: f(x) \to y$ be the arrow corresponding to the restriction of $\overline{r}$ to $\Delta^1 = \{l\}^{\triangleright} \subseteq L^{\triangleright}$. Since $f$ is $\kappa$-cofinal there exists an arrow $\beta: x \to z$ in $\mathcal{I}$ and a map $\eta:\alpha \to f(\beta)$ in $\mathcal{J}_{f(x)/}$. Since the inclusion $\{l\} \subseteq L$ is right anodyne we have that $L\coprod_{\{l\}} [\{l\} \star \Delta^0] \subseteq L \star \Delta^0$ is inner anodyne and so may now extend the map $L\coprod_{\{l\}} [\{l\} \star \Delta^0] \to \mathcal{I}$ determined by $\overline{q}$ and $\beta$ to a map $\overline{q}_{\beta}:L \star \Delta^0 \to \mathcal{I}$. Next since $L \coprod_{\{l\}} \{l\}^{\triangleright} \subseteq L^{\triangleright} \subseteq \Big[[K \star K']\coprod_{K'} L \Big]^{\triangleright} $ is a sequence of an inner anodyne map followed by a right anodyne map we may extend the map $$ \Big[[K \star K']\coprod_{K'} L \Big]^{\triangleright} \coprod_{L} [L \star \Delta^0] \coprod_{\{l\} \star \Delta^0} [\{l\}^{\triangleright} \star \Delta^0] = $$ $$ \Big[[K \star K']\coprod_{K'} L \Big]^{\triangleright} \coprod_{\big[L \coprod_{\{l\}} \{l\}^{\triangleright}\big]} \Big[\big[L \coprod_{\{l\}} \{l\}^{\triangleright}\big] \star \Delta^0\Big]\to \mathcal{J} $$ determined by $\overline{r}$, $f\overline{q}_{\beta}$ and $\eta$ to a map $$ \overline{r}_{\eta}:\Big[[K \star K']\coprod_{K'} L \Big]^{\triangleright} \star \Delta^0 \to \mathcal{J} .$$ The restriction of $\overline{r}_{\eta}$ to $K \star K'\star \Delta^0$ and the restriction of $\overline{q}_{\beta}$ to $K'\star \Delta^0$ now determine an extension of $\tilde{q}:K' \to \mathcal{I}_{p/}$ to a map $K'\star \Delta^0 \to \mathcal{I}_{p/}$, as needed.<|endoftext|> TITLE: Where to publish a long classification? QUESTION [11 upvotes]: Suppose that the classification of some mathematical (say algebraic) notions requires (say) 70 pages. Let clarify that (say) 90% of the pages are used to write the result itself, whereas only 10% are required to introduce the notions, to mention some theorems and to explain the computer programs used. Finally, assume that the data of this classification is useful to some (pure) mathematicians, and also to some physicists (like an atlas). Question: Where to published such a classification? (refereed) book or journal? Where? The classification I have in mind: multiplicity-free complex fusion categories up to rank $6$, and braiding structures. REPLY [10 votes]: Perhaps I can try a suggestion: a journal which accepts long papers on the topics you sketched, aimed towards pure and applied mathematicians (including physicists and engineers), which has also an illustrious history is the Journal published by Castelnuovo Institute of Mathematics of the Sapienza University of Rome, i. e. the "Rendiconti di Matematica e delle sue Applicazioni". On the homepage of the Journal, you can read the following statements: The Journal "Rendiconti di Matematica e delle sue Applicazioni" is regularly issued since 1914. The Journal traditionally publishes high-quality research articles in Pure and Applied Mathematics and adheres to the EMS Code of Practice. Articles of any length are considered for publication. Submission of surveys, of articles of foundational nature, of doctoral dissertations etc. is also encouraged. Every article submitted is subjected to a first screening by the Editorial board: if the manuscript meets the journal’s basic requirements, it will be sent to a referee for a single blind peer review process. Once a paper is accepted it goes immediately into production and published online shortly after the approval of galley proofs. I am almost sure that the impact factor of the journal is not "between the top ones": however in the past it has published several important papers and many illustrious mathematicians have been its directors (Volterra, Severi, Segre, Fichera, to cite a few names). Well, my two cents.<|endoftext|> TITLE: The first female algebraist in US/Britain? QUESTION [21 upvotes]: Recently I dug up some biographical details of Lindsay Burch, of Hilbert-Burch Theorem fame, whose few papers have had quite an impact on commutative algebra. This made me curious about the first women who obtained PhDs in abstract algebra in the US and Britain. Question 1: Who was the first woman to get a PhD in the US on a topic in algebra? Potential answer: Mildred Sanderson, who obtained her PhD from Dickson in 1913 (published in Annals of Mathematics), and tragically died one year later. I found her name in Bell's "50 years of algebra in America, 1888-1938". Is that correct? For Britain, I could not find any similar information. Burch got her PhD in 1967 under David Rees, but surely other women got British PhDs in algebra earlier. So: Question 2: Who was the first woman to get a PhD in Great Britain on a topic in algebra? REPLY [16 votes]: I think a very honourable mention is the American Ida May Schottenfels, who did not receive a PhD, but who nevertheless was very active in mathematical research: Ida May Schottenfels graduated from Northwestern University with a Bachelor of Arts degree in 1892. She then studied at Yale and the University of Chicago, earning a Masters degree in mathematics from Chicago in 1896. [...] In 1910 she was appointed head of the department of mathematics in Toledo University. In their study of "Women in the American Mathematical Research Community: 1891-1906" Fenster and Parshall cite Schottenfels as one of the two most "active" participants (along with Charlotte Angas Scott), listing her as giving 17 talks at mathematics conferences, publishing 3 papers, and attending 23 meetings and/or colloquia including those of the American Mathematical Society. She was also the first mathematician to prove that there are two finite simple groups of the same order (namely of order $20160$). So although she did not receive a PhD, she surely seems to have deserved one!<|endoftext|> TITLE: Infinitely long rods that touch one another QUESTION [5 upvotes]: Background: The basic question as given in 'Research Problems in Discrete Geometry' By Moser, Brass and Pach (page 98) is: What is the max number of congruent infinite circular cylinders that can be arranged in 3d space so that every pair is touching? Several partial results are known including a proof by Bezdek that this number cannot exceed 24. Questions: What if we relax this question to infinite circular cylinders with not necessarily equal radii? What happens if we look at congruent and infinitely long prisms all with same cross sections (triangular, square etc..)? If each infinitely long object only needs to be of uniform cross section but could have mutually differently shaped cross sections what can one say? Can one have an arbitrarily large number of convex and mutually congruent solid objects that can all touch one another? (Note: last question above is known to have "yes" answer as given in the comments below). REPLY [3 votes]: "an arbitrarily large number of objects that can all touch one another" I think one must be careful to define "object" and "touch." Otherwise, all these $n$ line segments (objects) intersect (touch) at the origin:       But this example violates the spirit of the OP's questions.<|endoftext|> TITLE: Does $0\to I\to\mathrm{Gal}_K\to\mathrm{Gal}_k\to 0$ always split? QUESTION [12 upvotes]: Let $K$ be a henselian valuation field with residue field $k$, then the decomposition group surjects onto Galois group of the residue field, with kernel the inertia subgroup, namely we have short exact sequence:$$0\to I\to\mathrm{Gal}_K\to\mathrm{Gal}_k\to 0$$ When $K$ is a local field, we can split the sequence by lifting the Frobenius; when $K=k((t))$, we can split the sequence by lifting the Galois action (with trivial action on $t$). But in general, do we know if the sequence always split? (The splitting of the sequence is mentioned as a well-known fact in Proposition A5 in "Exposant et indice d'algèbres simples centrales non ramifiées", but I couldn't find a reference.. Splitting of the sequence would imply that the restriction morphism $H^i_{et}(\mathrm{Spec}(V),M)\to H^i_{et}(\mathrm{Spec}(K),M)$ is injective for any locally constant sheave $M$ on the valuation ring $V$.) REPLY [13 votes]: Good question! Let me try to guess what Gabber had in mind there. (Note that he only says "known" (to him), not "well-known"...) The claim is that the extension splits. Note that to prove this, we are free to replace $K$ by any (algebraic) extension $K'$ whose residue field $k'$ is purely inseparable over $k$. By Zorn's lemma, we can choose $K$ so that it admits no further such extensions. In particular, $K$ is perfect and the value group is divisible. Let $p$ be the characteristic of $k$. Then it follows that $I$ must be pro-$p$, as the maximal unramified extension $K^{\mathrm{ur}}$ of $K$ will still have divisible value group, and so have no nontrivial tame extensions. Now, if $I$ was nonzero, there is a map $I\to \overline{I}$ where $\overline{I}$ is a nonzero $\mathbb F_p$-vector space (the quotient by the Frattini subgroup). By maximality of $K$, the induced sequence $$0\to \overline{I}\to \overline{G}\to \mathrm{Gal}_k\to 0$$ is nonsplit, so gives a nonzero class in $H^2(\mathrm{Gal}_k,\overline{I})$. But Galois groups in characteristic $p$ have $p$-cohomological dimension $\leq 1$ by Artin-Schreier theory.<|endoftext|> TITLE: Why does abelianization preserve finite products, really? QUESTION [19 upvotes]: The abelianization functor $(-)^{ab} : \mathrm{Grp} \to \mathrm{Ab}$ is left adjoint to the inclusion of abelian groups into groups. As such, it preserves all colimits, but it doesn't generally preserve limits (e.g. the mono $A_3 \hookrightarrow S_3$ is not preserved). Now curiously, it does preserve finite products, that is $$(G \times H)^{ab} \cong G^{ab} \times H^{ab}$$ I am trying to find an intuition why this is true from an abstract point of view: Which special properties of (abelian) groups really go into proving this? Before I generalize, let's recall one possible proof of product preservation: The equivalence relation for the abelianization of $G$ can be described as follows: Let $g \sim g'$ if for all homomorphisms $\phi : G \to A$ into abelian groups, we have $\phi(g) = \phi(g')$. Write $[g]$ for equivalence classes. This is a congruence and $G/\!\sim$ has the universal property of the abelianization. To show that the comparison morphism $(G \times H)^{ab} \to G^{ab} \times H^{ab}$ given by $[(g,h)] \mapsto ([g],[h])$ is an iso, we attempt to show that the obvious inverse $([g],[h]) \mapsto [(g,h)]$ is well-defined: That is, if $g \sim g', h \sim h'$ then $(g,h) \sim (g',h')$. Let $\phi : G \times H \to A$ be a homomorphism into an abelian group, then $\phi(-,1), \phi(1,-)$ are homomorphisms separately. Hence as desired $$\phi(g,h) = \phi((g,1)\cdot(1,h)) = \phi(g,1) \cdot \phi(1,h) = \phi(g',1)\cdot\phi(1,h') = \phi(g',h')$$ Note I had to use the neutral element as a way of relating homomorphisms out of a product to homomorphisms of the factors. Towards a generalization, let us look at a different example: A convex sets is a set equipped with convex-combination operations $x +_r y$ for $0 \leq r \leq 1$. A semilattice is a convex set where $+_{1/2}$ is associative. This forces all operations $+_r$ for $0 < r < 1$ to become equal, associative and a semilattice operation, jointly written $\vee$. Like abelianization, the inclusion of semilattices into convex sets has a left adjoint $(-)^{col} : \mathrm{Cx} \to \mathrm{Sl}$ which "collapses" probability to possibility so to speak, identifying all points which lie on the interior of some line segment. Again, this operation preserves products, but now for a slightly different reason: We don't have a neutral element, however for $\phi : X \times Y \to Z$ and fixed $y$, $\phi(-,y)$ is a homomorphism because all operations are idempotent ($y+_r y=y$). I arrive at my general question: Fix some signature. Let $T$ be an algebraic theory and $T' \supset T$ a super-theory over the same signature. The inclusion of $T'$-algebras into $T$-algebras has a left adjoint, freely enforcing the additional $T'$-equations. When does this left adjoint preserve products? From the above examples this seems to require particular properties of the respective theories. Yet, I don't know an example where product preservation does not hold, and would appreciate one if it exists. I'd also like other (category-theoretic) insights into the abelianization case. Possible lines of thought: The result doesn't seem to be a straightforward instance of the fact that reflexive coequalizers commute with finite products, though certainly coequalizers are involved. This proof to the abelianization case constructs the inverse $G^{ab} \times H^{ab} \to (G \times H)^{ab}$ by using the fact that in $\mathrm{Ab}$, the product $G^{ab} \times H^{ab}$ is actually a biproduct, by which it is enough to find morphisms $G^{ab},H^{ab} \to (G \times H)^{ab}$ separately. Again, this situation seems rather particular. REPLY [3 votes]: Here's an answer which doesn't really seem to lend itself to thinking about general algebraic theories, but does put this result in some larger context: think about it topologically! Review of the Hurewicz theorem for $n=1$: Consider the Hurewicz theorem for $n=1$, which says that for any based space $X$, we have $H_1(X) = \pi_1(X)^{ab}$. Working simplicially or cellularly, this case of the theorem is totally elementary to prove: given $X$, the van Kampen theorem implies that a presentation of its fundamental group is given by taking one generator for each nondegenerate 1-cell in the connected component of the basepoint, and a relation is imposed for each nondegenerate 2-cell between these. On the other hand, $H_1(X)$ is the first homology of the free simplicial abelian group on the simplicial set $X$, i.e. it has one generator for each 1-cell and a relation $d_1(x) = d_2(x) + d_0(x)$ for each 2-cell. That is, it is the free abelian group with the same presentation. Since abelianization is a left adjoint, it preserves the coproducts and coequalizers implicit in saying that these are "presentations", and so it carries the former to the latter. Now I claim that from this description, we can read off the fact that abelianization preserves finite products. For we are saying that if $G$ is a group, then $G^{ab} = H_1(BG) = \pi_1(F_{\mathbb Z}(BG))$. So we've broken $(-)^{ab}$ down as a composite of 3 functors: $B: Grp \to sSet$ really factors as the composite $B = N\mathbb B$, where $\mathbb B: Grp \to Cat$ carries a group $G$ to $\mathbb B G$, i.e. the same group considered as a 1-object category, and $N: Cat \to sSet$ is the nerve functor. Both of these functors are right adjoints, and in particular preserve finite products. $F_{\mathbb Z}: sSet \to sAb$ is the functor which carries a simplicial set $X$ to the free abelian group $F_{\mathbb Z}X$ on $X$ (viewing $X$ as a functor $\Delta^{op} \to Set$, this is just postcomposing with the free abelian group functor $Set \to Ab$). Since the free abelian group functor is a strong symmetric monoidal functor $(Set,\times) \to (Ab,\otimes)$, it follows that $F_{\mathbb Z}$ is a strong symmetric monoidal functor $sSet \to sAb$. At this point, one should worry a little bit -- the composite $F_{\mathbb Z} B: Grp \to sAb$ is strong symmetric monoidal, but with respect to a different monoidal structure on $sAb$ than the cartesian one ($\otimes \neq \times$). Never fear: The functor $\pi_1: sAb \to Ab$ doesn't carry tensor products to cartesian products in general. However, when $A = F_{\mathbb Z} X$, $B = F_{\mathbb Z} Y$ are the free simplicial abelian groups on simplicial sets $X,Y$ which each have a unique 0-cell, we get that $(A \otimes B)_0 = \mathbb Z$ and $(A\otimes B)_1 = A_0 \otimes B_1 \oplus A_1 \otimes B_0 = B_1 \oplus A_1$. Similarly, we have $(A\otimes B)_2 = A_0 \otimes B_2 \oplus A_1 \otimes B_1 \oplus A_2 \otimes B_0 = B_2 \oplus A_1 \otimes B_1 \oplus A_2$; differentials from the first and third summands enforce relations on $B_1,A_1$ to cut down to $H_1(B),H_1(A)$, while differentials from the middle term ensure that sums from the two factors commute with each other. The picture I want to get across is that very roughly (this is actually true over a field) we have $H_n(X \times Y) = \oplus_{i+j} = H_i(X) \otimes H_j(Y)$; when $n=1$ the only terms we have to pick up are $H_1(X) \otimes H_0(Y)$ and $H_0(X) \otimes H_1(Y)$, whose sum gives the desired cartesian product.<|endoftext|> TITLE: Which graphs on $n$ vertices have the largest determinant? QUESTION [34 upvotes]: This is a question that seems like it should have been studied before, but for some reason I cannot find much at all about it, and so I am asking for any pointers / references etc. The determinant of a simple graph is the determinant of its adjacency matrix which is, of course, a symmetric binary matrix with zero-diagonal. Question: Which graph has the largest value of $|\det(A)|$ over all simple undirected graphs on $n$ vertices? (We could ask separately for the graphs with minimum determinant, which will be some negative number, and those of maximum determinant, but by taking the absolute value, this just becomes a question of which graphs have the most extreme determinant.) For small numbers of vertices, we can just find the answer almost by hand: On $2$ vertices, the winner is the complete graph $K_2$ with spectrum $\{-1,1\}$, so determinant $-1$, of absolute value $1$. On $3$ vertices, the winner is $K_3$ which has absolute value of determinant equal to $2$ On $4$ vertices, the winner is $K_4$ with value $3$. On $5$ vertices, there are two winners (both with value $4$), namely $K_5$ and the "bowtie" graph (two triangles sharing a common vertex). Sequence so far is $1$, $2$, $3$, $4$ (don't bother looking this up in OEIS). For larger numbers of vertices, we turn to the computer and discover that on $6$ vertices, the maximum value is $7$, and this is realised by two graphs, namely a triangle and a $4$-clique sharing a vertex, and two $4$-cliques sharing an edge. From there, the sequence continues: 12, 28, 128, 256, 576 for 7, 8, 9, 10 and 11 vertices respectively, with between 2 and 7 graphs achieving the maximum for each of these values. So now I do have enough to look up in the OEIS, but there are no matches. The problem of finding the maximum determinant of a (0,1)-matrix, often called the Hadamard maximal determinant problem has been extensively studied, and there are lots of bounds, and constructions of extremal matrices etc. (There is a mapping between (0,1)-matrices and (-1,1)-matrices which changes the determinant predictably and when they exist, Hadamard matrices are the winners.) So lots is known, and it is summarised in sequence A003432 on the OEIS which gives exact values up to $n=20$. Just for comparison, for $n = 6$ to $n=11$, we have 7, 12, 28, 128, 256, 576 (for graphs) 9, 32, 56, 144, 320, 1458 (for (0,1)-matrices) From several sources, including the OEIS, I have been advised to check a website attributed to Will Orrick at http://www.indiana.edu/~maxdet/ but this appears to be offline or removed or something, and nor could I find his email address in the directory for that university. So my question remains: what is known about maximal determinant (in absolute value) of the adjacency matrix of a simple graph on $n$ vertices? REPLY [2 votes]: This may not be responsive, but I think it's amusing. I apologize in advance. To get an order 16 graph with determinant 327680 begin with (a) the line graph of $K_5$, (b) $K_5$, and (c) an isolated point. Join all vertices in (a) to (c) and no vertices in (b) to (c). Join each vertex in (b) to a maximum clique [of size 4] in (a). The graph6 format is: OV`vfmlJJNhRVDmdzkUVX<|endoftext|> TITLE: Positivity of Schur elements in Iwahori-Hecke algebras QUESTION [6 upvotes]: I'm interested in finite Iwahori-Hecke algebras. If $\mathcal{H}$ is such a Hecke algebra, defined over $\mathbb{Z}[q^{\pm 1/2}]$, and $\Lambda$ an irreductible representation, there is the notion of a Schur element $S_\Lambda$. Roughly speaking, to $\Lambda$ you can associate a central element whose matrix in $\Lambda$ is given by $S_\Lambda$ times the identity. I computed several of these Schur elements in type $A_n$ and I noticed a remarkable positivity property: all the coefficients in $q$ seem to be positive (see below)! I looked up in the literature, but I don't find explicitely this property. Is there a concrete reference for this positivity? Does it hold in other types? Here a small list of Schur elements: For $\mathfrak{sl}_2$: $1+q$, $1+q^{-1}$ For $\mathfrak{sl}_3$: $q+1+q^{-1}$, $1+2q+2q^2+q^3$, $1+2q^{-1}+2q^{-2}+q^{-3}$ For $\mathfrak{sl}_4$: $[4]!$ (quantum factorial), $q^2+3q+4+3q^{-1}+q^{-2}$, $q^{-1}+2+2q+2q^2+q^3$, ... REPLY [2 votes]: Together with Maria Chlouveraki, we determined the answer: The Schur elements are positive in all classical types $A_n, B_n$ and $D_n$ (use the formula of Theorem 4.3. in https://hal.archives-ouvertes.fr/tel-01411063/document, which uses generalized hook lengths). For dihedral groups $I_2(m)$, Theorem 8.3.4. in the book of Geck-Pfeiffer on characters of finite Coxeter groups and Hecke algebras shows that the Schur elements have negative coefficients for $m>4$. For the exceptional cases, by brute force, there are Schur elements with negative coefficients only in type $H_4, F_4$ and $E_8$.<|endoftext|> TITLE: What is the topological Hochschild cohomology of $\mathbb{F}_p$? QUESTION [6 upvotes]: Following the computation of the THH (topological Hochschild homology) of $\mathbb{F}_p$ as outlined in Krause-Nikolaus. We use the fact that $\mathbb{F}_p$ is initial $E_2$ ring with $0=p$ to compute $$\mathbb{F}_p \otimes_{\mathbb{S}} \mathbb{F}_p \cong \mathbb{F}_p[{\Omega^2 S^3}]$$ Then, $$THC(\mathbb{F}_p) \cong \mathrm{Hom}_{\mathbb{F}_p[{\Omega^2 S^3}]}(\mathbb{F}_p,\mathbb{F}_p)$$ Now using $$\mathrm{colim}_{BG} {G} = \{ \star\}$$ We have that $$\mathrm{colim}_{\Omega S^3} {\mathbb{F}_p[{\Omega^2 S^3}]} = \mathbb{F}_p$$ and hence, $$THC(\mathbb{F}_p) \cong \lim_{\Omega S^3}{\mathbb{F}_p} \cong \mathbb{F}_p^{\Omega^2 S^3}$$ where the action is trivial because $\mathbb{F}_p$ is discrete. Assuming the above is correct, I now am not sure how to compute this. REPLY [8 votes]: Let me write $HH^S(B) = THH(B) = B \wedge_{B^e} B$ for topological Hochschild homology, and $HH_S(B) = F_{B^e}(B, B)$ for topological Hochschild cohomology, where $B^e = B \wedge_S B^{op}$. For $B$ commutative the $B^e$-module action on $B$ factors through $\mu : B^e \to B$, so by adjunction we have $F_{B^e}(B, B) \cong F_B(B \wedge_{B^e} B, B)$. Hence $HH_S(B) \cong F_B(HH^S(B), B)$. For $B = H\mathbb{F}_p$ we have $\pi_* HH^S(B) = HH^S_*(B) = \mathbb{F}_p[x]$ with $|x|=2$ a primitively generated Hopf algebra, so dually $HH_S^{-*}(B) = \pi_* HH_S(B) = Hom_{\mathbb{F}_p}(\mathbb{F}_p[x], \mathbb{F}_p) = \Gamma_{\mathbb{F}_p}(\xi)$ is a divided power algebra with $\xi$ dual to $x$. Closer to the approach you outline: The final lim calculates the mod $p$ cohomology of $\Omega S^3$, which is the Hopf algebra dual to the mod $p$ homology of $\Omega S^3$. Either one can be calculated with the Wang sequence, or Serre spectral sequence, or by reference to the James construction.<|endoftext|> TITLE: Using HoTT, why is twisted cohomology of BG group cohomology? QUESTION [5 upvotes]: I've been reading Michael Shulman's blog posts defining cohomology in homotopy type theory, and I'd like to understand (using HoTT) why cohomology of BG is group cohomology. if I understand correctly, given a parametrized spectrum (i.e. a fibration by spectra) $E: X \to \mathsf{Spectra}$, we define the twisted cohomology of $X$ with coefficients in $E$ to be $H^n(X; E) \equiv \Vert \prod_{x:X} \Omega^{-n} E_0 \Vert_0$. In particular, if we have a parametrized family $V: X \to \mathsf{AbGroup}$ then we can compose with the Eilenberg-MacLane construction $H: \mathsf{AbGroup} \to \mathsf{Spectra}$ to get a parametrized family of Spectra $HV: X \to \mathsf{Spectra}$. The cohomology $H^n(X; HV)$ is cohomology with local coefficients, which is the twisted version of ordinary cohomology. Now if we consider the case $X = BG$ (i.e. $BG=K(G,1)$ ) for $G$ a set-group, then a parametrized family $V: BG \to \mathsf{AbGroup}$ is the same as a group representation of $G$, since functions are functorial on paths in HoTT. In other words, given $g: \bullet = \bullet$, we get a path $g_*: V(\bullet) = V(\bullet)$. Now, if we consider the corresponding twisted cohomology $H^n(BG; HV) \equiv \Vert \prod_{x:BG} K(V(x);n) \Vert_0$, why do we get group cohomology? For now let's just consider $H^0(BG; HV) \equiv \Vert \prod_{x:BG} V(x) \Vert_0 = \prod_{x:BG} V(x)$, where the second equality follows because $V(x)$ is a set. In order to get group cohomology, it should be the case that any $v: \prod_{x:BG} V(x)$ encodes a $G$-invariant element of the $G$-representation. But it isn't immediately obvious to me why this should be the case. Any help would be greatly appreciated! REPLY [3 votes]: There's not really much to say (which is why I'm marking this answer as CW): As Phil Tosteson commented, forming $\Pi$-types is right adjoint to the constant family map $\mathcal{U} \to \mathcal{U}^{BG}$ (and $\Sigma$-types gives the left adjoint). So that's the universal property of invariants (and co-invariants). Perhaps it would help to see what happens if $G$ is a $1$-group: We can present $BG$ as a HIT with constructors $$ \text{pt} : BG, \quad \text{loop} : G \to \text{pt}=\text{pt},\quad \text{loop-cmp} : \prod_{g,h:G}\text{loop}(g\cdot h) = \text{loop}(g)\cdot\text{loop}(h), $$ as well as a constructor forcing $BG$ to be a $1$-type. If $V : BG \to \text{Set}$ is any $G$-set, then by the universal property of $BG$ as a HIT, we get an equivalence $$ \prod_{t:BG}V(t) \simeq \sum_{v:V(\text{pt})}\prod_{g:G}g\cdot v = v.$$ (Since $V$ is a family of sets, there's nothing to do for the $\text{loop-cmp}$ constructor.)<|endoftext|> TITLE: Galois group of a polynomial modulo $p$ QUESTION [5 upvotes]: It is well known that if $f(x)$ is a polynomial over $\mathbb Z$ then for every prime $p$ (not dividing the discriminant of $f$ (thanks to KConrad)) the Galois group of that polynomial mod $p$ over $\mathbb{F}_p$ embeds into the Galois group of $f$ over $\mathbb{Q}$. Where can I find a (easy) proof of this fact? REPLY [2 votes]: I'm not a number theorist, but I did find the Tate proof enlightening because it avoids the machinery of Dedekind domains, although at heart the proof is the same as in @KConrad's linked note. To the best of my understanding, which might be faulty, the only thing it uses that is not typically seen before Galois theory is that the algebraic integers (complex numbers which are roots of monic integer polynomials) form a ring and that the only rational algebraic integers are the actual integers. The way I understand the proof is like this. Let $f$ be a monic integer polynomial which we can assume doesn't have repeated roots and suppose that $p$ is a prime not dividing the discriminant This is essential since you need $f$ to not have multiple roots modulo $p$. Let $\alpha_1,\ldots,\alpha_n$ be the roots of $f$ in $\mathbb C$, $L=\mathbb Q(\alpha_1,\ldots, \alpha_n)$ and let $G$ be the Galois group of $L$ over $\mathbb Q$ (i.e., of $f$). We can think of $G$ as a faithful group of permutations of $\alpha_1,\ldots,\alpha_n$. Now let $\mathcal O=\mathbb Z[\alpha_1,\ldots, \alpha_n]$. This is the one step where Tate differs from the classical proof using Dedekind domains I believe because $\mathcal O$ may be smaller than the full ring of integers in $L$ (a number theorist can correct me if I am mistaken). Note that $\mathcal O$ is $G$-invariant because $G$ permutes $\alpha_1,\ldots, \alpha_n$. Choose a maximal ideal $\mathfrak m$ of $\mathcal O$ containing $p\mathcal O$ (where $p$ is our prime). This exists since $\mathcal O$ consists of algebraic integers and hence $1/p\notin \mathcal O$, whence $p\mathcal O$ is a proper ideal. Then $\mathcal O/\mathfrak m$ is a finite extension of $\mathbb F_p=\mathbb Z/p\mathbb Z$ generated by the cosets $\overline{\alpha_1},\ldots, \overline{\alpha_n}$, which are distinct by the assumption that $p$ does not divide the discriminant of $f$. (This is where that assumption is used.) Let $H$ be the Galois group of $\mathcal O/\mathfrak m$ over $\mathbb F_p$ (which is the Galois group of the reduction $\overline{f}$ of $f$ modulo $p$). Then $H$ is a faithful permutation group of $\overline{\alpha_1},\ldots, \overline{\alpha_n}$. Since $\mathcal O$ is $G$-invariant, $G$ permutes the maximal ideals of $\mathcal O$. Let $D$ be the stabilizer of $\mathfrak m$. For Dedekind domains this is called the decomposition group so it is probably ok to call it that. Then $D$ acts on $\mathcal O/\mathfrak m$ and it acts faithfully because $\alpha_i\mapsto \overline{\alpha_i}$ is a bijection and the $D$ action on both rings is determined by its action on these finite sets. It then suffices to show that the map $D\to H$ (which we just saw is injective) is onto. This is more or less proved in greater generality in @KConrad's link. The basic idea is we can choose by the Chinese remainder theorem $\beta\in \mathcal O$ such that $\beta$ maps to a primitive element $\overline{\beta}$ of $\mathcal O/\mathfrak m$ and $\beta\in \sigma(\mathfrak m)$ whenever $\sigma\notin D$ (i.e., $\sigma(\mathfrak m)\neq \mathfrak m$). Consider the polynomial $g(x)=\prod_{\sigma\in G}(x-\sigma(\beta))$. This is a monic polynomial what is fixed by $G$ and hence has rational coefficients but it also has coefficients in $\mathcal O$ (which consists of algbraic integers) and so it has coefficients in $\mathbb Z$. So we can consider $\overline g$, the reduction of $g$ modulo $p$. Clearly $g(\overline{\beta})=0$ and so the minimal polynomial $m$ of $\overline{\beta}$ divides $g$. Also $\overline{g(x)} = x^k\cdot \prod_{\sigma\in D}(x-\sigma(\overline \beta))$ because if $\sigma\notin D$, then $\beta\in \sigma^{-1}(\mathfrak m)$, whence $\sigma(\beta)\in \mathfrak m$. It follows that $m$ divides $\prod_{\sigma\in D}(x-\sigma(\overline \beta))$. But if $\tau\in H$ (the Galois group of $\mathcal O/\mathfrak m$), then $\tau(\overline{\beta})$ is a root of $m$ and hence one of the $\sigma(\overline \beta)$ with $\sigma\in D$. Since $\overline{\beta}$ is a primitive element, we deduce that $\sigma=\tau$ on $\mathcal O/\mathfrak m$. This finishes the proof that $H\cong D\leq G$.<|endoftext|> TITLE: Is $\pi (x)=\operatorname{R}(x)-\sum_{\rho}\operatorname{R}(x^{\rho})$ correct at all? QUESTION [13 upvotes]: It should be the case that, in some appropriate sense $$\pi (x)\sim \operatorname{Ri}(x)-\sum_{\rho}\operatorname{Ri}(x^{\rho}) \tag*{(4)}$$ with $\operatorname{Ri}$ denoting the Riemann function defined: $$\operatorname{Ri}(x)=\sum_{m=1}^\infty \frac{\mu (m)}{m}\operatorname{li}\left(x^{\frac{1}{m}}\right). \tag*{(5)}$$ This relation $(4)$ has been called "exact" [in Ribenboim's The New Book of Prime Number Records], yet we could not locate a proof in the literature; such a proof should be nontrivial, as the conditionally convergent series involved are problematic. In any case relation $(4)$ is quite accurate, and furthermore the Riemann function $\operatorname{Ri}$ can be calculated efficiently (...) The sum in $(4)$ over critical zeros is not absolutely convergent, and furthermore the phases of the summands interact in a frightfully complicated way. —from Journal of Computational and Applied Mathematics by Borwein et al. Of profound importance, Bernhard Riemann proved that the prime-counting function is exactly $$\pi (x)=\operatorname{R}(x)-\sum_{\rho}\operatorname{R}(x^{\rho})$$ where $$\operatorname{R}(x)=\sum_{n=1}^\infty \frac{\mu (n)}{n}\operatorname{li}\left(x^{\frac{1}{n}}\right),$$ (...) $\rho$ indexes every zero of the Riemann zeta function, and $\operatorname{li}\left(x^{\frac{\rho}{n}}\right)$ is not evaluated with a branch cut but instead considered as $\operatorname{Ei}\left(\frac{\rho}{n}\ln x\right)$. Equivalently, if the trivial zeros are collected and the sum is taken only over the non-trivial zeros $\rho$ of the Riemann zeta function, then $\pi (x)$ may be written $$\pi (x)=\operatorname{R}(x)-\sum_{\rho}\operatorname{R}(x^{\rho})-\frac{1}{\ln x}+\frac{1}{\pi}\arctan\frac{\pi}{\ln x}.$$ —from Wikipedia's Prime counting function article (before 30/7/2021) Questions: According to Borwein and Ribenboim, the index in $(4)$ should run only over non-trivial zeros. According to Wikipedia, the index in $(4)$ should run over all zeros. Wikipedia states that if the sum runs only over non-trivial zeros, then we add the $\ln$ and $\arctan$ terms, which is even more confusing. So what's true? I'm pretty sure that Riemann did not prove the formula $(4)$. He only proposed a "weaker" form of it, namely $$\pi (x)=\sum_{m=1}^\infty \frac{\mu (m)}{m}J\left(x^{\frac{1}{m}}\right)$$ where $$J(x)=\operatorname{li}(x)-\sum_{\rho}\operatorname{li}\left(x^{\rho}\right)+\int_x^\infty \frac{dt}{t(t^2-1)\ln t}-\ln 2$$ and where $\rho$ runs over all non-trivial zeros. The formula was proven by Mangoldt, not Riemann. The Wikipedia article is wrong in that historical fact, isn't it? Even though the proof of $(4)$ is nowhere to be found in the literature, Raymond Manzoni provided a partial proof here. I call it partial because it is unknown whether the series converges at all: How could that be settled down? Note: When I refer to the formula $(4)$ in this question, I assume $=$ instead of $\sim$. Riesel and Gohl Riesel and Gohl show in Some Calculations Related to Riemann's Prime Number Formula that $$\sum_{n=1}^N \frac{\mu (n)}{n}\left(\int_{x^{\frac{1}{n}}}^\infty \frac{\mathrm dt}{t(t^2-1)\ln t}-\ln 2\right)=\frac{1}{2\ln x}\sum_{n=1}^N \mu (n)+\frac{1}{\pi}\arctan\frac{\pi}{\ln x}+\epsilon (x,N)$$ where $$\epsilon (x,N)=-\sum_{n=N+1}^\infty \frac{\mu (n)}{n}\left(\frac{1}{2}\ln\ln x+C\right)+\frac{1}{2}\sum_{n=N+1}^\infty \frac{\mu (n)\ln n}{n}+\sum_{n=N+1}^\infty \frac{\mu (n)}{n}\sum_{k=1}^\infty \frac{1}{\pi k}\arctan\frac{\ln x}{n\pi k}$$ where $$C=\sum_{k=1}^\infty \frac{1}{\pi k}\arctan\frac{1}{\pi k}+\int_1^\infty \frac{\mathrm du}{u(e^{2u}-1)}-\ln 2+\frac{1}{2}.$$ Now, $\epsilon\to 0$ as $N\to\infty$. If $\sum_{n=1}^\infty \mu (n)=-2$, then $$\sum_{n=1}^\infty \frac{\mu (n)}{n}\int_{x^{\frac{1}{n}}}^\infty \frac{\mathrm dt}{t(t^2-1)\ln t}=\frac{1}{\pi}\arctan\frac{\pi}{\ln x}-\frac{1}{\ln x}.$$ But this uses the zeta-regularized result $\sum_{n=1}^\infty \mu (n)=-2$. It is one way of assigning a finite value to a divergent sum, obtained by just plugging $s=0$ in $$\frac{1}{\zeta (s)}=\sum_{n=1}^\infty \frac{\mu (n)}{n^s}.$$ In general, this assigning of values is arbitrary. There are lots of ways to do that. I don't understand how can this arbitrary choice produce an "empirically-correct" result about the distribution of prime numbers. In fact, the paper of Riesel and Gohl implies the mysterious formula $$\pi (x)= \operatorname{R}(x)-\sum_{\rho}\operatorname{R}(x^{\rho})-\frac{1}{\ln x}+\frac{1}{\pi}\arctan\frac{\pi}{\ln x}.$$ For the empirical results, see here. Questions [continued]: What makes zeta-regularization (or other regularizations producing the same value) the only regularization that seems to produce the "empirically-correct" result? Or is it the case that the result is ultimately wrong? Note: $\operatorname{li}x$ is to be interpreted as the Cauchy principal value of $\operatorname{Ei}\ln x$. This question has been on MSE for some time, but it still doesn't have any answers, so I decided to post it here. Edit: The Wikipedia formula (before 30/7/2021) $$\pi (x)= \operatorname{R}(x)-\sum_{\rho}\operatorname{R}(x^{\rho})-\frac{1}{\ln x}+\frac{1}{\pi}\arctan\frac{\pi}{\ln x}$$ turned out to be a misinterpretation of Riesel and Gohl. REPLY [11 votes]: Preliminaries. We will denote the slight modifications of the prime counting function and the prime power counting function with $\pi^*$ and $J^*$ respectively, which assume a halfway step at discontinuities. Furthermore we will denote the non-trivial zeros of the zeta function with $\rho$ and all zeros i.e. the trivial and non-trivial zeros with $\varrho$ in questions two to four. Q1. It depends on what the meaning of $\sim$ in the statement $\pi^* (x)\sim \operatorname{Ri}(x)-\sum_{\rho}\operatorname{Ri}(x^{\rho})$ is. Putting convergence questions aside, which we will discuss in Q3, we indeed have an equality if the sum runs over all zeros of $\zeta$. If we omit the trivial zeros, the statement might still be correct, if $\sim$ denotes asymptotic equivalence since $\sum_n\operatorname{Ri}(x^{-2n})=o(1).$ This is because $$\left\lvert\lim_{x\to\infty}\sum_n\sum_m \frac{\mu(m)}{m}\operatorname{li}(x^{-2n/m})\right\rvert \leqslant \lim_{x\to\infty}\sum_n\sum_m \frac{1}{m}\left\lvert\operatorname{li}(x^{-2n/m})\right\rvert\overset{\text{(Tonelli)}}{=}0.$$ Q2. "Proposed" might not be the best fitting word. Riemann showed and proved essential ideas of his, while leaving out certain steps, which caused a certain lack of rigor. Hence, Riemann's proof was not complete. Quoting Edwards, p.38, 1.19 QUESTIONS UNRESOLVED BY RIEMANN: Riemann evidently believed that he had given a proof of the product formula for $\zeta(s)$, but, at least from the reading of the paper given above, one cannot consider his proof to be complete, and, in particular, one must question Riemann's estimate of the number of roots $\rho$ in the range $\{0 < \operatorname{Im} \rho < T\}$ on which this proof is based. It was not until 1893 that Hadamard proved the product formula, and not until 1905 that von Mangoldt proved the estimate of the number of roots in $\{0 < \operatorname{Im} \rho < T\}$. Q3. We want to prove that $\pi^*(x)=\operatorname{R}(x)-\sum_{\varrho} \operatorname{R}(x^{\varrho})$ follows from $\pi^*(x)=\sum_{m=1}^{\infty}\frac{\mu(m)}{m}J^*(x^{1/m})$. Define $\pi_X^*\colon \mathbb R_{>1}\to\mathbb N$ as $$x\mapsto\sum_{m\leqslant X}\frac{\mu(m)}{m}J^*(x^{1/m}),$$ where $X>1$. If $X>\log_2(x)$, then $\pi_X^*(x)=\pi^*(x)$, since $J^*(x)=0$ for $x<2$. After applying Möbius inversion on $J^*$ we get that for all $X>\log_2(x)$ $$\pi^*(x)=\sum_{m\leqslant X}\frac{\mu(m)}{m}\operatorname{li}(x^{1/m})-\sum_{\varrho}\sum_{m\leqslant X}\frac{\mu(m)}{m}\operatorname{li}(x^{\varrho/m})-\log 2\sum_{m\leqslant X}\frac{\mu(m)}{m}.$$ We were able to change the order of summation involving the conditional convergent series since the series over $m$ is finite. Now the idea is not to look at the case $X\to\infty$ seperately. There exists an $N\in\mathbb N$ for every $X>1$ so that $\pi^*_X(x)=\pi^*_N(x)$. The sequence $(\pi_N^*)_{N\in\mathbb N}$ converges since it is Cauchy. Q4. Riesel and Göhl objective was to find an approximation and not an exact form of the last two terms of $\pi_N(x)$, $\sum_{n=1}^N \frac{\mu (n)}{n}\left(\int_{x^{\frac{1}{n}}}^\infty \frac{\mathrm dt}{t(t^2-1)\ln t}-\ln 2\right)$. The integral is equivalent to $\sum_m\operatorname{li}(x^{-2n/m})$. They eventually showed that it is equal to $\frac{1}{2\ln x}M(N)+\frac{1}{\pi}\arctan\frac{\pi}{\ln x}+\epsilon (x,N),$ where $M$ denotes the Mertens function and $\epsilon$ is an error term, which tends to zero as $N\to\infty$. In order to make the calculation more efficient they replace the Mertens function with $-2$ and argue on page 979 as follows: It is thus advantageous to choose such a value of $N$ that the sums $g_k$ become comparatively small. It also is advantageous to have $\sum_1^N\mu(n) = -2$ at the same time, since (32) then has the order of magnitude only $= O((\log x)^{-3})$ instead of $O((\log x)^{-1})$. Here no zeta regularization was used. The triple sums in the error term are dependent on $g_k$. There exist infinitely many $N$ such that $M(N)=-2$. This immediatly follows from the bounds $\liminf~M(n)/\sqrt n < -1.009$ and $\limsup~M(n)/\sqrt n > 1.06$, which were used to disprove the Mertens conjecture.<|endoftext|> TITLE: Embedding an icosahedron QUESTION [5 upvotes]: A transitive set in $\mathbf{R}^n$ is a finite set with a transitive group of symmetries. I want to understand how subsets of a transitive set constrain the group. Let me start with the example of a tetrahedron. A tetrahedron has symmetry group isomorphic to $S_4$, which is of course nonabelian. However, it is possible to embed the tetrahedron into a larger set, the cube, which has a transitive abelian group of symmetries isomorphic to $C_2^3$. (Of course, the full symmetry group of the cube is $C_2 \times S_4$, which is nonabelian, but my point is there is an abelian subgroup which is still transitive.) So we made things simpler by going bigger. --> [pictures borrowed from Wikipedia] My specific question is the same with (tetrahedron, abelian) replaced with (icosahedron, soluble). --> ??? Is there a finite set in $\mathbf{R}^n$ (with possibly $n > 3$) with a transitive soluble group of symmetries containing the vertices of a regular icosahedron? I would also appreciate any references to related matters. EDIT: Achim Krause answered trivially the question I asked but not the question I intended. At the risk of making my question a bit like Douglas Adams's concept of the Universe in that, if ever anyone discovers exactly what it is for and why it is here it will instantly be replaced by something even more bizarre and inexplicable, I now wish to replace "icosahedron" with "snub dodecahedron". Now the group of rotational symmetries is $A_5$, which acts regularly on the vertices. The point is I meant to choose a transitive set for which the answer was "obviously" no, but such that it is not clear how to prove that, especially if we allow embedding into possibly higher dimensions. REPLY [8 votes]: The answer to your question is actually "yes", but maybe not in the way you wanted. Indeed, the full (oriented) symmetry group of the icosahedron is isomorphic to $A_5$. The stabilizer of a vertex is cyclic of order $5$, so the set of vertices with action by $A_5$ can be identified with the set of cosets $A_5/C_5$. The solvable subgroup $A_4\subseteq A_5$ acts freely and transitively on this: It has the right cardinality (12), and it acts freely, since if an $h\in A_4$ stabilizes a coset $gC_5$, it means that $h$ is conjugate to an element of $C_5$. But $A_4$ has not elements of order $5$, so $h=e$. So you can take the set of vertices to be just the vertices of the icosahedron!<|endoftext|> TITLE: Deligne Tensor Product of Categories, Explicit Equivalence of $A\otimes_\mathbb{C} B\text{-Mod} \cong A\text{-Mod}\boxtimes B\text{-Mod}$ QUESTION [5 upvotes]: $\newcommand\Mod[1]{#1\text{-Mod}}$Does any one have a reference on a explicit equivalence between $$\Mod{A\otimes_\mathbb{C} B} \cong \Mod A\boxtimes \Mod B?$$ The proof in "Tensor Categories EGNO" uses the universal property of $\boxtimes$, but I would like to see an explicit functor that defines an equivalence. In BaKi definition 1.1.15 there is a explicit description of $\mathcal{C}_1\boxtimes \mathcal{C}_2$ when $\mathcal{C}_1$, $\mathcal{C}_2$ are additive categories over $k$. Namely, $\operatorname{Ob}(\mathcal{C}_1\boxtimes \mathcal{C}_2)=$ finite sums of the form $\bigoplus X_i\boxtimes Y_i$, where $X_i\in \operatorname{Ob}(\mathcal{C}_1)$, $Y_i\in \operatorname{Ob}(\mathcal{C}_2)$. and $\operatorname{Hom}_{\mathcal{C}_1\boxtimes \mathcal{C}_2}(\bigoplus X_i\boxtimes Y_i,\bigoplus X_j'\boxtimes Y_j')= \bigoplus \operatorname{Hom}(X_i,X'_j)\otimes \operatorname{Hom}(Y_i,Y'_j)$. Using this description I wanted to construct the equivalence. I tried showing that \begin{align*} F:\Mod A\boxtimes \Mod B &\rightarrow \Mod{A\otimes_\mathbb{C} B}\\ X\boxtimes Y&\mapsto X\otimes_\mathbb{C}Y \end{align*} is full, faithul, and essentially surjective. But I got stuck on trying to show that this functor is essentially surjective. There is what seems to be a counter example, shown on MathStack (Link). Any help/suggestions would be greatly appreciated. I currently only need the case when the modules are finite dimensional. But dont want to be restricted to finite dimensional algebras. Namely I am working with finite dimensional modules over $U(\mathfrak{g}_1\oplus \mathfrak{g}_2)\cong U(\mathfrak{g}_1)\otimes U(\mathfrak{g}_2)$. Edit (More context): I was really trying to use the Relative Tambara Tensor product introduced in (Tam01), and saw a sentence in (DSS18) (First Paragraph), that when using $\mathcal{C}=\operatorname{Vect}$ it agrees with Deligne's Tensor Product (though I could have misunderstood). The Relative Tambara Tensor Product has a similar description of objects and morphisms, but also has much more relations. REPLY [6 votes]: The explicit definition you give is not the Deligne tensor product, it’s what’s often called the “naive” tensor product. The naive tensor product typically won’t be abelian. The Deligne tensor product is universal for right exact bifunctors, and there’s nothing about right exactness in the naive tensor product. There is no description of the Deligne tensor product that’s more explicit than the theorem you’re looking at, more general Deligne tensor products are explicitly constructed by taking colimits of this construction for subcategories.<|endoftext|> TITLE: Pointed versus unpointed maps into a topological monoid QUESTION [8 upvotes]: I've just stumbled on something that seems either too good to be true, or else too good for me not to have heard of it before. It has to do with the basepoint forgetting map $$ u: [A, M] \to \langle A, M \rangle, $$ where $A$ is a pointed space, $M$ is a topological monoid, and $\langle A, M\rangle$ indicates unpointed homotopy classes. For general $X$, I like to understand the map $u$ by writing $A_\circ$ for the unpointed version of $A$, and setting up the long cofiber sequence $$ S^0 \to (A_\circ)_+ \to A \to S^1 \to \Sigma((A_\circ)_+) \to \cdots $$ where $(A_\circ)_+$ indicates the basepoint free $A$ with a disjoint basepoint added, and $S^0\to (A_\circ)_+$ sends the nontrivial point to the basepoint we've forgotten about. Now we map this into $X$ to get the exact sequence $$ [S^0, X]\gets[(A_\circ)_+, X]\gets [A, X] \gets \pi_1(X) \gets [\Sigma((A_\circ)_+), X]\gets\cdots . $$ The first few terms are just pointed sets, and the map $[(A_\circ)_+, X]\gets [A, X]$ can be identified with $u$. It is easy to see that $\pi_1(X) \gets [\Sigma((A_\circ)_+), X]$ is surjective (it follows from naturality, using the composition $*\to A\to *$), so that the kernel of $u$ is trivial. However, it is not true that pointed homotopy and unpointed homotopy agree, and this is captured in the action of $\pi_1(X)$ on $[A,X]$ coming from the long cofiber sequence. We have $u(f) = u(g)$ if and only if $f$ and $g$ are in the same orbit of the action. The great thing is that when we take $X$ to be a topological monoid $M$, even the initial terms in the sequence are groups and homomorphisms, so the orbits are cosets of the image of $[A, M] \gets \pi_1(M)$, which is trivial. Consequently, pointed and unpointed homotopy classes into $M$ agree. QUESTION: This fact is not in my mental toolbox! Am I missing something? Alternatively, is there a nice reference to quote for this? REPLY [12 votes]: As is implicitly pointed out in the comments, you really want to assume that $X$ ($=M$) is path connected. And then your analysis is fine. Note that $M$ will then wish to be equivalent to $\Omega BM$, so those pointed homotopy classes then look like $\langle \Sigma A, BM \rangle$ and fundamental group issues will go away, as $BM$ will be simply connected.<|endoftext|> TITLE: So after all, what is this thing about topos theory and non-binary truth? QUESTION [12 upvotes]: Disclaimer. The question below is necessarily vague. I understand neither the subject matter topos theory nor the object about which my question is (the construction of a fractional / non-binary / elastic notion of truth). Thanks in advance for your patience. I've often heard whispers of vague statements like: "topos theory can be used to give continuous, subtle, and perhaps a much more interesting notion of truth (not just True or False, Right or Wrong, Correct or Incorrect)". For lack of better terminology, let us refer to this mysterious phenomenon / concept as fractional truth. Question. What is a minimal construction which already displays this phenomenon? I have essentially no knowledge of topos theory (but a decent handle on set theory, group theory, and topology, etc., and I'm naively assuming that this should be sufficient to understand the would-be construction, which I'm also assuming can be made so simple). REPLY [10 votes]: Suppose you always want to talk about two things simultaneously for some reason. When you say "a set $A$" you actually mean a pair of sets $(A_1, A_2)$, when you say "a function $f : A \to B$" you mean a pair of functions $f_1 : A_1 \to B_1$, $f_2 : A_2 \to B_2$. When you say a statement $p$ is true, you actually mean $p_1$ is true and $p_2$ is true. Then from an external perspective there obviously are truth values which "are neither true nor false", namely $(true, false)$ and $(false, true)$. (Note that the partial order of truth values is not a total order here, so it is not very similar to fractional numbers.) This is the internal language of the topos $\mathrm{Sh}(\{*\} \sqcup \{*\})$, the sheaf topos on the discrete two-point space. (A sheaf on this space is just a pair of sets.) If you take $\mathrm{Sh}(X)$ instead, for $X$ any topological space, there are as many internal truth values as $X$ has open subsets. Be careful, however, to distinguish between internal and external statements. The truth values $p = (\mathrm{true}, \mathrm{false})$ and $(\mathrm{true}, \mathrm{true})$ are different externally, but the internal statement "$p$ is different from $\mathrm{true}$" has truth value $(\mathrm{false}, \mathrm{true})$, so it is not valid (everywhere). In fact, we can prove in intuitionistic logic (which is valid in every topos) that there is no truth value which is neither true nor false. Also note that $\mathrm{Sh}(\{*\} \sqcup \{*\})$ is a boolean topos, that is, the internal language is in fact classical -- the law of omniscience (excluded middle) is valid internally. For example, $p \lor (\lnot p) = (\mathrm{true}, \mathrm{false}) \lor (\mathrm{false}, \mathrm{true}) = (\mathrm{true} \lor \mathrm{false}, \mathrm{false} \lor \mathrm{true}) = (\mathrm{true}, \mathrm{true})$.<|endoftext|> TITLE: Who started the "-oid" suffix fashion in math? QUESTION [16 upvotes]: There are lots of structures which have name suffixed by "oid". Off the top of my head, matroid, greedoid, perfectoid, causaloid... Who started this? AFAIK, "matroid", by Whitney, was a start, and led the way to several combinatorial oids. However, the Cardioid has had its name for some centuries now, so the use of the suffix is old. Still, it seems a bit different to name a family of specific objects, and to name some sort of abstract structure. REPLY [4 votes]: Though this might not be what you are expecting, I will explain you "oidification" or horizontal categorification as I understood (Experts are fell free to add or edit as necessary). This is the process that generalizes a "certain type of category with a single object" to "such type of categories with multiple objects". This is done mostly via "enriching" the initial category $\mathcal{C}$ over another monoidal category $\mathcal{K},$ which roughly says homsets (set of arrows two objects) of $\mathcal{C}$ are replaced by objects of $\mathcal{K}.$ Examples include X X-oid Enrichment monoid Category categories enriched over Set Category 2-Category categories enriched over Cat Group Groupoid Ring Ringoid category enriched in tensor category Ab Quantale Quantaloid category enriched in suplattices Algebr Algebroid category enriched in Vect or RMod C*-algebra C*-category *-category enriched in Ban You can find more details ringoid, and algebroid here. But as far as I know Hopf algebroids and Lie algebroids does not fit into this general definition of algebroids, but still multi-object generalizations of their counterparts. Also, it should worth remind that not every enrichment of a categories consider as an "odification". For example (Lawvere) generalized metric spaces is a category enriched in the monoidal poset category $([0,\infty],\ge),$ where the monoidal product is taken to be addition.<|endoftext|> TITLE: References for Hopf Galois module theory QUESTION [5 upvotes]: I am a first-year PhD student and I am really interested in Galois module theory, both in a "classical" and in a "nonclassical" sense. In the last months I have been reading about Hopf Galois theory, since it seems to be a nice way to find the structure of the ring of integers (or valuation ring, in the local case), when the classical approach fails. In particular, I am reading Childs's book "Taming Wild Extensions: Hopf Algebras and Local Galois Module Theory", where he makes a nice exposition of the known facts untile 2000. I was looking for some more recent results, but my search was not really fruitful. Could you please advise some more modern papers (or books, survey, notes, or anything you want) discussing about Hopf Galois module theory, so Galois module theory applied in the nonclassical setting of Hopf Galois extensions? REPLY [3 votes]: You could try this survey article: MR3415265 Crespo, T. ; Rio, A.; Vela, M. From Galois to Hopf Galois: theory and practice. Trends in number theory, 29–46, Contemp. Math., 649, Amer. Math. Soc., Providence, RI, 2015. You could also use MathSciNet to look at the list of publications that cite Taming Wild Extensions (there are currently 72 items on that list).<|endoftext|> TITLE: English translation of Schwartz's papers on vector-valued distributions QUESTION [5 upvotes]: I am interested in systematically studying the theory of vector-valued distributions. The original two papers due to Laurent Schwartz entitled Théorie des distributions à valeurs vectorielles. I & II (1957-58) are written in French. Occasionally I have read mathematics papers in French when needed, with the help of Google Translate. But these papers are to the tune of 350-odd pages making it infeasible for such an undertaking. Is there an English translation of these seminal papers by Schwartz? The closest I have seen to a systematic exposition of vector-valued distributions is the book Vector-valued distributions and Fourier multipliers (http://user.math.uzh.ch/amann/files/distributions.ps) by Herbert Amann. But here some results are presented without proof and the reader is referred to Schwartz's papers. This brings me to my next question. Is there a standard textbook (in English) which gives a self-contained account of the theory of vector-valued distributions? What I am looking for is something like a vector-valued version of the series of books on Generalized Functions by Gelfand and Shilov (which had been promptly translated from Russian to English). Thank you. REPLY [3 votes]: These papers have not been translated, as far as I know, however there exist lecture notes in english of courses by Schwarz on this topic: • Introduction to the Theory of Distributions • Lectures on Partial Differential Equations and Representations of Semigroups. Vector-Valued Distributions and Fourier Multipliers, by Herbert Amann, is freely available. A more extensive list of text books is at Good books on theory of distributions<|endoftext|> TITLE: Necessary and sufficient condition for tangential polygon to be cyclic QUESTION [6 upvotes]: Can you prove or disprove the following claim? Claim. Let $A_1,A_2, \ldots ,A_n$ be the vertices of an $n$-sided tangential polygon and let $B_1,B_2, \ldots ,B_n$ be the contact points of the inscribed circle and polygon sides such that $B_1$ lies on $A_1A_2$, $B_2$ lies on $A_2A_3$ ,etc. Denote by $H_1,H_2, \ldots,H_n$ the orthocenters of the triangles $\triangle A_1B_1B_n$, $\triangle A_2B_2B_1$,....,$\triangle A_nB_{n}B_{n-1}$ . Then the polygon is cyclic if and only if $H_1,H_2,\ldots ,H_n$ are concyclic. Picture for the case $n=6$: GeoGebra applets that demonstrate this claim can be found here , here and here. REPLY [8 votes]: $H_i$ lies on the ray $IA_i$ and $IH_i\cdot IA_i=2r^2$ (where $r=IB_i$), since the midpoint of $IH_i$ is the midpoint of $B_iB_{i-1}$. Hence $H_i$ is the image of $A_i$ under the inversion with respect to center $I$ and radius $\sqrt{2}r$. Thus the result.<|endoftext|> TITLE: Functorial kernel in derived category QUESTION [7 upvotes]: By the work of Verdier, we know that cones in a triangulated category $\mathcal{T}$ are functorial if and only if $\mathcal{T}$ is semisimple abelian. However, in these notes, it is said that In the context of triangulated categories, it is well known that cones are not functorial. However, we have just proven that if a triangulated category arises as the homotopy category of a stable $\infty$-category, cones in it are indeed functorial at the $\infty$-level. I must admit that I don't understand what it means that cones are functorial at the $\infty$-level. Since the derived category of an abelian category (with enough injectives) is the homotopy category of the derived infity category, which is a stable $\infty$-category, that would mean that we have functorial cones in derived abelian categories. What does that mean? Can I "take kernels"? REPLY [20 votes]: Let $\mathcal{C}$ be a stable $\infty$-category. Then $\mathcal{C}$ has a homotopy category $h \mathcal{C}$, which is triangulated. The collection of morphisms $f: X \rightarrow Y$ of $\mathcal{C}$ can be organized into an $\infty$-category $\mathrm{Fun}( \Delta^1, \mathcal{C} )$. The operation $f \mapsto \mathrm{Cone}(f)$ can be obtained from a functor of $\infty$-categories $\mathrm{Fun}( \Delta^1,\mathcal{C} ) \rightarrow \mathcal{C}$. You can pass to homotopy to get a functor of ordinary categories $$\mathrm{Cone}: h \mathrm{Fun}( \Delta^1, \mathcal{C} ) \rightarrow h \mathcal{C}.$$ The functor $\infty$-category $\mathrm{Fun}( \Delta^1, \mathcal{C} )$ is equipped with an evaluation functor $e: \Delta^1 \times \mathrm{Fun}( \Delta^1, \mathcal{C} ) \rightarrow \mathcal{C}$. You can take homotopy categories here, to get a functor of ordinary categories $[1] \times h\mathrm{Fun}( \Delta^1, \mathcal{C} ) \rightarrow h\mathcal{C}$, which can be identified with a functor (again of ordinary categories) $$ U: h \mathrm{Fun}( \Delta^1, \mathcal{C} ) \rightarrow \mathrm{Fun}( [1], h\mathcal{C} ).$$ Here $[1]$ denotes the category $\{ 0 < 1 \}$ consisting of two objects and one morphism between them. The phenomenon you're asking about is due to the fact that $U$ is not an equivalence of categories. Moreover, the functor $\mathrm{Cone}$ does not factor through $U$. If $f: X \rightarrow Y$ and $f': X' \rightarrow Y'$ are morphisms of $\mathcal{C}$, then morphisms from $f$ to $f'$ on in the $\infty$-category $\mathrm{Fun}( \Delta^1, \mathcal{C} )$ can be thought of triples $(u,v,h)$, where $u: X \rightarrow X'$ and $v: Y \rightarrow Y'$ are morphism of $\mathcal{C}$ and $h$ is a homotopy from $f' \circ u$ to $v \circ f$. In these terms, the functor $U$ is given on morphisms by the construction $[(u,v,h)] \mapsto ( [u], [v] )$ where $[s]$ denotes the homotopy class of a morphism $s$. In particular $U$ "forgets" the data of the homotopy $h$, and fails to be a faithful functor.<|endoftext|> TITLE: Get the commands history from GAP system QUESTION [5 upvotes]: I am not sure whether this was asked before, but I didn't find a reference in GAP system documentation on how to print the history of the command line (Ubuntu installation). For instance: gap> G := Group((1,2)(3,4),(1,2,3)); > irr := Irr(G); How to save or print the list of my previous commands (similar to history command on Ubuntu). Many thanks. REPLY [6 votes]: There are three commands that will do all variations of this: LogTo("filename.txt") will save all subsequent input and output to a file with the specified name. InputLogTo("filename.txt") will save all subsequent input to a file with the specified name. OutputLogTo("filename.txt") will save all subsequent output to a file with the specified name. In all three cases, running the command with no argument will stop the logging. More details can be found here in the GAP Manual.<|endoftext|> TITLE: Convex lattice polygons with equal area and perimeter QUESTION [5 upvotes]: A convex polygon all of whose vertices have integer coordinates is a convex lattice polygon. Do there exist mutually non-congruent convex lattice polygons which have the same area and same perimeter? If answer to 1 is yes, are there convex lattice polygons which can be cut into some integer number of convex lattice polygons which are not all congruent and all have same area and same perimeter? Note: The questions have natural analogs in higher dimensions. REPLY [2 votes]: My earlier comment "parallelogram and kite" was signaling an infinite family of examples of groups of $m$ convex lattice polygons where all of them from the same group have the same diameter, perimeter, and area (where finite $m$ can be arbitrarily large). There are even two such infinite families. The FIRST two power families: Let $\ d_1>d_2>\ldots>d_n>0,\,\ $ and $\,\ y_0 > 0\,\ $ and $\,\ y_k=y_{k-1}+d_k\ $ for $\ k=1\ldots n.$ Let $\ 0\le a_0 TITLE: Contemporary introduction to Godement-Jacquet "Zeta functions of simple algebras" QUESTION [6 upvotes]: The question is in the title: The book Godement-Jacquet "Zeta functions of simple algebras" is from 1971. Has there ever been a textbook introduction to this material, or at least part of it? (but beyond just Tate's thesis). REPLY [11 votes]: As you mention in your question, for $\mathrm{GL}_1$, this is just Tate's thesis revisited. The only exposition on this that I know of occurs in "Automorphic Representations and $L$-Functions for the General Linear Group" by Dorian Goldfeld and Joseph Hundley. In particular, Chapter 11 of volume 1 covers the theory of the Godement-Jacquet zeta integral for $\mathrm{GL}_2(\mathbb{A}_{\mathbb{Q}})$ in some detail. Chapter 15 of volume 2 covers the same theory for $\mathrm{GL}_n(\mathbb{A}_{\mathbb{Q}})$ with $n$ arbitrary. More generally, the two volumes are a detailed introduction to the theory of automorphic representations of $\mathrm{GL}_n(\mathbb{A}_{\mathbb{Q}})$ at a reasonable pace. There are other books that cover the $\mathrm{GL}_2$ theory at a little higher level (for example, "Automorphic Forms and Representations" by Daniel Bump or the original, "Automorphic Forms on $\mathrm{GL}(2)$" by Hervé Jacquet and Robert Langlands) or at a quicker pace ("Automorphic Forms on Adele Groups" by Stephen Gelbart). To my knowledge, however, Goldfeld-Hundley is the only textbook that covers the Godement-Jacquet zeta integral in detail. I should also mention that Hervé Jacquet has also written two short surveys on the Godement-Jacquet theory: "Principal $L$-Functions of the Linear Group", which appears in the famous Corvallis proceedings, and "Principal $L$-Functions for $\mathrm{GL}(n)$". Edit: per the review of Goldfeld-Hundley by Ramin Takloo-Bighash in the Bulletin of the American Mathematical Society, "The book under review is the first book since the appearance of [Godement-Jacquet] in 1972 that gives a thorough treatment of the principal $L$-function using matrix coefficients." Second edit: Google Scholar has informed me that there is a new book coming out soon, The Genesis of the Langlands Program, that includes a long chapter by Goldfeld and Jacquet called "Automorphic Representations and $L$-Functions for $\mathrm{GL}(n)$", which seems to be a detailed discussion of the theory of the Godement-Jacquet zeta integral. You can read some of this chapter on Google Books.<|endoftext|> TITLE: "Very lax" $2$-dimensional co/limits QUESTION [12 upvotes]: In the setting of $1$-categories, there are two (unweighted) variant notions of limits, namely limits and colimits. For bicategories, there are fourteen of them: Here $\mathsf{LaxCones}(\Delta_{X},D)\overset{\mathrm{def}}{=}\mathsf{LaxNat}(\Delta_{X},D)$, and similarly for the other entries in the last column. (Incidentally, these notions are also interrelated: for instance, passing to the weighted case, weighted lax co/limits can be expressed as weighted $2$-co/limits.) Each of these notions has an associated "strength", given by whether we require an equivalence of categories (e.g.) $$ \mathsf{Hom}_{\mathcal{D}}(X,\mathsf{bilim}^{\mathsf{lax}}(D)) \overset{\mathrm{eq}}{\cong} \mathsf{LaxCones}(\Delta_{X},D) $$ or an isomorphism of categories (e.g.) $$ \mathsf{Hom}_{\mathcal{D}}(X,\mathsf{lim}^{\mathsf{lax}}(D)) \cong \mathsf{LaxCones}(\Delta_{X},D). $$ Bicategorical adjunctions, on the other hand, are usually discussed using three different "strengths": equivalences, isomorphisms, and adjunctions. This leads to the following question: Question: Have the "very lax" notions of $2$-dimensional co/limits corresponding to weakening the equivalences of categories in the above table to be merely adjunctions been studied before? Are there any interesting/natural examples of them? REPLY [6 votes]: I believe these kinds of (co)limits are the ones defined in Gray's Formal Category Theory: Adjointness for $2$-categories I.7.9.1 (as quasi-(co)limits); some examples and computations of them are given further on in I.7.10, I.7.11 and I.7.12, and recover several classical constructions such as (co)comma objects or Eilenberg–Moore (or Kleisli) categories, as well as the usual description, when the target $2$-category is $\mathfrak{Cat}$, in terms of sections of an associated opfibration. Note that Gray defines them in the $3$-dimensional—or rather, (lax) Gray-categorical—setting, using his notion of quasi-adjunction, a.k.a. soft adjunction or lax adjunction or local adjunction ("adjunction up to adjunctions"), this last description (established in MacDonald–Stone's Soft adjunction between $2$-categories) being what recovers the definition you are asking for. However, I have not read this book in enough detail to be able to answer confidently on what exactly Gray's somewhat idiosyncratic terminology corresponds to (I am only writing this as an answer because it is slightly too long for a comment), though various flavours of laxity, not all of which have stood the test of time, are allowed—see the discussion at the end of Definition I.7.1 for a list; I think the relevant kind for your question (and in general use) should be "strict weak quasi-adjunction" or "transcendental quasi-adjunction".<|endoftext|> TITLE: Must the inclusion of an indecomposable module in the direct sum of two copies always split? QUESTION [5 upvotes]: We consider finitely generated modules over an Artin algebra. Let $X$ be an indecomposable module and let $f:X \longrightarrow X \oplus X$ a monomorphism. Must $f$ always be a split monomorphism? REPLY [11 votes]: Yes, it must be split. Since $M$ is an indecomposable module for an Artin algebra, its endomorphism ring $E$ is a local ring with nilpotent Jacobson radical $J(E)$. Say $J(E)^n=0$. Let the monomorphism $\varphi:M\to M\oplus M$ be given by $\varphi(m)=(\alpha(m), \beta(m))$. If either $\alpha$ or $\beta$ is an isomorphism, then $\varphi$ splits, so we shall assume that $\alpha,\beta\in J(E)$. Consider the sequence of monomorphisms $$M\xrightarrow{\varphi}M\oplus M\xrightarrow{(\varphi,\varphi)}M\oplus M\oplus M\oplus M\xrightarrow{(\varphi,\varphi,\varphi,\varphi)}\cdots.$$ Since $J(E)^n=0$, the composition of the first $n$ maps in the sequence is zero, contradicting the fact that they are all monomorphisms.<|endoftext|> TITLE: Tic-tac-toe with one mark type QUESTION [18 upvotes]: Parameters $a,b,c$ are given such that $c\leq\max(a,b)$. In an $a\times b$ board, two players take turns putting a mark on an empty square. Whoever gets $c$ consecutive marks horizontally, vertically, or diagonally first wins. (Someone must win because we use only one mark type.) For each triple $(a,b,c)$, who has a winning strategy? For $a=b=c=3$ (tic-tac-toe size), the first player can win by first going on the middle square and winning in the next turn. In the one-dimensional case ($a=1$), this may well be a known game, but I also cannot find a reference. I asked the question on math.SE but it has not been solved. REPLY [4 votes]: All I have are three small observations. For all $a,c\in \mathbb N^+$, $(2a,1,2c)$ is a second player win. Player two's winning strategy is to divide the board into adjacent pairs, and to respond with the square paired with player one's last move. I tried to find a similar pairing strategy for a $(2a,2b,2c)$ board, but failed. In the game where only orthogonal rows count as a win, there is a sort of pairing strategy for the second player: divide the board into $2\times 2$ blocks, and make a winning move if it exists, otherwise respond diagonally opposite in the same block. Unfortunately, this fails when diagonals are wins. The game $(a,1,3)$ is the octal game 0.11337, which is equivalent to 0.007 with an offset, as Timothy Chow said. That is, when $n\ge 2$, an empty $(n,1,3)$-board is equivalent to a 0.11337-heap of size $n$, which it turn equivalent to a 0.007-heap of size $n-2$ .<|endoftext|> TITLE: Core for a Sobolev space QUESTION [6 upvotes]: Let $D$ be a domain of $\mathbb{R}^d$. That is, $D$ is a connceted open subset of $\mathbb{R}^d$. The first-order Sobolev space $W^{1,2}(D)$ on $D$ is defined by \begin{align*} W^{1,2}(D)=\{f \in L^2(D,m) \mid \partial f/\partial x_i \in L^2(D,m),\, 1\le i \le d\}. \end{align*} Here, $m$ is the Lebesgue measure and $ \partial f/\partial x_i$ is the distributional derivative of $f$. It is well known that $W^{1,2}(D)$ becomes a Hilbert space. The norm is determined by $\|f\|_{W^{1,2}(D)}:=[\int_{D}\{f(x)^2+\sum_{i=1}^d (\partial f/\partial x_i)^2\}\,m(dx) ]^{1/2}$. When smooth functions $C^\infty_{c}({\overline{D}})(=C^\infty_{c}(\mathbb{R}^d)|_{\overline{D}})$ are dense in $W^{1,2}(D)$ ? If there is a bounded linear operator $T\colon W^{1,2}(D) \to W^{1,2}(\mathbb{R}^d)$ such that $Tf=f$, $m$-a.e. on $D$, we can easily check that $C^\infty_{c}({\overline{D}})$ becomes a dense subspace of $W^{1,2}(D)$ (because $C^\infty_{c}(\mathbb{R}^d)$ is a dense subspace of $W^{1,2}(\mathbb{R}^d)$). Such an operator is called an extension operator, and it seems that its existence is known even when the boundary of $D$ is very complicated [for example, the Koch snowflake domain]. Can $C^\infty_{c}({\overline{D}})$ become dense in $W^{1,2}(D)$ without the extension operator? I don't know such an example (of domains), so if anyone knows, please let me know. REPLY [2 votes]: (Too long for a comment.) I have just learned that my colleagues, Bartłomiej Dyda and Michał Kijaczko, wrote a paper [1] on that particular problem for fractional Sobolev spaces. In their work, they cite Theorem 3.25 in McLean's book [2], which reads as follows (with the original notation): Theorem: For any open set $\Omega$ and any real $s \geqslant 0$, the set $W^s(\Omega) \cap \mathcal E(\Omega)$ is dense in $W^s(\Omega)$. So at least smooth functions in $\Omega$ are always dense in $W^s(\Omega)$ (but of course they need not extend smoothly to the boundary). I did not have time to check carefully the two references mentioned above for an answer to your question — perhaps it is written somewhere in [2]. References: [1] B. Dyda, M. Kijaczko, On density of smooth functions in weighted fractional Sobolev spaces, Nonlinear Anal. 205 (2021): 112231, DOI:10.1016/j.na.2020.112231, arXiv:2009.00077 [2] W. McLean, Strongly elliptic systems and boundary integral equations. Cambridge University Press, Cambridge, 2000.<|endoftext|> TITLE: Making sense of "every non-commutative algebra has its own internal time evolution (aka a one-parameter group)"? QUESTION [12 upvotes]: I've listened to many interviews and lectures of Alain Connes, in which he says something which goes roughly as follows "Every non-commutative algebra has its own time (evolution of), by which I mean a one-parameter group." I find this statement somewhat mysterious and intriguing at the same time. Question. What is the precise statement of this result and how to can this proven explicitly even for a simple scenario like, say, the von Neumann algebra $M_2(\mathbb C)$ of $2 \times 2$ complex matrices ? Disclaimer. I have essentially no knowledge of non-comutative geometry, etc. I'd appreciate a very simple construction / proof; nothing too fancy. Thanks. REPLY [12 votes]: Given any von Neumann algebra $M$, we can define its noncommutative $\def\L{{\cal L}} \L^p$-spaces $\L^p(M)$ for any $\def\C{{\bf C}} p∈\C$ such that $\Re p≥0$. Here I use the notation $\L^p:={\rm L}^{1/p}$, where the right side is the usual notation from measure theory and real analysis. (This notation is explained in more detail in another answer.) We have $\L^0(M)≅M$, $\L^1(M)≅M_*$, and $\L^{1/2}(M)$ is the standard form of $M$ due to Haagerup. The spaces $\L^p(M)$ for all $p∈\C$ form a $\C$-graded *-algebra, where the involution is a $\C$-antilinear map $\L^p(M)→\L^{\bar p}(M)$. The multiplication is well-defined because of Hölder's inequality, which in this context says that we have a map $$\L^p(M)⊗_\C\L^q(M)→\L^{p+q}(M).$$ In fact, we can do better: the induced map $$\L^p(M)⊗_{\L^0(M)}\L^q(M)→\L^{p+q}(M)$$ is an isomorphism for any $p,q∈\C$ such that $\Re p≥0$, $\Re q≥0$. Here the tensor product on the left side is purely algebraic, it happens to be automatically complete. In particular, the bimodules $\L^p(M)$ are invertible for any $\def\I{{\bf I}} p∈\I=\{z∈\C\mid \Re z=0\}$, with the inverse being the bimodule $\L^{-p}(M)=\L^{\bar p}(M)$. Thus, we have a morphism of 2-groups $$\def\VNA{{\sf VNA}} \L(M)\colon\I→\VNA^⨯_M,$$ where $\VNA$ denotes the bicategory of von Neumann algebras, W*-bimodules, and (bounded) intertwiners, $\VNA^⨯$ denotes its maximal 2-subgroupoid, i.e., we only take invertible bimodules (alias Morita equivalences) and invertible intertwiners, and $\VNA^⨯_M$ denotes the 2-subgroupoid of $\VNA^⨯$ on a single object, namely, $M$, whose endomorphisms form a monoidal groupoid, i.e., a 2-group. The bicategory $\VNA^⨯$ is analogous to the bicategory of Lie groupoids, Morita equivalences (given by invertible bibundles), and invertible morphisms of bibundles. In particular, 1-morphisms in $\VNA^⨯$ can be seen as invertible maps of corresponding noncommutative measurable spaces in the same way as Morita equivalences of Lie groupoids can be seen as invertible maps of stacks corresponding to these Lie groupoids. Thus, the canonical homomorphism of 2-groups $\L(M)\colon\I→\VNA^⨯_M$ equips every von Neumann algebra $M$ with a canonical action of the Lie group $\I=i{\bf R}$ of imaginary numbers. This is precisely the Tomita–Takesaki modular flow, expressed in a canonical manner without arbitrary choices. If $M$ is a type III von Neumann algebra, this action is not isomorphic to the trivial action. This can be most easily seen as follows. The trivial action sends all $t∈\I$ to the $M$-$M$-bimodule $M$, with the obvious trivial coherence data. If α is an isomorphism from the trivial action to the action $\L(M)$, then for each $t∈\I$ we have an isomorphism of $M$-$M$-bimodules $$α_t\colon M→\L^t(M).$$ Such an isomorphism is uniquely characterized by the element $α_t(1)∈\L^t(M)$, which must be a central element of support 1. Now the relations imposed by the definition of a homomorphism of 2-groups tell us that $$α_s(1)α_t(1)=α_{s+t}(1)∈\L^{s+t}(M)$$ and $α_0(1)=1∈\L^0(M)=M$. Thus, $$t↦α_t(M)∈\L^t(M)$$ is a one-parameter group of unitary elements of $\L(M)$. Such one-parameter groups are in bijection with faithful semifinite normal weights $μ$ on $M$: given $μ$, we set $$α_t(1)=μ^t∈\L^t(M).$$ The elements $α_t(1)$ are central in $\L^t(M)$ if and only if $μ$ is a trace. (If $μ$ is finite, then $μ(ab-ba)=0$ can be rewritten as $(μa-aμ)(b)=0$, i.e., $aμ=μa$ for all $a∈M$. That is to say, $α_1(1)=μ^1=μ∈\L^1(M)$ is a central element.) Thus, the action $\L(M)$ is isomorphic to the trivial action if and only if $M$ admits a faithful semifinite normal trace, i.e., $M$ is a semifinite von Neumann algebra, equivalently, a direct integral of type I and type II factors. Hence, if $M$ is a type III algebra, the action is nontrivial. This approach also makes it clear the extent to which the traditional modular automorphism groups depend on the choice of a faithful semifinite normal weight. Given $t\in\I$ and a faithful semifinite normal weight $μ$, the traditional Tomita–Takesaki modular automorphism is $$σ_μ^t\colon M→M,\qquad x↦σ_μ^t(x)=μ^t x μ^{-t}.$$ We have a canonical homomorphism of 2-groups $$\def\Aut{\mathop{\rm Aut}} T\colon \Aut(M) → \VNA^⨯_M$$ that sends $σ∈\Aut(M)$ to the $M$-$M$-bimodule $T(σ)$. This is indeed a homomorphism because we have canonical isomorphisms $T(σ)⊗_M T(σ')≅T(σσ')$ that satisfy the relevant coherence conditions. Now consider the $M$-$M$-bimodule $M_{μ,t}=T(σ_μ^t)$, where the right action is twisted by $σ_μ^t$: $a⋅m⋅b=amσ_μ^t(b)=amμ^t b μ^{-t}$. We have a canonical isomorphism of $M$-$M$-bimodules $$ρ_{μ,t}\colon M_{μ,t}→\L^t(M), \qquad m↦mμ^t.$$ Furthermore, the isomorphisms $ρ_{μ,t}$ are compatible with tensor products: $$ρ_{μ,s}⊗_M ρ_{μ,t}≅ρ_{μ,s+t}.$$ This amounts to saying that for any faithful semifinite normal weight $μ$, the composition of the homomorphism of groups $$σ_μ\colon \I → \Aut(M)$$ with the homomorphism of 2-groups $$T\colon\Aut(M) → \VNA^⨯_M$$ that twists the right action is canonically isomorphic (via the isomorphism $ρ_μ$) to the homomorphism of 2-groups $$\L(M)\colon \I → \VNA^⨯_M.$$ Observe that both $σ_μ$ and the isomorphism $$ρ_μ\colon T∘σ_μ → \L(M)$$ depend on the choice of a faithful semifinite normal weight $μ$, but the homomorphism $\L(M)$ does not. Thus, $\L(M)$ expresses the modular automorphism group in a coordinate-free way.<|endoftext|> TITLE: Splitting of small primes in number fields generated by the torsion of elliptic curves QUESTION [7 upvotes]: Suppose $E/\mathbb Q$ is a non CM elliptic curve and we look at the number field $K_d$ generated by the $d$-torsion of $E$. What is known about the (complete) splitting of small primes in $K_d$? More precisely, since $|E(\mathbb F_p)| \sim p+1$, if $p$ were a completely split prime, we would need $d^2 < p+1$ (approximately). What can say about primes in, say, the range $d^2 < p < d^4$? What proportion of them split completely (with asymptotic dependence on $d$)? (I would also be interested in the analogous question for $\mathbb G_m$.) REPLY [8 votes]: Let's discuss the $\mathbb G_m$ question first. For $p$ to split completely in the field generated by the $d$-torsion of $\mathbb G_m$, i.e. the field generated by the $d$th roots of unity, a necessary and sufficient condition is that $\mathbb F_p$ contains the $d$th roots of unity, i.e. $p \equiv 1\mod d$. This requires $p>d$ so I guess the analogous question would be to count primes congruent to $1$ mod $d$ between $d$ and $d^2$. The usual heuristics suggest that this number should be roughly $d^2 / (2\phi(d) \log d)$. But a provable asymptotic is way too much to hope for. We don't even know a lower bound that this is nonzero - one would have to enlarge the range to $d < p < O(d^5)$ and apply Xylouris's strengthening of Linnik's theorem. Even under GRH the $d^2$ case is unknown. Probably one can get upper bounds that are reasonably close to the truth using sieve methods. For elliptic curves, the situation is similar, but more complicated. Some necessary conditions are that $p \equiv 1 \mod d$, since from two independent $d$-torsion points we can generate a $d$th root of unity by the Weil pairing, and $a_p \equiv p+1\mod d^2$, writing $E(\mathbb F_p) = p+1-a_p$. Are the sufficient? The right perspective is to think of Frobenius as a $2 \times 2$ matrix acting on the torsion points, $p$ as the determinant, and $a_p$ as the trace. Knowing the determinant $p$ is congruent to $1$ mod $d$ and the trace $a_p$ is congruent to $1+p$ mod $d^2$ does not suffice to guarantee that the matrix is congruent to the identity mod $d$ as the counterexample $\begin{pmatrix} 1 & 1 \\ 0 & p \end{pmatrix} $ shows. However, they do imply that the elliptic curve is congruent to an elliptic curve with full $d$-torsion, as any counterexample must be more-or-less congruent to that one modulo $d^2$. Because these conditions aren't quite sufficient, I don't know a criterion simpler then the claim that the Frobenius conjugacy class in $GL_2(\mathbb Z/d)$ is equal to the identity matrix. We expect this to hold for a proportion of primes equal to $1$ divided by the image of the Galois group in $GL_2(\mathbb Z/d)$. For a non-CM elliptic curve, this will typically be almost as large as $|GL_2(\mathbb Z/d)|\approx d^4$, and so we expect no or very few such primes $ TITLE: Can any $E_1$ algebra over $\mathbb{F}_p$ be modeled as a dg algebra? QUESTION [10 upvotes]: I'm not very familiar with dg algebras (not necessarily commutative) and I'm wondering if any $E_1$ algebra in the sense of infinity categories (i.e. monoid in the stable category of $R-Mod$) over $R$ some commutative ring (discrete) can be described by a dg algebra over R? Specifically I have some $E_1$ algebra in mind which has homotopy groups $\mathbb{F}_p$ in homological degrees $0$ and $-1$. I'm wondering if the $E_1$ structure has to be the "obvious" one. I think there may be other $E_1$ structures on it but I'm not sure how to see this. REPLY [9 votes]: Yes, this is precisely the content of Theorem 7.11 in arXiv:1410.5675, which should be combined with §7.4 of arXiv:1510.04969. In fact, the cited results prove this for any nonsymmetric operad in chain complexes over a commutative ring, and are also applicable to ∞-categories other than chain complexes. In the case of symmetric operads (such as ${\rm E}_∞$) the result continues to hold for characteristic 0, i.e., rational chain complexes. For symmetric operads in characteristic $p$ there are algebras that cannot be rectified as stated, and one must use other categories instead of chain complexes, e.g., simplicial modules over ${\bf F}_p$ (see §7.3 of arXiv:1510.04969). In all such results, there are two main ingredients. The first ingredient shows that any ∞-algebra over an ∞-operad $O$ can be rectified to a strict algebra over a strict operad $Q$, where the underlying ∞-operad of $Q$ is equivalent to $O$. This roughly amounts to the free strict $Q$-algebra computing the free ∞-algebra over $O$, which in its turn almost immediately boils down to the strict coinvariants of tensor products $Q_n⊗_{Σ_n}X^{⊗n}$ computing the corresponding homotopy coinvariants of derived tensor products, for any cofibrant object $X$. The main property required here is that $Q_n$ should be cofibrant (projectively cofibrant with respect to the $Σ_n$-action in the case of symmetric operads). The second ingredient shows that any strict $Q$-algebra can be rectified to a strict $R$-algebra, where $R$ is the operad we are interested in, e.g., the associative or commutative operad, and $f\colon Q→R$ is a weak equivalence of operads. The main property required here is that $f_n ⊗_{Σ_n} X^{⊗n}$ should be a weak equivalence for any cofibrant object $X$. This can happen for two reasons: either the maps $f_n$ are sufficiently nice (e.g., their source and target are $Σ_n$-projectively cofibrant) or the category in which we work is nice. The latter holds, for example, for rational chain complexes and for various categories of symmetric spectra. For nonsymmetric operads, such as the associative operad, we can drop $Σ_n$, which makes the condition much easier to satisfy in practice. In particular, it is always true for simplicial sets, chain complexes over any commutative ring, topological spaces, simplicial modules, simplicial presheaves, etc. In all these cases, ∞-algebras over nonsymmetric operads can always be rectified to strict algebras.<|endoftext|> TITLE: Constraints on the homology of amenable groups QUESTION [11 upvotes]: Edit (March 24): My first question has been answered nicely, but I am still looking for an answer to the second one. Due to the Kan–Thurston theorem, the homology of an arbitrary group can be anything you want. Using the Rips complex, we can see that hyperbolic groups are $F_{\infty}$ and, if torsion-free, even of type $F$, i.e., there exists a model of $BG$ that is a finite CW-complex. The other side of Bridson's universe of finitely presented groups is where amenable groups live. I wondered about constraints that amenablility imposes on the cohomology of a group. More precisely, let me ask the following two separate questions. Let $G$ denote a finitely presented torsion-free amenable group. Is $G$ always of type $F$? (True for nilpotent groups, see Brown's book. In general, this is probably false or open: Wikipedia taught me that it is an unresolved conjecture to prove that Thompson's group $F$ is not amenable.) Can it happen that $H_{j}(G;Z)$ is non-trivial in a single degree $d$? (For $d = 1$, we can obviously choose $G = \mathbf{Z}$, but I do not know any examples for bigger $d$.) REPLY [11 votes]: Is every finitely presented torsion-free amenable group of homotopical type F? It's false even for metabelian groups. For $i=1,2,3$, let $G_i$ be a copy of a solvable Baumslag-Solitar group, say $\langle t_i,x_i|t_ix_it_i^{-1}=x_i^2\rangle$. Start with $H=G_1\times G_2\times G_3$. It has a homomorphism $\phi$ onto $\mathbf{Z}$ mapping each $t_i$ to $1$ and each $x_i$ to $0$. Let $G$ be its kernel. Then $G$ is finitely presented, but not of type FP$_3$ / F$_3$. This follows from the "FP$_m$-conjecture" for metabelian groups. Namely $H$ is not finitely presented by the Bieri-Strebel criterion [3], and $H$ is not FP$_3$ / F$_3$, as the "FP$_m$-conjecture" is known for $m=3$ in a semidirect product of two abelian groups [2], and also known in general when the Prüfer rank is finite [1]. Note: the exponent $2$ could be replaced with any number $\ge 2$, possibly depending on $i$. Also, $\phi_i(t_i)$ could be defined as any positive integer $n_i$, with the same conclusion (while if $\phi(t_1),\phi(t_2)>0$ and $\phi(t_3)<0$ the resulting kernel has type F). [1] H. Åberg, Bieri–Strebel valuations (of finite rank), Proc. London Math. Soc. (3) 52 (1986) 269–304. [2] R. Bieri, J. Harlander, On the FP3-conjecture for metabelian groups, J. London Math. Soc. (2) 64 (2001) 595–610. [3] R. Bieri, R. Strebel, Valuations and finitely presented metabelian groups, Proc. London Math. Soc. (3) 41 (1980) 439–464. REPLY [10 votes]: Abels' groups provide simple examples of solvable groups that are finitely presented but not of type $F$. Given a ring $R$, define the group $$A_n(R):= \left\{ \left( \begin{array}{ccccc} 1 &&&& \\ & d_1 && \ast & \\ && \ddots && \\ & 0 && d_{n-1} & \\ &&&& 1 \end{array} \right) \in \mathrm{GL}(n+1,R), \ d_1,\ldots, d_{n-2} \in R^\times \right\}.$$ Extending Abels' result, Brown proved that, for every prime $p$ and every $n \geq 1$, the (virtually torsion-free) group $A_n(\mathbb{Z}[1/p])$ is of type $FP_{n-1}$ but not of type $FP_n$, and it is finitely presented whenever $n \geq 3$.<|endoftext|> TITLE: The core model and elementary embeddings QUESTION [7 upvotes]: Let $K$ be the core model (below a Woodin cardinal). Let $j \colon K \to M$ be an elementary embedding, where $M$ is well founded. Under which conditions can we conclude that $j$ is an iterated ultrapower of extenders in $K$ (possibly a branch in an iteration tree)? Are there generalizations of this result to larger inner models? REPLY [8 votes]: Some remarks: By Schindler's paper "Iterates of the core model", if $j:V\to N$ is elementary ($N$ transitive) and $N$ is closed under $\omega$-sequences, and $k:K\to K^N$ is the restriction of $j$, then $K^N$ is an iterate of $K$ and $k$ is the iteration map. Note that if $M$ is a mouse modelling ZFC + ``$\delta$ is Woodin'' and $(\delta^+)^M$ is countable, then letting $j:M\to U$ be a countable stationary tower embedding, $U$ is not an iterate of $M$, since $\omega_1^U=\delta$. But for example, assuming $M_1^\sharp$ exists and is fully iterable, if $U$ is already known to be a non-dropping iterate of $M_1$ and $j:M_1\to U$ is elementary, then $j=j_1\circ j_0$ where $j_0$ is the iteration map $M_1\to U$ and $j_1:U\to U$ has $\mathrm{crit}(j_1)$ above the Woodin of $U$. Given a mouse $M$ and $j:M\to N$ definable from parameters over $M$, with $N$ transitive, one can ask the same question, i.e. whether $N$ is an iterate and $j$ is an iteration map. There are various partial results on this (Schindler's mentioned above is relevant), but I think the question is open in general, even assuming that $M$ knows fully how to iterate itself.<|endoftext|> TITLE: When is a metric space a snowflake? QUESTION [5 upvotes]: Let $(X,d)$ be a metric space. For any $0<\epsilon<1$, we call the metric space $(X,d^{\epsilon})$; where $d^{\epsilon}(x,y)\triangleq (d(x,y))^{\epsilon}$ the $\epsilon$-snowflake of $(X,d)$. My question is, given $(X,d)$ and some $\epsilon\in(0,1)$, under what conditions can we deduce that there exists some other metric space $(Z,\rho)$ such that $(X,d)$ is the $\epsilon$-snowflake of $(Z,\rho)$; i.e.: $$ (X,d)=(Z,\rho^{\epsilon})? $$ REPLY [6 votes]: You may be interested in the following paper: Jeremy T. Tyson and Jang-Mei Wu, Characterizations of Snowflake Metric Spaces. Annales Academiae Scientiarum Fennicae Mathematica. Volume 30, 2005, 313-336. https://www.emis.de/journals/AASF/Vol30/tyson.pdf The paper contains a number of equivalent characterizations of the property that a metric space $X$ is bi-Lipschitz equivalent to a snowflake. Some of the characterizations in the paper require that the space is embedded in a certain Banach space, and some do not.<|endoftext|> TITLE: Basis of invariant tensors of rank n in three dimensions QUESTION [10 upvotes]: [This is a question motivated by theoretical physics, so apologies if the language is rough...] In three dimensions the spaces of invariant (or isotropic) tensors of rank $n$ have dimensions 1, 0, 1, 1, 3, 6, 15, 36, 91, 232, ... (see MathWorld). Equivalently, this is the number of trivial representations in the $n$-fold tensor product of the fundamental representation of $SO(3)$. Any invariant tensor can be expressed as a linear combination of products of Kronecker deltas and (for odd rank) one Levi-Civita symbol (see this note, which contains references to the original literture), but in general there will be linear independences. The dimensions are given by the Riordan numbers, of which there are many combinatoric interpretations. My question is: is there an explicit bijection between one of these interpretations and a linearly independent basis of invariant tensors? For example, in the case of $SU(2)$ there is such a bijection, expressed in terms of noncrossing pairings on a circle, with each pairing corresponding to $\epsilon_{ab}$ (in quantum mechanics we say a two-spin singlet state). This is originally due to Rumer, Teller, and Weyl, and generalized by Temperley and Lieb. See this paper for a more recent discussion REPLY [10 votes]: Planar partitions with no singletons works. You need to pick for each $n>1$ some map with certain properties. One way to do this is to just fix a preferred trivalent tree of each size and interpret each vertex as a cross product. For example, one arbitrary choice gives the map $V^{\otimes n} \rightarrow \mathbb{C}$ given by $$v_1 \otimes \ldots \otimes v_n \mapsto (((v_1 \times v_2) \times v_3) \ldots \times v_{n-1})\ldots) \cdot v_n.$$ It’s a little awkward that the above formula isn’t invariant under cyclic permutation of the variables, so you have to arbitrarily pick a "starting point" for each connected component. A better way is to write the non-crossing partition of 2n strands where each end connects to its neighbor and then attach a bunch of “2nd Jones-Wenzl projectors” (i.e. the projection from the square of the 2d to the 3d) along the boundary in an interleaving way (ie connecting each strand to its other neighbor). As discussed in comments, the JW2 part can be interpreted as rewriting the vector in terms of the corresponding sum of Pauli matrices. If we let $\sigma_v$ denote the sum of Pauli matrices corresponding to $v$, then the rotationally invariant map is just the trace of the product of Pauli matrices: $$v_1 \otimes \ldots v_n \mapsto \mathrm{Tr}(\sigma_{v_1} \sigma_{v_2} \cdots \sigma_{v_n}).$$ This was the formula for a single connected component, so the general formula is a product over connected components of the trace of the product of Pauli matrices for each component. Another approach, using a different set, is to take n pairs of boundary points and look at non-crossing partitions of 2n points such that no pair connects to itself. Then just interpret this in SU(2) by embedding each 3-dimensional irrep inside the square of the 2-dimensional irrep.<|endoftext|> TITLE: Why is the mapping class group of a surface with nonempty boundary torsion-free? QUESTION [6 upvotes]: On page 201 of Farb and Margalit's Primer on Mapping Class Groups, they explain why the mapping class group $\mathrm{Mod}(S)$ is torsion-free when $\partial S \neq \varnothing$. Here is my understanding of the argument: Let $S$ be a surface with a hyperbolic metric and let $\phi\colon S \to S$ be an isometry fixing $\partial S$ pointwise. Then $\phi$ fixes a frame (a basis for the tangent space) at every point of $\partial S$. Since isometries of surfaces are determined by their action on a frame, we must have that $\phi = \mathrm{id}$. The Nielsen realization theorem (Theorem 7.1) states that for every order $k < \infty$ element $f \in \mathrm{Mod}(S)$, there is an isometry $\phi \in \mathrm{Homeo}^+(S)$ of order $k$ representing $f$. However, there is no guarantee that $\phi$ will fix the boundary, we only know that $\phi$ is in the free homotopy class $f$. Up to here I understand all the points that have been made. What I don't understand is how they go from this, and the fact that Dehn twists about boundary components have infinite order, to conclude that $\mathrm{Mod}(S)$ is torsion-free whenever $\partial S \neq 0$. Is the point that if $f$ is a torsion element, then the isometry representative $\phi$ given by the NIelsen realization theorem must fix the boundary pointwise and therefore be the identity? I don't know how you would show this. I also don't understand why Dehn twists entered the argument. REPLY [8 votes]: I think the reason that Dehn twists enter is that we can take the differential of a (orientation-preserving) diffeomorphism $f$ of $S_g$ that fixes a chosen basepoint $\ast \in S_g$ at this point, and will get a map $d \colon \text{Diff}^+(S_g,\ast) \to \text{GL}_2^+(\mathbb R)$. The fiber of this fibration is then $\text{Diff}_{\partial}(S_{g,1})$. When you study the effect of the resulting fiber sequence $\text{Diff}_{\partial}(S_{g,1}) \to \text{Diff}^+(S_g,\ast) \to \text{GL}_2^+(\mathbb R)$ on homotopy groups, you obtain a short exact sequence $1 \to \mathbf{Z} \to \Gamma_{g,1} \to \Gamma_g^1 \to 1$ where $\Gamma_g^1$ stands for the mapping class group of a once-punctured surface of genus $g$, and the map $\mathbf{Z} \to \Gamma_{g,1}$ is given by Dehn twisting around the boundary curve. Ok, so far so good, but now let us prove that $\Gamma_{g,1}$ is torsion-free, by contradiction: suppose $f \in \Gamma_{g,1}$ has finite order $k > 1$. Then (as you remarked), we can find $\phi \in \text{Diff}(S_g,\ast)$ of order $k$ representing $f$. We can also find a compatible metric on $S_g$. Then $d(f)$ actually lands in $SO(2) \subset \text{GL}_2^+(\mathbb R)$ and is given by a rotation, say around angle $\theta$. Then $k\theta = \ell2\pi$ for some non-zero integer $\ell$, and you can convince yourself that $f^k$ represents a non-trivial Dehn twist in $\Gamma_{g,1}$, contradicting $f^k = 1$.<|endoftext|> TITLE: Universal enveloping algebra of Malcev Lie algebra associated to nilpotent group QUESTION [7 upvotes]: Let $G$ be a finitely generated torsion-free nilpotent group. The Malcev completion of $G$ is a nilpotent Lie group $N$ into which $G$ embeds as a lattice. One way to construct this is to take the completion $\widehat{\mathbb{R}[G]}$ of the real group ring with respect to the augmentation ideal. This is a Hopf algebra, and $N$ is the set of group-like elements in it. The Lie algebra of $N$ is the set of primitive elements $P(\widehat{\mathbb{R}[G]})$. Let $R$ be the subalgebra of $\widehat{\mathbb{R}[G]}$ generated by $P(\widehat{\mathbb{R}[G]})$. Question: is $R$ the universal enveloping algebra of $P(\widehat{\mathbb{R}[G]})$? If I understand the setup correctly, Quillen proved in his paper "On the associated graded of a group ring" that the associated graded of $R$ is the universal enveloping algebra of the associated graded of $N$, but I don't know if that is true before we take the associated graded. Of course, the field $\mathbb{R}$ is only playing a minor role here, and the natural question is to replace it by an arbitrary field of characteristic $0$. REPLY [4 votes]: Yes, equivalently, for $A = \widehat{kG}$ where $k$ is of characteristic zero, the map $U = \widehat{U(\mathrm{Prim }A)} \to A$ is injective. In fact, this holds whenever $A$ is complete with respect to its augmentation ideal. Proof: let $f: U \to A$ be the natural map, and let $J$ be the augmentation ideal of $U$. If $f$ is not injective, then there is a least $n$ such that $J^n \cap ker(f) \neq 0$. Let $x \in J^n \cap ker(f)$. Then $$y=\Delta(x) - x\otimes 1 - 1\otimes x \in \sum_{p=1}^{n-1} J^p \otimes J^{n-p}.$$ Since $ker(f)$ is a Hopf ideal, $0 = (f\otimes f)(\Delta(x)) - f(x)\otimes 1 - 1 \otimes f(x) = (f \otimes f)(y)$. But $f\otimes f$ is injective on $\sum_{p=1}^{n-1} J^p \otimes J^{n-p}$ by hypothesis on $n$, so $y = 0$ and $x$ is primitive. But the primitive elements of $U$ are exactly $\mathrm{Prim}\, A$, and the map $\mathrm{Prim}\, A \to A$ is injective. Hence $x = 0$, a contradiction. In fact, more is true when $A$ is cocommutative. Theorem (Milnor-Moore): if $A$ is a cocommutative Hopf algebra over a field $k$ of characteristic zero, complete with respect to its augmentation ideal $I$, then $A \cong U = \widehat{U(\mathrm{Prim}\, A)}$, the completion of the enveloping algebra of its primitive elements. See for instance the original paper of Milnor-Moore or the book Tensor Categories by Etingof et al.<|endoftext|> TITLE: Order from Coxeter-Dynkin diagram QUESTION [5 upvotes]: How is the order of a Coxeter group determined from its Coxeter-Dynkin diagram? REPLY [3 votes]: Several answers to this question are given in Section 12 of a paper called "Generalized cluster complexes and Coxeter combinatorics" (https://arxiv.org/abs/math/0505085). Specifically, there are four (more or less crazy) ways to determine invariants including exponents recursively from the diagram (if you know all invariants for connected subgraphs of the diagram). Once you know the exponents, you know the order of $W$. These methods are much too involved to describe here, but several of them have to do with the Fuss-Catalan number associated to $W$. This is a polynomial in a parameter $m$ and has a product formula in terms of the Coxeter number and exponents. You can factor the Fuss-Catalan number (polynomial) completely and (if you know the Coxeter number $h$) read off the exponents.<|endoftext|> TITLE: Approximation of a complex manifold by an algebraic variety QUESTION [5 upvotes]: What are some natural notions of distance $d$ between two complex manifolds of dimension $n$? For any of these notions what are the current best results on approximation of a complex manifold $M$ by a complex algebraic variety $V_M$? What happens if we impose the condition that $V_M$ must be non-singular? Can the precision of approximation be measured in terms of algebraic invariants of $M$ and $V_M$? REPLY [3 votes]: You ask several questions, but I will only really address the last, or at least my interpretation of it. In the real setting, any compact manifold is diffeomorphic to a real algebraic variety by work of Nash-Tognolli. In the complex setting, such a result is not possible. For example, any finitely presented group is the fundamental group of a compact complex manifold (Taubes), but this is far from true for nonsingular algebraic varieties. For instance, if $\Gamma$ is a nonvirtually nilpotent solvable group, then it won't be the fundamental group of a nonsingular algebraic variety. So I'm really sure what approximation would be mean in this case.<|endoftext|> TITLE: Why are characters orthogonal to cusp forms? QUESTION [9 upvotes]: Let $G = \operatorname{GL}_2$, and let $V = L^2(Z(\mathbb A)G(\mathbb Q) \backslash G(\mathbb A),\omega)$, for $\omega$ a character of the ideles $\mathbb A^{\ast}$, identified with a central character. For a character $\mu$ of $\mathbb A^{\ast}/\mathbb Q^{\ast}$ such that $\mu^2 = \omega$, let $\chi = \mu \circ \operatorname{det}$. Then $\chi$ is an element of $V$. How does one see (or intuit) that $\chi$ should be orthogonal to all cusp forms in $V$? Recall a cusp form is an element $f \in V$ such that $$\int\limits_{N(\mathbb A)/N(\mathbb Q)} f(ng) dn = 0$$ for almost all $g \in G(\mathbb A)$, where $P = TN$ is the usual Borel subgroup of $G$ with its Levi decomposition. My idea was to take the usual maximal compact subgroup $K$ of $G(\mathbb A)$ and say that, just as we have $\int\limits_{G(\mathbb A)} = \int\limits_{N(\mathbb A)} \int\limits_{T(\mathbb A)} \int\limits_K$, we should also have something like $$\int\limits_{Z(\mathbb A)G(\mathbb Q) \backslash G(\mathbb A)} f(g) \overline{\chi(g)}dg = \int\limits_{N(\mathbb Q) \backslash N(\mathbb A)} \space \int\limits_{ Z(\mathbb A) T(\mathbb Q) \backslash T(\mathbb A)} \space \int\limits_{[ Z(\mathbb A) G(\mathbb Q) \cap K] \backslash K} f(ntk) \overline{\chi(ntk)} \space dk dt dn$$ which should come out to $0$, using the fact that $\chi(ng) = \chi(g)$ for all $n \in N(\mathbb A)$. The problem is, I don't know if the integration over $Z(\mathbb A)G(\mathbb Q) \backslash G(\mathbb A)$ should decompose like this. I tried to prove this for some time, but I am not sure whether this step can be made sound. Is such a decomposition of measures ''legit?'' I would be grateful for any explanation or reference. REPLY [3 votes]: For the second measure theoretic question, I don’t know the answer, but I think, for the first question about the orthogonality is resolved as follows: We may assume that $\omega$ is unitary by tensoring a central character. By the Gelfand-Piatetski-Shapiro theorem, the space of L^2 cusp forms is decomposed to a direct sum of irreducible unitary representations with finite multiplicities (e.g. use Corvallis, Borel-Jacquet, 2.2 and 4.3, Wallach real reductive groups I 1.6.6). If $\pi$ is one of the irreducibles, then orthogonal projection from $\chi$ to $\pi$ is G-equivariant and not isomorphism, hence 0, by Schur’s lemma, so $\chi$ is orthogonal to L^2-cusp forms.<|endoftext|> TITLE: What is the indecomposable decomposition of holomorphic differentials of an Artin-Schreier curve C as a Z/p-representation? QUESTION [9 upvotes]: I am attempting to decompose the holomorphic differentials of an Artin-Schreier-Witt curve as a $\mathbb{Z}/p^n$-representation. This is done in Theorem 1 of Madan-Valentini Automorphisms and holomorphic differentials in characteristic p. However, I seem to be unable to understand their result even in the simplest case ($n=1, p=3$). First, I will define the quantities that show up in their theorem. $F$ is the function field of the curve with affine model $y^3-y = x^2$. $E$ is $K(x)$ the function field of $P^1$. They use the notation $\Omega_F:= H^0(C, \Omega_C)$. In our example, we have just one ramified prime $P'$ (the elliptic curves point at infinity) which is totally ramified over the point $P$ (infinity on the projective line). In our case, $r = 0$. [[This is because $r := 1 - max(e_i)$, where $e_i$ is the ramification index of all primes $P_i$ in $F$ which ramify over $P$ in $E$, and $e_i = 1$, thus, $r = 0$.]] Further, $\Gamma_k := v_{ik} := \lfloor \frac{d(P'|P) + kv_{P'}(x^2)}{p} \rfloor$, where $d(P'|P)$ is the exponent of the different of the point $P'$ over the point $P$. Their theorem is as follows: What I know is true: In our example, there is only one degree 1 component when considering $H^0(C, \Omega_C)$ as a $Z/3$-representation, and no other components. This degree 1 component is generated by $dy/x$. As a sanity check, we know apriori that the holomorphic differentials of C coincide with the cotangent space of the Jacobian at its origin, that is $H^0(C, \Omega_C) \simeq T^*_e(Jac(C))$, so it should indeed be 1-dimensional since $Jac(C)$ of an elliptic curve $C$ is 1-dimensional. Madan-Valentini result: The two values that go into the definition of $\Gamma_{k+1} := v_{1k}$ in Theorem 1 are (1) $d(P'|P)$ the exponent of the different of the point $P$ at infinity in $K(x)$ wrt the point $P'$ at infinity in the elliptic curve, and (2) the valuation of $x^2$ at the point $P'$ in the elliptic curve totally ramified above infinity. I calculated that (1) is 6, and (2) is 6 also. Plugging in my values of (1) and (2), I get $v_{11} = 0$ and, $v_{12} = -2$. But!! But, having only 1 degree one component in the representation forces the values of $\Gamma_0 := v_{11}$ and $\Gamma_1 := v_{12}$ in their theorem to have the relation $v_{11} = v_{12} + 1$. I can't find my error, however! It's driving me up the wall! Here's my question: What have I done wrong? Why am I getting a contradiction? I will explain how I calculated (1) and (2), I hope that someone can spot an error. The method of calculating (2) was explained to me by Jora Belousov, and all errors are mine and mine alone. (1) calculation of the exponent of the different $d(P'|P)$ The only ramification point is at infinity which I'll call $P := (1/x)$. We consider a valuation $v_P$ on $K(x)$. If the $x$-valuation of $w \in K(x)$ is nonpositive, then $v_P(x^2 - (w^p-w)) = v_P(x^2) = -2$. If the $x$-valuation of $w$ is positive, then $v_P (x^2 - (w^p-w)) = v_P(w^p) = p v_P (w)$ which is negative and less than $-p$, so the answer is $-2$ and it is well-defined. Thus, we have that our jump at $P$ is $m_P = 2$. Here I am using Stichtenoth, Algebraic Function Fields and Codes, Thm 3.7.8. We may then immediately calculate the exponent of the different $d(P'|P) = (p-1)(m_P + 1)$, i.e., 2*3 = 6. (2) calculation of $v_{P'}(x^2)$ (which is called $-\Phi(1, 1)$ in Valentini-Madan) Let $v_{P'}$ be the valuation extending the valuation at infinity of the projective line, which we've called $v_P$. First, we homogenize the equation $x^2 = y^3 - y$: and get $x^2z = y^3 - yz^2$. Then, we go to the chart $x = 1$, in this chart, the function $f=x^2$ becomes $f = 1/Z^2$. Here, our equation becomes $Z = Y^3-YZ^2$ which implies $$Z = Y^3/(YZ + 1).$$ We now localize the coordinate ring $K[Y,Z]/(Y^3 - YZ^2 - Z)$ at the point $(Y = 0, Z=0)$, let's call that ring $A_P$. Since $(YZ+1)$ doesn't vanish at this point, it is a unit in $A_P$. Thus, $Z$ is contained in the ideal generated by $Y$. Because the maximal ideal of the local ring $A_P$ is the principal ideal of $Y$, the valuation that we are interested in can be computed as the $Y$-degree: $v_(Y,Z) = v_Y$. Again using the elliptic curve equation, we may express our original $f = x^2$ in our new coordinates as $f = (1/Z)^2 = (YZ+1)^2/Y^6$. Thus, $v_{P'}(x^2) = v_{(Y,Z)}(Z^{-2}) = v_{Y} (Y^{-6}) = -6$. REPLY [3 votes]: Thanks to Jeremy Booher for explaining the following in private correspondence. My error was in the indexing in my definition of $v_{1k}$. In their paper, they define $$v_{1k} := \lfloor \frac{d(P'|P) - k\Phi(1,1)}{p} \rfloor.$$ I misunderstood the definition of $\Phi(1, 1)$ due to losing track of a sneaky index shift. My error was in thinking that $\Phi(1, j) = v_{P_j}(x^2)$. In their notation, $$\Phi(1,j) =v_{P_{j-1}}(x^2),$$ where $j$ is the point $P$ considered in the field $E_j$ (where we are considering the $\mathbb{Z}/p^n$ extension $F/E$ as a series of $n$ successive $\mathbb{Z}/p$ extensions $$E = E_0 \subset E_1 \subset \cdots \subset F=E_n.$$ Thus, $\Phi(1,1) = -v_{P}(x^2)= 2$, and $$v_{1k} := \lfloor \frac{d(P'|P) + kv_{P}(x^2)}{p} \rfloor.$$ This resolves the issue entirely.<|endoftext|> TITLE: About the number of fixed points of a torus action QUESTION [6 upvotes]: Suppose that $X$ is a complex, irreducible, projective variety with at most terminal singularities. Furthermore, assume that $\mathbb{C}^*$ acts on $X$ with exactly $k$ fixed points, where $k>0$. Question. Is it true that $k > \dim_{\mathbb{C}}(X)$? REPLY [11 votes]: This is true in a much more general setting. Let $X$ be any normal projective $T$-variety defines over an algebraically closed field. Then I claim that $\# X^T>\dim X$. We show this in two steps. (1) According to a theorem of Sumihiro, there is an equivariant embedding of $X$ into a projective space ${\bf P}^n$. Normality is used only in this first step. Often this step is superfluous because $X$ is sitting already in a projective space. Then all we need for the second step is that $X$ is closed. (2) Now let $T$ act linearly on ${\bf P}^n$ and let $X\subseteq{\bf P}^n$ be closed and $T$-stable. Then there are three substeps: (a) Assume $X={\bf P}^n$. Then ${\bf P}^n={\bf P}(V)$ where $V$ is an $n+1$-dimensional representation of $T$. This representation is diagonalizable. Hence $V=U_1\oplus\ldots\oplus U_n$ where each $U_i$ is $1$-dimensional with $T$ acting with a character $\chi_i$. Then $T$ has $n+1$ fixed points namely ${\bf P}(U_0),\ldots,{\bf P}(U_n)$. (b) Assume $X^T=({\bf P}^n)^T$. Then $\# X^T=\#({\bf P}^n)^T>n\ge\dim X$ by (a). (c) Finally assume $X^T\subsetneq ({\bf P}^n)^T$. Then there is a $T$-fixed point $x$ not lying in $X$. Let $\pi:{\bf P}^n\setminus\{x\}\to{\bf P}^{n-1}$ be the projection with center $x$. Then the restriction of $\pi$ to $X$ is both affine and projective hence finite. Let $\tilde X:=\pi(X)\subseteq{\bf P}^{n-1}$. Then $\#\tilde X^T>\dim\tilde X=\dim X$ by induction on $n$. We conclude with $\# X^T\ge\#\tilde X^T$ since the fibers of $\pi|_X$ are finite and $T$ is connected. Edit: Part (2) is Prop. IV.13.5 of A. Borel, Linear Algebraic Groups 2nd ed., Springer GTM 126.<|endoftext|> TITLE: Least prime in Artin's primitive root conjecture QUESTION [6 upvotes]: Let $a$ be an integer which is neither a square nor $-1$. Artin's conjecture states that there are infinitely many primes $p$ for which $a$ is a primitive root modulo $p$. My question is whether there is anything on the literature for (1) the (conjectural) size of the smallest such prime $p$. (2) Conditional or unconditional upper or lower bounds. Regarding (2), one can chase down the implied constants in Hooley's paper (Hooley, Christopher (1967). "On Artin's conjecture." J. Reine Angew. Math. 225, 209-220) to show that his asymptotic must kick in after $x\geq x_0=|a|^{C\log \log 3|a|}$ for some absolute $C>0$. Thus, under GRH the least prime is at most $$|a|^{C\log \log 3|a|}.$$ REPLY [8 votes]: One would guess much smaller. If p=2q+1 where p and q are prime, then a is a primitive root mod p IFF (a/p)=-1 and p does not divide a+1. One would guess from the usual heuristics that there is such a prime p once one has tested about c log |a| loglog |a| of them. Therefore we would expect to find such a p which is << log |a| (log log |a|)^3. For most a, one of the first 100 Sophie Germain primes will work; indeed I very much doubt you can compute an example where they will not.<|endoftext|> TITLE: Howson property of automorphism group of $F_2$ and of $F_3$ QUESTION [7 upvotes]: Is the intersection of any two finitely generated subgroups of $\operatorname{Aut}(F_2)$ (resp. $\operatorname{Aut}(F_3)$) again finitely generated? That is, does $\operatorname{Aut}(F_2)$ (resp. $\operatorname{Aut}(F_3)$) have the Howson property? Here $F_n$ is the free group of rank $n$. For $n \geq 4$, it follows by @YCor’s comment that $\operatorname{Aut}(F_n)$ does not have the Howson property. This question is maybe very simple (apologies if so!), but I have not been able to track down any references for this. I suppose my question extends quite naturally to the same question for the outer automorphism group $\operatorname{Out}(F_3)$ given the embedding $\operatorname{Aut}(F_n) \hookrightarrow \operatorname{Out}(F_{n+1})$. Note that $\operatorname{Out}(F_2) \cong \operatorname{GL}_2(\mathbb{Z})$, which is virtually free and hence has the Howson property. REPLY [7 votes]: According to wikipedia (https://en.wikipedia.org/wiki/Howson_property), any group of the form $F_2 \rtimes \mathbb{Z}$ fails to have the Howson property. Assuming we believe wikipedia, then since $Aut(F_2)$ contains subgroups of this form, it also fails to have the Howson property. (For example, take the subgroup generated by $Inn(F_2)\cong F_2$ together with any $\alpha\in Aut(F_2)$ whose image in $Out(F_2)\cong GL_2(\mathbb{Z})$ has infinite order.)<|endoftext|> TITLE: Categorification of determinant QUESTION [29 upvotes]: The notion of trace of a matrix can be generalized to trace of an endomorphism of a dualizable objects in a symmetric monoidal category. (See Ponto & Shulman for a nice description.) Is there a categorification of the notion of determinant as well? If it exists, where can I read about it? If it doesn't exist, what is the conceptual obstruction to it and what is special about the trace that makes it amenable to categorification in such generality? REPLY [8 votes]: Let me expand a bit on David C's answer. A Picard groupoid is a symmetric monoidal category in which every object is invertible (together with certain commutativity and associativity constraints), so one can apply the trace formalism there. Consider the Picard groupoid $\mathrm{Pic}^\mathbb{Z}(X)$ whose objects are graded lines, that is, $(\mathcal{L},\alpha)$ for $\mathcal{L}$ a line bundle on $X$ and $\alpha:X\to\mathbb{Z}$ a continuous function. For the morphisms, we set $\mathrm{Hom}((\mathcal{L},\alpha),(\mathcal{L}',\alpha'))$ to be isomorphisms $\mathcal{L}\to\mathcal{L}'$ if $\alpha=\alpha'$ and the empty set otherwise. Given any vector bundle $V$ on $X$, we can define an object $\mathrm{det}(V)\in\mathrm{Pic}^\mathbb{Z}(X)$. In particular, when $X=\mathrm{Spec}(k)$ and $V$ is a finite dimensional vector space, an automorphism $f:V\to V$ yields a map $\mathrm{det}(f):(\mathrm{det}(V),\dim(V))\to (\mathrm{det}(V),\dim(V))$ in $\mathrm{Pic}^\mathbb{Z}(\mathrm{Spec}(k))$, whose categorical trace is given by the usual determinant of $f$. Deligne constructed, for any exact category $\mathcal{E}$, a Picard groupoid $\mathcal{P}(\mathcal{E})$ such that $\pi_i(\mathcal{P}(\mathcal{E}))=K_i(\mathcal{E})$ for $i=0,1$, together with a universal determinant functor, and one could possibly play the same game as in the previous example. That is, to get a notion of the determinant of an automorphism $f:V\to V$ in $\mathcal{E}$ we can apply the universal determinant $\mathrm{det}(f):\mathrm{det}(V)\to\mathrm{det}(V)$ to obtain an endomorphism of $\mathrm{det}(V)\in\mathcal{P}(\mathcal{E})$, and we can look at $\mathrm{tr}_{\mathcal{P}(\mathcal{E})}(\mathrm{det}(V))$ which lives in $K_1(\mathcal{E})$.<|endoftext|> TITLE: Exact coverability of $\mathbb{Z}_n$ by cyclic shifts of a given set -- easy? NP-complete? QUESTION [14 upvotes]: Recently Ernest Davis asked me about the following computational problem: we're given as input a composite integer $n$, a divisor $k$ of $n$, and a subset $S \subset \mathbb{Z}_n$ of size k. The problem is to decide whether $\mathbb{Z}_n$ can be covered with $n/k$ cyclic translations of $S$, i.e. sets of the form $S+a_i$ for various $a_i \in \mathbb{Z}_n$. This is simply a special case of the NP-complete EXACT COVER problem---namely, where the available sets are all cyclic shifts of each other, and where all cyclic shifts are available. My suspicion is that the special case is already NP-complete, while Ernest suspects that an $n^{O(1)}$ time algorithm exists. I searched Google (and Garey&Johnson) and couldn't find leads -- would appreciate any thoughts or references! REPLY [4 votes]: E.D., I think the conjecture is not true. If one tries it on the smallest non-periodic tiling (see below) it fails. The example is from N.G. de Bruijn, On the factorization of cyclic groups, Indag. Math. 15, 4, 1953. The following Python program # A + B is a tiling of Z mod 72 A = [0, 8, 16, 18, 26, 34] B = [18, 54, 24, 60, 48, 12, 17, 41, 65, 45, 69, 21] m = len(A)*len(B) # Cheking tiling S = m*[0] for x in A: for y in B: S[(x+y) % m] += 1 if S == m*[1]: print("Yes, they tile") L = [x % len(B) for x in B] # The elements of B mod len(B) H = len(L)*[0] # Compute the histogram of L for x in L: H[x] += 1 print(f"Frequencies of B mod {len(B)}: {H}") prints Yes, they tile Frequencies of B mod 12: [4, 0, 0, 0, 0, 3, 2, 0, 0, 3, 0, 0] I also believe the problem is polynomial. I do not even know if it is in co-NP (are there certificates for non-tiling?).<|endoftext|> TITLE: Existence of point with zero mean curvature QUESTION [5 upvotes]: I'm a physicist studying differential geometry for my GR research, and I come up with the following claim (not sure if it's true or not): For any compact surface $S$ that's not homeomorphic to a sphere, $S$ must have a point with vanishing mean curvature, i.e $H=0$ Is this claim true or not? I know that by applying Gauss-Bonnet theorem, we can prove an analogous claim for the Gaussian curvature $K$ REPLY [7 votes]: A more pedestrian example: let $(u,v)\in [0,2\pi)^2$ and parametrize the torus $\mathbb{T}_{a,r} = \{ ( (a + r \cos u) \cos v, (a+r\cos u) \sin v, r \sin u) \}$. The mean curvature is given by $$ H = \frac1r + \frac{\cos u}{a + r \cos u} $$ When $r \ll a$ (this is something that looks like a bike tire: large overall radius and small cross section) then $\frac1r > \frac{1}{a - r} \geq |\frac{\cos u}{a + r \cos u}| $ and hence the mean curvature is never zero.<|endoftext|> TITLE: Realizability of a real representation using real cyclotomic coefficients QUESTION [10 upvotes]: Let $G$ be a finite group and $\rho: G \rightarrow GL(d,\mathbb{C})$ an irreducible representation with Frobenius-Schur indicator $\frac{1}{|G|}\sum_{g\in G} \operatorname{tr} \rho(g^2) = 1$. Thus $\rho$ is a real representation. A theorem by Brauer states that every irreducible representation over $\mathbb{C}$ can be written using coefficients taken from $\mathbb{Q}[\zeta_n]$ where $\zeta_n=\exp(\frac{2\pi i}{n})$ where $n$ is at most the exponent of the $G$. We also know that any real representation can be written with coefficients in $\mathbb{R}$ Can we satisfy both? For example, the irreducible representations for the dihedral groups can satisfy both conditions (real coefficients using cosines and sines of fractions of $\pi$). I can solve the problem over the algebraic integers by computing an antilinear equivariant involution $F$ such that $F(x) = J \overline{x}$ for a suitable matrix $J$, and such that $\rho(g) F(x) = F(\rho(g) x)$, which implies $\rho(g) J = J \overline{\rho(g)}$. Then I know that $J \overline{J} = \alpha \mathbb{1}$ for a positive scalar $\alpha$, and I set $J' = J/ \sqrt{\alpha}$, so that $J'$ represents the complex conjugation. Now, $J$ can be found with cyclotomic coefficients (one iterates over a basis of the space of matrices with $\{0,1\}$ coefficients, and averages over the group). But the square root operation is not necessarily closed over the cyclotomics, this is where I am currently stuck. REPLY [5 votes]: $\def\CC{\mathbb{C}}\def\RR{\mathbb{R}}\def\QQ{\mathbb{Q}}\def\ZZ{\mathbb{Z}}$The main result of this answer will be that, if $V$ is a representation defined over $\RR$ and over $\QQ(\zeta_M)$, then there is an integer $N$ such that $V$ is defined over $\RR \cap \QQ(\zeta_{MN})$. This proof will not give us good control over the size of $N$. We'll abbreviate $\QQ(\zeta_M)$ to $K_M$ and abbreviate $K_M \cap \RR$ to $R_M$. Start with a $K_M$-vector space $V$ and a representation $G \to GL(V)$ which is known to descend to $\RR$. Let $V_R$ be $V$ considered as an $R_M$-vector space and let $\Delta$ be the endomorphism of $V_R$. Then $E$ contains $K_M$. Also, since $V$ descends to $\RR$, we know that $V$ is isomorphic, over $K_M$, to the complex conjugate representation $\overline{V}$. We can compose the complex conjugate map $V \to \overline{V}$ with a $K_M$-linear isomorphism $\overline{V} \to V$ to get another element $\sigma$ of $E$. We have $\sigma \alpha = \overline{\alpha} \sigma$ for all $\alpha \in K_M$. Also, $\sigma^2$ commutes with $K_M$ so, by Schur's lemma applied to $V$, we have $\sigma^2 \in K_M$, say $\sigma^2 = a$. In other words, $E$ contains the quaternion algebra $( -1, a )$ for some $a \in K_M$ and some dimension counting shows that it is this quaternion algebra. We have $V$ defined over $R_M$ if and only if this quaternion algebra splits over $R_M$: If $V$ is defined over $R_M$, it is easy to see that $E \cong \mathrm{Mat}_{2 \times 2}(R_M)$ and conversely, if $E$ splits, the image of any rank $1$ idempotent is an $R_M$-form of $V$. So we want to show: If $E$ is a quaternion algebra over $R_M$ which splits over $\RR$ then there is an integer $N$ such that $E$ splits over $R_{MN}$. Quaternion algebras are order $2$ elements in the Brauer group. The Brauer group injects into the direct sum of all the local Brauer groups, which are $\ZZ/2 \ZZ$ for the Archimedean places and $\QQ/\ZZ$ for the non-Archimedean. Since we assumed that $E$ splits over $\RR$, there are no contributions from the Archimedean places. If $\lambda/\kappa$ is a degree $d$ embedding of local fields, then the corresponding map on Brauer groups is multiplication by $d$ from $\mathrm{Br}(\kappa) \cong \QQ/\ZZ$ to $\mathrm{Br}(\lambda) \cong \QQ/\ZZ$. In particular, in order to split our quaternion algebra $E$, we just need to find an extension $R_{MN}$ of $R_M$ such that every local field of $K_M$ at which $E$ is not split has even degree extensions in $R_{MN}$. Let $p_1$, $p_2$, \dots, $p_r$ be the primes of $\ZZ$ lying below the primes $R_M$ where $E$ is split. Choose $N$ relatively prime to $M$ and a quadratic nonresidue modulo every $p_j$. Then $R_M[\sqrt{N}] \subset R_{MN}$ has the desired property. $\square$. Obviously, this is not the answer you are looking for, but it is the best I can do.<|endoftext|> TITLE: Does homeomorphism between cones imply homeomorphism between sections QUESTION [8 upvotes]: For any topological space $A$, the cone $C(A)$ is defined to be $A \times [0,\infty)$ with $A \times 0$ identified to a point (cone point). Let $X$ and $Y$ be two compact Hausdorff spaces such that there is a homeomorphism between $C(X)$ and $C(Y)$ which preserves the cone points. Can we prove that $X$ and $Y$ are homeomorphic? REPLY [14 votes]: The answer is no. Let $X$, $Y$ be any two smoothly h-cobordant closed manifolds of dimension $\ge 4$ that are non-homeomorphic. For example, we can take $X=S^2\times L(7,1)$ and $Y=S^2\times L(7,2)$ where as usual $L(p,q)$ is a 3-dimensional lens space. By the weak h-cobordism theorem the interior of the h-cobordism is diffeomorphic to $X\times\mathbb R$ and also to $Y\times \mathbb R$. The interior is two-ended, and we can compactify it by adding two points (= the ends). The "identity" homeomorphism $X\times\mathbb R\to Y\times \mathbb R$ extends to a homeomorphism of the compactifications that takes ends to ends. Removing the points corresponding to one end gives a homeomorphism from the cone over $X$ to the cone over $Y$ that preserves the cone point. Conversely, if $X$, $Y$ are closed manifolds and $X\times\mathbb R$ is diffeomorphic to $Y\times \mathbb R$, then $X$, $Y$ are h-cobordant. The same trick of course works when $X$, $Y$ are compact non-homeomorphic spaces and $X\times\mathbb R$ is homeomorphic to $Y\times \mathbb R$.<|endoftext|> TITLE: reverse KL-divergence: Bregman or not? QUESTION [6 upvotes]: I am having a little trouble getting my head around the two "directions" of the Kullback-Leibler divergence: Definition (Kullback-Leibler divergence) For discrete probability distributions $P$ and $Q$ defined on the same probability space, $\chi$, the Kullback-Leibler divergence from $Q$ to $P$ is defined to be $$D_{KL}(P||Q) := \sum_{x \in \chi}P(x)\log\bigg(\frac{P(x)}{Q(x)}\bigg). $$ Note that The KL-divergence is not really a true distance measure, since it is does not satisfy the Triangle Inequality and $D_{KL}(P||Q)$ does not in general equal $D_{KL}(Q||P)$. Hence the need to distinguish the KL-divergence from its dual, the so-called "reverse KL-divergence". Fact. Both the KL-divergence and the reverse KL-divergence are examples of f-divergences. Fact. The KL-divergence is an example of a Bregman divergence. Question: Is the reverse Kullback-Leibler divergence also a Bregman divergence? This is not obvious to me. Also, I have read conflicting information, with Amari (2009) arguing that the KL-divergence and its dual are are unique divergences belonging to the $f$-divergence and Bregman divergence classes, and others (e.g., Wang et al. 2019) saying that the reverse KL-divergence is no longer a Bregman. Which is correct? Any pointers would be greatly appreciated. References: Wang, Q., Li, Y. and Xiong, J. (2019) Divergence-Augmented Policy Optimization link Amari, S. (2009) $\alpha$-Divergence Is Unique, Belonging to Both $f$-Divergence and Bregman Divergence Classes link REPLY [9 votes]: Define the KL convergence as in the Amari's paper linked by you: $$KL(x||y):=D_{KL}(x||y):=\sum(y_i-x_i+x_i\ln\frac{x_i}{y_i}).$$ Then $$KL(x||y)=F(x)-F(y)-\nabla F(y)\cdot(x-y)$$ if $F(x):=\sum(x_i\ln x_i-x_i)$. So, the KL-divergence is a Bregman one. On the other hand, the dual divergence, defined by $$LK(x||y):=KL(y||x)=\sum(x_i-y_i+y_i\ln\frac{y_i}{x_i}),$$ is not a Bregman one. Indeed, if it were a Bregman one, then for some appropriate function $G$ we would have $$LK(x||y)=G(x)-G(y)-\nabla G(y)\cdot(x-y)$$ and hence $$\nabla_x(LK(x||y))=\nabla G(x)-\nabla G(y),$$ whereas in fact $$\nabla_x(LK(x||y))=(1-y_i/x_i),$$ which cannot be of the form of a difference $\nabla G(x)-\nabla G(y)$ -- because otherwise we would have $(1-v/u)+(1-w/v)=(1-w/u)$ for all positive real $u,v,w$. (Amari's proof contains formula (52), which contains functions $\psi_\alpha$ and $\psi_{-\alpha}$, supposedly defined by formula (49). However, the expression for $\psi_\alpha$ in (49) is undefined for $\alpha=-1$, whereas it is both of the values $\pm1$ of $\alpha$ that are needed in (52) for $LK$.)<|endoftext|> TITLE: Who was H. Vogt? QUESTION [28 upvotes]: In Chapter I.9 of Chandler-Magnus "The History of Combinatorial Group Theory", a number of important mathematicians in the early history of the development of group theory and sources for their obituaries are given. For example, we certainly find an entry Dehn, Max, 1878-1951. For other names, less information is known, such as Pick, G., 1859-1943(?). This latter question mark reflects the fact that Georg Pick died in the Theresienstadt concentration camp in 1942, and finding this information might have been difficult at the time of the writing of the book (1982). All names in this list have a source for where their obituary may be found, and at least one of a birthyear or death year is present -- except for one name. This name is listed simply as H. Vogt: ?-?., with no further information. Curiosity piqued, this gives my question: Who was H. Vogt? What were his mathematical contributions? Here's the clues I've got this far. The most relevant piece of information is the following paper: Vogt, H., Sur les invariants fondamentaux des équations différentielles linéaires du second ordre. , Ann. Sci. École Norm. Sup. (3) 6 (1889), 3–72. (Thèse, Paris), This paper is the only paper cited by Chandler and Magnus for Vogt, and is hence the only publication I am certain is by the desired H. Vogt. It also appears to have been his Ph.D. thesis. No result can be found on Mathematics Genealogy matching this. There are a number of matches on MathSciNet for publications by an H. Vogt; the earliest is from 1879, by a Heinrich Vogt, and this could in principle be the same H. Vogt as above. The latest that could conceivably be by our H. Vogt is from 1923 -- this is again on differential equations, so seems very likely to be by the same author! This would give a (very!) rough idea of (1860-1930) as the lifespan of our dear H. Vogt -- perhaps this helps the search. One idea is that H. Vogt could possibly be related to (father of?) Wolfgang Vogt, a young German mathematician whose last paper was in 1914, and who may well have perished, as did so many other young German academics at the time, in World War 1, such as Werner Boy, of Boy's surface fame, and Hugo Gieseking. The topic of his 1906 Ph.D. thesis seems -- at least on a surface level -- somewhat related to what H. Vogt did, especially if some of the other publications on MathSciNet were by the same H. Vogt. Note: there is a 1932 paper by someone called H. Vogt, namely Vogt, H., Max Wolf., Astronomische Nachrichten 247, 313-316 (1932). ZBL59.0039.09. However, this seems to be by the nazi astronomer Heinrich Vogt (1880-1968), who seems unrelated (and likely did not write an article about differential equations at the age of 1). REPLY [20 votes]: According to Zentralblatt (which is freely accessible on Internet, since Jan 1, 2021, btw) Henri Gustave Vogt was a mathematician, apparently French, since he wrote in French, published in French journals, and his first and second names use French spelling, though his last name hints German origin. In Zentralblatt, we find 28 publications, including 14 books, beginning in 1889. His by far most famous paper is the one you mention. See also William Goldman, An exposition of results of Fricke and Vogt, arXiv:math/0402103. for a modern exposition of this paper.<|endoftext|> TITLE: Positivity of Iwahori–Hecke algebra characters on the Kazhdan-Lusztig basis QUESTION [10 upvotes]: $\DeclareMathOperator\tr{tr}$I'm interested in the irreducible characters of a finite Iwahori–Hecke algebra evaluated at the Kazhdan–Lusztig basis. These are Laurent polynomials. Are the coefficients of these polynomials always positive? In type $A_n$, the irreducible representations are given by left cells. Denote by $(C_w)$ the Kazhdan-Lusztig basis and by $(C^w)$ its dual basis with respect to the standard trace $\tr$. Take $\Lambda$ such a left cell and denote by $\chi_\lambda$ the associated character. It is known that $\chi_\lambda(h) = \tr(\sum_{x \in \Lambda}C_xC^xh)$. Hence $$\chi_\lambda(C_w) = \sum_{x\in\Lambda} \tr(C_xC^xC_w)$$ which is positive since the expression equals a sum of structure constants in the KL-basis. In the literature, one finds character tables, but always on the standard basis, not the KL-basis. Is there a reference where the character tables on the KL basis are computed? Finally, most of the positivity properties in the Hecke algebras can be proven using the categorification with Soergel bimodules. Is there a categorification of the irreducible characters? REPLY [3 votes]: I checked using the Chevie package of Gap3. It seems that for all types apart $A_n$, there are negative coefficients in some values of characters on the KL-basis. The smallest example is type $B_2$. Some character value of $C'_{rsr}$ has negative coefficients. So the first question seems only to work in type $A_n$ for which everything is understood.<|endoftext|> TITLE: The Einstein minus convention, lost QUESTION [7 upvotes]: In his milestone paper on general relativity, Einstein not only introduces the Einstein summation convention, but also (formula (45) in [1]) abbreviates a minus at the Christoffel symbols away by introducing the Gamma notation for the connection coefficients of his variant of the covariant derivative, constructed on cotangent space first from given geodesics. In later standard literature the minus is gone. (E.g. Schrödinger p.66 [3]) Why, and who dropped it? Addendum/scholium: Thanks for linking from a related question on https://hsm.stackexchange.com/questions/7974/notation-for-christoffel-symbols ! The physical relevance of that sign might explain why it is gone, and I first thought about not asking on mathoverflow. But Einstein's introduction of the covariant derivative strikes me as masterful abstract differential geometry (of his time). Plus, it can be simplified and abstracted to general symmetric connections with given geodesics without Christoffel symbol stuff (if I'm not mistaken). I have not yet found it in other literature. (Already the standard geodesic equation with Christoffel symbols looks like an ugly case of superfluous zero to me. Like Einstein I prefer an own side for the 2nd derivative.) [1] A.Einstein: Die Grundlage der allgemeinen Relativitätstheorie (1916) p.802 https://web.archive.org/web/20060830030952/http://www.alberteinstein.info/gallery/pdf/CP6Doc30_pp284-339.pdf [2] Manuscript of [1], translation and more: H.Gutfreund, J.Renn: The Road to Relativity (2015) p.92 (German manuscript facsimile) p.209 (English translation). https://press.princeton.edu/books/hardcover/9780691162539/the-road-to-relativity [3] E.Schrödinger: Space-Time Structure (1950/4) https://www.cambridge.org/core/books/spacetime-structure/554B50728DF38139E42E60BBED654D85 REPLY [3 votes]: Since the question has now narrowed down to "who lost the minus sign" in the Christoffel symbol, let me start a new thread: The OP asks for a reputable source, later than Einstein's 1916 paper, in which the minus sign is abandoned. I propose that it was Einstein himself who dropped it. Below I copy from his 1921 lectures at Princeton University, page 46:<|endoftext|> TITLE: Is an interpretation mathematics (fit for publication)? QUESTION [23 upvotes]: Background I am a mathematician with two published papers. The first is based on my PhD thesis and generalised a tool to a more general setting. The thesis was cited a number of times by the time the paper was submitted and I think that helped get it published in a top journal. The citing papers solved many of the open problems in the thesis but one eluded the experts but I managed to solve it in my second paper. What I did in the thesis could probably have been done by any expert in the field if they found the time, but the elusive open problem had been open for a few decades, and a number of experts clearly attacked the problem without success. I was delighted to solve the problem and submitted to a top journal. The referee took particular issue with the length of the paper and felt that I elaborated too much and included too much elementary material. This is kind of a side issue but an expert in the field warned me that the paper might fall down on these grounds. I found this quite deflating - I am no expert nor star, and struggle sometimes with the brevity employed in papers. I find authors could be allowed explain a lot more in papers and find far too much is in the minds of the experts and not committed to paper. I never have any issue with overly long papers, any parts that are familiar can easily be skimmed. So the background is that I find myself at odds with convention in terms of what I consider to be the good mathematical writing. The second paper did get published by a good journal. Now I have written a third paper which I love but I am worried it is moving further away from what journals might want. Or indeed mathematicians. The Paper under Discussion I work in a field that has a level of abstraction inbuilt it where intuitions are very difficult to come by. For a mere mortal like me it provided a ferocious learning curve. Last year I emailed an expert on a technical point and they included in the reply a little bit of intuition that they use. I was not satisfied one bit with the intuition - it did not capture at all the nature of the objects we study. Since then I have developed what I consider to be an comprehensive and consistent interpretation rather than mere intuition. I have a paper explaining the background to the area, the motivations, the conventional interpretations, and have presented this new interpretation. In terms of theorems and proofs I have some rather elementary but most of the paper is dealing with examples explaining how the interpretation works. It has inspired easier proofs of known results and I have some new counterexamples. What the paper possibly does is make the whole subject far more accessible to even undergrads where formerly it is a subject that is firmly of the grad level. The paper touches on heavy technical machinery but mostly works in easier examples. The interpretation brings forth streams of conjectures, etc and I have included a number of these ideas in the paper. The second half of the paper as such comprises a miscellany of ideas and conjectures and I believe can help people understand these mathematics better than before. I think it makes the whole area easier to understand but because it doesn't really contain any new big results I am worried whether or not it actually comprises mathematics. It is of an expository nature but not traditionally because it does not quite survey known results (although naturally there is an element of survey to it). It is a paper I wish I could have read at the start of my PhD but I am worried it is inappropriate for publication. Question I intend to put it on the arXiv and send it to some correspondents for feedback. However I am a little bit nervous of doing even that. I really would like advice here but need to ask a concrete question. Is a paper whose purpose is to introduce a new interpretation of given results mathematics fit for publication. PS: The irony about the second paper is that after the negative review I trimmed the elaboration and elementary material. The referee who approved it for another journal asked me to elaborate further in spots. EDIT: I have suggested in comments that I would submit the paper as is to the arXiv and then spawn a note and an article. On reflection I have decided to host the long version on my website and liberally edit down to one solid expository article for the arXiv and submission to a journal. Thanks for your help. REPLY [19 votes]: Why don't you make it a book? I know of several quite good books based on theses which were not really teeming with groundbreaking results. It does not even have to be that long as there is a niche now (absent until fairly recently) for something in the range of 80-100 pages (too long for an article and too short for a "real" book). Books are much easier to publish than papers, and no one will be picking on you for indulging in "interpretation" and "philosophy" (alas, there is a lot of people who think that there is no place for them in what they call "mathematics").<|endoftext|> TITLE: Nonconvexity and discretization QUESTION [62 upvotes]: Edit: Here's a more down-to-earth, and somewhat weakened, but I believe still nontrivial, version of the main theorem. Prototypical nonconvex spaces are $\ell^p$-spaces for $00$. The Yoneda-Ext group $\mathrm{Ext}^i$ classifies exact sequences with $i+2$ terms $$0\to V\to W_1\to\ldots\to W_i\to \ell^{p'}(\mathbb N)\to 0$$ where all $W_n$ are quasi-Banachs (but possibly only $q$-Banach for $q1$, and can be done by a direct attack.) Below is the long, rambling, original question. I recommend reading the paragraph starting with "Say you want to travel from A to B" to get some intuition for what happens in the proposition. Three months ago, I wrote the blog post Liquid Tensor Experiment, posing a challenge to verify a certain proof. The computer formalization is proceeding extremely well, I am really impressed by the efforts. But this question is about the "human" part of that same proof: I think I still don't understand (or am just beginning to understand, with writing this question a key part of this) what is happening in our proof. I can roughly follow the proof line-by-line, but I don't have a real understanding for why it works, or am only slowly getting it. (I realize that this is an awkward thing to say about one's own proof. There's a reason I want a verification...) Let me state the theorem in question. There are three inputs: The category of condensed abelian groups --- a version of topological abelian groups, but forming an abelian category --- in which this computation takes place. For an introduction to this, see the blog post. For the kind of "big picture" question I want to ask, it is however not very relevant. A $p$-Banach space $V$, for some $00$? It turns out that this is false: One can show that there are explicit $\mathrm{Ext}^1$-classes, first constructed by Ribe, coming from the entropy function. (I hadn't expected to run into the entropy functional any day soon...) In a first try, we then passed to measures of "bounded entropy", on which these extensions split; but they acquire new extensions of their own, leading to infinite regress. So one needs to replace $\mathcal M_p(S)$ by some space of measures distinctly smaller than $\mathcal M_p(S)$; for example $\mathcal M_{p'}(S)$ for $p'0$. This theorem is the basis for the "$p$-liquid analytic ring structure" on the reals; I refer you again to the blogpost for some discussion about this. Basically, the intuition here is that if you have a $p'$-measure $\mu$, then you can split it into $N$ summands that are either Dirac measures, or are of $p'$-norm roughly $\tfrac 1N$ of the $p'$-norm of the original sequence; mapping to $V$, we know where the Dirac measures go, and the other terms have very small image in $V$, as the norm scales differently on $V$. Any such decomposition of $\mu$ can be thought of as a path connecting a given measure to Dirac measures. To see that higher Ext-groups vanish, one needs to see that for any two such paths, there is some homotopy connecting them, also being sufficiently small; and higher homotopies between those. Unfortunately, this seems to be a bit tricky, as in a nonconvex shape, it is hard to find enough free room to move around: You cannot simply take the average of two maps if you want to keep track of bounds. So roughly, any such decomposition is like some valley in a function you would like to minimize, but now you have to connect different valleys without climbing too high; that's tricky. The way we (inadvertently -- see the following comments) solved this issue is by passing to a discretization of the problem, where the discretized problem is convex. (Our original motivation for the discretization was in some sense of a purely technical sort, having to do with the nature of the category of condensed abelian groups; this motivation is explained in the blog post. Only recently, when in the formalization process a significant amount of convex geometry has to be formalized, did I realize that we're using convex geometry in a critical way, and that this is a much more profound reason for the discretization.) More precisely, for some fixed radius $r$ with $0 1$ (so that $1/p'$ is positive) the passage to the major subset is not needed and one can just take $E'=E$, making the right-hand side completely convex. But for $p \leq 1$ the need to remove an exceptional minor subset is needed (consider for instance the basic example $S = {\bf N}$, $f(n) = n^{-1/p}$). This characterisation of weak $\ell^p$ interacts very well with Calderon-Zygmund theory (in which one is always discarding various small exceptional sets) and also with real interpolation methods which can often upgrade weak $\ell^p$ estimates to strong $\ell^p$ estimates (at the cost of modifying $p$ slightly).<|endoftext|> TITLE: Is the harmonic series worse than any summable series? QUESTION [5 upvotes]: It is well-known that the harmonic series is not summable. In some sense this means that it takes a lot of rather large values. We define the operator $F_{\varepsilon}: \ell^{\infty}(\mathbb N) \rightarrow [0,\infty]$ by $$F_{\varepsilon}(x) = \sum_{i=1}^{\infty} 2^{-\varepsilon \vert x_i \vert^{-1}} \text{ for }\varepsilon>0.$$ Now consider a positive summable sequence $x$ and the harmonic sequence $(1/n)_n$. Intuitively, the slow decay of the harmonic series should imply that it converges slower than anything summable (for most of it). Therefore, I ask: Is it true that for any positive summable sequence $x$ $$\limsup_{\varepsilon \downarrow 0} \frac{F_{\varepsilon}(x)}{F_{\varepsilon}((1/n))} \le 1?$$ Please just let me know if you have any questions. REPLY [6 votes]: $\newcommand\ep\varepsilon\newcommand\de\delta$ Let us show more: \begin{equation*} \frac{F_{\ep}(x)}{F_{\ep}((1/n))}\to0\tag{$*$} \end{equation*} (as $\ep\downarrow0$). Indeed, \begin{equation*} F_{\ep}((1/n))=\sum_{i=1}^\infty 2^{-\ep i}=\frac{2^{-\ep}}{1-2^{-\ep}}\asymp\frac1\ep. \tag{0} \end{equation*} On the other hand, take any positive $x_n$'s such that $\sum_1^\infty x_n<\infty$. For each natural $k$, let $J_k$ denote the set of all natural $n$ such that $\frac1k\le x_n<\frac1{k-1}$, where $\frac1{k-1}:=\infty$ for $k=1$: \begin{equation*} n\in J_k\iff\frac1k\le x_n<\frac1{k-1}. \end{equation*} Then the $J_k$'s partition the set of all natural numbers. Moreover, the condition $\sum_1^\infty x_n<\infty$ implies \begin{equation*} \sum_{k=1}^\infty|J_k|/k<\infty, \tag{1} \end{equation*} where $|J_k|$ is the cardinality of $J_k$. In particular, it follows that $|J_k|<\infty$ for all $k$. Further, $2^{-\ep/x_n}<2^\ep\times2^{-\ep k}$ for $n\in J_k$. So, \begin{equation*} F_{\ep}(x)<2^\ep\sum_{k=1}^\infty 2^{-\ep k}|J_k|. \end{equation*} Take now any real $\de>0$ and, in view of (1), let $k_\de$ be a natural number such that \begin{equation*} \sum_{k\ge k_\de}|J_k|/k<\de. \end{equation*} Let $c_\de:=\sum_{k=1}^{k_\de-1}|J_k|$. Then \begin{align*} F_{\ep}(x)&0$. Now ($*$) follows, in view of (0).<|endoftext|> TITLE: Checking exactness of a triangle on stalks QUESTION [8 upvotes]: Suppose I have a triangle $$A \to B \to C \to A[1]$$ in $D(Ab(X))$, the derived category of abelian sheaves on some topological space $X$. For each $x \in X$, there is an exact functor $D(Ab(X)) \to D(Ab)$ that takes a complex of sheaves to the complex of stalks at $x$. Is my triangle an exact triangle if the image under the stalk functor is exact for all $x \in X$? I suspect that the triangle need not be exact. REPLY [4 votes]: The answer is no, but we can say a bit more : it can become true if you pass to the derived $\infty$-category and replace the words "distinguished triangle" with "cofiber sequence" (modulo the choice of a nullhomotopy) I'll assume we already know that the composite $A\to C$ vanishes - this cannot be deduced from a stalkwise assumption, as we will see later. Complete the morphism $A\to B$ to an exact triangle $A\to B\to \mathrm{cofib}\to A[1]$, then there exists a map of sequences of composable morphisms (not triangles a priori) $\require{AMScd}\begin{CD}A@>>> B@>>> \mathrm{cofib}\\ @VVV @VVV @VVV \\ A @>>> B @>>> C\end{CD}$ (it's not unique, in fact there will be different ones depending on a chosen nullhomotopy of $A\to C$ in the derived $\infty$-category) I then claim that $\mathrm{cofib}\to C$ is an equivalence: because taking stalks is exact, your assumption implies that it is an equivalence stalkwise. Further note that $H_n(A_x)\cong H_n(A)_x$ naturally in $A$, so that the assumption implies $H_n(A)_x\cong H_n(B)_x$ for all $n$, and so $H_n(A) \cong H_n(B)$ for all $n$, i.e. the map $\mathrm{cofib}\to C$ is an equivalence. What we have proved is : Under your assumptions, there exists a morphism $C\to A[1]$ such that the triangle $A\to B\to C\to A[1]$ is distinguished. However, as is maybe clear from the proof, this morphism $C\to A[1]$ need not be the one you started with, and in particular it is not clear that your original triangle will be distinguished. Here's a counterexample to that effect: over $X= Spec(\mathbb Z)$ for instance, find an extension of abelian groups $0\to A\to B\to C\to 0$ which splits when localized at each prime (and thus rationnally), but doesn't split integrally. Then $A\to B\to C\to A[1]$ where we put the $0$ map as the map $C\to A[1]$ is an exact triangle at each stalk, but is not an exact triangle. The original triangle might not be distinguished. However, as explained before, there is a map $C\to A[1]$ making it into an exact triangle. So that's the best we can hope for, and it is true. (for an explicit counterexample, on can use an extension of $\mathbb Q$ by $\mathbb Z$ given by the following composite morphism $\mathbb{Q\to Q/Z \to Z}/p^\infty \to \mathbb{Q/Z}$ where we use the decomposition of $\mathbb{Q/Z}$ as the sum of its local torsions; where we fix a prime $p$, e.g. $p=2$)<|endoftext|> TITLE: What is meant by this notation of the real forms of $E_6$? QUESTION [5 upvotes]: There are five real forms of the exceptional Lie group, $E_6$. Four of them are notated as in the following: The split form as EI or $E_{6(6)}$ The quasi-split form as EII or $E_{6(2)}$ EIII or $E_{6(-14)}$ EIV or $E_{6(-26)}$ What do the annotations to $E_6$ actually indicate and are they also used for real forms for Lie groups in general? REPLY [8 votes]: The notation is a bit complicated to make precise, but the number in parentheses is the character, which is defined on page 353 section C of Helgason, Differential Geometry, Lie Groups and Symmetric Spaces. The character is the difference $\dim \mathfrak{p}_0 - \dim \mathfrak{k}_0$ in dimensions in a Cartan decomposition of a real form. As Helgason explains, the character of an exceptional real simple Lie algebra determines the Lie algebra, but for the classical real simple Lie algebras, it doesn't, so the notation is only used for exceptionals. REPLY [7 votes]: $E_{6(n)}$ means that $n$ is the dimension of the group $E_6$ minus twice the dimension of a maximal compact subgroup. example: $E_{6(6)}$ of dimension 78 has maximal compact subgroup ${\rm Sp}(4)/\mathbb{Z}_2$ of dimension 36.<|endoftext|> TITLE: Why is the automorphism group of the octonions $G_2$ instead of $SO(7)$ QUESTION [14 upvotes]: I can calculate the derivation of the octonions and I clearly find the 14 generators that form the algebra of $G_2$. However, when I do the same calculation for the quaternions, I end up with the three generators of $SO(3)$, which basically tells me that I can rotate the set of imaginary units anyway I like. Intuitively, I don't understand, why it is not possible for the octonions to be rotated in the same way with an arbitrary rotation of $SO(7)$. Instead, the calculations show, that the possible transformations are restricted to the $G_2$ subgroup. Is there way to understand this geometrically or algebraically? Is it related to the non-associative property of the octonions? REPLY [24 votes]: The quaternions are generated by any two imaginary elements $x$ and $y$ that are orthonormal, i.e., $\bigl(1,\, x,\, y,\, xy\bigr)$ is an orthonormal basis of the quaternions. Moreover, the multiplication table using such a pair does not depend on which pair you choose. That's why the automorphism group of the quaternions acts transitively on orthonormal pairs in the imaginary quaternions. Meanwhile, the octonions are generated algebraically by any three imaginary elements, say, $x$, $y$, and $z$ that are orthonormal and $z$ is perpendicular to $xy$. In fact, $\bigl(\,1,\, x,\, y,\, xy,\, z,\, zx,\, zy,\, z(xy)\,\bigr)$ is an orthonormal basis for the octonions. Moreover, as was shown by Dickson, if $x'$, $y'$, $z'$ are another orthonormal triple of imaginary octonions such that $z'$ is perpendicular to $x'y'$, there is a unique automorphism of the octonions that carries $(x,y,z)$ to $(x',y',z')$. Note that the set of orthonormal triples $(x,y,z)$ in the imaginary octonions is the Stiefel manifold $V_{7,3}$, which has dimension $6+ 5+ 4 = 15$, and the single extra relation $z\cdot xy = 0$ cuts out a submanifold of dimension $14$. Hence the automorphism group of the octonions is a Lie group of dimension $14$. Note that, any automorphism of the octonions that fixes three such elements is the identity. Thus, $\mathrm{SO}(7)$ (which has dimension $21$) is too large to be the automorphism group of the octonions, not the least because it acts transitively on the set of oriented orthonormal bases of the imaginary octonions. (Sadly, when $x$, $y$, $xy$, and $z$ are imaginary orthogonal octonions, the defining property $uv\cdot uv = (u\cdot u)(v\cdot v)$ turns out to imply that $(xy)z = -x(yz)$, so the octonions are not associative. However, this is not really the reason that $\mathrm{SO}(7)$ is not the automorphisms of the octonions.)<|endoftext|> TITLE: Is there a Grothendieck correspondence for sheaves/stacks? QUESTION [6 upvotes]: Given a category $\mathcal{C}$, the category of elements functor sets up an equivalence of categories $$ \mathsf{DFib}(\mathcal{C}) \cong \mathsf{PSh}(\mathcal{C}), $$ whereas the Grothendieck construction sets up a $2$-equivalence $$ \mathsf{CartFib}(\mathcal{C}) \cong \mathsf{PseudoPSh}(\mathcal{C}). $$ Question: If one puts a Grothendieck topology $\mathcal{T}$ on $\mathcal{C}$ and replaces the right sides of the above ($2$-)equivalences with $\mathsf{Shv}_{\mathcal{T}}(\mathcal{C})$ and $\mathsf{Stacks}_{\mathcal{T}}(\mathcal{C})$, what should the corresponding categories on the left be? REPLY [3 votes]: The essential image of the Grothendieck construction from (weak) functors $C\to\operatorname{Cat}$ which are sheaves with respect to a Grothendieck topology $\mathcal T$ is described in Section 8.4 of the Stacks project, with the full subcategories of sheaves of groupoids and sets (more precisely, "setoids", i.e. groupoids such that all morphism sets have at most one element) described in the following sections Section 8.5 and Section 8.6, respectively. To be precise, these sections define stacks, but you can probably compare this definition with your favourite one using the description of the Grothendieck construction in Section 4.36.<|endoftext|> TITLE: Independence of $\Pi^1_1$-induction from ATR$_0$ QUESTION [7 upvotes]: Is it known that $\Pi^1_1$-induction is independent of ATR$_0$? Simpson's book shows this for $\Pi^1_1$ transfinite induction ($\Pi^1_1$-TI), but I'm only interested in inducting on $\omega$. I can show that ATR$_0$ + $\Pi^1_1$-induction implies $\Sigma^1_1$-TI, but unlike simpler inductions, it's not clear that the $\Pi$ and $\Sigma$ forms are equivalent here. REPLY [3 votes]: Let me sketch the proof that $\mathsf{ATR}_0+\Pi^1_1\textsf{-Ind}\vdash\mathsf{Con}(\mathsf{ATR}_0)$ (by Gödel's 2-nd incompleteness this implies that $\mathsf{ATR}_0$ doesn't prove $\Pi^1_1\textsf{-Ind}$). We reason in $\mathsf{ATR}_0+\Pi^1_1\textsf{-Ind}$. Assume for a contradiction that $\mathsf{ATR}_0$ isn't consistent. Recall that $\mathsf{ATR}_0$ could be axiomatized over first-order logic by a single $\Pi^1_2$ sentence (i.e. sentence of the form $\forall \vec{X}\exists \vec{Y}\;\varphi$, where $\varphi$ doesn't contain second-order quantifiers). By our assumption there is a FOL proof of $\lnot \forall \vec{X}\exists \vec{Y} \;\varphi$. So we have a cut-free Tait calculus proof $P$ of the sequent $\exists \vec{X}\forall \vec{Y} \;\lnot \varphi$. We prove by $\Pi^1_1$-induction on the $P$-subproofs $$\displaystyle\frac{\ldots}{\Gamma}$$ that the universal closure of $\bigvee (\Gamma\cap \Pi^1_1)$ is true. The main point in this proof by induction is that when some new non-$\Pi^1_1$ arise the last rule should have been of the form $$\displaystyle \frac{\Delta,\forall \vec{Y}\;\lnot \varphi}{\Delta,\exists X_n\forall \vec{Y}\;\lnot \varphi}$$ and since we know that actually $\forall \vec{X}\exists \vec{Y} \varphi$, in fact $\forall \vec{Y}\;\lnot \varphi$ was false on any value of parameters $\vec{X}$ and hence the universal closure of $\bigvee (\Delta \cap \Pi^1_1)$ was true. For the case of the whole derivation $P$ we see that the universal closure of the empty disjunction is true, contradiction. Note that the same argument shows that over $\mathsf{RCA}_0$ the principle $\Pi^1_1\textsf{-Ind}$ is equivalent to uniform $\Pi^1_3$-reflection for $\mathsf{RCA}_0$.<|endoftext|> TITLE: Largest set of $k$-wise linearly independent vectors in $\mathbb F_q^n$? QUESTION [8 upvotes]: What is known about the largest set of $k$-wise linearly independent vectors in $\mathbb F_q^n$? I am especially interested when $q=2$, and in the regime where $k$ is fixed an $n\to\infty$. Here are some partial results: The largest pairwise independent set has $\frac{q^n-1}{q-1}$ vectors. The largest size of a $3$-wise independent set is at most $$\frac{q^{n-1}-1}{q-1}+1.$$ To see why, if $S$ is $3$-wise independent, and ${\bf v}_0\in S$, then the set of vectors of the form $a{\bf v}_0+b{\bf v}$ with $a\in \mathbb F_q, b\in \mathbb F_q^{\times}, {\bf v}\in S\setminus\{v_0\}$ are pairwise distinct. This is achieved when $q=2$ by the set of vectors with an odd number of ones. The next simplest case is $q=3$, where I am stuck. It seems to be that finding a large independent set is related to finding a large cap-set in $\mathbb F_3^{n-1}$, which is a hard open problem. Specifically, by multiplying all the vectors in $S$ with an appropriate scalar, you can assume their leftmost nonzero coordinate is $1$, and then if $u,v,w$ all have their first coordinate equaling $1$, then $\{u,v,w\}$ being linearly independent is equivalent $\{u',v',w'\}$ being non-colinear, where $u'$ is obtained by deleting the initial $1$ from $u$, and similarly for $v',w'$. Any $4$-wise independent set $S$ must satisfy $$ (q-1)^2\binom{|S|}2\le q^n, $$ which follows by realizing that the vectors of the form $a{\bf v}_1+b{\bf v}_2$, with $a,b\in \mathbb F_q^\times$ and ${\bf v}_1,{\bf v}_2\in S$, are pairwise distinct (for this to work, each pair $\{{\bf v}_1,{\bf v}_2\}$ must appear at most once). This gives $|S|\lesssim \sqrt{2}q^{n/2}/(q-1)$. I am not sure if this the true growth rate, even for $q=2$. It seems straightforward to generalize these upper bound arguments to show that the size of a $k$-wise independent set is $\mathcal O(q^{n/\lfloor k/2\rfloor})$. To summarize, I am wondering if the upper bounds I gave are tight, and if any other general constructions are known. (The reason for the pr.probability is that $k$-wise independent vectors give rise to $k$-wise independent sets of random variables, which is how I came across this problem). REPLY [9 votes]: $\newcommand{\F}{{\mathbb F}}$ Let's focus on the case $q=2$ which you are primarily interested in. Suppose that $v_1,\dotsc,v_m\in\F_2^n$, and let $V$ be the matrix over $\F_2$ of size $n\times m$ with $v_1,\dotsc,v_m$ (written in the standard basis) as its columns. It is readily seen that for $v_1,\dotsc,v_m$ to be $k$-wise linearly independent, it is necessary and sufficient that the kernel of $V$ did not contain any vector of weight $k$ or less. You are thus asking how large can the length $m$ of a linear code be given that the minimum distance of the code is at least $k+1$, and that the code has co-dimension $n$. This is essentially equivalent to the central problem of coding theory: what is the largest possible dimension of a linear code of given length and minimum distance ($m$ and $k+1$, respectively, in our case)?<|endoftext|> TITLE: Distance of low-rank matrices to the identity for the $\infty$-norm QUESTION [5 upvotes]: I am trying to get a lower bound (or even the exact value) of $$ \min_{X \in \mathbb{R}^{n\times n}} \|X - I_n\|_{\infty} \enspace \text{s.t.} \enspace \mbox{Rank}(X) = m $$ where $m \leq n$, and the infinity norm is $$ \| X \|_{\infty} := \max_{ij}|X_{ij}| $$ I have a very simple lower bound, obtained with norm equivalence: $$\|X - I_n\|_{\infty} \geq \frac{1}{n} \|X - I_n\|_F\geq \frac{\sqrt{n - m}}{n}$$ but this is obviously not tight, and experiments suggest that the scaling is wrong. Thanks a lot ! :) REPLY [4 votes]: Theorem 1.1 from Alon's paper "Perturbed identity matrices have high rank: proof and applications" says that if $\|X-I_n\|_\infty TITLE: Largest Hopfian quotient QUESTION [9 upvotes]: Let $\Gamma$ be a group, say finitely generated if it helps. Does $\Gamma$ admit a largest Hopfian quotient? That is, does there exist a Hopfian quotient $H$ of $\Gamma$, such that every surjective homomorphism from $\Gamma$ onto a Hopfian group factors through $H$? I first thought to define $H = \Gamma / K$, where $K$ is the intersection of all kernels of surjective endomorphisms of $\Gamma$, but I cannot see why this should be Hopfian, or why it should have the factorization property. Then I thought to define $H = \Gamma/K$, where $K$ is the intersection of all kernels of surjective homomorphism from $\Gamma$ onto a Hopfian group, and this clearly has the factorization property but I don't see why it should be Hopfian. Does such a quotient always exist? Does it exist under some additional conditions? Do you have some explicit example where it exists, and it is not the same as the largest residually finite quotient? In particular I would like to know this for the Baumslag-Solitar group $BS(2, 3)$. REPLY [9 votes]: The answer is no even for finitely generated groups. Here's a construction of a finitely generated residually Hopfian, non-Hopfian group. It is even solvable (actually center-by-metabelian). Denote by $M(u,v,x,y,z)$ the matrix $$\begin{pmatrix}u & x & z\\ 0 & v & y \\ 0 & 0 & 1\end{pmatrix}.$$ Fix a prime $p$; let $G$ be the group of matrices $M(0,t^n,x,y,z)$ for $(n,x,y,z)\in\mathbf{Z}\times\mathbf{F}_p[t,t^{-1}]^3$. Note that $G$ is finitely generated (by $\{M(0,1,0,0,0),M(0,0,1,0,0),M(0,0,0,1,0)\}$). Let $Z$ be the central subgroup of $G$ of those matrices of the form $M(0,0,0,0,z)$ for $z\in\mathbf{F}_p[t]$. Let $\alpha$ be the matrix $M(t,0,0,0,0)$ (i.e., diagonal $(t,1,1)$). Then $\alpha Z \alpha^{-1}$ is contained in $Z$ with index $p$. In particular, $\alpha$ induces a surjective, non-injective endomorphism of $G/Z$, so $G/Z$ is non-Hopfian. Proposition. $G/Z$ is residually Hopfian. I will indeed prove that it has two normal subgroup with trivial intersection, each quotient being Hopfian. Let us "classify" quotients of $G$. Let $Z'$ be its center (the subgroup of those $M(0,0,0,0,z)$ for $z\in\mathbf{F}_p[t,t^{-1}]$. Claim 1: every normal subgroup $N$ of $G$ is either contained in the center $Z'$, or contains some finite index subgroup of $Z'$. Proof: suppose that $N$ contains an element not in $Z'$. Write $N'$ for the set of elements in $N$ of the form $M(0,0,x,y,z)$; then $N'$ is also a normal subgroup of $G$. Performing a commutator if necessary, we see that $N'$ is not contained in $G$: it contains an element of the form $M(0,0,x,y;z)$ with $(x,y)\neq (0,0)$; by symmetry we can suppose $x\neq 0$. Write $p_1,p_2,p_3$ for the projections $M(0,0,x,y;z)\mapsto x,\mapsto y,\mapsto z$. Since $N'$ is a normal subgroup, $p_1(N')$ is a nonzero $\mathbf{F}_p[t,t^{-1}]$ submodule of $\mathbf{F}_p[t,t^{-1}]$, that is, an ideal. Taking commutators with $M(0,0,0,1,0)$, we deduce that $p_3(N')$ also contains this finite index ideal. So $N$ contains some finite index subgroup of $Z'$.$\Box$ Claim 2: every finite-by-metabelian f.g. group $G$ is Hopfian. Proof: every such group satisfies the max-n property (max condition on chains of normal subgroups), an old result of Ph. Hall, and max-n immediately implies Hopfian (for a surjective non-injective endomorphism, the sequence of kernels of iterates is strictly increasing). $\Box$ If $N$ is a normal subgroup of $G$ above, if $N$ contains a finite index subgroup of $Z'$, then $G/N$ is finite-by-metabelian, hence Hopfian by Claim 2. Otherwise, $N$ is contained in $Z'$, i.e., is central. We now assume so. Since the metabelian group $G/Z'$ has trivial center [this has index 2 in a lamplighter group $C_p\wr\mathbf{Z}$, and is a lamplighter $C_p^2\wr\mathbf{Z}$], the center of $G/N$ is reduced to $Z'/N$. Let $f$ be a surjective endomorphism of $G/N$. By surjectivity, $f$ maps center into the center, and hence induces an endomorphism of the metabelian quotient $G/Z'$, which is Hopfian. Hence the inverse image of the center equals the center, and $f$ has kernel contained in the center $Z'/N$. Up to replace $f$ with its square, we can suppose that the action of $f$ on $\mathbf{Z}$ (which appears modding out by the set of torsion elements, which is a subgroup) is trivial. Hence the action of $f$ on "$(x,y)$" (formally speaking, its action on $\mathbf{F}_p[t,t^{-1}]^2$ obtained modding out the center and restricting to the torsion subgroup) is by $\mathbf{F}_p[t,t^{-1}]$-module automorphism: it is given by some matrix $q\in\mathrm{GL}_2(\mathbf{F}_p[t,t^{-1}])$. Passing to some proper power if necessary, we can suppose that the determinant $\det(q)$ equals $t^m$ for some $m\in\mathbf{Z}$. Shaking hands, the action of $q$ on the center $Z'$ (this doesn't make sense!) should be given by multiplication the determinant $t^m=\det(q)\in\mathbf{F}_p[t,t^{-1}]^\times$ and this should restrict the possibilities for $N$. More precisely, for $s\in\mathbf{F}_p[t,t^{-1}]$, write $X_s=M(0,0,s,0,0)$, $Y=M(0,0,0,1,0)$, so $X_sYX_s^{-1}Y^{-1}=M(0,0,0,0,s)=Z_s$. Let $X'_s$, $Y'$ and $Z'_s$ be their images in $G/N$. So $f(Z'_s)=Z'_{F(s)}$ for some $F(s)$ which is well-defined up to addition by an element of $I=p_3(N)$. Then computing $f$ on this commutator shows that $\det(q)s-F(s)\in I$ for every $s$. Prescribing $F$ on the canonical basis, we can suppose that $F$ is an additive homomorphism. Let $K$ be the kernel of $f$ and $J=p_3(K)$. Then $I$ is strictly contained in $J$, and $F^{-1}(I)=J$ (essentially by definition). If $m=0$ by contradiction, we get that $F$ is identity modulo $I$, which contradicts $f(J)\subset I$. Now suppose that $I$ is has basis $(t^n)_{n\in A}$ for some negatively unbounded subset $A\subset\mathbf{Z}$ containing $\mathbf{N}$ which in the negative is "more and more sparse" (in the sense that $A=\mathbf{N}\cup\{a_k:k\in\mathbf{N}\}$ with $a_k-a_{k-1}\to - \infty$). We combine the conditions $F(I)\subset I$ and $t^ms-F(s)\in I$, to get $t^{m+n}\in I$ for all $n\in A$, which contradicts the sparse condition. Hence if $N=N_I$ is the group just chosen, then $G/N_I$ is Hopfian. Taking $I,I'$ sparse and with intersection reduced to $\mathbf{N}$, we get $N_I\cap N_{I'}=Z$. So $G/Z$ is residually Hopfian.<|endoftext|> TITLE: How to structure a proof by induction in a maths research paper? QUESTION [13 upvotes]: I am 16 years old at the time of writing (so I have no supervisors to seek advice from) and I have written a mathematics research paper, which I plan on submitting to a journal for publication. I asked an identical question on Academia.SE and I was advised to ask the question here. For a couple of the assertions that I make, I use proofs by induction. Now, in school we're encouraged to write proofs by induction in the following (rigid) format: Base case:... Assumption(s): …. Inductive step: …. Conclusion: …. I have noticed that no research articles that I have seen have written proofs by induction using this sort of format. The authors usually make it flow much more smoothly, eg 'For the base case, the result is trivial. Now assume the result holds for some $n=k$, so that …. Now consider the expression for $n=k+1$ … and by the inductive hypothesis this equals … hence the result is true by mathematical induction.' So, is it good practice to write proofs by induction in the pretty rigid structure I first outlined or is it ok/better to write the proofs more naturally so that it flows better? REPLY [5 votes]: Welcome to math overflow! There is no need to list up the names of the steps of the induction (Base case, Inductive step etc.) if it clear from context and/or obvious what you're doing. This will likely only make the argument take up more physical space on the page than strictly necessary. It should however be clear that you are doing an argument by induction, what the base case is, and how the induction step is performed. If a step of the argument is a little bit difficult to comprehend mentally, then I would say it's no loss in making it explicitly clear why the argument holds, without drowning in details, of course. I had a teacher once that used the term "Weierstrass-rigorous" for certain arguments / persons. This refers to the ideal standard that your proofs should be logically consistent (not contradict them selves) and "water-proof" , in other words correct, as others have stated. When attempting to confine to such a standard, it often happens that there are steps used in a proof that actually assume more than they seem. For example, something as innocent as $$"\text{Let} \hspace{2mm}(x_n)_{n=1}^{\infty}\hspace{2mm} \text{be a sequence of elements in} \hspace{2mm} X", $$ may actually require the assumption of the Axiom Of Choice to obtain the sequence. The proof should also be readable. This means, among other things, that all symbols, functions, etc. that you utilize throughout the proof are defined ether a priori, or on the fly. It is also a bonus if the proof follows a natural progression, if this is possible. In summary: Too rigorous is better than to little rigorous, and write it in away that makes it readable (or perhaps also enjoyable). And don’t forget to "get to the point" in the proof. [Ps. Writing style is very much a personal preference, as long as it is rigorous.]<|endoftext|> TITLE: Fair cutting of the plane with lines QUESTION [29 upvotes]: An infinite countable family $\cal{L}$ of lines in the plane $\mathbb{R}^2$ forms a fair cutting of the plane if the following conditions are satisfied: $\bullet$ No circle intersects infinitely many lines from the family $\cal{L}$; $\bullet$ Let $L$ denote the union of the lines in $\cal{L}$. Each connected component of the complement of $L$ is a bounded set; the closure of each of these sets will be called a cell (observe that each cell is a convex polygon). This condition requires that the diameters of the cells formed by the cutting have a common upper bound; Finally, $\bullet$ All cells are of the same area. We say that two fair cuttings are affinely equivalent if there exists an affine transformation of $\mathbb{R}^2$ sending the lines of one family onto the lines of the other. Also, we say that a fair cutting $\cal{L}$ is extra-fair, if, in at least one of its affine equivalents, all cells are congruent. Four examples of mutually non-equivalent extra-fair cuttings are shown below. Question 1. Is there a fair cutting whose at least one cell has more than four sides? Question 2. Must all cells of a fair cutting have the same number of sides? Question 3. Is every fair cutting extra-fair? Question 4. Does there exist another example of a fair cuting, non-equivalent to any of the four pictured here? REPLY [7 votes]: The answer to question 4 is no - every fair cutting is affinely equivalent to one of your four. This implies the answers to questions 1, 2, and 3 are no, yes, and yes, respectively. (Question 1 was already answered by Yaakov Baruch.) My argument will use Yaakov Baruch's partial answer. Beyond what he does, there is one tricky algebraic calculation, and the rest is reasonably nice geometric arguments. The general principle is that when one has inside a given fair cutting a large enough piece of one of the four cuttings on the list in the original post (i.e. one knows that certain cells from the cutting on the list are cells of the cutting, thus the lines through their edges are lines of the cutting and intersections of these lines are vertices of the cutting), one can extend this to get more of the cutting on the list. This gets easier the more one has, so the greatest difficulty is in starting out. However, finishing the argument requires getting to the point where the cutting repeats, allowing an inductive argument to fill the whole space. This is proportionally more difficult as the cutting gets less periodic (say, as the fundamental domain for the translation symmetries contains more and more cells), which is in increasing order from (2) to (3) to (4) to (6). As a balancing force, as we get further in the argument we know certain configurations expand to a full cutting of the plane and therefore we can assume that those configurations don't appear, making the argument easier. Thus we want to go in increasing order - we proceed from proving that certain configurations produce (2), to proving that some other configurations produce (3), to proving that still other configurations produce (4), and finally prove that all other configurations produce (6). I apologize for not including diagrams, which I am bad at drawing, in my answer, and trying to argue verbally / in coordinates as clearly as possible. Step 1: Every fair tiling which contains a cell with four or more sides is equivalent to the square grid (2) This was proven by Yaakov Baruch in his answer. It reduces us to considering tilings where every cell is a triangle. Step 2: If two triangles in a fair tiling share an edge, their vertices are equidistant from that edge. This follows from the area formula for triangles as one half base times height, i.e. one half the length of the edge times the distance of the vertex to the edge. Step 3: A triangle in a fair tiling where every cell is a triangle such that every vertex is the intersection of three or more lines in the tiling, together with its three adjacent triangles, are affinely equivalent to four triangles from the equilateral triangular tiling (3). This is a mouthful, and is the hardest step. Let $P,Q,R$ be the vertices of the triangle in clockwise order. Because every cell is a triangle, each edge is shared with one other triangle. The triangle that shares the edge $PQ$, say, has a vertex $A_R$ whose distance from the line $PQ$ is equal to the distance of $R$ from $PQ$. This vertex $A_R$ therefore lies on the line $L_R$ parallel to $PQ$ and of that distance from $PQ$. The vertex $A_R$ must lie between the intersection of $PR$ and $L_R$ and the intersection of $QR$ and $L_R$, since otherwise those lines would intersect the triangle. Let $a_R$ be the distance from $A_R$ to to $PR \cap L_R$ divided by the distance from $PR \cap L_R$ to $QR \cap L_R$, so that $0 < a_R <1$. (If $a_R$ were exactly $0$, then $PR$ and $PQ$ would be the only lines through $P$.) Define $a_P$ and $a_Q$ the same way, but rotated. Let us now calculate in coordinates. After an affine transformation, we can take $P = (0,-1)$, $Q = (-1,0)$, $R = (0,0)$. Thus $L_P = \{x,y \mid y=1 \}$ and $L_Q = \{ x,y \mid x=1\} $. We have $A_P = ( 2 a_R - 2, 1)$ and $A_Q = (1, - 2 a_Q)$. The triangle $Q R A_P$ contains the line $R A_P$ between $ ( 2 a_R - 2, 1)$ and $(0,0)$ given by the equation $x = (2a_R -2) y$. If $A_Q$ lies to the right of this line then this line intersects the triangle $PR A_Q$, which is impossible, so we must have $$1 \leq (2 a_R -2) ( -2 a_Q)$$ or $$ a_Q \geq \frac{1}{ 4 -4 a_R} = a_R + \frac{ (2 a_R -1)^2}{4 -4 a_R} \geq a_R $$ with equality only if $a_R =\frac{1}{2}$. Symmetrically, we have $$ a_P \geq a_Q \geq a_R \geq a_P$$ so $$a_P = a_Q = a_R =\frac{1}{2}$$ so $A_P = (-1,1)$, $A_Q= (1,-1)$, $A_R = (-1,-1)$, which together with $P,Q,R$ are indeed affine equivalent to four triangles from the equilateral triangular tiling (3). Step 4: A fair tiling where every cell is a triangle, containing one triangle such that every vertex is the intersection of at least three lines, is affinely equivalent to (3). From Step 3, we may assume, up to affine transformation, that the tiling contains four triangles of the triangular tiling (3). For example, we can say it contains the vertices $(-1,1)$, $(0,0)$, $(1,-1)$, $(-1,0)$, $(0,-1)$, $(-1,-1)$, the lines $x+y=0$, $x+y=-1$, $x=0$, $x=-1$, $y=0$, $y=-1$, and the four triangles bounded by those lines. We will show any fair tiling, where every cell is a triangle, containing those, contains also the reflection of those four triangles around the line $x=y=0$. Iterating this, reflecting in any desired direction, we get all the triangles of the triangular tiling. To show this, consider the triangle which shares the edge between $(1,-1)$ and $(0,0)$ with the triangle $(1,-1),(0,0), (0,-1)$. By step 2, its third vertex $V$ must satisfy $x+y=1$. If the edge connecting $V$ to $(0,0)$ is not one of our existing lines, then there is another line through $(0,0)$, which intersects one of our existing triangles, contradicting the assumption that these are cells of the tiling. Therefore, $V$ must lie on one of the lines through $(0,0)$ - either $x=0$ or $y=0$. If $V$ lies on $x=0$ then this triangle is intersected by the line $y=0$ so $V$ lies on $x=0$, that is $V = (0,1)$. Thus our tiling contains the edge $x=1$ between $(1,-1)$ and $(0,0)$. Symmetrically, our tiling contains the vertex $(1,0)$ and the edge $y=1$. The edges $x=0$, $x=1$, $y=0$, $y=1$ bound a square. Since our cells have area $1/2$, we must have another edge bisecting the square into two triangles. This can be either $x=y$ or $x+y=1$, but $x=y$ would bisect the triangle with vertices $(0,0)$, $(-1,0)$, $(0,-1)$ already assumed, so it must contain $(x+y=1)$. It follows that the tiling contains the vertices $(-1,1), (0,0), (1,-1) , (0,1), (1,0), (1,1)$, the edges $x=0$, $x=1$, $y=0$, $y=1$, $x+y=0$, $x+y=1$, and the four triangles bounded by these edges. This is affine-equivalent to our starting configuration, so we may iterate, covering the plane. Thus, we may assume every triangle has a vertex which is the intersection of only two lines. Step 5: In a fair tiling where every cell is a triangle, a vertex which is the intersection of only two lines, together with the four triangles adjacent to it, form a parallelogram. Let $L_1$ and $L_2$ be the two lines intersecting at the vertex $V$. Let $A$ and $C$ be the two vertices on the line $L_1$ closest to $V$, and let $B$ and $D$ be the two vertices on $L_2$ closes to $V$. Then $AVB, BVC, CVD, DVA$ must be the four triangles adjacent to $V$. By step 2, $A$ and $C$ are equidistant from $L_2$. Because they each lie on $L_1$, they are equidistant from $V$ - $V$ is the midpoint of $AC$. Similarly $V$ is the midpoint of $BD$. Thus ABCD is a parallelogram. Step 6: The parallelograms constructed in Step 5 can't overlap each other. In a fair tiling where every cell is a triangle and every triangle has one of its four vertices the intersection of only two lines, the plane is tiled by such parallelograms. If two parallelograms of this form overlapped, an internal edge of one would be an external edge of the other. The internal edges all touch the center vertex, which is the intersection of two lines, but the external edges only touch the external vertices, which are each the intersection of at least three lines - the two sides of the parallelogram plus the line through the center. So this is impossible. By step 5, every triangle lies in a parallelogram, and by step 6, the parallelograms overlap, so indeed they cannot tile the plane. Step 7: A fair tiling where every cell is a triangle and every triangle has a vertex which is the intersection of only two lines, which contains two parallelograms of step 5 which are equivalent under translation and share an edge, must be affine-equivalent to the quadrisected square tiling (4). Every parallelogram is affine-equivalent to any given square, say the square with vertices $(0,0),(2,0), (0,2), (2,2)$. Without loss of generality, our two adjacent parallelograms are that one and the square with vertices $(2,0), (2,2), (4,0), (4,2)$. Thus our tiling contains the vertices $(0,0), (0,2), (1,1) , (2,0) ,(2,2), (3,1), (4,0), (4,2)$, the lines $x=0, x=2, x=4, y=0, y=2, x+y=2, x+y=4, x=y, x=y+2$ and the eight triangles with vertices among those eight vertices bounded by those nine lines. Using this, we will show that our tiling contains the same configuration with $y$ shifted vertically by $2$. Iterating, and rotating, we get the whole square lattice. To do this, note that the triangles in our tiling have area $1$, and the lines $y=2$, $x=0$, $x+y=4$ of our tiling bound the triangle $(0,2),(0,4), (2,2)$ with area $2$, so this triangle must be the union of two triangular cells of the tiling. One of them must share the edge edge $(2,0),(2,2)$ with the triangle $(2,0),(2,2),(1,1)$. By step 2, its third vertex $V$ must be on the line $y=3$, hence be either $(1,3)$ or $(0,3)$, since when cutting a triangle into two triangles the only new vertices are on the edges of the triangle. If it's $(0,3)$ then the line $y = 3 -x/2$ lies in the tiling and intersects the already-existing triangle $(2,2), (4,2), (2,0)$, which is a contradiction, so the last vertex must be $(1,3)$. Thus the edge $y=x+2$ lies in our tiling. The edges $x=2, x=0, y=2, y=x+2, y+x=2$ define the triangles $\{(0,4), (1,3), (0,2)\}$, $\{(0,2), (1,3), (2,2)\}$, and $\{(1,3), (2,2), (2,4)\}$, which all must be cells of our tiling. Any lines other than these which intersect the triangle $(0,4), (1,3), (2,4)$ would intersect one of those three tiles, contradicting the fact that they are cells of our tiling. Thus the triangle $(0,4), (1,3), (2,4)$ is contained in a cell of the tiling, and since it has area $1$, must be a cell of the tiling, so the edge $y=4$ lies in our tiling. Symmetrically, the vertex $(3,3)$ and the edge $x+y=6$ lies in our tiling. So our tiling contains the edges $y=2, y=4, x=0, x=2,x=4, x+y=4, x+y=6, y=x, y=x+2$, the vertices $(0,2), (0,4), (1,3), (2,2), (2,4), (3,3), (4,2),(4,4)$, and the eight triangles bounded by those lines and with vertices among those vertices, which indeed is the translation of our original configuration. It also contains our original configuration with $x$ and $y$ swapped, allowing us to travel in the $x$ direction. Iterating, we fill out the tiling (4). Step 8: In a fair tiling where every cell is a triangle and every triangle has a vertex which is the intersection of only two lines, if one of the parallelograms constructed in step 5 is $(0,0),(1,0),(0,1),(1,1)$, then the tiling contains the lines $x=n$ and $y=n$ for every integer $n$ By symmetry, it suffices to check this for the lines $y=n$ for positive integers $n$. Because, by step 6, the plane is tiled by parallelograms, the edge $(1,0), (1,1)$ is shared by another parallelogram. Since every parallelogram contains four cells, they all have the same area $1$. The parallelogram must have another edge parallel to this one, and because it has area $1$, that edge must lie on the line $y=2$. That edge is then shared by a third parallelogram, which must have a third parallel edge, which for the same reason must lie on the line $y=3$. Iterating, we have the claim. Step 9: A fair tiling where every cell is a triangle and every triangle has a vertex which is the intersection of only two lines, which does not contain two parallelograms which are translates of each other and share an edge is affine-equivalent to the tiling (6). Without loss of generality, one of the parallelograms is $(0,0),(1,0),(0,1), (1,1)$, so by step 8 the tiling contains the lines $x=n$ and $y=n$ for integer $n$. Now consider the parallelogram adjacent to the edge $(0,1), (1,1)$. Because it also has area $1$, its other vertices have the form $(x,2), (1+x,2)$ for some real $x$. By assumption, $x \neq 0$. By a reflection symmetry in $y$, we may assume $x \neq 0$. The line $x=1$ lies in our tiling and intersects this parallelogram. It therefore must pass through the center of the parallelogram, which implies $x=1$. So our tiling contains the parallelogram $(0,1), (1,1), (1,2), (2,2)$. This one is affine-equivalent to $(0,0),(1,0),(0,1), (1,1)$ under the transformation $(x,y) \mapsto ( x+y, y+1)$ so by an affine-equivalent form of step 8, the tiling contains the edges of the form $x+y=n$ for an integer $n$. The edges $x=n, y=n, x+y=n$ divide the plane into triangles of area $1/2$, each of which contains two cells of the tiling, and therefore each of which is bisected into two triangles by an edge. Thus each vertex of the tiling is either a vertex of this triangular grid or the midpoint of an edge of the triangular grid. the vertices of the triangular grid lie on at least three lines, so only the midpoints of edges may be centers of parallelograms - thus each parallelogram consists of two triangles from this grid. There are three types of parallelograms formed by three triangles of this grid, up to translation. No two congruent ones can be adjacent. For each triangle adjacent to a fixed parallelogram, there are two possible parallelograms it can be contained in, and one is congruent to the original, so it must be the other. Iterating this, our one starting parallelogram forces a tiling of the other grid by parallelograms, which gives a fair tiling of the plane, which indeed is (6).<|endoftext|> TITLE: Comparing divergent and convergent sums QUESTION [7 upvotes]: Let $(x_n)$ be a monotonically decreasing sequence of positive real numbers that is also summable. Let $(y_n)$ be a sequence of positive real numbers such that $\sum_n x_n y_n$ converges. Let $(z_n)$ be a monotonically increasing sequence of positive real numbers such that $\sum_n x_n z_n =\infty.$ Assume that the sequences $y_n$ and $z_n$ are such that $2^{-\varepsilon y_n}$ and $2^{-\varepsilon z_n}$ are summable for every $\varepsilon>0.$ Does it follow that there is some $\delta>0$ such that $$ \sum_n \Big(2^{-\varepsilon y_n}-2^{-\varepsilon z_n}\Big) \ge 0 \text{ for all } \varepsilon \in (0,\delta)?$$ The motivation for this statement to be true is that $z_n$ should be larger most of the time than $y_n$ and we capture this most of the time by taking $\varepsilon$ small. Please let me know if you have any comments, questions or remarks. REPLY [4 votes]: No. For example $x_n=2^{-n}$, $y_n=n$, and we now keep $z_n$ constant on long intervals. More precisely, consider first $\epsilon=1$, and set $z_1= \ldots = z_{N_1}=c_1$, with $N_1$ taken so large that $N_1 2^{-1\cdot 1}>\sum 2^{-1\cdot n}$. We then continue in the same way: let $z_{N_1+1}=\ldots = z_{N_2}= c_2$, and again take $N_2$ so large that the inequality you want fails for $\epsilon=1/2$ etc. Here we must choose the $c_n$ such that $\sum 2^{-n}z_n=\infty$ and $\sum 2^{-\epsilon z_n}<\infty$. I'll take $c_n=2^{N_{n-1}}$ to guarantee the first property. If the $n$th step deals with $\epsilon=1/n$, then, since $\sum n2^{-\epsilon n} \simeq 1/\epsilon^2$, I can take $N_n=n^3 2^{N_{n-1}/n}$ to make sure your inequality fails. With these choices in place, the final condition becomes $$ \sum N_n 2^{-\epsilon 2^{N_{n-1}}} =\sum n^3 2^{(-\epsilon+1/n)2^{N_{n-1}}}< \infty $$ for all $\epsilon >0$, and this holds because the $N_n$ increase rapidly.<|endoftext|> TITLE: Finite $\Gamma$-modules with trivial $H^2$, where $\Gamma$ is a group of order 2 QUESTION [6 upvotes]: Let $\Gamma=\{1,\gamma\}$ be a group of order 2. Let $A$ be a finite $\Gamma$-module, that is, a finite abelian group on which $\Gamma$ acts. It is a hopeless problem to classify finite $\Gamma$-modules. We consider $A^\Gamma=\{a\in A\mid{}^\gamma a=a\}$ and $$H^2(\Gamma,A)=A^\Gamma/\{a'+{}^\gamma a'\mid a'\in A\}.$$ Question 1. Is it possible to classify finite $\Gamma$-modules $A$ with $H^2(\Gamma,A)=0$ ? Remark. If $A$ is of odd order, then $H^n(\Gamma,A)=0$ for $n>0$. Reduction 1. Write $A=A_{\rm odd}\oplus A_2$, where $|A_{\rm odd}|$ is odd and $|A_2|=2^m$. Then $H^2(\Gamma,A)=H^2(\Gamma,A_2)$. Therefore, from now on we assume that $A$ is a 2-group. Remark. Let $B$ be an induced $\Gamma$-module, that is, $B=C\oplus C$, where $C$ is an abelian group and $\Gamma$ acts on $B$ by $$^\gamma(c,c')=(c',c)\quad\text{for $(c,c')\in C\oplus C=B$.} $$ Then $H^n(\Gamma,B)=0$ for $n>0$. Reduction 2. Let $B\subset A$ be a $\Gamma$-submodule that is induced. Then the canonical epimorphism $A\to A/B$ induces an isomorphism $H^n(\Gamma,A)\cong H^n(\Gamma,A/B)$ for $n>0$. Therefore, from now on we assume that $A$ has no induced $\Gamma$-submodules. Question 2. Does there exist a finite $\Gamma$-module $A$ of order $2^m$ for some $m>1$ without induced $\Gamma$-submodules and such that $H^2(\Gamma,A)=0$ ? REPLY [6 votes]: The answer to question 2 is yes. Take $A=\mathbb Z/8\mathbb Z$ where $\gamma$ acts by multiplication by $5$. Then $A^\Gamma= \ker( \cdot 4)= 2\mathbb Z/8\mathbb Z$ but $(\mathrm{id} +\gamma)(3)=6\cdot 3=2$, thus $H^2(\Gamma,A)$=0 (and obviously $A$ doesn't contain any induced module). For the question 1 I do not really know what to say except that if $A$ is killed by 2, then $H^2(\Gamma,A)=0$ forces $A$ to be induced. However as you see in the example above, the "$H^2(\Gamma,A)=0$"-property does not pass to the 2-torsion subgroup $A[2]$ or the subquotients $A[2^i]/A[2^{i+1}]$, so I do not think this gives you anything when considering a general $A$. I also think that asking $H^2(\Gamma,A)=0$ doesn't really simplify the classification (so if we restrict to $2$-groups we just get finite $\mathbb Z_2[x]/(x^2-1)$-modules for which $\ker(x-1)=\mathrm{im}(1+x)$).<|endoftext|> TITLE: What is the Möbius function of substrings? QUESTION [6 upvotes]: Define a poset on the set of all finite binary strings, defined by $a \le b$ whenever $b = uav$ for (possibly empty) binary strings $u, v$. What is the Möbius function of this poset? REPLY [9 votes]: This problem was solved by Anders Björner, The Möbius function of factor order, Theoretical Computer Science 117 (1993), 91-98. In particular, the Möbius function assumes only the values $-1,0,1$. It is also Exercise 3.134(b) in Enumerative Combinatorics, vol. 1, second edition.<|endoftext|> TITLE: Why do Chern classes and Stiefel-Whitney classes satisfy the "same" Whitney sum formula? QUESTION [12 upvotes]: The Whitney sum formula for Stiefel-Whitney classes, $w_n(V \oplus W) = \sum w_i(V) w_{n-i}(W)$, looks a lot like the one for Chern classes $c_n(V \oplus W) = \sum c_i(V)c_{n-i}(W)$. But I don't know a way to prove both formulae "at once". Question: Is there an abstract computation from which both Whitney sum formulae follow? For instance, constructing cell structures on $BO$ and $BU$ and then just reading off what the relevant maps do on cohomology doesn't "explain" why the formulas "look the same" in both cases, so doesn't fit the bill for me. Notes: The Whitney sum formula for Stiefel-Whitney classes is also related to the Cartan formula for Steenrod squares, $Sq^n(xy) = \sum Sq^i(x)Sq^{n-i}(y)$, since Stiefel-Whitney classes can be defined in terms of Steenrod squares. I don't know a corresponding formula in integral cohomology operations related to Chern classes. An answer which sheds light on this connection would be most welcome. These formulae are equivalent to working out the comultiplication on $H^\ast(BO;\mathbb F_2)$ and $H^\ast(BU;\mathbb Z)$ respectively. It appears to me to be forced algebraically that e.g. the pullback of $c_n \in H^\ast(BU(n);\mathbb Z)$ to $H^\ast(BU(1)^n;\mathbb Z)$ is some scalar multiple of $c_1^{\otimes n}$, but I can't seem to rule out that the scalar multiple is zero except by invoking the splitting principle (which I'd ideally like to avoid -- I'd really like a computation of this comultiplication which implies the splitting principle!), and I'm also not sure I can show that if nonzero, the scalar is a unit. EDIT: I did eventually arrive at a way to compute this coproduct without constructing complete cell structures: In the fibration $\mathbb{CP}^{n-1} \to BU(n-1) \times BU(1) \to BU(n)$, observe that the Serre spectral sequence for integral cohomology must collapse because we know the rank of all the groups involved, which are free. It follows that $H^\ast(BU(n)) \to H^\ast(BU(n-1) \times BU(1))$ is injective; iterating gives the splitting principle and the coproduct formula. The fibration $\mathbb{RP}^{n-1} \to BO(n-1)\times BO(1) \to BO(n)$ works similarly in the real case. I still find this argument unsatisfactory because one must still treat the two cases "in parallel" rather than proving one theorem and deducing both results from it. There might be some sort of argument which deduces the comultiplication on $H^\ast(BO)$ from the one on $H^\ast(BU)$ or vice versa -- this isn't precisely what I'm looking for, but I'd be interested to see this worked out. It would be very nice if there were an argument which were to abstractly construct an $E_\infty$ map $MU \to H\mathbb Z[t^\pm]$ (where $|t| = 2$) corresponding to the total Chern class $c = \sum_n c_n t^{-n}$, and deduce the comultiplication formula by computing that the map does this on the homology of $\Sigma^{\infty-1} BU \subset MU$ and using that it is multiplicative. I haven't quite been able to see through such an argument though. (As pointed out by Oscar Randal-Williams below, Totaro has shown that this does not work!) REPLY [13 votes]: I prefer this approach which I believe is due to Grothendieck. (I haven't checked how this compares with the sources cited by Nick Kuhn.) Let $(\mathbb{K},R,d)$ be $(\mathbb{R},\mathbb{Z}/2,1)$ or $(\mathbb{C},\mathbb{Z},2)$. Let $V$ be a $\mathbb{K}$-linear vector bundle over $X$. Over the associated projective bundle $PV$ we have a $\mathbb{K}$-linear tautological line bundle $T$ classified by a map $PV\to \mathbb{K}P^\infty$. I'll assume that we know that $H^*(\mathbb{K}P^\infty;R)=R[x]$ with $|x|=d$. Pulling back $x$ gives a class $x\in H^d(PV;R)$. Induction over the cells of $X$ shows that $H^*(PV;R)$ is a free module over $H^*(X;R)$ with basis $\{x^i\mid 0\leq i<\dim(V)\}$. Thus, there is a unique monic polynomial $f_V(t)\in H^*(X;R)[t]$ of degree $\dim(V)$ such that $H^*(PV;R)=H^*(X;R)[x]/f_V(x)$. The characteristic classes of $V$ are just the coefficients of $f_V(t)$ (possibly with an extra $\pm$-sign, according to conventions). The cofibre of the inclusion $PV\to P(V\oplus W)$ is the Thom space of a $\mathbb{K}$-linear vector bundle over $PW$, and it follows that $f_V(x)f_W(x)$ annihilates $H^*(P(V\oplus W);R)$, and thus that $f_{V\oplus W}(t)$ must be equal to $f_V(t)f_W(t)$. By comparing coefficients we get the standard formula for characteristic classes of $V\oplus W$.<|endoftext|> TITLE: Check if a polygon has an axis of symmetry in $O(n)$ time QUESTION [5 upvotes]: Question: Is it possible to check if an $n$-gon has an axis of symmetry in $O(n)$ time? Note: An $O(n^2)$ algorithm is easy to see: it is easy to check if any given line is an axis of symmetry of the $n$-gon in $O(n)$ time and for any $n$-gon, there are only $2n$ lines that are candidates for being an axis of symmetry — the angular bisectors of the n angles and the perpendicular bisectors of the $n$ sides. Further question: If an $n$-gon has more than one axis of symmetry, is there an $O(n)$ algorithm to find them all? REPLY [9 votes]: Indeed that paper I cited in the comments describes how to determine all symmetries of a polygon $P$ in $O(n)$ time. The polygon is first translated so that its centroid is at the origin. Then a "representation" $L(P)$ of $P$ is constructed. $L$ is an $n$-tuple of pairs $(d_i,\alpha_i)$, where $d_i$ is the length of polygon edge $i$, and $\alpha_i = \theta_{i+1} - \theta_i$, where $\theta_i$ is the angle the edge makes with the positive $x$-axis. So $\alpha_i$ is essentially the angle between adjacent edges. A rotation by $2 \pi/k$ about the origin induces a rotation of $L$. All matching rotations are found using the Knuth–Morris–Pratt linear-time string-matching algorithm. Reflective symmetries are found by equating reflections with reversal followed by rotation. Then again string-matching may be employed. This algorithm finds all symmetries in $O(n)$ time. I believe the algorithm as described is due to Glenn Manacher. Peter Eades, "Symmetry Finding Algorithms." Machine Intelligence and Pattern Recognition, North-Holland, Volume 6, 1988, Pages 41-51. DOI<|endoftext|> TITLE: Is there a nullstellensatz for trigonometric polynomials? QUESTION [45 upvotes]: Let $$ f(x) = \sum_{j=1}^n c_j e^{2\pi i\alpha_j x}, g(x) = \sum_{k=1}^m d_k e^{2\pi i\beta_k x}$$ be two (quasi-periodic) trigonometric polynomials, where the coefficients $c_j, d_k$ are complex and the frequencies $\alpha_j,\beta_k$ are real (but are not necessarily commensurable). Suppose that $f,g$ have infinitely many zeroes in common, thus $f(x)=g(x)=0$ for infinitely many reals $x$. Does this imply that $f,g$ have a non-trivial common factor (in the ring of trigonometric polynomials)? In the commensurable case (so that $f,g$ have a common period) this is clear from the factor theorem, but my interest is in the incommensurable case. I would also be interested in the special case when $f,g$ are both real-valued, though I suspect that this restriction does not materially impact the question. Here is a closely related question. Let $G = \{ (e^{2\pi i \alpha_1 x}, \dots, e^{2\pi i \alpha_k x}): x \in {\bf R} \}$ be a one-parameter subgroup of the torus $(S^1)^k$ for some reals $\alpha_1,\dots,\alpha_k$ that are linearly independent over ${\bf Q}$ (so that $G$ is dense in $(S^1)^k$). Is it true that every codimension two real subvariety of $(S^1)^k$ intersects $G$ in at most finitely many points? Note that if one specialises to the subvarieties $\{ z \in (S^1)^k: P(z)=Q(z)=0\}$ for two polynomials $P,Q$ with no common factor one basically obtains a version of the first question. I'm somewhat aware of the literature on exponential polynomials (e.g., Ritt's theorem), but I couldn't find quite the appropriate tool; results in transcendental number theory (e.g., the six exponentials theorem) also seem vaguely relevant, but again it's not quite a perfect fit. REPLY [31 votes]: I'm converting the comment above to an answer: Shapiro made the following conjecture in the paper "The expansion of mean-periodic functions in series of exponentials" Comm. Pure and Appl. Math. 11 (1958) 1–22: If two exponential polynomials have infinitely many zeros in common they are both multiples of some third (entirely transcendental) exponential polynomial. In the paper "On common zeros of exponential polynomials" Enseignement Math. (2) 21 (1975), no. 1, 57–67, van der Poorten and Tijdeman prove a special case of Shapiro's conjecture for the case when one of the polynomials has rational frequencies. They observe that in a certain sense this statement is equivalent to the Skolem–Mahler–Lech theorem. More recently, in the paper "From Schanuel’s Conjecture to Shapiro’s Conjecture", D'Aquino, Macintyre, and Terzo show that the general conjecture can be proven assuming Schanuel's conjecture.<|endoftext|> TITLE: Relating different constructions of the universal compact quantum group QUESTION [5 upvotes]: Before asking my question, let me give the necessary background. Readers that are comfortable with the language of universal and reduced compact quantum groups may skip the following two sections. Definition: A pair $(A, \Delta)$ is called $C^*$-algebraic compact quantum group if $A$ is a unital $C^*$-algebra and $\Delta: A \to A \otimes A$ is a unital $*$-morphism that is coassociative and satisfies the quantum density rules. Definition: A pair $(A, \Delta)$ is called an algebraic compact quantum group if $(A, \Delta)$ is a Hopf$^*$-algebra together with a positive functional that is both left and right invariant. Some constructions. Full details can be found in e.g. Timmerman's book "An invitation to quantum groups and duality". (1) Given a $C^*$-algebraic compact quantum group $(A, \Delta)$, we can define the space of matrix coefficients of unitary finite-dimensional corepresentations. This becomes an algebraic compact quantum group for the induced comultiplication and the Haar state. Constructing the antipode and the counit takes some work. (2) Given an algebraic compact quantum group $(A, \Delta)$, one can associate two C*-algebraic compact quantum groups: (i) The universal one: This is the $C^*$-envelope of the space of matrix coefficients. (ii) The reduced one: One forms the associated universal $C^*$-algebraic compact quantum group $(A_u, \Delta_u)$ as in (i) and considers the Haar state $h_u:A_u \to \mathbb{C}$ on this quantum group. One then considers the associated GNS representation $\pi_u: A_u \to B(H_u)$ with respect to $h_u$ and gives $A_r := \pi_u(A_u)$ the structure of a compact quantum group by extending the comultiplication $\Delta$ on $A$. Now, let $(A, \Delta)$ be a $C^*$-algebraic compact quantum group and let $(A_0, \Delta_0)$ be the space of matrix coefficients, which has the structure of an algebraic compact quantum group. With this algebraic compact quantum group, we can associate $(A_r, \Delta_r)$ as in $(ii)$ (thus first consider the universal compact quantum group and consider the GNS-image). However, in the literature, the following construction also occurs: One considers the Haar state on $(A, \Delta)$ and one associated the GNS representation $\pi_h: A \to B(H_h)$ to it. One then defines $\widetilde{A}_r = \pi_h(A)$ and shows that the comultiplication $\Delta:A_0\to A_0\odot A_0$ extends to a comultiplication on $\widetilde{A}_r$ (note that the Haar measure is faithful on $A_0$, so that $\pi_h$ is injective on $A_0$). Thus, this construction "skips the universal compact quantum group". I have the impression that $A_r$ and $\widetilde{A}_r$ are used interchangeably, so they must be somehow related. Question: How are these $C^*$-algebraic compact quantum groups related? Are they canonically isomorphic? I.e. do we have $\pi_u(A_u)\cong \pi_h(A).$ REPLY [4 votes]: Yes, $\widetilde{A}_r$ and $A_r$ are canonically isomorphic. This follows from the following result in which you could take as $(B,\Delta_B)$ the universal C$^*$-algebraic compact quantum group $(A_u,\Delta_u)$ associated with $(A_0,\Delta_0)$. Proposition. Let $(B,\Delta_B)$ and $(A,\Delta_A)$ be C$^*$-algebraic compact quantum groups and let $\pi : B \to A$ be a surjective unital $*$-homomorphism satisfying $(\pi \otimes \pi) \circ \Delta_B = \Delta_A \circ \pi$. Let $B_0 \subset B$ be a dense Hopf $*$-subalgebra. Assume that the restriction $\pi|_{B_0}$ is injective. Denote by $h_B$ and $h_A$ the Haar states on $(B,\Delta_B)$ and $(A,\Delta_A)$, respectively. Denote by $\pi_B$ and $\pi_A$ the corresponding GNS-representations. Then, $h_B = h_A \circ \pi$ and there is a unique $*$-isomorphism $\theta : \pi_B(B) \to \pi_A(A)$ satisfying $\theta \circ \pi_B = \pi_A \circ \pi$. Proof. Since $\pi|_{B_0}$ is injective, the restriction of $h_A \circ \pi$ to $B_0$ is an invariant state on the Hopf $*$-algebra $B_0$. By uniqueness of the invariant state, $h_A \circ \pi$ and $h_B$ are equal on $B_0$. By density of $B_0 \subset B$, we get that $h_B = h_A \circ \pi$. Since $\pi$ is assumed to be surjective, the rest follows immediately.<|endoftext|> TITLE: Proposition 5.13 (ii) in Scholze's Perfectoid Spaces QUESTION [6 upvotes]: In Proposition 5.13 (ii) in Scholze's Perfectoid Spaces, we have $R \to S$ a morphism of $\Bbb F_p$-algebras and the assumption that the relative Frobenius $\Phi_{S/R}$ induces an isomorphism $R_{(\Phi)} \otimes_R^{\Bbb L} S \to S_{(\Phi)}$ in $D(R)$, the derived category of $R$-modules. Here $R_{(\Phi)}$ is the ring $R$ with the $R$-algebra structure given by the Frobenius $R \to R$, and $S_{(\Phi)}$ is similarly defined. Then they go on to claim, in the proof of (ii), that this assumption says that $\Phi_{S^\bullet/R}: R_{(\Phi)} \otimes_R S^\bullet \to S^\bullet_{(\Phi)}$ induces a quasi-isomorphism of simplicial algebras, where $S^\bullet$ is a simplicial resolution of $S$ by free $R$-algebras. I do not understand why. My guess is that this is because the complex $R_{(\Phi)} \otimes_R^{\Bbb L} S \in D(R)$ in the assumption is constructed by taking such a resolution of $S$, so $R_{(\Phi)} \otimes_R^{\Bbb L} S \triangleq R_{(\Phi)} \otimes_R S^\bullet \in D(R)$, and then $S^\bullet_{(\Phi)} \to S_{(\Phi)}$ is a quasi-isomorphism because $S^\bullet_{(\Phi)}$ is a resolution of $S_{(\Phi)}$, so in $D(R)$ we have $R_{(\Phi)} \otimes_R S^\bullet \triangleq R_{(\Phi)} \otimes_R^{\Bbb L} S \cong S_{(\Phi)} \cong S^\bullet_{(\Phi)}$? After that, they say that this implies that $\Phi_{S^\bullet/R}: R_{(\Phi)} \otimes_R S^\bullet \to S^\bullet_{(\Phi)}$ gives an isomorphism $R_{(\Phi)} \otimes_R^{\Bbb L} \Bbb L_{S/R} \cong \Bbb L_{S_{(\Phi)}/R_{(\Phi)}}$. I also do not understand why. According to the same paper, the complex $\Bbb L_{S/R}$ is defined to be $\Omega^{1}_{S_\bullet/R} \otimes_{S_\bullet} S$, but I do not see a relation. The reference for this part is Lemma 6.5.9 of Gabber–Romero, which references Proposition II.1.2.6.2 of Illusie's Complexe Cotangent et Déformations I, but I do not see how to apply the proposition to this situation. Finally, both Scholze and Gabber–Romero claim that $\Bbb L_{S_{(\Phi)}/R_{(\Phi)}} \cong \Bbb L_{S/R}$, but I do not know why. I think it is because $R_{(\Phi)}$ is defined to be $R$ as a ring, and the map $R_{(\Phi)} \to S_{(\Phi)}$ as rings should be the same as the map $R \to S$ as rings, so their modules of Kähler differentials should be the same. P.S. I don't understand why Scholze uses $S_\bullet$ in the definition of $\Bbb L_{S/R}$ when he uses $S^\bullet$. Indeed, Gabber–Romero uses $P^\bullet$ in Lemma 6.5.9. REPLY [9 votes]: Regarding the first and third question, what you say is correct. For the second question, you are looking for the base change compatibility of the cotangent complex: If $R\to R'$ is any map of rings and $S$ is an $R$-algebra such that $S'=S\otimes^L_R R'$ sits in degree $0$, then $$R'\otimes^L_R \mathbb L_{S/R}\cong \mathbb L_{S'/R'}.$$ To see this, use any simplicial resolution $S_\bullet$ of $S$ by free $R$-algebras; then $S_\bullet\otimes_R R'$ is such a resolution of $S'$. The left-hand side is computed by the simplicial $R'$-module $R'\otimes_R(\Omega^1_{S_\bullet/R}\otimes_{S_\bullet} S)$, and the right-hand side by $\Omega^1_{S'_\bullet/R'}\otimes_{S'_\bullet} S'$; these two simplicial $R'$-modules agree. Implicit here is that one can use any free resolution to compute the cotangent complex, not necessarily the standard functorial one. But I think there must be some more basic confusion on cotangent complexes, as you say that you do not see the relation between $\mathbb L_{S/R}$ and $\Omega^1_{S_\bullet/R}\otimes_{S_\bullet} S$. To address this: What do you take as the definition of the cotangent complex?<|endoftext|> TITLE: How Dirichlet proved Dirichlet's unit theorem for general number fields? QUESTION [13 upvotes]: For a general number field $K$, Dirichlet's unit theorem states that the unit group of the ring of integers of $K$ is a finitely generated group of rank $r_1+r_2-1$. It seems that standard algebraic number theory textbooks usually prove this theorem by using Minkowski's theory or using the Blichfeldt-Minkowski Lemma to show that the group of norm 1 ideles modulo $K^\times$ is compact. Since both Minkowski and Blichfeldt were born significantly later than Dirichlet, I wonder that Dirichlet's proof may be different from the above two? I should say that, for real quadratic fields, or equivalently, Pell's equation $x^2-dy^2=1$, I have saw an argument in Dirichlet's book "Lectures on number theory" where the author first shows that, for the irrational number $\sqrt{d}$ there are infinitely many rational numbers $\frac{p}{q}$ such that $$ |\frac{p}{q} -\sqrt{d}| \leq \frac{1}{q^2}. $$ Using this result, one can easily produce infinitely many principal ideals with bounded norms of the ring of integers $\mathbb{Q}(\sqrt{d})$ whence produce a nontrivial unit. Done! To me this argument is distinct from the Minkowski's method in some sense and I am unable to generalize the above argument to prove the Dirichlet's unit theorem for general number fields. So I wonder that how Dirichlet proved his theorem for general number fields. Thanks. REPLY [13 votes]: Dirichlet's proof is described in Number Theory: Algebraic Numbers and Functions (starting on page 48). Dirichlet did not use Minkowski’s theorem; he proved the unit theorem in 1846 while Minkowski’s theorem appeared in 1889. Dirichlet’s substitute for the convex-body theorem was the pigeonhole principle. Dirichlet did not state the unit theorem for all orders, but only those of the form $\mathbf{Z}[\alpha]$, since at the time these were the kinds of rings that were considered. [source] There is an oft-repeated story that the idea for the proof came to Dirichlet while he was attending Easter mass in the Sistine Chapel in Rome. Attempts to document that story are described in this MO question.<|endoftext|> TITLE: Uniform spaces as condensed sets QUESTION [10 upvotes]: $\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Unif{Unif}\DeclareMathOperator\CHaus{CHaus}\DeclareMathOperator\Set{Set}\DeclareMathOperator\op{op}\DeclareMathOperator\Ind{Ind}\DeclareMathOperator\Fin{Fin}\DeclareMathOperator\Cond{Cond}\DeclareMathOperator\Top{Top}$In Barwick–Haine - Pyknotic objects, I. Basic notions Example 2.1.10, they showed that the functor $\Hom_{\Unif}(-,X)\colon\CHaus^{\op}\to\Set$ is a pyknotic set, i.e., a sheaf on the site $\CHaus$ of compact Hausdorff spaces equipped with the coherent topology (Lurie - Ultracategories). I did not check whether the sheaf above is accessible (update: it is just the underlying topological space, see the Update part). However, I am slightly skeptical about this approach. It seems to me that the "correct" realization of a uniform space should be a condensed set which records its underlying topological space, along with an extra structure which records the uniform structure. Question: I wonder how much is known about any kind of realization of uniform spaces as condensed sets. I suppose that this extra structure would be described by a certain kind of enriched groupoid. Indeed, the uniform structure on a topological space could be understood as a groupoid enriched in filters. See nLab page for a description of this sort. This is motivated by an attempt to eliminate the restriction that the adjective "solid" only applies to condensed abelian groups. Let $M$ be a topological abelian group. I was about to understand what it means for $M$ that the condensed abelian group $\underline M$ is solid. Following Lecture II of Scholze - Lectures on Analytic Geometry, for any sequence $(m_n)_{n\in\mathbb N}\in M^{\mathbb N}$ convergent to $0$, we associate a (continuous) map from the profinite set $S\mathrel{:=}\mathbb N\cup\{\infty\}$ to $M$ which maps $n$ to $m_n$ and $\infty$ to $0$, or equivalently, a map $S\to\underline M$ of condensed sets by, say, Yoneda's lemma. Suppose that $\underline M$ is solid, then this map extends uniquely to a map $\mathbb Z[S]^\blacksquare\to\underline M$ of condensed abelian groups. If I am not mistaken, $\mathbb Z[S]^\blacksquare\to M$ further factors through $\underline{\mathbb Z[[t]]}$, the condensed abelian group associated to the topological abelian group $\mathbb Z[[t]]$ with $(t)$-adic topology, and by full faithfulness, we get a factorization $S\to\mathbb Z[[t]]\to M$ where the second map is additive. Consequently, for every sequence $(a_n)_{n\in\mathbb N}\in{\mathbb Z}^{\mathbb N}$ of integers, the series $\sum_na_nt^n$ converges in $\mathbb Z[[t]]$, therefore the series $\sum_na_nm_n$ converges in $M$, which should imply, if I am not mistaken, that the uniform structure on $M$ is non-archimedean and complete, at least when $M$ is first countable (by the way, I don't understand why it is claimed that “it is not directly as any kind of limit of finite sums”). So the non-archimedean nature is rooted in the formalism. I suppose that a natural approach is to generalize the uniform structure to condensed sets, and to generalize the classical Cauchy-completeness. I don't know whether it is convinced that this does not work. The current presentation separates non-archimedean and $\mathbb R$-case, which covers neither non-abelian groups nor the general completeness of topological abelian groups. Update: When discussing the functor $\Hom_{\Unif}(-,X)\colon\CHaus^{\op}\to\Set$, I missed the simple fact that $\Hom_{\Unif}(-,X)=\Hom_{\Top}(-,X)\mathrel{=:}\underline X$ by the Heine–Cantor theorem. In other words, when the uniform space in question is $T_0$ (therefore $T_1$), $\underline X$ is thus a condensed set. Update2: It seems to me that I did not phrase the question unambiguously. In fact, I wanted to ask for any realization of unform spaces as condensed sets with some extra data, which should be recorded in the data of a condensed abelian group when it comes from a topological abelian group. At the time that I posed the question, I did not realize that Barwick–Haine's approach only records the data about the underlying space, but just doubted that it is a "correct" approach to record all data. Update3: We say that a uniform space is non-archimedean if the uniformity is generated by equivalence relations. REPLY [2 votes]: Here is an essentially tautological answer. The notion of uniformity makes sense also for condensed sets -- it is a condensed set $X$ together with certain subsets $U\subset X\times X$ termed entourages satisfying all the usual axioms. Moreover, one can also define the completion of $X$ with respect to its uniform structure "by interpreting everything internally in condensed sets". If $X$ is a condensed abelian group with a nonarchimedean uniformity -- meaning say that the $U$'s above can be chosen to be subgroups, and all $X/U$ are discrete (as condensed sets) -- then the completion of $X$ is the inverse limit of the $X/U$, and hence is solid.<|endoftext|> TITLE: Abelianization of monoids in arbitrary (symmetric) monoidal categories QUESTION [7 upvotes]: Under which 'minimal' conditions on a symmetric monoidal category does an abelization functor from its monoids to its commutative monoids exist? More precisely: let $(\mathcal{C},\otimes,1)$ be a symmetric monoidal category, and let $U : \mathrm{CMon}(\mathcal{C}) \rightarrow \mathrm{Mon}(\mathcal{C})$ be the forgetful functor from its category of commutative monoids to its category of monoids. Define the abelianization functor $(-)_{\mathrm{ab}}$, if it exists, as the left adjoint of the forgetful functor ($(-)_{\mathrm{ab}} \dashv U$). It seems to me that this functor does not necessarily exist in general, and one has to assume some extra conditions on the symmetric monoidal category $\mathcal{C}$. Which 'minimal' such extra conditions do you know? (Of course this 'minimal' is not well defined, as some conditions might not be weaker than others, just different, that's why I put it in quotes and by that I mean a general idea of 'as weak conditions as possible'). The page abelianization in nLab, under Section 3, mentions "we can do abelianization of monoid objects in many monoidal categories", the word many (and not all) suggesting that it is indeed not possible to do it in general, but it does not give any further hints on what kind of conditions one needs. Thanks! REPLY [3 votes]: Here is an approach that has the advantage of avoiding to go in the details of a concrete construction. More explicit constructions of the abelianization might give even more general results, in terms of corollary 2 below they would corresponds to more explicit description of the object $X_{ab}$, but I think corollary 2 below already cover most of the applications. Theorem: Let $C$ be a symmetric monoidal locally presentable category in which the tensor product is accessible. Then the categories of (commutative) monoids in $C$ is locally presentable. Proof: The categories of monoids or commutative monoids can be expressed as Cat-enriched limits of powers of $C$ and map induced by the tensor product between them. So as long as the tensor product is accessible, these are accessible categories because Cat-enriched limits of accessible categories and accessible functors between are accessible. As these category have all limits (they are created by the forgetful functor to $C$) they are locally presentable categories. Corollary 1: Let $C$ be a locally presentable symmetric monoidal category in which the tensor product is accessible. Then the abelianization functor exists. Proof: This follows from the special adjoint functor theorem applied to the forget full functor from commutative monoids to moinods. Fix $\kappa$ such that $\otimes$ is $\kappa$-accessible. The forgetfull functor commutes to all limits and $\kappa$-directed colimits, i.e. it is an accessible right adjoint , so it has a left adjoint by the special adjoint functor theorem. Exploiting this, one can do much better and drop the local presentability assumption by observing that "the universal example" is itself locally presentable: Corollary 2: Let $C$ be a symmetric monoidal category with all colimits in which the tensor product preserves $\kappa$-directed colimits for some regular cardinal $\kappa$. Then the abelianization functor for $C$ exists. Proof: Let $M_\kappa$ be the "free symetric monoidal category with all $\kappa$-small colimits generated by a monoid object $X \in M_\kappa$". It exists, because all the structure I've put on $M_\kappa$ can be encoded as a $\kappa^+$ essentially algebraic theory (let says in a bicategorical sense for simplicity, though it would also work in a $1$-categorical sense). Let $V_\kappa$ be the $\kappa$-ind completion of $M_\kappa$. Because Ind completion commutes to products, the tensor product on $M_\kappa$ induces a $\kappa$-accessible tensor product on $V_\kappa$, making $V_\kappa$ a symmetric monoidal category (if you prefer, ind completion being monoidal for the product, it sends monoids object to monoids object). It follows that $V_\kappa$ is a locally presentable category with a $\kappa$-accessible tensor product, so one can apply the previous result and the universal monoid object $X \in M_\kappa$ admit a symmetrisation $X_{ab}$ in $V_\kappa$. If now $C$ is a general symmetric monoidal category with all colimits and such that $\otimes$ preserves $\kappa$-directed colimits in each variables. Then given a monoid object $T \in C$ you have an essentially unique (strong) symmetric monoidal functor $F: M_\kappa \to C$ that preserves $\kappa$-small colimits and sends $X$ (as a monoid) to $T$. It extends to a monoidal functor preserving all colimits $V_\kappa \to C$, in fact (as $V_\kappa$ is locally presentable), it is a left adjoint functor. It then follows from a general argument that the image of $X_{ab}$ by this functor is an abelianisation of $X$ (one just check the universal property in $C$ using the universal property of $X_{ab}$ in $V_\kappa$ and the adjunction).<|endoftext|> TITLE: Closure of the diagonal is an equivalence relation QUESTION [5 upvotes]: Let $X$ be a topological space and $\overline{\Delta_X} \subseteq X \times X$ the closure of its diagonal. Then $\overline{\Delta_X}$ is the graph of an equivalence relation on $X$. This statement can be found in Borceux's Handbook of Categorical Algebra I, chapter 3, page 96, as he introduces adjoint functors. I'm currently trying to prove this. Reflexivity is clear, and so is symmetry, for $(x,y) \mapsto (y,x)$ is an automorphism of $X \times X$. However, I'm not sure how to prove transitivity. Here's what I found. Not belonging to $\overline{\Delta_X}$ is the same as being separable by open subsets of $X$ : indeed, $(a,b) \notin \overline{\Delta_X}$ iff there exists $U,V$ open subsets of $X$ such that $(a,b) \in U \times V \subseteq (X \times X) \setminus \overline{\Delta_X}$, that is iff there exists $U,V$ open subsets of $X$ such that $a \in U$,$ b \in V$ but $U \cap V = \varnothing$. Thus it seems to me that the statement is false : taking $X = \{x,y,z\}$ with basis $\{x\},\{z\}$, we have $(x,y),(y,z) \in \overline{\Delta_X}$ and yet $(x,z) \notin \overline{\Delta_X}$. Thank you for clearing my mind. Edit : It seems that this statement really is false, as Eric Wofsey suggests in this answer at math.SE. REPLY [4 votes]: The set $\{1,2,3\}$ for which the open subsets are $\emptyset$, $\{1\}$, $\{3\}$, $\{1,3\}$, $\{1,2,3\}$, is a topological space for which the closure of the diagonal is $\{1,2,3\}^2\smallsetminus\{(1,3),(3,1)\}=\{1,2\}^2\cup\{2,3\}^2$, which is not an equivalence relation. Possibly more intuitively it can be viewed as the quotient of $\mathbf{R}$ by the equivalence relation $x\simeq y$ if $\exists t>0$: $x=ty$.<|endoftext|> TITLE: A road map through group cohomology QUESTION [13 upvotes]: I am a PhD student in algebraic topology, and I would like to learn something about group cohomology. The final goal would be to present one or two seminars on this topic, in order to give my mates a gently introduction to this subject and at the same time showing them some striking result/application of this theory. Ideally, my plan for the seminar is: Introduce group cohomology, with a lot of motivations and examples Explain what makes group cohomology awesome Focus on a specific result, and showing some pretty applications of it (something that could be interesting to an algebraic topologist if possible) in order to strenghthen point 2 I am not looking for books, which are already given in these questions: https://math.stackexchange.com/questions/2697778/reference-for-group-cohomology https://math.stackexchange.com/questions/695613/reference-request-introduction-to-finite-group-cohomology?rq=1 So my questions are: Does anyone know any introductory papers/lecture notes where I can find a concise introduction to group cohomology? I am looking for something which do not contains all the details but which gives me a general view of the main results and applications of the theory. Youtube videos/lecture series are also very welcome. Of course if you want to mention book that are not in the previous answers it is fine aswell. Are there any suggestions about results/applications that I can put in points 2 and 3 of the seminar? As I said before the idea is to present this material to other students of algebraic topology, so I would prefer theorems/applications that will appeal to this kind of audience. EDIT: an answer of this kind References and resources for (learning) chromatic homotopy theory and related areas is also very welcome and pertinent! Thank you in advance, Tommaso REPLY [3 votes]: Brown: Lectures on the cohomology of groups. Adem: Lectures on the cohomology of finite groups. Carlson: The cohomology of groups (from Handbook of Algebra, Vol.1, 1996). Rotman: Homology of groups (chapter 9 from one of his algebra books that I forget the name). REPLY [2 votes]: Chapter 2 of these notes by Milne have been helpful to me.<|endoftext|> TITLE: Criterion for Kuratowski Limit Inferior QUESTION [7 upvotes]: Let $(X,d_X)$ be a compact metric space and let $\{K_n\}_{n=1}^{\infty}$ be a collection of non-empty compact subsets. Let $K\subseteq X$ be compact. Then, if for every $x_n \in K_n$ we have $$ d_X(x_n,K)\leq \frac1{n}, $$ does this imply that $K$ is the Kuratowski lower limit ($\mathop{\mathrm{Li}}_{n \to \infty} K_{n}$) of the $K_n$, where the Kuratowksi limit limit is defined by $$ \mathop{\mathrm{Li}}_{n \to \infty} K_{n} = \left\{ x \in X \left| \limsup_{n \to \infty} d(x, K_{n}) = 0 \right. \right\} \;? $$ REPLY [6 votes]: The answer is no, or what am I missing. Let $X = K = \{0,1\}$ and $K_n = \{0\}$, $n \in \mathbb{N}$ with $d_X(0,1) = 1$. Then $\text{LI}_{n \to \infty} K_n = \{0\} \not= K$.<|endoftext|> TITLE: Taylor series with coefficients in $\mathbb{Q}$ QUESTION [10 upvotes]: Is there a sequence of rational numbers $a_0, a_1, \dotsc$ such that $\sum\limits_{i\geq 0}a_i x^i$ converges absolutely to $2^x$ for every $x\in \mathbb{Z}$? REPLY [13 votes]: Yes. We construct inductively a sequence $b_0,b_1,\ldots$ of real numbers such that $|b_k|\leqslant 1/k!$ and all coefficients of the entire function $$ 2^z+(b_0+b_1z+\ldots)\sin \pi z $$ are rational. If $b_0,\ldots,b_{m-1}$ are already defined and the coefficients of $1,z,z^2,\ldots,z^m$ are rational, we may choose appropriate $b_m$ so that the coefficient of $z^{m+1}$ (which is $\pi b_m$ plus something known) is rational. REPLY [10 votes]: Yes. You can construct such a series of the form $$ 1 + \sum_{i=0}^\infty (a_i + b_i x) \left( \frac{x^2}{(i+1)^2}\right)^{e_i} \prod_{j=-i}^i (x-j) $$ for some sequence $a_i, b_i$ of rational numbers and some sequence $e_i$ of natural numbers (converging to $\infty$.) This certainly defines a power series with rational coefficients since the $i$th term is divisible by $x^{2 e_i}$ and $e_i$ goes to $\infty$, so the coefficient of each power of $x$ is a finite sum of rational numbers. The $\prod(x-j)$ term ensures that the $i$th term vanishes at every integer from $-i$ to $i$. Thus the function has the value $2^x$ at $x=i+1$ and $x=-i-1$ if and only if the sum of the first $i$ terms has the value $2^x$ at those points. For any values of $a_1,\dots, a_{i-1}, b_1,\dots, b_{i-1}, e_1,\dots, e_{i-1}$, there is a unique rational $a_i,b_i$ which ensures this power series takes the correct value at these $x$. (The denominator $i+1$ is so that the value of $e_i$ does not affect this.) Now we can choose $e_i$ sufficiently large depending on $a_1,\dots, a_{i-1}, b_1,\dots, b_{i-1}, e_1,\dots, e_{i-1},a_i,b_i$ so that the version of the $i$th term with absolute values everywhere $$(|a_i| + |b_i| |x|) \left( \frac{|x|^2}{(i+1)^2}\right)^{e_i} \prod_{j=-i}^i (|x|+|j|) $$ is as small as desired at the points $-i,\dots, i$. For example, we can ensure it is at most $2^{-i}$. Having done this, the series is certainly absolutely convergent at all integer points (and thus at all points), because all but finitely many of the terms have absolute value at most $2^{-i}$ at any given point. Choosing $a_i,b_i,e_i$ according to this rubric, our series satisfies all the desiderata.<|endoftext|> TITLE: An integration identity on $\mathbb{P}^{n-1}$ QUESTION [8 upvotes]: Let $\omega_{\text{FS}}$ denote the Fubini–Study metric on $\mathbb{P}^{n-1}$ with unit volume, and let $[w_1 : \cdots : w_n]$ be standard unitary homogeneous coordinates. On page 5 of Yang–Zheng's paper On real bisectional curvature for Hermitian manifolds they claim the following is well-known: $$\int_{\mathbb{P}^{n-1}} \frac{w_i \overline{w_j} w_k \overline{w_{\ell}}}{\lvert w \rvert^4} \omega_{\text{FS}}^{n-1} = \frac{1}{n(n+1)}(\delta_{ij} \delta_{k \ell} + \delta_{i \ell} \delta_{kj}).$$ Does anyone have a reference for this "well-known fact"? Perhaps is so well-known that nobody knows it? REPLY [6 votes]: If I remember correctly, you can find this in the book "Complex Differential Geometry" by Fangyang Zheng in Chapter 7. The analogous result "with real coefficients" (i.e. for real vectors and with $S^{n-1}$ instead of projective space) is originally due to Marcel Berger in Lemma 7.4 here.<|endoftext|> TITLE: Is this set theory used by Gandy first-order with signature $(\in, \lambda)$? QUESTION [5 upvotes]: In On the Axiom of Extensionality, Part II, The Journal of Symbolic Logic, Vol. 24, No. 4 (Dec., 1959), https://doi.org/10.2307/2963897, pp. 287-300, R. O. Gandy shows that a class theory X containing NBG minus extensionality is not weaker than NBG; X includes a use of class-abstraction denoted with $\lambda$, so that $\lambda x(x=x)$ is a universal class. Gandy does have identity. It is worth noticing that one can show (use A2 p. 289 and I29 p. 290) that $x\in \lambda z\phi(z)\leftrightarrow \exists y(x\in y)\wedge\phi(x)$. Is X a first-order theory with signature $(\in, \lambda)$? REPLY [3 votes]: A signature of first order logic is usually taken to be a list of extra-logical symbols that range over specific elements (for constants) or over specific subsets (for n+1 ary functions, predicates) of the universe of discourse. The class-abstraction symbol $\lambda$ here doesn't fit into any of those. I'm not sure if it can be considered among symbols of the underlying logic, but by then that kind of logic won't be called just first order, one may call it first order logic with class-abstractions, or something to that effect. That said, I think the signature of Gandy's theory if described in terms of first order logic then it would be very extensive (in agreement with comment by James Hanson), so if $ \{x_i: i \in \mathbb N \}$ is the set of all variable symbols in a langauge, and $\{\phi_j(x_i): i,j \in \mathbb N \}$ is the set of all formulas in one free variable in the language then the signature would be something like: $(=,\in, \lambda x_i \phi_j(x_i): i,j \in \mathbb N)$, a countably infinite signature! Where each $\lambda x_i \phi_j(x_i)$ is a constant (zero place function) symbol, i.e. an argumentless (doesn't take an element of the universe of discourse as argument) expression that range over a single element of the universe of discourse. However, if $n$ many free variables other than $x_i$ are allowed to occur in $\phi_j(x_i)$, then the expression $\lambda x_i \phi_j(x_i)$ would become an $n$-ary function symbol. In nutshell X is a first order theory with signature $(=,\in, \lambda x_i \phi_j(x_i): i,j \in \mathbb N)$<|endoftext|> TITLE: Multiplicative Brown representability? QUESTION [15 upvotes]: The Brown representability theorem can be convenient way to construct a spectrum. But to get a ring spectrum of even a very unstructured form seems to be harder. There's even currently a statement on the nlab to the effect that any multiplicative cohomology theory is represented by a ring spectrum, but my strong suspicion is that this statement was made in error. Assumption: Let $E^\ast: Spaces^{op} \to GrAb$ be a cohomology theory. To avoid subtleties with the unpointed version of Brown's theorem, let's just assume that $E^\ast$ is represented by a spectrum $E$. Now when we impose "multiplicative" structure on $E^\ast$, the question is whether the "multiplication" is also "representable". I can think of at least two possible ways to make this precise: Question 1: Suppose that $E^\ast: Spaces^{op} \to GrAb$ admits the structure of a lax (symmetric) monoidal functor. Then does there exist a map $\mu: E \wedge E \to E$ inducing the (symmetric) monoidal structure of $E^\ast$ in the appropriate sense? I think that's the cleanest form of the question, but I think it would be more classical to ask it in the following form: Question 2: Suppose that $E^\ast: Spaces^{op} \to GrAb$ admits a lift to a functor $E^\ast: Spaces^{op} \to GrRing$. Then does there exist a map $\mu: E \wedge E \to E$ such that the multiplication $$E^\ast(X) \otimes E^\ast(X) = \pi_{-\ast} F(\Sigma^\infty_+,E) \otimes \pi_{-\ast}F(\Sigma^\infty_+,E) \to \pi_{-\ast}F(\Sigma^\infty_+ X, E) = E^\ast(X)$$ is always given by the formula $$f\otimes g \mapsto \mu_\ast \circ (f \wedge g) \circ \Delta^\ast$$ where ($\Delta: X \to X \times X$ is the diagonal)? Remark: The category "$GrRing$" could be interpreted in multiple ways -- if our rings are associative, we could ask for a homotopy associative $\mu$; if our rings are graded-commutative we could ask for a homotopy commutative $\mu$, etc. Note: The Yoneda lemma does allow us to give an "$E_0$ version" of Brown representability. That is, from the structure of unit elements $u_X \in E^\ast(X)$ we can construct a unit map $u: \mathbb S \to E$, just by the Yoneda lemma. That's why the question is about the multiplication. Remark: Actually -- we might even just as well have our cohomology theory be defined as a functor $E^\ast: Spectra^{op} \to GrAb$, and the question still seems to be interesting. REPLY [13 votes]: I'll write $h_E$ for the functor from finite spectra to abelian groups given by $h_E(X)=\pi_0(E\wedge X)=[DX,E]$. This is an object of the category $\mathcal{A}$ of all additive functors from finite spectra to abelian groups. Given $A,B,C\in\mathcal{A}$, a pairing from $A$ and $B$ to $C$ means a natural map $A(X)\otimes B(Y)\to C(X\wedge Y)$. By the usual Day construction, there is a symmetric monoidal product $\boxtimes$ on $\mathcal{A}$ such that $\mathcal{A}(A\boxtimes B,C)$ is the set of pairings from $A$ and $B$ to $C$. This preserves colimits in both variables. We can use Spanier-Whitehead duality to identify $\mathcal{A}$ with a category of contravariant functors, and then use standard properties of Day convolution to see that $h_E\boxtimes h_F=h_{E\wedge F}$ when $E$ and $F$ are finite. We can then take colimits over finite subspectra to see that $h_E\boxtimes h_F=h_{E\wedge F}$ in all cases. Any of the kinds of input data specified in the question will give rise to a pairing from $h_E$ and $h_E$ to $h_E$, and thus to a map $h_{E\wedge E}=h_E\boxtimes h_E\to h_E$. By Brown representability, this will arise from a map $\mu\colon E\wedge E\to E$ of spectra, which will be unique mod phantoms. If the original pairing is commutative or associative, then $\mu$ will be commutative or associative mod phantoms. Corollary 2.15 of "Morava K-theories and localisation" gives a commonly satisfied criterion under which the group of phantom maps in degree zero is trivial. I was surprised to find that Dan Christensen and I did not discuss this monoidal structure in our 1998 paper on phantom maps and homology theories, as it would fit very naturally there. Probably someone else has published this at some point, but I don't know where.<|endoftext|> TITLE: Where do some "energy identities" in PDE theory come from? QUESTION [15 upvotes]: There are a lot of very complicated expression that helps us obtain useful estimates for PDEs. To just give one example, the following is one of the Virial identities: $$ \frac12\frac{d}{dt} \Im \left(\int (x \cdot \nabla u) u^* \right)dx =\int|\nabla u|^2 dx -\epsilon\left(\frac{n}{2}-\frac{n}{p+1}\right)\int|u|^{p+1} dx, $$ where $u\in H^1, xu\in L^2$ solves the non-linear Schrödinger equation $i\partial_t u + \Delta u = -\epsilon u|u|^2$ in $\mathbb R^n.$ One way to come up with complicated "energy" or "momentum" expressions is to use Noether's theorem, but apparently many identities similar to the one above cannot be easily derived by Noether's theorem. In particular, the above expression can be derived without using any Lagrangians at all. I often see such complicated identities in texts about PDE theory, stated and proven without any clues about where they come from. My question is: how do people come up with them in the first place? It would be very helpful if there were an explanation for at least one or two identities like this so that I can see what type of reasoning is involved when we discover them. REPLY [6 votes]: Another construction when your PDE has a Hamiltonian structure, but the Poisson structure has a non-trivial kernel. Roughly speaking, this means that the phase space is foliated by submanifolds, and each of these varieties is equiped with a non-degenerate Poisson structure, this one varying smoothly along the foliation. Functions that are constant along each leaf are called Casimir's. These are conserved quantities along the flow of the PDE because this flow stay on the leaf associated with the initial data. A typical example arises in the Euler equation governing the velocity field $u$ of an incompressible fluid. In $2$ space dimensions, the functional $$\int_\Omega f(\partial_1u_2-\partial_2u_1)\,dx$$ are Casimir's ($f$ smooth but arbitrary). In $3$ space dimensions, in a torus or in ${\mathbb R}^3$, the helicity is a Casimir: $$\int u\cdot{\rm curl}\,u\,dx.$$ Both examples have natural generalizations to higher dimensions. They also have counterparts in the case of compressible fluids. This is quite important because despite the family of Casimir's can be quite large, it does not contribute to the complete integrability of a Hamiltonian system. In some sense, these are trivial invariants, because they depend only upon the Poisson structure, but not on the choice of the Hamiltonian.<|endoftext|> TITLE: CFSG-free bound for the number of generators of a finite simple group QUESTION [7 upvotes]: We know that every finite simple group can be generated by $2$ elements. This (correct me if I'm wrong) was proved, as far as I know, by Steinberg (Steinberg, R. (1962). Generators for Simple Groups. Canadian Journal of Mathematics, 14, 277-283) for Chevalley groups and by Aschbacher and Guralnick (M. Aschbacher, R. Guralnick, Some applications of the first cohomology group, Journal of Algebra, Volume 90, Issue 2, 1984, Pages 446-460, Theorem B) for the other simple groups, using the classification of the finite simple groups (CFSG). We also know, by an easy application of Lagrange's theorem, that a group of order $n$ can be generated by at most $\log_2(n)$ elements. My question is the following. Are there results proving that every simple group of finite order $n$ can be generated by $d$ elements, where $d$ is some quantity between $2$ and $\log_2(n)$ (for example $\log_2 \log_2 (n)$ or - more optimistically - an unspecified constant) without using the classification of finite simple groups? REPLY [7 votes]: Still elementary, but marginally stronger than the $\log_{2}(n)$ bound is: every non-Abelian finite simple group $G$ of order $n$ can be generated by fewer than $\log_{p}(n)$ elements, where $p$ is the largest prime divisor of $\lvert G\rvert$. In particular (using Burnside's $p^{a}q^{b}$-theorem), such a group $G$ can be generated by fewer than $\log_{5}(n)$ elements, since we always have $p \geq 5$. This improves the $\log_{2}(n)$ bound by a factor of at least $2$, but is still very weak. The proof is easy and elementary— $G$ is generated by its elements of order $p$. Let $S$ be a set of elements of order $p$ of $G$ which generates $G$ with $\lvert S\rvert$ minimal. Then for each positive integer $j$, any $j$ elements of $S$ generate a subgroup of order $p^{j}$ or greater, so $\lvert S\rvert < \log_{p}(\lvert G\rvert)$. I don't know what the best known "elementary" bound is (or even if this is a well-defined quantity). I seriously doubt whether there is a constant-type bound known without using CFSG.<|endoftext|> TITLE: The (current) obstructions for a cohomological interpretation of the Riemann zeta function QUESTION [9 upvotes]: I am interested in the idea of a cohomological interpretation of the Riemann hypothesis (suggested by Deninger/Connes). I am a beginner in étale cohomology, and I would like to ask the following Question. Why does étale cohomology not offer a cohomological interpretation of the local zeta function of the scheme $\operatorname{Spec} \mathbb{Z}$ in terms of the (étale?) cohomology of the associated topological space? I would appreciate any answer as well as a reference. I understand this question may be a bit annoying since the obstructions should supposedly be immediate for if otherwise this would probably be well known. Nonetheless, I am curious and do not know who to ask. REPLY [17 votes]: One can't give a complete answer to this question without first understanding how etale cohomology does give a cohomological interpretation of the zeta function in the function field case. Let $X$ be a smooth projective curve over a finite field $\mathbb F_q$. Then $$ \zeta_{X}(s) = \frac{ \det (1 - \operatorname{Frob}_q q^{-s}, H^1(X_{\overline{\mathbb F_q}}, \mathbb Q_\ell))}{\det (1 - \operatorname{Frob}_q q^{-s}, H^0(X_{\overline{\mathbb F_q}}, \mathbb Q_\ell)) \cdot \det (1 - \operatorname{Frob}_q q^{-s}, H^2(X_{\overline{\mathbb F_q}}, \mathbb Q_\ell))}$$ It is crucial that, to get this formula, one passes to the curve $X_{\overline{\mathbb F_q}}$ obtained by base-changing to the algebraic closure. The reason this is important to note is that there is no analogue of "the algebraic closure of the base finite field" after replacing $X$ with $\operatorname{Spec} \mathbb Z$, at least not in the usual world of schemes. An optimistic guess is that it suffices to take the etale cohomology of $\operatorname{Spec} \mathbb Z$. If that were true, then it would presumably likewise suffice in the function field world to take the etale cohomology of $X$, without base-changing to the algebraic closure. What happens when we do this? We can see this using the long exact sequence $$ \to H^i(X, \mathbb Z_\ell) \to H^i (X_{\overline{\mathbb F_q}}, \mathbb Z_\ell) \to H^i (X_{\overline{\mathbb F_q}}, \mathbb Z_\ell) \to$$ where the third arrow denotes the map $\operatorname{Frob}_q-1$. Examining this exact sequence and the known description of the action of Frobenius on $H^0$ and $H^2$, we see that $$H^0 (X, \mathbb Z_\ell) = \mathbb Z_\ell$$ $$ H^1 (X, \mathbb Z_\ell)= \mathbb Z_\ell$$ $$H^2(X, \mathbb Z_\ell) = H^1(X_{\overline{\mathbb F_q}}, \mathbb Z_\ell)/ (1 - \operatorname{Frob}_q $$ $$H^3(X, \mathbb Z_\ell) = \mathbb Z_\ell/ (q-1)$$ The only interesting cohomology group here is $H_2$, which is a finite abelian $\ell$-group. Its order is the maximum power of $\ell$ dividing the determinant of $\operatorname{Frob}_q-1$ acting on $H^1$. Equivalently, this is the maximum power of $\ell$ dividing the residue of the zeta function of $X$ at $s=1$. Combining this information for all primes $\ell$, we have a cohomological interpretation of the residue of the zeta function at $s=1$ (at least up to a power of $p$). Similarly, over rings of integesr of number fields, not necessarily $\mathbb Z$, we can obtain a cohomological intepretation of the residue of the zeta function. This cohomology group turns out to be dual to the $\ell$-part of the class group, so this is just the Dirichlet class number formula. There is one last thing you can do here. The field of functions on $X_{\overline{\mathbb F_q}}$ is a Galois extension of the field of functions on $X$, with Galois group $\prod_p \mathbb Z_p$. We can just pick an extension of $\mathbb Q$ with Galois group $\prod_p \mathbb Z_p$, or even just $\mathbb Z_p$, pretend that this is $X_{\overline{\mathbb F_q}}$, and take etale cohomology of its ring of integers. This, modulo technical details, is the approach of Iwasawa theory, which gives a cohomological interpretation of $p$-adic $L$-functions. Some $p$-adic $L$-functions are related to special values of the Riemann zeta function at negative integers, so there is, in a sense, a cohomological interpretation of these specific values. You will note some synchrony here with Peter Scholze's point in the comments that etale cohomology is only suitable for constructing $\ell$-adic $L$-functions, and is only able to construct complex-analytic ones in the function field setting by a trick using the fact that the $L$-functions are polynomials.<|endoftext|> TITLE: Testing whether $e^x+ax^2+bx+c$ has a zero QUESTION [12 upvotes]: What is the simple test with exponential polynomials to determine whether $$f(x)=e^x+ax^2+bx+c$$ has a positive zero? This was prompted by the question about discriminants here. We have an ineffective result, a partial result, and a result for a broader class of algorithms. As an ineffective result: According to Wilkie's 1996 paper, there must be a list of exponential polynomials in $a,b,c$ whose signs will determine whether $f$ has a positive zero. However, that paper does not provide an effective algorithm. It might be possible to get a similar ineffective result more simply using compactness and some of the arguments below. As a partial result: If $a=0$, then $f$ has a zero iff one of the following holds: \begin{align} &b>0 \\ &b=0\ \ \&\ \ c<0 \\ &b<0\ \ \&\ \ b + e^{1-c/b}<0 \end{align} The last of these comes from using calculus to find the minimum of $f$, evaluating $f$ there, and replacing the logs in the expression with exponentials. As a result with a broader class of algorithms: We can test this for any algebraic $a,b,c$. Check for each positive rational $q$ if $$e^q+aq^2+bq+c<0$$ and check for each natural $n$ if $$\forall x(x>0\implies ax^2+bx+c+\sum^n_{k=0}x^k/k!\ge0)$$ We can test the first piece using the proof of the transcendence of $e$ (like this answer); if $f$ goes negative we will eventually verify that. We can test the second piece using Tarski-Seidenberg quantifier elimination; if $f$ is always positive we will eventually verify that. The third possibility, that $f$ is tangent to the $x$-axis, would imply that $f$ and $f'$ have a simultaneous zero, and we can rule that out by the Lindemann-Weierstrass theorem. However, this is an algorithm without an obvious time bound. And even if we convert each of the subtests above into a test via exponential polynomials, this doesn't give a single set of exponential polynomials which can be used for all $a,b,c$. What is the single set of exponential polynomials in $a,b,c$ whose signs determine if $f$ has a positive zero? REPLY [5 votes]: First let's consider a simpler problem on testing whether $g(x) := e^x + ux + v$ has a zero in the given interval $(L,U]$. It does when one of the following cases takes place: $g(L)<0$ and $g(U)\geq 0$; $g(L)>0$ and $g(U)\leq 0$; $g(L)\geq 0$ and $g(U)\geq 0$ and $u<0$ and $L<\log(-u)\leq U$ and $u(\log(-u)-1)+v \leq 0$. Let $\text{TEST}(u,v,L,U)$ denotes testing these conditions and giving True (pass) / False (no pass) answer. To be on a safe side, we also let it return False when $L\geq U$ (i.e., when the given interval is empty). Back to the original problem, we consider the case $a\ne 0$. If $c<-1$, then $f(0)=1+c<0$, so the answer is Yes. If $c\geq -1$ and $\neg \text{TEST}(2a,b,0,+\infty)$, then $f’$ has no positive zeroes, so $f$ must be increasing for $x\ge 0$, so the answer is No. If $c\geq -1$ and $\text{TEST}(2a,b,0,+\infty)$ then let $z$ be a positive zero of $f’$. It remains to check whether $f(z)\leq 0$. Since $0=f'(z)=e^{z} + 2az + b$, we have $$f(z) = -(2az+b) + az^2 + bz + c = az^2 + (b-2a)z + (c-b).$$ Let $D:=4a^2 + b^2 - 4ac$ be the discriminant of this quadratic, and let $s := \frac{2a-b-\sqrt{D}}{2a}$ and $t:= \frac{2a-b+\sqrt{D}}{2a}$ be its roots. We use these roots to teat whether $f’$ has a zero in the interval(s) where $z$ is positive and this quadratic is negative; if so, that root is $z$ and therefore $f(z)$ is negative, leading to a Yes. Now there are four subcases: If $D<0$ and $a<0$, then the quadratic is always negative, so the answer is Yes. If $D<0$ and $a>0$, then the quadratic is always positive, so the answer is No. If $D\ge 0$ and $a<0$, then let $ s_0:=\max(s,0)$. The answer is Yes iff $\text{TEST}(2a,b,0,t)$ or ($s_0>0$ and $f'(s_0)=0$) or $\text{TEST}(2a,b,s_0,+\infty)$. If $D\geq 0$ and $a>0$, then let $s_0:=\max(s,0)$. The answer is Yes iff ($s_0>0$ and $f'(s_0)=0$) or $\text{TEST}(2a,b,s_0,t)$. Remark. If numbers $a,b,c$ are algebraic (in particular, rational), then $f'(x)$ cannot have positive algebraic zeros, and thus conditions ($s_0>0$ and $f'(s_0)=0$) never hold and can be safely removed from the last two cases.<|endoftext|> TITLE: Units of group algebra of dihedral group QUESTION [8 upvotes]: Question: Can we fully describe the group of units (=invertible elements) $(KG)^\times$ of the group algebra $KG$ for $K=\mathbf{F}_2$, $G=D_\infty=\langle s,t|s^2=t^2=1\rangle$, the infinite dihedral group? I'm also interested in other fields $K$ (in which case one can focus on describing the quotient by its central subgroup $K^*$). If $A=G_{\mathrm{ab}}$ is the Klein group, I'd already be happy in a description of the kernel of $(KG)^\times \to (KA)^\times$, which has finite index (for a finite field $K$). This question is motivated by this question concerning a more complicated (but torsion-free) virtually abelian group. I'd also be interested in the same question replacing $G$ with the Klein bottle group, which is a non-abelian semidirect product $\mathbf{Z}\rtimes\mathbf{Z}$. Still, in a sense I'd like to take advantage of the semidirect product decomposition $\mathbf{Z}^d\rtimes$(finite). Note that when $G$ is torsion-free abelian, the group of units in $KG$ is reduced to $K^\times\times G$. "Fully describe" is a bit unclear a priori, but I'd like a characterization of its elements within the group ring, if possible for which we can deduce the answer about the simplest natural questions about its structure (is it virtually abelian? solvable? etc.) REPLY [13 votes]: A description of the group of units of the group algebra $\mathbb{F}_2 D_\infty$ can be found in Theorem 4.1 of the following paper: M. Mirowicz: Units in group rings of the infinite dihedral group, Can. Math. Bull. 34 (1991), 83-89 DOI link (the paper is accessible with no restriction at the moment)<|endoftext|> TITLE: Why is the regulator of the degree 11 extension of $\mathbb{Q}$ so large? QUESTION [22 upvotes]: Let $K_\infty/\mathbb{Q}$ denote the $\hat{\mathbb{Z}}$-extension of $\mathbb{Q}$. Then for each $n\geq1$, $K_\infty$ has a unique subfield $K_n$ of degree $n$ over $\mathbb{Q}$. The fields $K_n$ are rather nice: $K_n$ is Galois over $\mathbb{Q}$, and $\mathrm{Gal}(K_n/\mathbb{Q})$ is cyclic of order $n$ (so Kronecker-Weber applies), The subfields of $K_n$ are exactly $K_m$ for $m\mid n$, $K_n$ is totally real, It is conjectured that $K_{p^j}$ always has class number 1 (see this answer). I was looking through these fields in the LMFDB (links at the end of this question), and I noticed something peculiar: $$\begin{array}{l|l} n&\lim_{s\to1^+}(s-1)\zeta_{K_n}(s)\\\hline 1&1\\ 2&0.623225240141\\ 3&0.3774610891760\\ 4&0.431652451862\\ 5&0.411382062671\\ 6&0.26281913742\\ 7&0.29967691809\\ 8&0.3399574212\\ 9&0.231073814089\\ 10&0.59812750863\\ 11&\mathbf{\color{red}{10.9924154253}}\\ 12&0.272503409905\\ 13&0.480494415585\\ 14&0.303247583959919\\ 15&0.126218760817\\ 16&0.271689000529\\ 17&0.149703203263054\\ 18&0.121386452739\\ \end{array}\qquad\qquad\qquad\begin{array}{l|l} n&\lim_{s\to1^+}(s-1)\zeta_{K_n}(s)\\\hline 19&0.127769546503240\\ 20&0.148293571484876\\ 21&0.325798468160489\\ 22&0.481687105754706\\ 23&0.123796267561194\\ 24&0.139575702501298\\ 25&0.167117342071058\\ 26&0.522724794223398\\ 27&0.174105094869797\\ 28&0.448262864859088\\ 29&0.239734187517100\\ 30&0.106243397680971\\ 31&\\ 32&0.278099976864148\\ 33&\mathbf{\color{red}{10.8513191982336}}\\ 34&\\ 35&\\ 36&0.121939637724450 \end{array}$$ For these values of $n$, the limit always lies in the interval $[0.1,1]$, except for $n=11$ and $n=33$ (but not $n=22$), where the limit is over $10$ times larger! Why is the residue of the pole of $\zeta_{K_n}(s)$ so large for $n=11$? (and also $n=33$, but not $n=22$) Here are two potential ways of approaching the question: If $p$ is an odd prime then $\Delta_{K_p}=p^{2(p-1)}$ and conjecturally $h_{K_p}=1$ so we have $$\lim_{s\to1^+}(s-1)\zeta_{K_p}(s)=(2/p)^{p-1}\mathrm{Reg}_{K_p}.$$ Thus, the question could be phrased as: Why is the regulator of $K_{11}$ so large? We have $\lim_{s\to1^+}(s-1)\zeta_{K_n}(s)=\prod_{\chi\neq1}L(1,\chi)$. Here are pictures of $L(1,\chi)$ for $n=7,11,13$. Of course, these pictures don't really explain anything by themselves. $p=7$ (mostly in the unit circle so the product is small, $\approx0.3$) $p=11$ (almost entirely outside of the unit circle so the product is large, $\approx11$) $p=13$ (mostly in the unit circle so the product is small, $\approx0.5$) Here are the LMFDB links: $K_1$ $K_2$ $K_3$ $K_4$ $K_5$ $K_6$ $K_7$ $K_8$ $K_9$ $K_{10}$ $K_{11}$ $K_{12}$ $K_{13}$ $K_{14}$ $K_{15}$ $K_{16}$ $K_{17}$ $K_{18}$ $K_{19}$ $K_{20}$ $K_{21}$ $K_{22}$ $K_{23}$ $K_{24}$ $K_{25}$ $K_{26}$ $K_{27}$ $K_{28}$ $K_{29}$ $K_{30}$ $K_{31}$ $K_{32}$ $K_{33}$ $K_{34}$ $K_{35}$ $K_{36}$ REPLY [19 votes]: A prime $\ell$ splits in $K_p$ for $p$ odd if and only if $$\ell^{p-1} \equiv 1 \mod p^2.$$ We can express $\lim_{s \to 1} (s-1) \zeta_{K_n(s)}$ as the product of $\frac{1}{ 1- |\mathfrak a|^{-1} }$ over primes $\mathfrak a$ of $K_n$, regualarized by dividing by the product of $\frac{1}{1-\ell}$ over primes $\ell$ of $\mathbb Q$. From this perspective, large values of $\lim_{s \to 1} (s-1) \zeta_{K_p(s)}$ are "explained" by the existence of small primes that split in $K_p$, which give a large contribution to this product. For fixed $\ell$, the $p$ for which $\ell$ splits in $K_p$ are the generalized Wieferich primes. Examining that table, we see that for $\ell=3$, $11$ is the smallest Wieferich prime. It is exceptionally small. In particular, if we ignore $\ell> p$ as that is the range where we start expecting to have a split prime for random reasons, then the next-smallest Wiefierch prime for any $\ell<11$ is $p=1093$ for $\ell=2$. The contribution of the primes lying over $3$ to the zeta function is $(3/2)^{11}\approx 86.5$. If we divide by the Euler factors for primes $<11$ of the zeta function of $\mathbb Q$, or $ 2 \cdot (3/2) \cdot (5/4) \cdot (7/6) $, we get $19.8$, which is on the same order of magnitude as your limit. Thus, this exceptional split prime gives an explanation for the unusually large value. Examining the table linked above, we might suspect that exceptionally large values might occur for $K_p$ when $p=43, 71, 137 863, 1093, 3511, \dots $ owing to the split primes $\ell=19, 11,19, 13, 2,2,\dots$ respectively. The distinct behavior of $K_{22}$ and $K_{33}$ is probably related to what happens to these small primes in those two extensions. In $K_{22}$ they are inert, taking the local factor of the zeta function lying over $3$ from $(3/2)^{11}$ to $(9/8)^{11}$, a loss of a factor of $\approx 23.7$ (matching the ratio of the residues in your table), while in $K_{33}$ they are ramified, preserving their norms, and preserving the local factor $(3/2)^{11}$ (explaining why the residue barely changes).<|endoftext|> TITLE: Integer valued polynomials and polynomials with integer coefficients QUESTION [15 upvotes]: It is well known that the subring $S$ of integer valued polynomials ${\mathbb Q}[x]$ is generated by the binomial functions $P_n={x \choose n}$. One can ask a dual question: how to characterize the polynomial functions ${\mathbb Z} \to {\mathbb Z}$ which come from an element in ${\mathbb Z}[x]$. I understand one can write down derivative as a (terminating) series of difference derivatives and thus express each coefficients in terms of values of the polynomial but does this (or another) procedure lead to a neat answer? There is a necessary condition for a polynomial function $f:{\mathbb Z} \to {\mathbb Z}$ to come from an element in ${\mathbb Z}[x]$, namely for every $n$ the residue of $f(x) \mod n$ depends only on the residue of $x \mod n$. This necessary condition is not sufficient but I am interested in the subring of elements in $S$ satisfying that necessary condition. Is there a nice set of generators and/or a basis? REPLY [14 votes]: In the second paragraph of the post you ask about those integer-valued polynomials $f(x)$ for which $\frac{f(x)-f(y)}{x-y}$ is an integer for integers $x\ne y$. The basis in this $\mathbb{Z}$-module $M$ may be described as follows. For each $n=1,2,\ldots$ define the minimal positive rational $\alpha_n$ such that $\alpha_nx(x-1)\ldots(x-n+1)\in M$. I claim that $\alpha_n={\rm lcm}(1,2,\ldots,n)/n!$ The polynomials $1,\alpha_1x,\alpha_2x(x-1),\ldots$ form a basis of $M$. You may read the proof here (Problem 4, solution 1). Now about characterizing those polynomials which have integer coefficients via the values. This may be done if we generalize the necessary condition $$\frac{f(x)-f(y)}{x-y}\in \mathbb{Z}\quad\quad\quad\quad\quad\quad\quad\,\,\,\,\,\,\,(1)$$ as follows: $$ \sum_{i=0}^k \frac{f(x_i)}{\prod_{j:j\ne i} (x_i-x_j)}\in \mathbb{Z} \quad\quad\quad\quad\quad(2) $$ for all distinct integers $x_0,x_1,\ldots,x_k$ ((2) reduces to (1) when $k=1$). By Lagrange interpolation formula, the expression in (2) is the coefficient of $x^k$ in the polynomial $g(x)$ which has degree at most $k$ and takes the same values as $f$ at points $x_0,\ldots,x_k$. In other words, $g$ is the remainder of $f$ modulo $(x-x_0)(x-x_1)\ldots (x-x_k)$. So, if $f(x)\in \mathbb{Z}[x]$, we have $g(x)\in \mathbb{Z}[x]$ and in particular (2) holds. On the other hand, if (2) holds for $k=\deg f+1$, we see that the leading coefficient of $f$ is integer, so we may subtract the leading term from $f$ and induct, proving that $f(x)\in \mathbb{Z}[x]$.<|endoftext|> TITLE: Smallest $3$-regular graph with a unique perfect matching QUESTION [5 upvotes]: What is the smallest 3-regular graph to have a unique perfect matching? With a large enough number of nodes, it is possible for a 3-regular graph to have no perfect matching (example can be seen in this question Cubic graphs without a perfect matching and a vertex incident to three bridges ). So I believe 3-regular graphs with a unique matching likely exists, but I am unsure how to go about constructing and proving what the smallest one is. Likely there is no better answer than to brute force check all the possibilities, so I am hoping someone happens to know what this graph looks like. Even better: Does anyone know of an online searchable graph database that allows searching for small graphs with certain properties? REPLY [11 votes]: There is no such graph. I have some reading to do as my intuition is off, but the details and a related question are available here: Does there exist an r-regular graph (r≥2) with a unique maximum matching? Akbari, Ghodrati, Hosseinzadeh (2017), On the structure of graphs having a unique k-factor, Aust. J. Combin. (pdf) show: ... we prove that there is no r-regular graph (r≥2) with a unique perfect matching. REPLY [4 votes]: Regarding your second even better question, I warmly suggest the Brendan McKay page on combinatorial objects, that gives many kinds of graph examples.<|endoftext|> TITLE: The surface area measure in terms of support functions QUESTION [7 upvotes]: $\def\RR{\mathbb{R}}$Let $K$ be a closed bounded convex body in $\RR^n$. The support function $h_K$ on $\RR^n$ is defined by $$h_K(v) = \max_{w \in K} \langle v,w \rangle.$$ Let $S^{n-1}$ be the unit sphere. The surface area measure is the measure on $\sigma$ such that, for an open set $U$ in $S^{n-1}$, the measure $\sigma(U)$ is the $(n-1)$-dimensional Lebesgue measure of the set of $w \in \partial K$ where at which supporting hyperplanes normal to $v$ make contact. (I believe this is fairly standard terminology; I am reading Schneider's "Convex Bodies The Brunn-Minkowski Theory" as my reference.) Now, if $K$ is smooth, then $\sigma$ is a smooth $(n-1)$-form, and we can compute $\tfrac{\sigma}{\mathrm{Area}}$ in terms of the Hessian of $h$. Namely, restrict $h$ to the affine hyperplane $v+v^{\perp}$. (In other words, the tangent plane to $S^{n-1}$ at $v$.) Then $\tfrac{\sigma}{\mathrm{Area}}$ is the determinant of the Hessian of this restricted function. On the other hand, suppose that $K$ is a polytope, so $h$ is piecewise linear. Then $\sigma$ is an atomic measure, concentrated on the normals to the facets of $K$. Then we can also compute $\sigma$ in terms of $h$ restricted to $v+v^{\perp}$: For $u$ in $v^{\perp}$, take the directional derivative $\tilde{h}(u) := \lim_{t \to 0^+} \tfrac{h(v+tu)-h(v)}{t}$. Then $\tilde{h}(u)$ is (I believe) the support function of the facet normal to $v$, and $\sigma(v)$ is the volume of that facet. We can recover the facet, and hence its volume, as the dual of $\tilde{h}$. What I am trying to understand is how to interpolate between these formulas. In general, is $\sigma$ something in terms of the restriction of $h$ to $v+v^{\perp}$. And why am I seeing second derivatives in the smooth case and first (directional) derivatives in the polytopal case? REPLY [5 votes]: I came to convex geometry from differential geometry and had many similar questions, too. One way to view surface area measure is the following: The normal to a supporting hyperplane is essentially the Gauss map from $\partial K$ to $S^{n-1}$. This is a multi-valued map that is differentiable whenever it is single-valued. The points where it is multi-valued has measure zero on $\partial K$. The surface area measure is the measure $dA$ on $\partial K$ pushed forward to $S^{n-1}$ by the Gauss map. At a point on $\partial K$, where the Gauss map has invertible Jacobian (i.e., $\partial K$ has positive Gauss curvature), the change of measure is given in terms of the the determinant of the Jacobian, which is Gauss curvature. In particular, Gauss curvature $\kappa$ satisfies $du = \kappa\,dA$. Therefore, if you pull back $dA$ by the inverse Gauss map (which is well-defined here), you get $dA = f\,du$, where $f$ is what convex geometers call the curvature function and is the reciprocal of the Gauss curvature as a function of the unit normal. On the other hand, the Gauss map is constant along a flat face, So if you push forward $dA$, you get an atomic measure at the normal to the face with mass equal to the area of the face. As for how all this is connected to the support function, the observation is that, wherever $h$ is differentiable, its gradient is the inverse Gauss map. So the Jacobian of the inverse Gauss map is the Hessian of $h$. It's also worth noting that the support function of the polar body is the reciprocal of what is called the radial function. And the gradient of the polar support function, if normalized appropriately, is really the Gauss map itself. What's cool about all of this is that, even though the above mentions the unit sphere and the Gauss map, all of these concepts do not require the Euclidean structure of $\mathbb{R}^n$. The only structures needed are the vector space structure and Lebesgue measure. A lot of what's in Schneider's book is invariant or equivariant under the action of $SL(n)$. But, at least for me, this is hard to see using the standard approach. So I worked out my own way of doing things. The first thing I always do is view the body $K$ as a subset of an abstract vector space $V$, the polar body is then in $V^*$, and the domain of the support function is $V^*$. Figuring out how to describe surface area measure in this setting took a bit more work. I developed my own formulation of all of this, avoiding any use of the dot product or the unit sphere, and wrote it up in a survey paper, Affine Integral Geometry from a Differentiable Viewpoint.<|endoftext|> TITLE: Do there exist positive definite matrices $A$ and $B$ satisfying this condition? QUESTION [5 upvotes]: Denote by $\mathbb{P}_n$ be the set of real symmetric positive definite $n \times n$ matrices. In $\mathbb{P}_n$, we define an partial order as follow: for $X, Y \in \mathbb{P}_n$, we say $X \prec Y$ if $Y-X$ is also a positive definite matrix. Let $\lambda = 2^{1-1/p}$ with $p>1$ be a real number. Question: For $0\prec X \prec Y \prec \lambda X$, do there exist positive definite matrices $A, B$ such that $$X = \dfrac{A+B}{2}, \qquad Y = \left(\dfrac{A^p+B^p}{2}\right)^{1/p}?$$ Remark: The answer in the case of a scalar is Yes, and $\lambda = 2^{1-1/p}$ comes from this case to ensure that the function $$f(a) = \left(\dfrac{a^p + (2x-a)^p}{2}\right)^{1/p}$$ is surjective from $[0,2x]$ to $[x, \lambda x]$. REPLY [10 votes]: Edit: The original answer only dealt with $p=2$, now the case $1 TITLE: Filling square to push-out in abelian category QUESTION [8 upvotes]: Let $\mathcal{C}$ be an abelian category. In $\mathcal{C}$ we consider the diagram \begin{array}{ccc} A&&\\\ \downarrow&&\\\ C&\rightarrow&D \end{array} with arrows being monomorphism. Is it possible to say when there is a $B$ such that we have a push-out square \begin{array}{ccc} A&\rightarrow&B\\\ \downarrow&&\downarrow\\\ C&\rightarrow&D \end{array} and if possible how to get $B$ (up to isomorphism)? REPLY [11 votes]: If there is such a pushout, then $B\to D$ is also a monomorphism, i.e. $B$ is a subobject of $D$. Phrased more concretely, you're asking when there is a subobject $B$ of $D$ such that $B+C = D$ and $B\cap C = A$. Let us mod out $D$ by $A$: we get a monomorphism $C/A\to D/A$, and a pushout square $\require{AMScd}\begin{CD} C @>>> D \\ @VVV @VVV \\ C/A @>>> D/A\end{CD}$ which we can glue to the previous one to get a pushout square $\require{AMScd}\begin{CD} A @>>> B \\ @VVV @VVV \\ C/A @>>> D/A\end{CD}$ This means that $D/A\cong C/A\oplus B/A$ Conversely, assume $C/A \to D/A$ has a summand $X$, and consider its pullback $B$ to $D$. Then clearly $B\cap C = A$ and $B+C = D$. So there is such a $B$ if and only if $C/A$ has a summand in $D/A$; and in fact if you're more careful with the above constructions, it seems like $\{$ subobjects $B\to D$ realizing such a pushout $\}\to \{$ summands of $C/A$ in $D/A \}$ is a bijection, with inverse given by the pullback; this can be seen as a relativization of the claim that subjects of $D$ containing $A$ are the same thing as subobjects of $D/A$.<|endoftext|> TITLE: Ramanujan and his influence on others QUESTION [5 upvotes]: A few years ago I saw a paper where a few important researchers were asked which theorem of Ramanujan impressed them most. I don't remember details, but I would like to see this paper again. Details, please! REPLY [6 votes]: I believe the request of the OP matches closely with the article Your Hit Parade: The Top Ten Most Fascinating Formulas in Ramanujan's Lost Notebook by George Andrews and Bruce Berndt in the January 2008 Notices of the AMS, which involved taking a survey of 34 "renowned experts" to appropriately rank the formulas.<|endoftext|> TITLE: Lower eigenvectors of nonnegative matrices with zero trace QUESTION [9 upvotes]: Let $A$ be an $N\times N$ nonnegative matrix with all diagonal entries equal to zero and such that there is $n_0$ such that all entries of $A^{n_0}$ are strictly positive. Let $\lambda_1,\ldots, \lambda_N$ be its eigenvalues ordered in the decreasing order with respect to their real parts, and $v_1,\ldots, v_N$ be the corresponding (left) eigenvectors. Perron and Frobenius tell us that $\lambda_1$ is a strictly positive real number and therefore (since the sum of eigenvalues must be zero) there must also be eigenvalues with strictly negative real part; let $\lambda_{k_0},\ldots, \lambda_N$ be those. Questions: (1) is it true that the "smallest" (with respect to the real part of the corresponding eigenvalue) eigenvector $v_N$ can be chosen in such a way that all of its entries are nonzero? (2) if the above doesn't hold, is it at least true that for any $j\in \{1,\ldots,N\}$ we can find $m\geq k_0$ such that $v_m$ has nonzero $j$th component (that is, the set of eigenvectors corresponding to eigenvalues with negative real part cannot have a common all-zero entry index)? REPLY [3 votes]: Partial answer: For the special case of self-adjoint matrices, the answer to (2) is yes. Funnily enough, this has nothing to do with the non-negativity of the matrix: Proposition. Let $A \not= 0$ be a self-adjoint complex $N \times N$ matrix with all diagonal entries equal to $0$, let $\lambda_1 \ge \dots \ge \lambda_N \in \mathbb{R}$ be its eigenvalues, and let $v_1, \dots, v_n \in \mathbb{C}^N$ be an orthonormal basis of eigenvectors. Assume that every entry of $v_1$ is non-zero. Then for each $j \in \{1,\dots,N\}$ there exists an eigenvector for a negative eigenvalue whose $j$-the component is non-zero. Proof. Fix $j$ and write the $j$-th canonical unit vector $e_j$ as $$ e_j = \sum_{k=0}^N \alpha_k v_k, $$ where $\alpha_k = \langle v_k, e_j \rangle$ for each $k$ (here, I use the "physical" convention that the inner product be linear in the second component). We have $$ 0 = \langle e_j, A e_j \rangle = \sum_{k,\,\ell=1}^N \overline{\alpha_k} \alpha_\ell \langle v_k, Av_\ell \rangle = \sum_{k=1}^N \lvert \alpha_k \rvert^2 \lambda_k. $$ Since $\lambda_1 > 0$ and $\alpha_1 \not= 0$ by assumption, it follows that there exists $k_0$ such that $\lambda_{k_0} < 0$ and $\alpha_{k_0} \not= 0$. $\square$ Remark. The assumption that every entry of $v_1$ be non-zero can be weakened; without this assumption, the following is still true: Fix $j$. If there exists an eigenvector for a positive eigenvalue whose $j$-th component is non-zero, then there also exists an eigenvector for a negative eigenvalue whose $j$-th component is non-zero. The proof is the same.<|endoftext|> TITLE: Using algebraic geometry to understand class field theory QUESTION [33 upvotes]: In Algebraic Number Theory, S. Lang says "[a geometrical approach] allows one to have a much clearer insight into the whole class field theory, since the existence theorem and the reciprocity law become obvious once the machinery of algebraic geometry is available." Inspired by this, I wonder if there is some (preferably modern) reference for class field theory using algebraic geometry. REPLY [3 votes]: I would like to point out a letter from A Grothendieck to L Breen [pdf] In which he explains some details about it. See for example page 45<|endoftext|> TITLE: Various authors of the Bourbaki's books QUESTION [13 upvotes]: As far as I understand, each chapter of the Bourbaki's collection was written by one (or two?) specific authors. The book itself was reviewed, corrected and after all approved by the whole Bourbaki assembly (if I may call them this way). But I have the feeling one (or two) specific person was in charge of the writting of any given chapter. In my opinion, the style and the clarity of the exposition really varies from one book to the other. I personally found the Lie Algebras book much more readable than the Algebra one. I haven't read any other volume of the Bourbaki's series, so I don't have an opinion on the other chapters. In any case, I'd like to know if one has any idea (or convictions) on the identity of the authors of each volume. Let me start filling the boxes where I am almost convinced I am correct: $\bullet$ Algebra (at least chapter 4 to 7?) : Adrien Douady (may be wrong as he is said to have joined after the beginning of the writting of Algebra 4-7)? $\bullet$ Espaces vectoriels topologiques (the whole book?) : Alexandre Grothendieck. Possible guess following suggestions in the comments: $\bullet$ Lie Algebras : François Bruhat (in connection with his co-author Jacques Tits)? $\bullet$ Functions of a real variable : Jean Delsartes? Any guess for the other chapters/books? EDIT : Following @abx remark (who actually was one of the later Bourbaki's member, so he can be trusted on that point), one should rather ask for the author of the preliminary draft of each chapter. This, in any case, is what I was looking for. I am definitely not interested in the various identities of the members who polished the initial version of the draft. EDIT BIS : Matthieu Latapy suggests a very interesting approach using comparison of n-gram distributions between the Bourbaki's chapters and some contemporary papers of the suspected authors. I must confess I don't have any idea on how to implement that effectively. But I'd very interested to know if it can be used to match François Bruhat with some chapters in Lie-Algebras, Adrien Douady with some chapters in Algebra and Commutative Algebra with Samuel/Chevalley/Cartier. REPLY [32 votes]: I think the whole idea is wrong. As you can see from the Bourbaki archives, Bourbaki worked as follows: one member would write a rough first draft, this would be read in the next meeting, criticized, heavily modified, then another member would write a new version, and so on. It is impossible to attribute any of the books to one author. By the way, all guesses in the post or in the comments are wrong: Algebra 4-7 was started much before Douady joined Bourbaki, Tits was never a member, etc.<|endoftext|> TITLE: Morphisms of Hochschild (or cyclic) homology induced by homotopic maps QUESTION [5 upvotes]: Let $f$ and $g$ be two maps between DG algebras $A$ and $B$, and assume that $f$ and $g$ are homotopic as chain maps, hence they induce the same map on the level of homology. Moreover, $f$ and $g$ induce morphisms between $\mathrm{HH}_\bullet(A)$ and $\mathrm{HH}_\bullet(B)$ and $\mathrm{HC}_\bullet(A)$ and $\mathrm{HC}_\bullet(B)$. What can we say about those morphisms between Hochschild or cyclic homology? Everything is over a field of characteristic zero. Any suggestions and helps would be really appreciated. REPLY [5 votes]: $\newcommand{\dd}{\mathrm d}\DeclareMathOperator{\Sym}{Sym}\DeclareMathOperator{\id}{id}$If $f,g:A\to B$ are dga morphisms which are homotopic as chain maps, the maps induced by $f$ and $g$ on Hochschild homology and its variants can be different. I give an example above the line; the easiest way to construct it, at least for me, is to systematically investigate the (homotopy) functoriality of Hochschild homology, which I sketch below the line. Let $A = \mathbb k[\{x_k\}_{k\ge 1}]$ be the free nonunital graded algebra on generators $x_k$ in (cohomological) degree $1$, equipped with the differential $\dd x_k = \sum_{i+j=k} x_ix_j$. In other words, $A = \Omega(B(\mathbb k[x]/x^2))$ is the cobar-bar resolution of the free nonunital graded commutative algebra on one generator in degree $1$; in particular there are chain maps \begin{align*} r: A &\leftrightarrows H_*(A)\cong \mathbb k[x]/x^2:i\\ x_k &\mapsto \begin{cases}x&\text{ for }k=1\\ 0&\text{ else}\end{cases}\\ x_1 &\leftarrow\!\shortmid x \end{align*} such that $r$ is a dga morphism, $r\circ i = \id_{H_*(A)}$ and there is a chain homotopy $h$ between $\id_A$ and $i\circ r$. Let $f:A\to H_*(A)$ be the zero map, and let $g:A\to H_*(A)$ be the (nonunital) algebra map with $g(x_k) = \begin{cases}x&\text{ for }k = 2\\0&\text{ else}\end{cases}$. Both are chain maps since the target is concentrated in degree $1$, and since $g\circ i = 0$, the map $g\circ h$ is a chain homotopy between $f$ and $g$. Let $\alpha = x_1\otimes x_1\in C_1(A)$. Since $x$ is in odd degree, the Koszul sign rule gives $\dd\alpha = 2x_1^2$, so that $\beta = x_1\otimes x_1 - 2x_2\in C_1(A)$ is closed. Its image under $f$ vanishes, while its image under $g$ gives the cycle $2x\in HH_1(H_*(A))$, which is nonzero since it maps to the nonzero element $2x$ under the trace map $HH_1(H_*(A))\to H_*(A)$ arising from the commutativity of $H_*(A)$. Thus $f$ and $g$ do not induce the same map on Hochschild homology. Hochschild homology is a functor for $A_\infty$-morphisms, i.e. a dga morphism $\Omega(B(A))\to B$ (explicitly, a collection of maps $f_n:A^{\otimes n}\to B[1-n]$ satisfying certain quadratic equations) induces a chain map $HH_*(A)\to HH_*(B)$. Said differently, they are Maurer-Cartan elements of the dgla $\operatorname{Hom}(B(A),B)$ (the Maurer-Cartan equation is exactly the quadratic equation satisfied by the components $f_n$). Now for any (nilpotent) dgla $\mathfrak g$, the set of Maurer-Cartan elements form the $0$-simplicies of a Kan complex/homotopy type with $n$-simplices given by Maurer-Cartan elements of $\mathfrak g\otimes\Omega_{PL}^*(\Delta^n)$, and we can actually replace the cdga $\Omega_{PL}^*(\Delta^n)$ by a smaller $C_\infty$-algebra, compare Robert-Nicoud--Vallette's Higher Lie Theory. In particular, we now have a straightforward definition of a homotopy between two $A_\infty$-morphisms: It is essentially a family of $A_\infty$-morphisms $f_t$, parametrized by $t\in [0,1]$, together with a "coherent nullhomotopy" of $\partial_t f_t$, such that $f_0 = f,f_1 = g$. Very explicitly, given the two morphisms $f,g$, it is a collection of maps $h_n:A^{\otimes n}\to B[-n]$ which satisfies a quadratic identity involving the components of $f$ and $g$, which for small $n$ and strict morphisms $f,g$ (i.e. $f_n = g_n = 0$ for $n>1$) becomes $h_1$ is a chain homotopy between $f_1$ and $g_1$ $h_2$ is a chain homotopy between $\mu_B\circ(f_1\otimes h_1 + h_1\otimes g_1)$ and $h\circ \mu_A$ $h_3$ is a chain nullhomotopy of a map $A^3\to B$ of degree $-2$, defined in terms of the maps $f_1,g_1,h_1,h_2$ Explicit formulas and a generalization to higher homotopies can be found in Mazuir's Higher algebra of A∞ and ΩBAs-algebras in Morse theory II. Your assumption that $f$ and $g$ are chain homotopic is the first step in this series of equations; each subsequent step asks you to find a nullhomotopy for a chain map defined in terms of the previous choices. In particular, this is impossible if this chain map is not nullhomotopic. This makes it easy to find an example of such a map; it is slightly more work to find one such that the induced maps on Hochschild homology are different: Let $V$ be a graded vector space and $A_V = \Omega((\Sym^* V[1])[-1])$ be the Koszul resolution of the free graded commutative algebra $\Sym^*V$, i.e. $A_V$ is the free algebra on $\Sym^* V[1]$ with the differential on generators defined via the coalgebra structure. By the HKR theorem, we have $HH_*(A_V)\cong HH_*(\Sym^*V)\cong \Sym^* (V\oplus V[1])$. The Hochschild cohomology of $A_V$ is quasiisomorphic to $\Sym^* (V\oplus V^\vee[-1])$; essentially, it gives infinitesimal deformations of the identity as a $A_\infty$-automorphism of $\Sym^* V$. In particular, any polyvector field $X\in\Sym^* V\otimes\Sym^{\ge 2}V^\vee[-1]$ of cohomological degree $0$ integrates to a $A_\infty$-automorphism of $\Sym^* V$ whose linear component is the identity, which can then be lifted to a dga automorphism of $A_V$ whose underlying chain map is homotopic to the identity. Its action on Hochschild homology is (the exponentiation of) the "Lie derivative" action of polyvector fields on differential forms, and thus usually not equal to the identity map. The example above essentially arises in this way, except for the fact that I took a deformation of the zero map. The polyvector field is $\partial_x\wedge\partial_x$, which defines an $A_\infty$-morphism from $H_*(A)$ to itself with nonvanishing degree $2$ component, and the differential form is $x\dd x$. My impression is that the explicit computation does not do much to explain why the example works. If you are interested in the functoriality properties of Hochschild and cyclic homology from a more abstract point of view, I can recommend Krause--Nikolaus's lecture series on topological cyclic homology.<|endoftext|> TITLE: Why are these sets divisible by n? QUESTION [5 upvotes]: Suppose we have a polynomial $z \to f_c(z)$ defined over $\mathbb Z$ with a free parameter $c$, for instance $z \to z^2 + c$ and we consider the iterates $z \to f_c^{(n)}(z)$ and define the polynomials $g_n(c) = f_c^{(n)}(0)$. That is, the roots of $g_n$ correspond to those $c$ for which $0$ has a period of size dividing $n$. For example, with $f_c(z) = z^2 + c$: \begin{gather*} g_2(c) = c^2 + c \\ g_3(c) = (c^2 + c)^2 + c \end{gather*} and so on. We also define $h_n(x) \mid g_n(x)$ to be the polynomial with roots corresponding to $c$ so that $0$ has period exactly $n$. Then: \begin{gather*} h_2(c) = c+1 \\ h_3(c) = c^3 + 2c^2 + c + 1 \end{gather*} and so on. In particular, if $f(z)$ has degree $d$, then $g_n(z)$ has degree $d^{n-1}$ and by möbius inversion: $$\deg(h_n) = \sum_{m\mid n}\mu\left(\frac{n}{m}\right)d^{m-1}.$$ Now it turns out to be true that we have the following congruence: $$\sum_{m\mid n}\mu\left(\frac{n}{m}\right)d^m \equiv 0 \pmod{n}$$ and therefore, at least when $\gcd(d,n) = 1$, we have that $\deg(h_n)$ is divisible by $n$. Is there a natural way to partition the roots of $h_n$ (corresponding to $c$ so that $0$ has an orbit of period exactly $n$) into sets of size $n$ (or perhaps $n$ sets?)? I don't see a straightforward way of doing this, especially because $h_3(c) = c^3 + 2c^2 + c + 1$ in the example above turns out not to generate a Galois extension and so we cannot write all 3 roots as algebraic expressions in one of the roots. (I believe it is a conjecture that the $h_n$ are irreducible over $\mathbb Q$ at least for $f(z) = z^2 + c$. I would be very happy if someone could provide a reference for this conjecture and also perhaps what we expect to happen in general.) REPLY [2 votes]: As Asvin mentioned, the polynomials that you're looking at are often called Gleason polynomials, although sometimes people only require that the critical point 0 have finite orbit, rather that being periodic. (Sometimes those preperiodic versions are called "Misiurewicz polynomials".) I'm not sure who first asked (conjectured) that the Gleason polynomials are irreducible, but it was listed as a problem worth studying at an AIM workshop on Postcritically finite maps in complex and arithmetic dynamics: http://aimpl.org/finitedynamics/3/ As for what's true in general, it depends on what you mean. Do you want to restrict to 1-parameter families of polynomials, in which case you'll get analogous Gleason-polynomials that are polynomials in the parameter. Or you could take higher dimensional families and get multi-variable Gleason polynomials. All of these lead to interesting, and mostly open, questions. For recent preprints on irreduciblity of some Misiurewicz polynomials in a family of rational maps that aren't polynomials, see: Misiurewicz polynomials for rational maps with nontrivial automorphisms Misiurewicz polynomials for rational maps with nontrivial automorphisms II<|endoftext|> TITLE: Relation between $\xi$-cohomological and discrete series QUESTION [9 upvotes]: Sometimes, results on automorphic representations are available only under local assumptions. Typically, one could require the representation to be a $\xi$-cohomological cuspidal representation, and I would like to understand the meaning of it. Let $G$ be a connected reductive group over $\mathbb{Q}$. Let $\xi$ be an irreducible (finite dimensional) algebraic representation of $G$ over $\mathbb{C}$. An (isomorphism classes of) irreducible admissible representations $\pi_\infty$ of $G(\mathbb R)$ is $\xi$-cohomological if it has the same central character and infinitesimal character as $\xi$. I would like an automorphic/classical interpretation of this condition: is it related to being infinitesimally equivalent of a certain discrete series? or to being in a certain $L$-packet? The reason why it is termed cohomological condition also puzzles me (and this may be part of my lack of understanding). REPLY [6 votes]: This condition comes up because of $(\mathfrak{g}, K)$-cohomology, which is an extremely important invariant of automorphic representations. If $\xi$ is an algebraic rep, then $\xi$ defines a locally-constant sheaf on the locally-symmetric space $Y_G(U)$ for any open compact $U \subset G(\mathbf{A}_f)$. The Betti cohomology $H^\star_B(Y_G(U), L_\xi)$ can be computed in terms of automorphic forms. If $G$ is anisotropic over $\mathbf{Q}$ (so $Y_G(U)$ is compact, and all automorphic reps of $G$ are cuspidal) then Matsushima's formula says that this Betti cohomology is given by $$\bigoplus_\pi m(\pi) \cdot \pi_f^U \otimes H^*(\mathfrak{g}, K; \pi_\infty \otimes \xi^\vee),$$ where the sum is over cuspidal automorphic reps of $G$. There is a generalisation of this to arbitrary $G$ by Franke (but you have to be careful to isolate the "cuspidal part" of the cohomology). So the correct definition of "cohomological" should really be "$H^*(\mathfrak{g}, K; \pi_\infty \otimes \xi^\vee)$ is non-zero for some $\xi$"; these are the representations which contribute to the Betti cohomology of symmetric spaces. However, an obvious necessary condition for this is that $\pi_\infty$ have the same central and infinitesimal character as $\xi$, since otherwise the cohomology vanishes for trivial reasons. Discrete series representations are always cohomological. However, there are more cohomological representations which are not discrete series (fortunately, since lots of groups, e.g. GL(n) for n > 2, don't have any discrete series representations).<|endoftext|> TITLE: What is known about the discrete group cohomology $H^2(\mathrm{SL}_2(\mathbb C), \mathbb C^\times)$? QUESTION [5 upvotes]: The cohomology ring of $\mathrm{SL}_2(\mathbb C)$ as a topological group is straightforward (it's generated by a Chern class), but what is known in the discrete case? I'm particularly interested in $H^2$ with coefficients in $\mathbb C^\times = \mathbb C \setminus \{0\}$, especially if there are explicit formulas for the classes. REPLY [8 votes]: There is a paper by Milnor called On the homology of Lie groups made discrete which contains many results on this kind of problem, as well as references to related literature. However, I don't think that it directly answers your question.<|endoftext|> TITLE: Finite compact quantum groups QUESTION [7 upvotes]: Let $(A, \Delta)$ be a $C^*$-algebraic compact quantum group (in the sense of Woronowicz). It is called finite if $A$ is a finite-dimensional $C^*$-algebra. By elementary $C^*$-algebra theory, we known that $$A\cong M_{n_1}(\mathbb{C}) \oplus \dots \oplus M_{n_k}(\mathbb{C})$$ as $C^*$-algebras. If $X$ is a finite (⇒ compact) group, then we can consider $C(X)$ with comultiplication $$\Delta(f)(x,y) = f(xy).$$ Then $(C(X), \Delta)$ is a finite compact group. Further, we can consider $C^*(X)= C_r^*(X)= \mathbb{C}[X]$, the group algebra, with comultiplication uniquely determined by $$\Delta(x) = x \otimes x, \quad x \in X.$$ Question: What are other examples of finite compact quantum groups? Are the finite compact quantum groups completely classified? REPLY [8 votes]: Another example apart from the example of Kac & Paljutkin (reference below) are the quantum groups of Sekine: Y. Sekine, An example of finite-dimensional Kac algebras of Kac-Paljutkin type, Proc. Amer. Math. Soc. 124 (1996), no. 4, 1139-1147. The title confused many authors into thinking that these are generalisations of the finite quantum group of Kac & Paljutkin. In fact Kac & Paljutkin (whose paper I would recommend getting your hands on) set out a general framework to construct finite quantum groups. Their dimension eight example involves a particular choice (of matrix with complex entries), but while the quantum groups of Sekine follow the general framework, the particular choice is different (the matrix has real entries). It is worth commenting that the construction of Sekine holds at $k=2$ and $k=1$. The case of $k=2$ has been mistaken for the Kac Paljutkin quantum group but is in fact the dual of the dihedral group, $\widehat{D}_4$. The case $k=1$ gives $\mathbb{Z}_2$. If you Google Sekine quantum groups you will find some recent studies of this family of finite quantum groups related to random walks. I. Baraquin, Random Walks on Finite Quantum Groups, J. Theoret. Probab. 33 (2019), 1715-1736. U. Franz and A. Skalski, On Idempotent States on Quantum Groups, Journal of Algebra 322 (2009), no. 5, 1774-1802. H. Zhang, Idempotent states on Sekine quantum groups, Comm. Algebra 47:10 (2019), 4095--4113: For more finite quantum groups there are some constructions such as the tensor product which can make new finite quantum groups out of old. A nice example is to take the tensor product of quantum groups with commutative and cocommutative algebras of functions, for example $C(S_3)\otimes C(\widehat{S_3})$. While all commutative and cocommutative algebras of functions on finite quantum groups are quantum permutation groups, the referenced paper of Banica, Bichon, and Natale gives a finite quantum group which is not a quantum permutation group. Another place where you might find more examples is the paper of Banica and Bichon on the quantum symmetries of four points where all the quantum subgroups of $S_4^+$, the quantum permutation group on four symbols, are found: T Banica and J. Bichon, Quantum groups acting on 4 points, J. Reine Angew. Math. 626 (2009), 74-114. The papers of Masuoka referenced might give some more examples. We see the phenomenon of quasi-subgroups of finite quantum groups more commonly referred to as non-Haar idemptotents. These can be viewed as subsets of the state space that are closed under the "quantum group law", "contain the identity" and are "closed under inverses". However they are not quantum groups and this can be understood on the level of measurement: there is a projection $p$ in the algebra of functions such that measurement of a state with $p$ will see conditioning in the sense of quantum probability to a state no longer in the quasi-subgroup. There are cocommutative examples given by non-normal subgroups. I am getting rather away from your question. You have asked in a comment to another question about whether all these examples are of Woronowicz type. The answer is yes, because they are Hopf*-algebras with an integral, and therefore algebraic compact quantum groups. For the existence of the integral, see: A. Van Daele, The Haar Measure on Finite Quantum Groups, Proc. Amer. Math. Soc. 125 (1997), no. 2, 3489-3500. My understanding is that a classification result is far beyond any current technology. See here. As a final comment, while every finite quantum group with commutative algebra of functions is indeed a finite group, and so finite quantum groups are a generalisation of finite groups, it might be tenable or rather more natural to instead go down the line of Cayley's Theorem and say instead that (possibly infinite) quantum permutation groups are a more correct generalisation of finite groups. Then however the wallpaper gets bubbly again when we consider that not all finite quantum groups are quantum permutation groups. G. I. Kac and V.G. Paljutkin, Finite Group Rings, Trudy Moskov. Mat. Obsc 15 (1966); English transl., Trans. Moscow Math. Soc. (1967), 251-284.<|endoftext|> TITLE: Godbillon–Vey invariant and leaf space of foliations QUESTION [7 upvotes]: I recently got to know about the existence of the so-called Godbillon–Vey invariant, and I am interested in its relationship with foliation theory in 3-manifolds. I briefly recall here the definition: Suppose that $\mathcal{F}$ is a codimension-$1$ foliation of a $3$-manifold $M$, and suppose that $\mathcal{F}$ is defined as the kernel of a $1$-form $\alpha$. We have that $d\alpha\wedge\alpha=0$ and this implies that there exists a $1$-form $\theta$ such that $d\alpha=\theta\wedge\alpha$ and now one can define the $3$-form $\theta\wedge d\theta$. This form is closed and its cohomology class does not depend on the choice of $\theta$, and it is called the Godbillon–Vey invariant of $\mathcal{F}$. I got the idea that this invariant should tell us something about the transverse topology of the foliation, and I have some questions about it. Does the vanishing of the Godbillon–Vey invariant imply something about the leaf space of $\mathcal{F}$? More precisely, can we say whether this space is Hausdorff or not? Is there some known technique to calculate this invariant? Is there some generalization of this invariant for foliation that are only $C^{\infty,0}$? That is to say, the leaves are smooth but the tangent plane bundle to the foliation is only continuous. Any reference is appreciated. REPLY [2 votes]: There's a discussion of the history of the Godbillon-Vey invariant in Hurder's Dynamic's & the Godbillon-Vey Class. He recalls: Reinhart and Wood ... gave a formula expressing the the Godbillon-Vey class of a foliation of a Riemannian 3-manifold in terms of the curvatures of the leaves and normal bundle, which can be interpreted as [Thurston's] helical wobble. He doesn't expressly detail it, but there is a reference. The early history of the subject suggested that when the invariant did not vanish there were leaves of exponential growth. This was conjectured by Moussu & Pelletier and Sullivan and was proved for increasingly broader classes of foliations. Finally Duminy proved in 1986 in a unpublished manuscript that: Theorem: If $F$ is a codimension one, $C^2$-foliation of a compact manifold $M$ with non-trivial Godbillon measure $g_F$ , then $F$ must have a hyperbolic resilient leaf, and hence there is an open subset of $M$ consisting of leaves with exponential growth. Here, Duminy 'factored' the Godbillon-Vey invariant into the Godbillon measure measure constructed from the leaf cohomology class $[\eta] \in H^1(M, F)$. The other 'factor' is the Vey class, $[d\eta] \in H2(M/F)$. The theorem was generalised to higher codimension: THEOREM: Let $F$ be a $C^1$ foliation of codimension $q ≥ 1$ such that almost every leaf has subexponential growth rate. Then the Godbillon measure $g_F = 0$. If $F$ is $C^2$ then the Godbillon-Vey class $GV (F)=0$. (See Kotschick's paper, Godbillon-Vey invariants for families of foliations for the generalisation of the Godbillon-Vey invariant to higher codimension). The results above are for $C^2$ foliations with smooth leafs but they do mention an unproven conjecture (as of '00) which Hurder terms 'very surprising if true' is that the pullback of the invariant by a continuous map should be invariant. But they note it was proven under additional regularity: by Raby under $C^1$-diffeomorphism whilst Hurder and Katok showed independently that if the 'conjugancy map' and its inverse are absolutely continuous, then the conclusion is also true. Finally Connes has shown that the Godbillon-Vey class can be calculated from the flow of weights for a natural von Neumann algebra associated to the foliation. In fact, if the class does not vanish then this algebra has factor of type III.<|endoftext|> TITLE: Does the axiom schema of collection imply schematic dependent choice in ZFCU? QUESTION [9 upvotes]: This question was asked on math.stackexchange and didn't receive an answer. But I think it's interesting, and I at least would love to know the answer. Let ${\sf ZFCU}$ be the axioms of ${\sf ZFC}$ modified to allow for urelements in the usual way. We do not assume that the urelements form a set. The question is whether ${\sf ZFCU}$ plus the axiom schema of collection---i.e.: (Collection) $\forall x\exists y \phi(x, y) \to \forall z\exists w\forall x\in z\exists y\in w \phi(x, y)$ implies schematic dependent choice---i.e.: (SDC) $\forall x\exists y\phi(x, y) \to \forall z\exists f(f(0) = z \wedge \forall n\phi(f(n), f(n+1)))$ REPLY [5 votes]: (Remark: Sam Roberts pointed out to me an error in the first version of the proof I posted earlier -- there was no reason that the desired automorphisms exist, as there was no specification of them on the atoms outside of the sets of atoms under consideration. The following is a significantly modified version, and I think this time it works. One should also compare with Emil Jeřábek's and Sam Roberts' comments regarding DC$_{\omega_1}$ in the original question and below this answer.) Assuming foundation is included (in the sense that for every non-empty set $x$ with no atomic elements, there is $y\in x$ such that $y\cap x=\emptyset$), SDC follows: Write $U$ for the entire universe (of all sets and atoms). If there is only a set $A$ of atoms, then $U=V(A)$, i.e. $U$ is the union of the cumulative hierarchy $\left_{\beta\in\mathrm{Ord}}$ above $V_0(A)=A$. In this case there's an easy argument by recursively collecting witnesses according to rank. That is, let $\beta_0$ be the least ordinal $\beta$ such that there is $y\in V_\beta(A)$ with $\varphi(z,y)$ (where $z$ was the given parameter). Then by collection, we can take $\beta_1$ least such that for all $x\in V_{\beta_0}(A)$ there is $y\in V_{\beta_1}(A)$ such that $\varphi(x,y)$. Etc, producing $\left<\beta_n\right>_{n<\omega}$. From this information we can build a set relation and apply DC to it to get the DC-branch $f$, i.e. such that $f(0)=z$ and $\varphi(f(n),f(n+1))$ for each $n<\omega$. So suppose $A$ is a proper class. We will adapt the argument just mentioned. Remark: I use parametrized collection in the proof, but this follows from the axioms in a routine way, by folding parameters into the ``$x$'' in unparametrized collection and modifying $\phi$ appropriately. Lemma 1: For every set $x$, the transitive closure of $x$ exists. Proof: Just the usual thing, taking the union of $\bigcup^nx$ over all $n<\omega$. Given a set of atoms $a$ and ordinal $\beta$, let $V_\beta(a)$ be the cumulative hierarchy built above $V_0(a)=a$. Lemma 2: For every set $x$, there is a set $a$ of atoms and ordinal $\beta$ with $x\in V_\beta(a)$. Proof: Let $t$ be the transitive closure of $\{x\}$ and $a=t\cap A$. So it suffices to see $t\subseteq V(a)$. If not let $t'=t\backslash V(a)$, so $t'\neq\emptyset$ and $t'$ is not an atom, and note that $t'$ contains no atoms. By foundation there is (a set) $y\in t'$ such that $y\cap t'=\emptyset$. So since $y\subseteq t$, we have $y\subseteq V(a)$. But then by collection there is $\beta$ with $y\subseteq V_\beta(a)$, but then $y\in V_{\beta+1}(a)$, so $y\in V(a)$, a contradiction. We now split into cases. Case 1: For all cardinals $\kappa$ there is a set of atoms of cardinality $\kappa$. Claim 1.1: Let $a,a'$ be sets of atoms of the same cardinality and $\pi:a\to a'$ a bijection, and $\pi^+:V(a)\to V(a')$ the resulting isomorphism. Then for all formulas $\varphi$ and all $x\in V(a)$, we have $\varphi(x)\Leftrightarrow\varphi(\pi^+(x))$ (with truth of $\varphi$ evaluated in $U$). Proof: By (meta-)induction on formula complexity. If $\varphi$ is $\Sigma_0$ then it is just absoluteness of $\Sigma_0$ between $V(a)$ and $U$ and since $\pi^+$ is an isomorphism (the language can include a predicate interpreted as $A$, which is equivalent to the sets $a/a'$ when interpreted over $V(a)/V(a')$). Suppose it holds for $\Sigma_n$ formulas, let $\varphi$ be $\Sigma_n$, let $x\in V(a)$, and suppose there is $y\in U$ such that $\varphi(x,y)$. Let $b$ be some set of atoms which is disjoint from $a$ and such that there is such a $y\in V(a\cup b)$. Let $b'$ be a set of atoms of the same cardinality as $b$, which is disjoint from $a'$. This exists by Case 1 hypothesis. Then $a'\cup b'$ has the same cardinality as $a\cup b$. Let $\sigma:a\cup b\to a'\cup b'$ be a bijection extending $\pi$ (which bijected $a$ with $a'$ already). Let $\sigma^+:V(a\cup b)\to V(a'\cup b')$ be the resulting isomorphism. Then $\pi^+\subseteq\sigma^+$, and by induction, since $\varphi(x,y)$ is true, so is $\varphi(\sigma^+(x),\sigma^+(y))$, which suffices, proving Claim 1.1. Now suppose that for all $x$ there is $y$ such that $\varphi(x,y)$, and let $z$ be given. Let $t_z$ be the transitive closure of $\{z\}$ and $a_z=A\cap t_z$. Let $(\kappa_0,\beta_0)$ be the lexicographically least pair $(\kappa,\beta)$ of ordinals such that for some set $a$ of atoms of cardinality $\kappa$, with $a$ disjoint from $a_z$, there is $y\in V_\beta(a_z\cup a)$ such that $\varphi(z,y)$. Now suppose we have defined $(\kappa_0,\beta_0),\ldots,(\kappa_n,\beta_n)$. Let $(\kappa_{n+1},\beta_{n+1})$ be the lex least pair $(\kappa,\beta)$ of ordinals such that for all sets $a_0,a_1,\ldots,a_n$ of atoms, if $a_z,a_0,\ldots,a_n$ are pairwise disjoint and each $a_i$ has cardinality $\kappa_i$, then there is a set $a'$ of atoms of cardinality $\kappa$ such that $a'$ is disjoint from $a_z\cup a_0\cup\ldots\cup a_n$, and for all $x\in V_{\beta_n}(a_z\cup a_0\cup\ldots\cup a_n)$, there is $y\in V_{\beta}(a_z\cup a_0\cup\ldots\cup a_n\cup a')$ such that $\varphi(x,y)$. (Note that we allow $\kappa_n=0$.) (Such a pair $(\kappa,\beta)$ exists. For let $(a_0,\ldots,a_n)$ and $(a'_0,\ldots,a'_n)$ satisfy these conditions. Using collection, there is a set $Y$ such that for all $x\in W=V_{\beta_n}(a_z\cup a_0\cup\ldots\cup a_n)$, there is $y\in Y$ such that $\varphi(x,y)$. From $Y$ we can compute a $(\kappa,\beta)$ which works for $W$. But then the same $(\kappa,\beta)$ works for $W'=V_{\beta_n}(a_z\cup a'_0\cup\ldots\cup a'_n)$, by Claim 1.1 and since by case hypothesis, we can always find a disjoint set of atoms of a given cardinality.) Now fix a sequence $\left_{n<\omega}$ of pairwise disjoint sets of atoms, disjoint from $a_z$, with each $a_i$ of cardinality $\kappa_i$. (Exists by case hypothesis.) Note that there is $y\in V_{\beta_0}(a_z\cup a_0)$ such that $\varphi(z,y)$ and for each $n<\omega$, for each $x\in V_{\beta_n}(a_z\cup a_0\cup\ldots\cup a_n)$ there is $y\in V_{\beta_{n+1}}(a_z\cup a_0\cup\ldots\cup a_{n+1})$ such that $\varphi(x,y)$. So we have reduced the whole issue down to set-size, so applying choice, we get our DC-branch $f$, as desired. Case 2: Otherwise. (There is some cardinal $\kappa$ such that there is no set of atoms of cardinality $\kappa$.) Note that there are infinite sets of atoms, because for each $n<\omega$ there is a set of atoms of cardinality $n$, and applying collection. And given any set $a$ of atoms, there is an infinite set $b$ of atoms which is disjoint from $a$, for similar reasons and because $A$ is not a set. Claim 2.1: There is a largest cardinal $\kappa$ such that there is a set of atoms of size $\kappa$. Proof: Let $\kappa$ be the sup of all $\kappa'$ such that there is a set of atoms of size $\kappa'$. So $\kappa\geq\omega$. By collection, we can find a set $Y$ which contains a set of atoms of size $\kappa'$, for each $\kappa'<\kappa$. Let $\kappa$ be the cardinality the set of atoms in the transitive closure of $Y$. Then $\kappa$ works. Let $\kappa_0=$ this largest $\kappa$. Subcase 2.1: For every set $c$ of atoms of cardinality $\kappa_0$, there is a set $b$ of atoms of cardinality $\kappa_0$ which is disjoint from $c$. This subcase is just a simplification of the subcase below. Subcase 2.2: Otherwise. (There is a set $c$ of atoms of cardinality $\kappa_0$ such that there is no set $b$ of atoms of cardinality $\kappa_0$ with $c\cap b=\emptyset$.) Let $c$ be a set as in Subcase 2.2 hypothesis. Then define $\mu_c$ as the largest cardinality of a set of atoms $b$ which is disjoint from $c$. This largest cardinality exists like in the proof of Claim 2.1. Now let $\mu_0$ be the least value of $\mu_c$, ranging over all such $c$. Let $c_0$ witness this choice. Note that $\mu_0\geq\omega$, since $A$ is not a set. Claim 2.2: For every set $c$ of atoms disjoint from $c_0$, there is a set $b$ of atoms disjoint from $c\cup c_0$, such that $b$ has cardinality $\mu_0$. Proof: If not, let $c'=c\cup c_0$ and observe that $\mu_{c'}<\mu_0$, a contradiction. From here we can run a simple variant of the proof from Case 1, but starting with $c_0\cup a_z$ as our base set of atoms. That is: Claim 2.3: Let $a,a'$ be sets of atoms of the same cardinality, which are disjoint from $c_0$. Let $\pi:a\to a'$ a bijection, and $\sigma:V(c_0\cup a)\to V(c_0\cup a')$ the resulting isomorphism. Then for all formulas $\varphi$ and all $x\in V(c_0\cup a)$, we have $\varphi(x)\Leftrightarrow\varphi(\sigma(x))$ (with truth evaluated in the entire universe). Proof: This is now proved like its version in Case 1, but using Claims 2.2 and 2.3 (the properties of $c_0$ and $\mu_0$) to see that we can find appropriate disjoint sets of atoms (of cardinality at most $\mu_0$). The rest of the proof is like before; given a sequence $a_0,\ldots,a_n$ of sets of atoms disjoint from $c_0\cup a_z$, they each have cardinality $\leq\mu_0$, and we can always find another set $a$ of atoms of cardinality $\mu_0$ disjoint from the rest. (So in fact, since $\mu_0\geq\omega$, one could just fix a sequence $\left_{n<\omega}$ of sets $a_i$ of atoms, each of cardinality $\mu_0$, disjoint from $c_0\cup a_z$ and pairwise disjoint, and then argue that given an ordinal $\beta$, there is an ordinal $\beta'$ such that for every $x\in V_{\beta}(c_0\cup a_z\cup a_0\cup\ldots\cup a_n)$ there is $y\in V_{\beta'}(c_0\cup a_z\cup a_0\cup\ldots\cup a_n\cup a_{n+1})$ such that $\varphi(x,y)$. Then we can define a resulting sequence $\left<\beta_n\right>_{n<\omega}$, and apply DC to sets as before.) Remark: Considering @SamRoberts' and @EmilJeřábekand's comments below, we might have $\kappa_0=\mu_0=\omega$, hence no $\omega_1$-sequence of atoms. In this case the proof wouldn't work to give DC$_{\omega_1}$, since it would rely on having a sequence $\left_{\beta<\omega_1}$ of pairwise disjoint (presumably non-empty) sets of atoms. We got a sequence like this above because we had the set of cardinality $\mu_0$ of atoms to work with, and since (in the argument) $\mu_0\geq\omega$, we can actually partition it as desired. But if $\mu_0=\omega$, this clearly breaks when we want an $\omega_1$-sequence.<|endoftext|> TITLE: Braided monoidal category, example QUESTION [5 upvotes]: Let $H$ be a cocommutative hopf algebra. Let $M$ be the category of $H$-bimodules. Does the category $M$ form a braided monoidal category with tensor product $\otimes_{H}$ ? REPLY [2 votes]: The answer is no in general. Here is a counter example. Let us work over a ground field $k$, and let $ H = \oplus_n k$ be the direct sum of $n$ copies of $k$, with $n \geq 2$. This is a commutative, cocommutative Hopf algebra. Of course, as others have commented, the monoidal category $({}_H Mod_H, \otimes_H)$ only depends on the algebra structure of $H$. The category of $H$-$H$-bimdoules with $\otimes_H$ monoidal structure has a nice interpretation. An $H$-$H$-bimodule $M$ can be thought of as an $n \times n$-matrix of vector spaces. The $(i,j)$th entry of this matrix is obtained by multiplying $M$ by the $i$th minimal idempotent on one side and the $j$th minimal idempotent on the other side. Under this identification the monoidal structure $\otimes_H$ is easy to describe: it is "matrix multiplication", but where you replace the usual addition and multiplication operations with direct sum and tensor product of vector spaces. However, now we see the problem emerge. For $n \geq 2$ matrix multiplication is not commutative, and it is impossible to equip $({}_H Mod_H, \otimes_H)$ with a braiding. For a concrete example we can set $$ M = \begin{bmatrix}0 & V\\0 & 0\end{bmatrix}$$ $$ N = \begin{bmatrix}0 & 0\\0 & W\end{bmatrix}$$ Then $N \otimes_H M = 0$, while $$M \otimes_H N = \begin{bmatrix}0 & V \otimes_k W \\0 & 0\end{bmatrix}$$ There can be no isomorphism between $N \otimes_H M$ and $M \otimes_H N$.<|endoftext|> TITLE: Elliptic curves of high rank over quadratic extensions QUESTION [7 upvotes]: Are there examples of elliptic curves which has rank 0 over $\mathbb{Q}$, but acquires a high rank ( $\geq 2$) over some quadratic extension? More generally, are there known bounds for a given extension of degree $n$, how big can the rank be if you start with a rank 0 curve ( over $\mathbb{Q}$ )? REPLY [4 votes]: Following the comments, we start with the $\mathbb{Z}/2\mathbb{Z}$ curve $$ E=[ 0, 0, 0, -15650411093524454493683423178813553233328484724599030637228, 492395613211229713687666165349208268497505680211462943972719303000171030884176146690352 ] $$ of rank $0$ and extend it over $\mathbb{Q}(\sqrt{2})$ to obtain the $\mathbb{Z}/2\mathbb{Z}$ curve found by Elkies - Klagsbrun (2020) of rank $20$. The use of Magma's TwoPowerIsogenyDescentRankBound significantly speeds up the process. SetClassGroupBounds("GRH"); E := EllipticCurve([ 0, 0, 0, -15650411093524454493683423178813553233328484724599030637228, 492395613211229713687666165349208268497505680211462943972719303000171030884176146690352 ]); Coefficients(E); TwoPowerIsogenyDescentRankBound(E); QT := MinimalModel(QuadraticTwist(E, 2)); Coefficients(QT); QT eq EllipticCurve([1,-1,1,-244537673336319601463803487168961769270757573821859853707,961710182053183034546222979258806817743270682028964434238957830989898438151121499931]); [ 0, 0, 0, -15650411093524454493683423178813553233328484724599030637228, 4923956132112297136876661653492082684975056802114629439727193030001710308841761\ 46690352 ] 0 [ 11, 5, 2, 2, 1 ] [ 7, 5, 2, 2, 1 ] [ 1, -1, 1, -244537673336319601463803487168961769270757573821859853707, 9617101820531830345462229792588068177432706820289644342389578309898984381511214\ 99931 ] true A great collection of references on the topic is provided on Andrej Dujella's page High rank elliptic curves with prescribed torsion over quadratic fields. Some of the papers also discuss the extensions over cubic and/or quartic number fields, e.g. J. Bosman, P. Bruin, A. Dujella and F. Najman, Ranks of elliptic curves with prescribed torsion over number fields, Int. Math. Res. Not. 2014 (11) (2014), 2885-2923, doi:10.1093/imrn/rnt013, arXiv:1201.0252, PDF. D. Jeon, C. H. Kim, and Y. Lee, Families of elliptic curves over cubic number fields with prescribed torsion subgroups, Mathematics of Computation, Volume 80, Number 273, January 2011, pp. 579–591, doi:10.1090/S0025-5718-10-02369-0.<|endoftext|> TITLE: Motivation for Henselian rings in algebraic geometry QUESTION [11 upvotes]: In Andrew Kobin's script on Algebraic Geometry I found on page 355 a comment I would like better understand. It states Another way to view formal smoothness is as an abstraction of Hensel's Lemma. Formal smoothness of a scheme $X \to S$ is characterized by the property that for every $S$-scheme $Y$ and every infinitesimal subscheme $Y_0 \subset Y$ (that is defined by a nilpotent ideal sheaf in $Y$) , the canonical morphism $$Hom_S(Y,X) \to Hom_S(Y_0, X)$$ is surjective. Well, why can this property be regarded as abstraction of Hensels lemma? The Hensel's lemma I am familar with on lifting polynomials under certain conditions from $(R/m)[X]$ to $R[X]$, where $A$ is local complete with maximal ideal $m$, that doesn't involve any assumptions that $m$ is nilpotent. In which sense can the above be regarded as an abstraction? Another question on general properties of Henselian rings. I heard fleetingly (but forgot the concrete context) that schemes over henselian local rings have generally a rich divisor theory that on one hand somehow allows in certain way to "reduce" the analysis of divisors on $R$-scheme $X$ ($R$ local hensel with maximal ideal $m$ and residue field $\kappa=R/m$) to that on the special fiber $X \times_R \kappa$ in the sense that a "lot of information" of the theory of divisors on $X$ maybe already be extracted from the study of divisors on $X \times_R \kappa$ using formal techniques, by lifting results from $X \times_R R/m^i$ to $X \times_R R/m^{i+1}$ in much more "fruitful way" than if we work without Henselian context. On the other hand if that's true, then what I heard for example in the case of $X$'s surface there is much more "flexibility" in the constructions of divisors with desired intersection behaviour. At least it seems that henselian assumption "guarantees" that much more information is conserved by passing to the special fiber as without this assumption. Could somebody give short insight into this correspondence principle and motivate how exactly henselianess enters fruitfully into the game? Or give a recomendable reference where these ideas are made precise and explaned in details? Sorry if the formulation is too vague, I've only heard it once about this ideas, but unfortunately couldn't find anything about it and still quite curious what I can take from it. REPLY [4 votes]: First about the ralation between henselian and formal smoothness property, I think a good idea is to look at what was the first version of Hensel lemma: it says that if $f(\bar{a})=0,f'(\bar{a})\, modp=0$ then there is a lift of $\bar{a}$ in $\mathbb{Z}_p$ such that $f(a)=0$. this is basically says that if you define a curve as $C=V(f)$ then $f'(a)\not = 0$ implies that $C$ has the formal smoothness property with respect to thickening of the form $F_p\to \mathbb{Z}_p/P^n$ over the point $\bar{a}$. so Hensel lemma for complete ring is a consequence of formal smoothness. but why we define the Henselian rings and don't only work with complete rings? I think the answer is the important concept of Henselisation: if you have a local ring $(O,m)$ you can consider its completion $(\bar O,\bar{m})$ or you can consider its Henselisation $O^{h}$. there are two ways to define $O^h$ you can define it as the smallest Henselian extension of $(O,m)$ inside $\bar{O}$ or as the limit of all etale extension $O'$ of $O$ with an isomorphism $O/m\to O'/m'$. you can also consider the strict Henselisation $O^{sh}$ as the limit of all etale extensions. there are two reasons why we are interested in $O^h,O^{sh}$ instead of $\bar{O}$. the first is that in non-Notherian setting $\bar{O}$ is not faithfully flat over $O$ but $O^h,O^{sh}$ are almost by definition always faithfully flat. the second reason is that $O_x^{sh}$ works in etale topology like $O_x$ in Zariski topology. but how we study Henselian rings? the Henselian property itself is very powerful and you can deduce a lot of things as a consequence of that property but there is another important tool: Artin approximation property and its consequences. First, let us look at the simplest case: the Hensilsation of $\mathbb{C}[x]$ at $x$ consists of algebraic power series: power series the satisfy a polynomial equation then it is not hard to deduce from Henselian property that if you have a system of equations $f^1=0,...f^r=0$ with invertible jacobian then any solution $y$ of this system in $\mathbb{C}[[x]]$ and any constant $c$ there is a solution $y'$ in $\mathbb C[[x]]^{alg}$ such that $y=y' mod x^c$. in summary, you can approximate the solution in $\bar{O}$ with the solutions in $O^h$. now back to your question let $A$ be a henselian ring, there is general version of Artin theorem: consider any "finitely presented" functor from the $A-algebras$ to $Sets$ for example the functor that sends $B$ to $Div(X_B)$. then Artin theorem says that for each $y\in F(\bar{A})$ and constant $c$ there is a $y'\in F(A)$ such that $y=y' mod\, m^c$. This is important because the are powefull tools in deformation theory to relate $F(\bar{A})$ to $F(\bar{A}/m)=F(A/m)$. in the case of divisors $Div(X_\bar{A})=Div(X_{A/m})$ so you get the desired relation between $Div(X)$ and $Div(X_k)$ in your language.<|endoftext|> TITLE: Nonvanishing section of infinite-dimensional tautological bundle QUESTION [6 upvotes]: Let $H$ be a real or complex Hilbert space. In the case where $H$ is infinite-dimensional, let us define a half-dimensional subspace as a subspace $W \subset H$ such that both $W$ and $W^\perp$ have infinite dimension. Fix one half-dimensional subspace $W_0$. The Grassmannian of $H$ is $$\mathrm{Gr}(H, W_0) = \{W \subset H ~|~ W \text{ is half-dimensional}, P_W - P_{W_0} \text{ is Hilbert-Schmidt}\}. $$ Here for $W \subset H$ a subspace, $P_W$ denotes the orthogonal projection onto $W$. $\mathrm{Gr}(H, W_0)$ can be given the structure of a Hilbert manifold in a natural way (see e.g. the book "Loop Groups" of Pressley and Segal). The space $\mathrm{Gr}(H, W_0)$ has a tautological vector bundle $\tau$ over it, where the fiber is given by $\tau(L) = L$. Question: Does $\tau$ have a nowhere vanishing section? I believe that in the case that $H$ is finite-dimensional (say of dimension $2n$), the answer is no, as one can show that the Euler class of $\tau$ is non-zero. But how would one proceed in the infinite-dimensional case? REPLY [8 votes]: Let $X$ be any paracompact space. Then Hilbert vector bundles over $X$ are classified by homotopy classes of maps $[X, BU(\mathcal H)]$. But when $\mathcal H$ is infinite-dimensional, the group $U(\mathcal H)$ is contractible (this is Kuiper's theorem), and hence every infinite-dimensional Hilbert bundle over a paracompact space is trivializable. Hilbert manifolds modelled on a separable Hilbert space are metrizable. Separable metric spaces are paracompact. As long as your $H$ is a separable Hilbert space, the above argument implies that your bundle is trivializable. There is probably a straightforward extension of this argument in the non-separable case but I didn't think about it. In particular, your bundle has infinitely many linearly independent sections. REPLY [5 votes]: As Mike notes Hilbert bundles are trivial over most spaces. The paper MR2481802 (2010e:46083) Reviewed Abbondandolo, Alberto (I-PISA); Majer, Pietro (I-PISA) Infinite dimensional Grassmannians. (English summary) J. Operator Theory 61 (2009), no. 1, 19–62. 46T05 (47A53 58B15) https://arxiv.org/pdf/math/0307192.pdf Studies the homotopy type of the relative Grasmannian when the projector is compact. It has been too long since I've thought about Hilbert Schmidt operators to see if this is the same answer.<|endoftext|> TITLE: sl(2)-reps categorifying q-binomials QUESTION [8 upvotes]: Recall that the $q$-binomial coefficient $\big[\begin{smallmatrix}a\\b\end{smallmatrix}\big]$ is the Laurent polynomial in $q$ given by $$ \big[\begin{smallmatrix}a\\b\end{smallmatrix}\big]=\frac{[a]!}{[b]![a-b]!} $$ where $[n]!=[1][2][3]...[n]$, and $[i]=\tfrac{q^i-q^{-i}}{q-q^{-1}}=q^{i-1}+q^{i-3}+q^{i-5}+...+q^{-(i-1)}$. Let $V_n$ be the $n$-dimensional irrep of $\mathfrak{sl}(2)$. It is easy to check that there exists an isomorphism of $\mathfrak{sl}(2)$-representations $$ S^k V_{n+1} \cong \textstyle \bigwedge^k V_{n+k} $$ between the $k$-th symmetric power of $V_{n+1}$ and the $k$-th exterior power of $V_{n+k}$, as both have $\big[\begin{smallmatrix}n+k\\k\end{smallmatrix}\big]$ as their character Proof: We exhibit a basis of weight vectors of $S^k V_{n+1}$, a basis of weight vectors of $\bigwedge^k V_{n+k}$, and a bijection between these bases which respects the weights. Let $x_0,\ldots, x_n$ be the standard basis of $V_{n+1}$, and let $y_1,\ldots y_{n+k}$ be the standard basis of $V_{n+k}$. Then $x_{i_1}{\cdot}\, x_{i_2} \cdot... \cdot x_{i_k}\mapsto y_{i_1+1}\wedge y_{i_2+2}\wedge...\wedge y_{i_k+k}$ is the desired bijection. $\square$ The problem with the above proof is that it only proves the existence of an isomorphism, without actually constructing one. Question: It there a natural isomorphism of $\mathfrak{sl}(2)$-representations $S^k V_{n+1} \cong \textstyle \bigwedge^k V_{n+k}$ that one can write? REPLY [4 votes]: Yes there is. In my article with Chipalkatti "On the Wronskian combinants of binary forms" in J. Pure Appl. Algebra, we gave an explicit construction in Section 2.5. We called it the Wronskian isomorphism because when followed (on the sym-sym side) by complete symetrization, we get the usual Wronskian (in homogenized form). Now, I'm sure the map was known a long time before. The earliest reference I found is the Thesis by Cyparissos Stephanos from 1884. Essentially, the map is nothing more than dividing by the Vandermonde in order to turn antisymmetric functions into symmetric ones, a kind of Boson-Fermion correspondence. You might be interested in a new development about this map regarding characteristic $p$. See the article "Koszul modules and Green’s conjecture" by Aprodu et al., Sec 3.4.<|endoftext|> TITLE: Nonvanishing section of infinite-dimensional tautological bundle II QUESTION [5 upvotes]: This is a follow-up question on a previous question of mine, which ended up to be trivial, because I overlooked the obvious problem with Hilbert space bundles, which I fix here. Let us write $E$ for the space of complex-valued sequences $\{\dots, a_{-2}, a_{-1}, a_1, a_2, \dots\}$, labeled by $(\mathbb{Z}\setminus\{0\})$, such that only finitely many $a_n$ are non-zero. For a subset $I \subset \mathbb{Z}$, write $$E_I = \mathrm{span}\{ e^i ~|~ i \in I\} \subseteq E,$$ where $e^i$, $i \in \mathbb{Z}$ are the canonical basis vectors of $E$. Let us call a vector subspace $W \subset E$ half-dimensional if there exists $N \in \mathbb{N}$, such that $$ W = W^\prime \oplus E_{[N, \infty)},$$ where $W^\prime$ is a half-dimensional subspace of the (finite-dimensional) space $E_{[-N, N]}$. Consider the following space $$\mathrm{Gr}_\infty = \{ W \subset E ~|~ W \text{ is half-dimensional} \}.$$ This has a natural direct limit topology: $ \mathrm{Gr}_\infty = \lim_{n \to \infty} \mathrm{Gr}_n(E_{[-n, n]}),$ where the map $\mathrm{Gr}_n(E_{[-n, n]}) \to \mathrm{Gr}_{n+1}(E_{[-(n+1), n+1]})$ is given by sending $W \to W \oplus \mathbb{C} e^{n+1}$. There is a tautological bundle $\tau$ over $\mathrm{Gr}_\infty$, with fibers $\tau(W) = W$. Q: Does this bundle have a non-vanishing section? In the previous version of the question, the answer was trivially yes: the bundle was a bundle of Hilbert spaces, which has contractible automorphism group. Here, we have a bundle with typical fiber $\mathbb{C}^\infty$, the automorphism group of which is $U_\infty = \lim_{n \to \infty} U_n$, which has lots of trivial topology. REPLY [2 votes]: $\DeclareMathOperator{\Gr}{Gr}\newcommand{\C}{\mathbb{C}}\DeclareMathOperator*{\colim}{colim}$The total space of your bundle is the colimit of the spaces $\tau_n\times\C^{[N,\infty)}$, where $\tau_n\to \Gr_n(E_{[-n,n]})$ is the tautological bundle over $\Gr_n(E_{[-n,n]})$ and the transition functions are the product of the bundle map $\tau_n\oplus \C\to \tau_{n+1}$, which exhibits the left-hand side as the pullback of $\tau_{n+1}$ along the inclusion $\Gr_n(E_{[-n,n]})\hookrightarrow \Gr_{n+1}(E_{[-(n+1),n+1]})$, and the identity of $\C^{[N+1,\infty)}$. In particular, it receives a map from $\colim_n \tau_n$, and the induced map from the complements of zero sections $\colim_n \tau_n^\times$ lands in the complement of the zero section. Now the map $\tau_n^\times\to \Gr_n(E_{[-n,n]})$ is a fiber bundle, in particular a Serre fibration, with fiber $S^{2n-1}$. Since a filtered colimit of Serre fibrations is a Serre fibration, the map $\colim_n \tau_n^\times\to \Gr_\infty$ is a Serre fibration with fiber $\colim_n S^{2n-1}\simeq *$, i.e. an acyclic Serre fibration. Since the base $\Gr_\infty$ has the homotopy type of a CW complex, there is a section $\Gr_\infty\to\colim_n\tau_n^\times$, which then maps to a nonzero section of your bundle. It is probably also possible to explicitly construct such a section by using that the shift map on sequences is homotopic to the identity.<|endoftext|> TITLE: Construction of a model of $ZFC+\neg Con(ZFC)$ QUESTION [7 upvotes]: By Gödel's second incompleteness theorem, the following assertion is true in ZFC: $$ Con(ZFC)\rightarrow Con(ZFC+\neg Con(ZFC)) $$ Considering the completeness theorem, this assertion is equivalent to that one can always construct a model of $ZFC+\neg Con(ZFC)$ from a given model of $ZFC$. Is there any concrete construction of the new model, just like the constructible universe and forcing method? REPLY [3 votes]: As has been discussed in the comments, there is no inner model or forcing-like construction that will give you a model, since the collection of natural numbers needs to grow. You'll have to tell me if this is 'natural' enough, but by modifying the construction mentioned by Joel David Hamkins here, we can get the following: Proposition. There is a formula $\varphi(x)$ in the language of set theory such that for any $V \models \mathsf{ZFC}$, there is a unique $a \in V$ satisfying $\varphi(x)$ and this $a$ is always (externally) a model of $\mathsf{ZFC}+ \neg\mathrm{Con}(\mathsf{ZFC})$. Proof: As Noah Schweber mentioned in the comments, by standard (computable) procedure, for any theory $T$, we can find an expansion $H_{T} \supseteq T$ which is term complete, i.e., for every formula $\varphi(x,\bar{y})$ and tuple of closed terms $\bar{t}$, there is a closed term $s$ such that $H_{T}\vdash \exists x \varphi(x,\bar{t}) \to \varphi(s,\bar{t})$. The two most common ways to do this are adding Henkin constants and Skolemizing the theory. This implies that for any completion $T'$ of $H_T$, the term structure of $T'$ is a model of $T'\supseteq T$, where the term structure is the structure whose universe is the collection of all closed terms in the language of $T'$ with the interpretations of the atomic relations dictated by $T'$. Another thing to note is that the construction of the language of $H_T$ can be done in such a way that if the language of $T$ is countable with a given enumeration, then the language of $H_T$ is countable with a given enumeration which is uniformly computable from the enumeration of the language of $T$. Also note that this language really only depends on the language of $T$, not on $T$ itself. $\mathsf{ZFC}$ proves that for any theory $T$, $\mathrm{Con}(T) \to \mathrm{Con}(H_T)$. (This is also provable in significantly weaker theories for the standard definitions of $H_T$.) Let $\{\psi_i\}_{i<\omega}$ be a fixed computable enumeration of the axioms of $\mathsf{ZFC}+\neg\mathrm{Con}(\mathsf{ZFC})$, and let $T_i$ be $\{\psi_j\}_{i TITLE: Smooth projective variety with no second homotopy group QUESTION [9 upvotes]: I am looking for an example (if such exist) of a smooth projective variety $X$ whose $\mathbb{Q}$-homology $H_*(X,\mathbb{Q})$ is generated by algebraic cycles, and yet does not have a second homotopy group, $\pi_2(X)=0.$ Thus, algebraic cycles that span $H_2(X,\mathbb{Q})$ are coming from some non-rational curves. REPLY [18 votes]: Fake projective planes have $H_2(X,\mathbb{Z}) \cong \mathbb{Z}$. They have metrics of pinched negative curvature, so they have $\pi_2(X) \cong \{0\}$. Thus $H_2(X,\mathbb{Q}) \cong \mathbb{Q}$ is generated by a hyperplane section, and this is not a rational curve. A Picard maximal fake quadric (a surface of general type with the same rational cohomology as $\mathbb{P}^1 \times \mathbb{P}^1$) with universal cover the product of two hyperbolic planes will also have the desired property with $H_2(X,\mathbb{Q}) \cong \mathbb{Q}^2$. They have metrics of nonpositive curvature, so again $\pi_2$ is trivial. For example, there are "product quotient" examples; see The classification of surfaces with pg=q=0 isogenous to a product of curves, Pure Appl. Math. Q. 4 (2008), no. 2, Special Issue: In honor of Fedor Bogomolov. Part 1, 547–586 by Bauer, Catanese, and Grunewald. You should also look into other surfaces with $p_g = q = 0$. You certainly want $p_g = 0$, and so $q = 0$ if the surface is minimal of general type (you certainly want minimal for $\pi_2(X)$ to be $\{0\}$). For example, see Bauer, Catanese, and Pignatelli's Surfaces of general type with geometric genus zero: a survey, Complex and differential geometry, 1-48, Springer Proc. Math., 8, Springer, Heidelberg, 2011.<|endoftext|> TITLE: Grand tour of the special orthogonal group QUESTION [6 upvotes]: Is there a continuous function $f:[0,+\infty) \to \operatorname{SO}(n)$ whose image is dense in $\operatorname{SO}(n)$ and that is well behaved in certain ways? For each $\varepsilon>0$ it doesn't take longer than necessary, or not much, to come within distance $\varepsilon$ of every point. It is not too hard to compute. One can write software for it without being a genius. REPLY [6 votes]: [EDIT: Dan Asimov notified me that this construction is similar to a construction in his 1985 paper entitled "The Grand Tour: a Tool for Viewing Multidimensional Data". The construction in the 1985 paper is somewhat more elegant than this one, avoiding the use of the exponential map and the sine function.] We'll describe such a function $f$ as the composition of three continuous maps: $h : [0, +\infty) \rightarrow [-1,1]^{\binom{n}{2}}$; $g : [-1,1]^{\binom{n}{2}} \rightarrow \mathcal{A}$; $j : \mathcal{A} \rightarrow SO(n)$; where $\mathcal{A}$ is the space of antisymmetric matrices with entries in $[-1, 1]$. Each of these three maps, and thus their composition $f$, is not only continuous but is in fact Lipschitz-continuous (unlike a spacefilling curve). In reverse order: $j(A) := \exp((\pi \sqrt{n}) A)$, where $\exp$ is the matrix exponential; $g^{-1}(A)$ is the vector $v$ obtained by 'flattening' the entries in the upper triangle of $A$ into a vector of $\binom{n}{2}$ elements, and $g$ is the inverse of the function $g^{-1}$ just described; $h(t) := (\sin(c_1 t), \sin(c_2 t), \dots, \sin(c_{\binom{n}{2}} t))$, where $c_1, c_2, \dots, c_{\binom{n}{2}}$ are a set of $\binom{n}{2}$ irrationals that are linearly independent over $\mathbb{Q}$. The image of $h$ is dense in the hypercube as mentioned here: https://en.wikipedia.org/wiki/Linear_flow_on_the_torus and, because $g$ is a homeomorphism, it follows that $g(h(t))$ is dense in $\mathcal{A}$. As such, it remains to show that $j$ is surjective onto $SO(n)$; that would establish that $f(t) = j(g(h(t)))$ is dense in $SO(n)$. Claim: Every matrix $R \in SO(n)$ is expressible as the matrix exponential of an antisymmetric matrix $A$ with Hilbert-Schmidt norm $\operatorname{tr}(A^T A) \leq \pi^2 n$. Proof: By an orthogonal change of basis, we can assume that the matrix $R$ is block-diagonal, consisting of $1 \times 1$ blocks of the form: $$ \begin{pmatrix} 1 \end{pmatrix} $$ and $2 \times 2$ blocks of the form: $$ \begin{pmatrix} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta} \end{pmatrix} $$ where $\theta \in [-\pi, \pi]$. As such, $R$ is the matrix exponential of a block-diagonal antisymmetric matrix with blocks of the form: $$ \begin{pmatrix} 0 \end{pmatrix} $$ and: $$ \begin{pmatrix} 0 & \theta \\ -\theta & 0 \end{pmatrix} $$ The result follows. Corollary: Every orthogonal matrix is the matrix exponential of an antisymmetric matrix with entries in $[-\pi \sqrt{n}, \pi \sqrt{n}]$. As such, $j$ is indeed surjective onto $SO(n)$.<|endoftext|> TITLE: Perfectoid approach to resolution of singularities in char $p$ QUESTION [19 upvotes]: Since perfectoid techniques have built a bridge between char $0$ and char $p$ worlds, it is conceivable that they can be applied to resolution of singularities in char $p$ using their successful resolution in char $0$, or offer an alternative view of alterations. Has this been attempted, and if not, why is it not a plausible line of approach? REPLY [11 votes]: Somehow that question slipped my radar, sorry! The truth is that shamefully I'm not able to say much, as I don't have a strong knowledge of resolution of singularities. But at least so far, the flow of information has been from characteristic $p$ to characteristic $0$, or more specifically to mixed characteristic. So maybe I wouldn't be surprised if once resolution of singularities has been proved in characteristic $p$, one can use some perfectoid(-inspired) things in order to generalize these results to mixed characteristic. But I would be very surprised if the known results in characteristic $0$ can be used to make progress in characteristic $p$. As Hailong Dao hints in the comments, this scenario has recently played out in the somewhat related field of the homological conjectures, and related questions in birational geometry. There, "everything" was known in characteristic $p$ building on the work of Hochster, Huneke, etc. An interesting feature here is that one measures singularities in characteristic $p$ not by comparison to a resolution of singularities, but by using Frobenius (in particular, by the relation to their perfection -- for example, by a theorem of Kunz a noetherian $\mathbb F_p$-algebra is regular if and only if its Frobenius morphism is flat, if and only if the map to its perfection is flat). Only very recently all of this has been generalized to mixed characteristic, by Andre, Bhatt, etc., using critically perfectoid methods. One can then measure singularities by comparing to perfectoid algebras: for example, by a theorem of Bhatt-Iyengar-Ma, a noetherian $p$-complete $\mathbb Z_p$-algebra is regular if and only if it admits a faithfully flat map to a perfectoid ring. The most recent results of Bhatt, on Cohen-Macaulayness of absolute integral closures and Kodaira vanishing up to finite covers, have in turn been used to establish the minimal model program for arithmetic threefolds, adapting previous work of Hacon--Xu in characteristic $p$.<|endoftext|> TITLE: Symplectic resolutions amongst cotangent bundles QUESTION [7 upvotes]: It is known that a generalized flag variety $X=T^*(G/P)$ is a (symplectic) resolution of singularities of its affinization $X^\text{aff}\mathrel{:=}\operatorname{Spec}(H^0(X,\mathcal{O}_X))$. In type A, this affinization is the closure of a nilpotent orbit, and in other types, it is finite over it. Moreover, it is conjectured in Kaledin's paper (Geometry and topology of symplectic resolutions, Conj. 1.3) that these are the only cotangent bundles of smooth projective varieties which are resolutions of their affinization. As this paper was written some time ago, I am interested in whether any progress has been made towards proving this conjecture? NB Everything is assumed under complex numbers, the flag variety, and the symplectic structure. REPLY [2 votes]: I think @Joel is saying the following (this was too long for a comment, so I put it as an answer): Given a conical symplectic resolution $X\mathrel{:=}T^*M \rightarrow X^\text{aff}$ whose $\mathbb{C}^*$-action contracts the fibres of $X$ with weight 1, the ring of global functions $R=H^0(X,\mathcal{O}_X)$ has a set of homogenous generators $\{x_1,\dotsc,x_n\}$ with maximal weight 1. (**) The last fact, together with the normality of the ring $R$, is precisely saying that $X^\text{aff}$ is a conical symplectic singularity with maximal weight 1, thus the Namikawa's theorem applies, stating that $X^\text{aff}$ must be a closure of a normal nilpotent orbit in a semisimple $\mathfrak{g}$. Then, by Fu_Symplectic Resolutions for Nilpotent Orbits (Theorem 0.1), we indeed have $X \cong T^*(G/P),$ for some parabolic subgroup $P$ of $G$ (and $\operatorname{Lie}(G)=\mathfrak{g}$). Moreover, $X^\text{aff}=\overline{O}_P\mathrel{:=}\operatorname{Im}(\mu),$ where $$\mu:T^*(G/P)\rightarrow \mathfrak{g}^*\cong \mathfrak{g}$$ is the moment map of the $G$-action (and $\mathfrak{g}^*\cong \mathfrak{g}$ is obtained via Killing form). Though, what still worries me is that the argument above does not cover resolutions $$X=T^*(G/P) \rightarrow X^\text{aff},$$ where the induced map $X^\text{aff}\rightarrow \overline{\mathcal{O}_P}$ is not an isomorphism, but rather a finite cover. This is a general setup, and in some cases (e.g. when $G=\operatorname{SL}_n$), this map is indeed an isomorphism. Having this in mind, perhaps the statement (**) about the weights of global functions is wrong?<|endoftext|> TITLE: How many people have the same exact number of hairs? QUESTION [14 upvotes]: Assume we look at $n\in\mathbb N$ people that can have anywhere between $1$ to $k\in\mathbb N$ hairs on their head. Formally, I look at $n$ independent (in fact, this is not really true in real life because of inheritance etc., but bear with me) uniformly (might also not be true in real life) distributed random variables $$X_1, X_2, \dots, X_n \sim \text{Uniform}(\{1, 2, 3, \dots, k\}).$$ My question: How many people am I expected to be able to find that have the same number of hairs? Formally, what is, where $\lvert\cdot\rvert$ denotes cardinality, $$\mathsf E\left(\max_{i\in\{1, 2, \dots, k\}} \lvert\{j\in\{1,2,\dots, n\}: X_j = i\}\rvert\right).$$ Remark. According to the Pigeonhole Principle, we have $$\max_{i\in\{1, 2, \dots, k\}} \lvert\{j\in\{1,2,\dots, n\}: X_j = i\}\rvert\geq\frac nk.$$ For example, if we assume that there are $n=82$ million people living in Germany, and everyone has at most $k=1$ million hairs, then we can always find at least $82$ people that have the exact same number of hair. (Note that the Pigeonhole Principle doesn't need the uniform distribution nor the independence of the $X_j$ !) Note that the random variables $\lvert\{j\in\{1,2,\dots, n\}: X_j = i\}\rvert$ for $i\in\{1,2,\dots, k\}$ are not independent. REPLY [15 votes]: This question has been studied extensively in the computer science literature under the name "balls in bins"; see [1] which gives quite tight bounds in Theorem 1, page 161 and also describes prior work before 1999. [1] Raab, Martin, and Angelika Steger. "“Balls into bins”—A simple and tight analysis." In International Workshop on Randomization and Approximation Techniques in Computer Science, pp. 159-170. Springer, Berlin, Heidelberg, 1998. Available on page 159 of https://link.springer.com/content/pdf/10.1007%2F3-540-49543-6.pdf<|endoftext|> TITLE: Convergence speed of a random dyadic rational generator QUESTION [5 upvotes]: We are given a multiset $M$ of real numbers which initially is equal to $\{0,1\}$. In a sequential fashion, at each round $r\in\mathbb{N}$ two distinct instances $x_r$ and $y_r$ of $M$'s numbers are selected uniformly at random from $M$ (which implies that they cannot be the same instance of any number contained in $M$, viz., $x_r$ is selected and temporarily removed from $M$, thereafter $y_r$ is selected from $M\setminus\{x_r\}$ without removing it, and finally $x_r$ is added back to $M$), and $z_r=\frac{x_r+y_r}{2}$ is added to $M$. Question: What is the probability $p_{r,{\epsilon}}$ that we have $\left|z_r-\frac{1}{2}\right|\le\epsilon$ for a given $\epsilon\in\left(0,\frac{1}{2}\right)$? Edit: I show below the experimental results for different simulations of the random process run with an increasing total number of rounds, and for a single simulation of the random process keeping track of the evolution over time of the total average and the last element added - For the sake of convenience, by writing "rounds", here I am counting the initial insertion of both $0$ and $1$ (simultaneously) as the very first round: New simul. with 2^1 rounds - Avg: 0.5 | Last added: 0.5 New simul. with 2^2 rounds - Avg: 0.5 | Last added: 0.5 New simul. with 2^3 rounds - Avg: 0.472222 | Last added: 0.5 New simul. with 2^4 rounds - Avg: 0.430147 | Last added: 0.375 New simul. with 2^5 rounds - Avg: 0.413826 | Last added: 0.40625 New simul. with 2^6 rounds - Avg: 0.40012 | Last added: 0.40625 New simul. with 2^7 rounds - Avg: 0.38313 | Last added: 0.46875 New simul. with 2^8 rounds - Avg: 0.377516 | Last added: 0.378906 New simul. with 2^9 rounds - Avg: 0.366866 | Last added: 0.378906 New simul. with 2^10 rounds - Avg: 0.362607 | Last added: 0.342743 New simul. with 2^11 rounds - Avg: 0.360595 | Last added: 0.358353 New simul. with 2^12 rounds - Avg: 0.359471 | Last added: 0.343569 New simul. with 2^13 rounds - Avg: 0.364962 | Last added: 0.336161 New simul. with 2^14 rounds - Avg: 0.500135 | Last added: 0.497771 New simul. with 2^15 rounds - Avg: 0.49995 | Last added: 0.488623 New simul. with 2^16 rounds - Avg: 0.602851 | Last added: 0.590848 New simul. with 2^17 rounds - Avg: 0.376087 | Last added: 0.372888 New simul. with 2^18 rounds - Avg: 0.655107 | Last added: 0.62898 New simul. with 2^19 rounds - Avg: 0.182425 | Last added: 0.201142 New simul. with 2^20 rounds - Avg: 0.709139 | Last added: 0.713385 New simul. with 2^21 rounds - Avg: 0.219937 | Last added: 0.220374 New simul. with 2^22 rounds - Avg: 0.112707 | Last added: 0.112427 Same simul. r=2^1 - Avg: 0.5 | Last added: 0.5 Same simul. r=2^2 - Avg: 0.5 | Last added: 0.75 Same simul. r=2^3 - Avg: 0.545139 | Last added: 0.46875 Same simul. r=2^4 - Avg: 0.625 | Last added: 0.28125 Same simul. r=2^5 - Avg: 0.60393 | Last added: 0.59375 Same simul. r=2^6 - Avg: 0.568329 | Last added: 0.71875 Same simul. r=2^7 - Avg: 0.57769 | Last added: 0.697266 Same simul. r=2^8 - Avg: 0.573474 | Last added: 0.631714 Same simul. r=2^9 - Avg: 0.575036 | Last added: 0.576538 Same simul. r=2^10 - Avg: 0.577153 | Last added: 0.47583 Same simul. r=2^11 - Avg: 0.578355 | Last added: 0.617221 Same simul. r=2^12 - Avg: 0.576684 | Last added: 0.57461 Same simul. r=2^13 - Avg: 0.576757 | Last added: 0.581285 Same simul. r=2^14 - Avg: 0.577305 | Last added: 0.546254 Same simul. r=2^15 - Avg: 0.577683 | Last added: 0.592735 Same simul. r=2^16 - Avg: 0.577662 | Last added: 0.56319 Same simul. r=2^17 - Avg: 0.577692 | Last added: 0.576607 Same simul. r=2^18 - Avg: 0.577675 | Last added: 0.571428 Same simul. r=2^19 - Avg: 0.577657 | Last added: 0.572818 Same simul. r=2^20 - Avg: 0.577655 | Last added: 0.579482 Same simul. r=2^21 - Avg: 0.577652 | Last added: 0.575974 Same simul. r=2^22 - Avg: 0.577654 | Last added: 0.5777 Same simul. r=2^23 - Avg: 0.577659 | Last added: 0.585123 Same simul. r=2^24 - Avg: 0.577657 | Last added: 0.571693 Same simul. r=2^25 - Avg: 0.577659 | Last added: 0.579782 Same simul. r=2^26 - Avg: 0.577659 | Last added: 0.574194 Same simul. r=2^27 - Avg: 0.577659 | Last added: 0.579098 REPLY [3 votes]: This is just a comment, but I wanted to include a graph. There are are $2^r-1$ possibilities for $z_r$. $z_r=\frac{p}{2^r}$ with$1 \le p \le 2^r-1.$ An interesting observation is that the denominator (in lowest terms) is unlikely to be close to $2^r$. If $z_r$ happens to have denominator $2^r$ then it is the only one with that denominator in (the current version of) $M$ and , aside from $\frac01$ and $\frac11$, each of $2,2^2,\cdots, 2^r$ is used as a denominator once. Then the denominator of $z_{r+1}$ is $2^s$ with probability $\frac{2s}{r^2+r}.$ so on average $s \approx \frac{2r}3.$ Here is the probability distribution for $z_6.$ Just under half the time it is $\frac12,\frac38,\frac58,\frac14,$ or $\frac34$ with frequencies roughly $12\%,10\%,10\%,8.5\%,8.5\%. $ That may not have any effect on the average behavior for very large $r$.<|endoftext|> TITLE: Matrix coefficients of a compact quantum group QUESTION [5 upvotes]: Let $(A, \Delta)$ be a $C^*$-algebraic compact quantum group (in the sense of Woronowicz). Definition: A corepresentation matrix of $(A, \Delta)$ is a matrix $a=(a_{i,j}) \in M_n(A)$ such that $$\Delta(a_{i,j}) = \sum_k a_{ik}\otimes a_{kj}$$ for all $i,j$. The matrix $a$ is called non-degenerate (unitary) when $a$ is invertible (unitary) in the $C^*$-algebra $M_n(A)$. Question: If $A_0$ is the set of matrix entries of unitary corepresentation matrices of $(A, \Delta)$ and $A_1$ is the set of corepresentations (with no restrictions) of $(A, \Delta)$, then do we have $A_0 = A_1$? Clearly $A_0 \subseteq A_1$, but I'm not sure if the converse inclusion holds. Any help will be appreciated! REPLY [4 votes]: In general, the converse inclusion $A_1 \subseteq A_0$ does not hold. As the counterexample below shows, it is not a good idea to define a corepresentation matrix as in the question. To really be considered as a corepresentation matrix, one should require $a$ to be invertible as an element in $M_n(A)$. Below is an example of a C$^*$-algebraic compact quantum group $(A,\Delta)$ containing an orthogonal projection $p \in A$ that is nontrivial ($p \neq 0$ and $p \neq 1$) and that satisfies $\Delta(p) = p \otimes p$. This pathological behavior can only happen in situations where the Haar state is not faithful. Let me therefore start with a positive result showing that in the reduced case (i.e. the case where the Haar state is faithful), the $*$-algebra $A_0$ of coefficients of finite dimensional unitary corepresentations is highly canonical and, in particular, contains all coefficients of corepresentation matrices as in the question, by applying the following proposition to the $*$-algebra generated by the elements $a_{ij}$. Proposition. Let $(A,\Delta)$ be a C$^*$-algebraic compact quantum group and assume that the Haar state on $A$ is faithful. Denote by $A_0$ the $*$-algebra of coefficients of finite dimensional unitary corepresentations. Let $A_1 \subseteq A$ be any $*$-subalgebra satisfying $\Delta(A_1) \subseteq A_1 \otimes_{\text{alg}} A_1$. Then $A_1 \subseteq A_0$ and equality holds if and only if $A_1 \subseteq A$ is dense. Proof. Denote by $h$ the Haar measure on $A$. By the Schur orthogonality relations, we can fix a complete set of irreducible and inequivalent unitary corepresentations $u_\alpha$ and bases for their underlying Hilbert spaces such that the matrix coefficients $u_{\alpha,i,j}$ satisfy $$h(u_{\alpha,i,j}^* u_{\beta,k,l}) = \begin{cases} F_{\alpha,i} > 0 &\;\;\text{if $\alpha = \beta$, $i = k$, $j = l$,} \\ 0 &\;\;\text{otherwise.}\end{cases}$$ Define the map $\Phi : A \to A \otimes A \otimes A$ by $\Phi = (\Delta \otimes id) \circ \Delta$. We thus have $$(h \otimes id \otimes h)((u_{\alpha,i,k}^* \otimes 1 \otimes u_{\alpha,l,j}^*) \Phi(a)) = F_{\alpha,l} \, h(u_{\alpha,i,j}^* a) \, u_{\alpha,k,l}$$ for all $a \in A$. Denote by $V(\alpha) \subseteq A$ the linear span of $u_{\alpha,i,j}$. The previous formula shows that either $A_1$ is orthogonal to $V(\alpha)$ (w.r.t. the scalar product given by $h$), or $V(\alpha) \subseteq A_1$. Given $a \in A_1$, we can take a finite-dimensional vector space $V \subseteq A$ such that $\Phi(a) \in V \otimes V \otimes V$. So if $a$ is not orthogonal to $V(\alpha)$, we have $V(\alpha) \subseteq V$. Therefore, $a$ is orthogonal to all but finitely many $V(\alpha)$. It follows that $a$ is contained in the linear span of finitely many $V(\alpha)$. This means that $a \in A_0$. Combined with the previous paragraph, we also find that $A_1 = A_0$ if we moreover assume that $A_1 \subseteq A$ is dense. Counterexample. Let $G$ be any nonamenable group. Denote by $A_u = C^*(G)$ the universal C$^*$-algebra and denote by $A_r = C^*_r(G)$ the reduced C$^*$-algebra. Denote by $\lambda : A_u \to A_r$ the regular representation and by $\varepsilon : A_u \to \mathbb{C}$ the trivial representation. Write $\pi = \lambda \oplus \varepsilon$. Put $A = \pi(A_u)$. Since the trivial representation is not weakly contained in the regular representation, $A = A_r \oplus \mathbb{C}$. Since $\pi \otimes \pi$ is weakly contained in $\pi$ (actually, contained in a multiple of $\pi$), there is a unique comultiplication $\Delta$ on $A$ such that $\Delta \circ \pi = (\pi \otimes \pi) \circ \Delta_u$. Then, $(A,\Delta)$ is a C$^*$-algebraic compact quantum group. Denote by $p = (0,1) \in A$ the natural projection. I prove that $\Delta(p) = p \otimes p$. Denote by $\pi_\lambda : A \to A_r$ and $\pi_{\varepsilon} : A \to \mathbb{C}$ the homomorphisms satisfying $\pi_\lambda \circ \pi = \lambda$ and $\pi_\varepsilon \circ \pi = \varepsilon$. Then, $$(\pi_\lambda \otimes \pi_\lambda) \circ \Delta \circ \pi = \Delta_r \circ \lambda \quad , \quad (id \otimes \pi_\varepsilon) \circ \Delta \circ \pi = \pi = (\pi_\varepsilon \otimes id) \circ \Delta \circ \pi \; .$$ Viewing $A_r \subseteq A$ (and noting that this inclusion is not unital), it follows that $$\Delta(a) = \Delta_r(a) + a \otimes p + p \otimes a \quad , \quad \Delta(p) = p \otimes p \;\;,$$ for all $a \in A_r$. The $*$-algebra $A_0$ of coefficients of finite dimensional unitary corepresentations of $(A,\Delta)$ is given by the group algebra $\pi(\mathbb{C}[G])$. In particular, the restriction of $\pi_\lambda$ to $A_0$ is injective. But $\pi_\lambda(p) = 0$. Thus, $p \not\in A_0$.<|endoftext|> TITLE: Which averages of products of a function give a norm? QUESTION [5 upvotes]: Let $f: [0,1] \rightarrow \mathbb{R}$ be a bounded measurable function. For some real non-negative numbers $a_1, a_2, b_1, b_2$ with $a_1+b_1=a_2+b_2=1$ consider the quantity $$N(f)=\int_{[0,1]} \int_{[0,1]} f(a_1x+b_1y)f(a_2x+b_2y) \, dx \, dy.$$ If $a_1=a_2=1$ and $b_1=b_2=0$ then $N(f)^{1/2}$ is just the $L_2$ norm of $f$. I am interested to understand when the quantity $N(f)^{1/2}$ defines a norm in other situations. Is anything like that known? Maybe at least some clear necessary conditions for the thing to be a norm? Naturally, if this is known and easy, the general question would for a $k$-wise product of $f$ with some affine combination of the integrating variables plugged in each $f$ and then the question would be whether or not $N(f)^{1/k}$ is a norm. REPLY [4 votes]: It is a norm when $a_1=a_2$, $b_1=b_2$. Otherwise making the change of variables $u=a_1x+b_1y, v=a_2x+b_2y$ we get the integral of $f(u)f(v)$ over certain parallelogram $P$ with a diagonal joining $(0, 0) $ and $(1, 1) $. Assume that $f$ has a small support $\Delta$ so that $\Delta^2\subset P$. Then the integral is just $(\int f)^2$ and this may be equal to 0.<|endoftext|> TITLE: Are there any "simple" monoids with intermediate growth? QUESTION [17 upvotes]: The discovery of the Grigorchuk group which has intermediate growth caused a number of other such groups to be found, but they are all fairly complicated, and as far as I know none of them are finitely presented. Are there simpler examples of intermediate growth if we drop the requirement that there exists an inverse? REPLY [22 votes]: Yes. Jan Okninski showed that $$\begin{bmatrix} 1 & 1 \\ 0 &1\end{bmatrix}\ \text{and}\ \begin{bmatrix} 1 & 0\\ 1 & 0\end{bmatrix}$$ generate a semigroup of intermediate growth. Details can be found in Nathanson. The growth was estimated there to be like the Hardy-Ramanujan estimate of the partition function. An exact asymptotic growth rate of $e^{\sqrt{n/\log n}}$ was obtained for this semigroup by Lavrik-Mannnlin. There are also finitely presented examples. The simplest example I know is Yuji Kobayashi. A finitely presented monoid which has solvable word problem but has no regular complete presentation. Theoret. Comput. Sci., 146(1-2):321–329, 1995.. This example is almost cancellative. It has a zero but you can cancel whenever neither product is zero. The growth is essentially the same as Okninski’s example. The first finitely presented example is due to James Shearer (James B. Shearer. A graded algebra with a nonrational Hilbert series. J. Algebra, 62(1):228– 231, 1980) and in a note added in proof he gives essentially the same presentation as Kobayashi but without proof. The generators are $a,b,c,0$ and the relations are $0x=0=x0$ for $x$ any generator and $ab=ba$, $bc=aca$ and $acc=0$. One can also build monomial examples easily that are not finitely presented. Take any infinite word $w$ over a finite alphabet $A$ whose factor complexity has intermediate growth (these exist for example here) and take the quotient of the free semigroup on $A$ by the ideal of all words not appearing as a factor in $w$. This is I believe the oldest construction due to Govorov I believe.<|endoftext|> TITLE: Is the Euler–Mascheroni constant an EL-number? QUESTION [7 upvotes]: This question is based on Chow - What is a closed-form number?. The author of the linked paper had proposed a plausible definition of "elementary numbers" (which he calls "EL-numbers") concept, building it analogously to the concept of elementary function (author admits, "elementary number" would be a better name for his proposal, but the term is already occupied). So, the author defines a set of $\mathbb{E}$ of "EL numbers" which stands for "elementary" and well as "exponentially-logarithmic". He defines the set as any numbers that can be produced by applying finite number of field operations, exponential and logarithmic functions to the number $0$. For instance, in his system \begin{gather*} 1=\exp(0) \\ e=\exp(\exp(0)) \\ i=\exp\left(\frac{\log(-1)}2\right)=\exp\left(\frac{\log(0-\exp(0))}{\exp(0)+\exp(0)}\right) \\ \pi=-i\log(-1)=-\exp\left(\frac{\log(0-\exp(0))}{\exp(0)+\exp(0)}\right)\log(0-\exp(0)). \end{gather*} It turns out that any root of a polynomial with rational coefficients, expressible in the radicals, is also in $\mathbb{E}$. So, my question is, does the Euler–Mascheroni constant $\gamma$ belong to the EL-numbers? I think no, but that it is "nearly-elementary" in the same way as the digamma function is a nearly-elementary function. My thoughts on this revolve around these points: There is symmetry between $\pi/4$ and $e^{-\gamma}$ $\gamma=\psi(1)$, but $\psi(x)$ is antidifference of $1/x$ while logarithm is antiderivative. $\psi(x)$ relates to $\log x$ the same way as Bernoulli polynomials relate to monomials (both Bernoulli polynomials and $\psi(x)$ are slices of Hurwitz Zeta function). Many divergent integrals and their logarithms regularize to $\gamma$, particularly, $\operatorname{reg} \int_0^1 \frac1x dx=\gamma$ (which makes it in some sense the regularized value of logarithm at zero). REPLY [10 votes]: Despite the simplicity and elegance of the definition of EL numbers, it is very hard to prove that an explicit number is not EL, as Chow points out throughout the paper. In particular, I am certain that as of right now, nobody knows if the Euler-Mascheroni constant is or isn’t EL. It most likely isn’t, but we can’t even prove that it is irrational.<|endoftext|> TITLE: Dehn surgery along primitive knot in 3-dimensional handlebody QUESTION [6 upvotes]: I'm studying the article "An alternative proof of Lickorish–Wallace theorem" (doi link) and I got stuck in a problem. Let $H_g$ be a 3 dimensional handlebody of genus $g$, a primate curve in $H_g$ is a knot in $\partial H_g$ that intersects an essential disk of $H_g$ in a single point. Let $c$ be a primitive curve, pushing $c$ in the interior of $H_g$ we obtain the knot $c'$. Now consider a spanning annulus $A$ in $H_g \setminus \eta(c')$ with $c \subset \partial A$, and the other boundary component of $A$ is called $c''$ and lies in $\partial \eta(c')$. How can I prove that if I perform a surgery on $c'$ along $c''$ I obtain a genus $g$ handlebody? According to my notations, a surgery on $c'$ along $c''$ means glueing the meridian $\{x\} \times \partial D^2 \subset S^1 \times D^2$ on $c''$. I found a similar question (Dehn surgery on handlebody), the answers (in particular the one by Ian Agol) seems to confirm that my statement is true, but there are no details. REPLY [9 votes]: Since $c\subset H_g$ intersects an essential disc $D$ in a single point, the boundary of a regular neighbourhood of $D\cup c$ is another disc $D'$, which splits $H_g$ into a solid torus containing $D\cup c$ and the rest. You can forget about the rest (this is a $\partial$-connected sum) and consider the solid torus alone. Here, if you push $c$ inside the solid torus, the complement will be diffeomorphic to $T \times [0,1]$ for a torus $T$, hence any Dehn surgery on one component will give you a solid torus back.<|endoftext|> TITLE: Morphisms between compact quantum groups QUESTION [6 upvotes]: Let $(A, \Delta_A)$ and $(B, \Delta_B)$ be two compact quantum groups (in the sense of Woronowicz). I would be tempted to define a morphism $(A, \Delta_A) \to (B, \Delta_B)$ to be a unital $*$-morphism $$\pi: B \to A$$ such that $$(\pi \otimes \pi)\circ \Delta_B = \Delta_A \circ \pi.$$ However, in the literature, I read that this definition is not the 'right' one because there are analytical subtleties. Instead, such a morphism is defined to be a $*$-morphism $\pi: B_0 \to A_0$ satisfying $(\pi \otimes \pi)\Delta_B = \Delta_A \pi$ where $A_0$ and $B_0$ are the associated spaces of matrix coefficients of finite-dimensional unitary representations. Equivalently , we can translate this to a $*$-morphism $B_u \to A_u$ between the associated universal compact quantum groups in the sense that I proposed above. What can go wrong with the definition I propose? Why is the definition with matrix coefficients the 'correct one'? Also, for most purposes, having a $*$-morphism $B\to A$ respecting the comultiplications seems good enough, so I am confused why one chooses to work with the matrix coefficients instead. For practical purposes, this seems bad as one has to know a lot of information about the spaces of matrix coefficients involved. REPLY [6 votes]: When $\pi : B \to A$ is a unital $*$-homomorphism respecting the comultiplication, then automatically $\pi(B_0) \subseteq A_0$, because $\pi$ maps a corepresentation of $(B,\Delta_B)$ to a corepresentation of $(A,\Delta_A)$. The converse need not hold: if $\pi : B_0 \to A_0$ is a unital $*$-homomorphism respecting the comultiplication, it need not be bounded for the C$^*$-norms on $B$ and $A$. So, by only considering $*$-homomorphisms $B \to A$, you may ``miss'' some quantum group morphisms. The required C$^*$-boundedness would be automatic if $B=B_u$ is the universal enveloping C$^*$-algebra of $B_0$. All this is best illustrated when $(B,\Delta_B)$ and $(A,\Delta_A)$ are C$^*$-algebras of discrete groups $\Lambda$ and $\Gamma$. Every group homomorphism $\delta : \Lambda \to \Gamma$ gives rise to a natural $*$-homomorphism $C^*(\Lambda) \to C^*(\Gamma)$, but not necessarily from $C^*_r(\Gamma)$ to $C^*_r(\Lambda)$.<|endoftext|> TITLE: Is the bitranspose continuous for the $\sigma$-strong topology? QUESTION [7 upvotes]: Let $\varphi\colon A\to B$ be a bounded, linear map between C*-algebras. Is the bitranspose $\varphi^{**}\colon A^{**}\to B^{**}$ continuous when the von Neumann algebras $A^{**}$ and $B^{**}$ are equipped with their $\sigma$-strong topologies? Motivation/Background: Note that $\varphi^{**}$ is clearly continuous when $A^{**}$ and $B^{**}$ are equipped with their $\sigma$-weak topologies, since these agree with the weak${}^*$-topology from the preduals, and $\varphi^{**}$ is weak${}^*$-continuous (that is, $\sigma(A^{**},A^*)-\sigma(B^{**},B^*)$-continuous). If $\varphi$ if completely positive, then it follows that $\varphi^{**}$ is a completely positive, normal map, and therefore is continuous for the $\sigma$-strong topologies. Thus, the question is only interesting if $\varphi$ is not completely positive. On bounded sets of a von Neumann algebra, the $\sigma$-strong topology agrees with the strong (operator) topology (SOT). If $M\subseteq B(H)$ is a von Neumann algebra, then a net $(a_j)_j$ in $M$ SOT-converges to $a\in M$ if $\|a_j\xi-a\xi\|\to 0$ for every $\xi\in H$. REPLY [6 votes]: I think the transpose map on the compacts gives a counterexample. Let $K(H)$ be the compacts on a separable infinite dimensional Hilbert space with orthonormal basis $\{ e_n \}.$ Let $T:K(H)\rightarrow K(H)$ be the transpose map (i.e. $T(e_{n,m})=e_{m,n}$ on matrix units). Then $T$ is $\sigma$-weakly continuous so $T^{**}:B(H)\rightarrow B(H)$ will also be the transpose map. Let $S$ be the shift $Se_n=e_{n+1}.$ Then $T(S^{*n})=S^n$ for all $n.$ Finally $S^{*n}\rightarrow 0$ strongly while $T(S^{*n})=S^n$ does not converge strongly to anything. Since all of these maps $S^n, S^{*n}$ are in the unit ball, as you mention the strong and $\sigma$-strong topologies coincide.<|endoftext|> TITLE: Brauer-Manin obstruction on an open subset of an elliptic curve QUESTION [9 upvotes]: First a disclaimer. This is an old question that I considered years ago and that I recently remembered. Since I am no longer in active research it may be considered as 'idle curiosity', although I feel I could probably learn a good deal from an answer (and maybe others could find it instructive as well). Now the question. Are there any examples of elliptic curves $E/\mathbb{Q}$ (or failing that over some other number field) of rank $0$, and an open subvariety $X\subset E$, such that $$ X(\mathbb{A}_\mathbb{Q})^{\operatorname{Br}(X)} = \varnothing? $$ Here the left-hand side denotes the Brauer-Manin set of $X$. Or more generally still, is it at all possible to use Brauer-Manin obstructions to prove the non-existence of rational points on curves outside of some closed (proper and non-empty) subset? The main problem that I recall having is that if one deals with Brauer classes on $X$ that are ramified (i.e. that do not come from $\operatorname{Br}(E)$), the evaluation map $$ \operatorname{ev}_X : X(\mathbb{A}_\mathbb{Q}) \times \operatorname{Br}(X) \to \mathbb{Q}/\mathbb{Z} $$ is hard to describe, since for $\mathscr{A} \in \operatorname{Br}(X)$ with $X$ non-proper there are in general infinitely many primes such that the local evaluation map $\operatorname{ev}_{\mathscr{A},p} : X(\mathbb{Q}_p) \to \mathbb{Q}/\mathbb{Z}$ is not identically zero (this in contrast to the case where $X$ is projective). The usual way of computing Brauer-Manin obstructions is that one exhibits a set of Brauer classes $\mathscr{A}_i$ and shows that the images of the maps $\operatorname{ev}_{\mathscr{A}_i} :U_i \to \mathbb{Q}/\mathbb{Z}$ are disjoint from $0$, for some partition $U_i$ of $X(\mathbb{A}_{\mathbb{Q}})$. But this becomes difficult when $X$ is non-proper, since one needs to determine the images of an infinite set of local evaluation maps $\operatorname{ev}_{\mathscr{A}_i,p}$ (for each $i$). (I don't even know whether it somehow follows from this that the examples I am asking for simply do not exist.) REPLY [7 votes]: Yes you can get: $$ X(\mathbb{A}_\mathbb{Q})^{\operatorname{Br}(X)} = \varnothing $$ in fact we can prove $$ X(\mathbb{A}_\mathbb{Q})^{\operatorname{Br}(E)} = \varnothing $$ and we can even make this (possibly larger) set to be empty. The Brauer Manin obstruction for Abelian varieties is nicely explained in section 4 of this article by Poonen en Voloch. And I will use the notation from this section without further explanation in this answer. Where it is useful to know that $A$ is an Abelian variety and $K$ a number field. The key ingredients are the exact sequence $$0 \to \widehat{A(K)} \to \widehat {\operatorname{Sel}} \to TШ \to 0$$ and the inclusions $$\widehat{A(K)} = \overline{A(K)} \subseteq A(\mathbb{A}_K)^{\operatorname{Br}(A)}_{•} \subseteq \widehat {\operatorname{Sel}}$$ of Theorem $E$. If we take an $A=E$ an elliptic curve over $\mathbb{Q}$ of analytic rank 0. Then we know by work of Kolyvagin that it also has algebraic rank 0 and that $Ш$ is finite. If $Ш$ is finite then $TШ$ is trivial and the exact sequence and the above inclusions imply: $$E(\mathbb Q) = \widehat{E(\mathbb Q)} = E(\mathbb{A}_\mathbb{Q})^{\operatorname{Br}(E)}_{•}$$ If we take $S=E(\mathbb{Q})$ and $X$ to be the open subvariety $E \setminus S$ then I claim $$ X(\mathbb{A}_\mathbb{Q})^{\operatorname{Br}(E)} = \varnothing $$ Indeed suppose for contradiction that $x \in X(\mathbb{A}_\mathbb{Q})^{\operatorname{Br}(E)}$ and let $x^f$ denote the finite part of $x$. Then $E(\mathbb Q) = E(\mathbb{A}_\mathbb{Q})^{\operatorname{Br}(E)}_{•}$ implies that $x^f$ comes from an element $s \in S$. However by definition of $X$ we have that the $x_v = s_v$ for at most finitely many places $v$ which is clearly incompatible with $x^f = s^f$. For explicitness taking $X$ to be the open part obtained from the curve https://www.lmfdb.org/EllipticCurve/Q/11/a/1 by removing the origin gives an example. This construction works more generally for $E$ over a number field $K$ whenever the rank is $0$ and $Ш$ is finite.<|endoftext|> TITLE: How does Mathematica do symbolic integration? QUESTION [30 upvotes]: I suppose there was at least once in our lifetime the point where we resorted to mathematica for help with an integral.-Unless you chose not to have the pleasure of using the continuum in your mathematical field of research. I am wondering however whether behind a computer algebra system like mathematica that does symbolic computations there is some high-level idea how it would approach an awkward integral and try to find an anti-derivative?-I focus here on the integration part, since I think I would understand how to implement symbolic differentiation, at least conceptually, but integration seems to be rather mysterious to me. REPLY [17 votes]: People usually mention the Risch algorithm first, as other answers have. Another approach, which is surprisingly successful, is to do what you or I would when solving integrals: look for patterns for which we have known-to-be-useful transformations. RUBI (which is usable from Mathematica as well as from other systems) does precisely this. According to its benchmarks, it works better than Mathematica's built-in Integrate. This is why I was not surprised to see that machine-learning methods can also work very well. While I did not read the paper, I imagine that the inherent unpredictability/unreliability of such methods is not an issue: the result can always be verified by differentiation. (Note: All I wrote above applies to indefinite integration. Definite integration is a different animal, and much harder problem, mostly because it is necessary to verify that the function behaves nicely in the middle of the integration interval.)<|endoftext|> TITLE: Reference for the Swan-Serre theorem as a monoidal equivalence QUESTION [15 upvotes]: Let $X$ be a compact Hausdorff The well-known Swan--Serre theorem gives an equivalence between the continuous vector bundles over a compact Hausdorff space $X$, and finitely-generated projective $C(X)$-modules. Both the category of vector bundles and the category of projective modules have evident monoidal structures. With respect to these structures, is the Swan--Serre equivalence a monoidal equivalence? I would guess that this is the case but I cannot find a reference. REPLY [12 votes]: Recall that the equivalence in question is given by taking global sections of a vector bundle, $V\mapsto \Gamma(X,V)$. Now if you have a section $s$ of $V$ and a section $t$ of $W$, this gives you a section $s\otimes t$ of $V\otimes W$ so this gives you a natural morphism $\Gamma(X,V)\otimes_{C(X)}\Gamma(X,W)\to \Gamma(X,V\otimes W)$ (note that if $f$ is a function on $X$, then $((f\cdot s)\otimes t)_x = f(x)s_x \otimes t_x = s_x \otimes f(x)t_x = (s\otimes (f\cdot t))_x = (f\cdot (s\otimes t))_x$, so this does descend to the $C(X)$-tensor product, and is indeed $C(X)$-linear) Furthermore, this morphism is an isomorphism when $V=W$ is the trivial line bundle over $X$, and both sides are additive functors in each variable, so it is an isomorphism whenever $V$ and $W$ are summands of trivial bundles, i.e. for all $V,W$ because any vector bundle is a summand of a trivial vector bundle (equivalently any finitely generated projective $C(X)$-module is a summand of a finitely generated free $C(X)$-module). So the answer is yes, $\Gamma(X,-)$ has a natural symmetric monoidal structure. I don't know the references though, so if this is what you're looking for then this doesn't quite answer.<|endoftext|> TITLE: Hurwitz numbers and $t$-cores QUESTION [7 upvotes]: For integers $k \geq 0$ and $d \geq 1$ let $H(k,d)$ be the Hurwitz number which, for the purposes of this posting, will be defined by: \begin{equation} H(k,d) \, := \ d! \, \sum_{\lambda \, \vdash d} \, \nu_{\scriptscriptstyle T}^k(\lambda) \ \ \text{where} \ \nu_{\scriptscriptstyle T}(\lambda) := \binom{d}{2} \cdot {{\chi^\lambda_{\scriptscriptstyle T}} \over {\dim(\lambda)} } \end{equation} and where $\chi^\lambda$ is the character value of the irreducible representation $V_\lambda$ of the symmetric group $S_d$ corresponding to the partition $\lambda \vdash d$ evaluated at any representative transposition (taken from the conjugacy class $T$ of all transpositions) and where $\dim(\lambda)$ is the dimension of $V_\lambda$. The Hurwitz number $H(k,d)$ can be interpreted, using the Verlinde formula, as counting the the number of homomorphisms (up to conjugation in $S_d$) \begin{equation} \rho : \pi_1 \Big( \Bbb{T}^2_k , \, \mathrm{base \, point}\Big) \longrightarrow S_d \end{equation} where $\Bbb{T}^2_k$ is the 2-torus with $k$ punctures. We can assemble these Hurwitz numbers into the following bivariate generating function \begin{equation} \begin{array}{ll} H(x;q) &\displaystyle = \ 1 \ + \ \sum_{d \geq 1} \, \sum_{k \geq 0} \, H(k,d) \, {x^k \over {k!}} \, q^d \\ &\displaystyle = \ 1 \ + \ \sum_{\lambda \ne \emptyset} \, q^{|\lambda|} \, \exp \big\{ x \, \nu_{\scriptscriptstyle T}(\lambda) \big\} \end{array} \end{equation} whose logarithm has a "genus" expansion \begin{equation} \log H(x;\tau) = F_1(\tau) \ + \ \sum_{g \geq 2} \, F_g(\tau) \, {x^{2g-2} \over {(2g-2)!}} \end{equation} where we set $q = e^{2\pi i \tau}$ and each $\tau$-series $F_g(\tau)$ is known to be a quasi-modular form. Now let $t \geq 2$ be an integer and let us consider the following $t$-core analogues: \begin{equation} \begin{array}{l} \displaystyle H_t(x;q) \, := \ 1 \ + \ \sum_{\stackrel{\scriptstyle \text{$t$-cores}}{\lambda \,\ne \, \emptyset}} \, q^{|\lambda|} \, \exp \ \big\{ x \, \nu_{\scriptscriptstyle T}(\lambda) \big\} \\ \displaystyle F_{g; \, t}(\tau) \, := \ \text{ the coefficient of} \ {x^{2g-2} \over {(2g-2)!}} \ \text{in} \ \log H_t(x;\tau) \end{array} \end{equation} Question 1: Does the generating function $H_t(x;q)$ have a nice closed expression, e.g. some sort of product formula? Question 2: Does the $\tau$-series $F_{g; \, t}(\tau)$ have any kind of modular property? thanks, ines. Post Script: As a kind of stupid example, consider the case of $2$-cores, which are precisely the stair-case partitions. The Murnaghan-Nakayama rule tells us that $\chi^\lambda_{\scriptscriptstyle T}$ can be evaluated recursively as the (signed) sum of dimensions $\dim(\mu)$ of partitions $\mu \vdash |\lambda| -2$ obtained by removing skew-hooks of size $2$ from the border of $\lambda$, i.e. \begin{equation} \chi^\lambda_{\scriptscriptstyle T} \ = \ \sum_{\stackrel{\scriptstyle \lambda \, = \, \mu + \sigma}{\sigma \, \vdash \, 2}} \, \big(-1 \big)^{\#(\sigma)-1} \, \dim(\mu) \end{equation} where $\#(\sigma)$ is the number of parts of $\sigma$. Of course there are no skew-hooks of size $2$ which can be excised from a stair-case partition, so $\chi^\lambda_{\scriptscriptstyle T} = 0$ for any $2$-core partition $\lambda$ and consequently \begin{equation} \begin{array}{ll} H_2(x \, ;q) &\displaystyle = \ 1 \ + \ \sum_{\stackrel{\scriptstyle \text{$2$-cores}}{\lambda \, \ne \, \emptyset}} \, q^{|\lambda|} \\ &\displaystyle = \ 1 \ + \ \sum_{d \geq 1} \, q^{{1 \over 2}d(d+1)} \\ &\displaystyle = \ \prod_{d \geq 1} \big(1 - q^{2d}\big) \cdot \big(1 + q^d \big) \end{array} \end{equation} Furthermore \begin{equation} \begin{array}{ll} \log H_2(x \, ; q) &\displaystyle = \ \sum_{d \geq 1} \, \log(1 - q^{2d}) \, + \, \log(1 + q^d) \\ &\displaystyle = \ F_{1 ; 2}(\tau) \end{array} \end{equation} The interesting computation begins with $3$-cores. Post-Post Script: One possible approach to the problem may be to take advantage of the Garvan-Kim-Stanton correspondence (GKS for short) which is a bijection \begin{equation} \lambda \stackrel{\phi}{\Longleftrightarrow} \vec{n} \end{equation} between $t$-cores $\lambda$ and integer vectors $\vec{n} = \big(n_0, n_1, \dots, n_{t-1} \big)$ with zero coordinate sum $n_0 + \cdots + n_{t-1} = 0$ such that \begin{equation} |\lambda| \ = \ {t \over 2} \| \vec{n} \|^2 \, + \, \vec{b} \cdot \vec{n} \end{equation} where $\vec{b} = \big(0 ,1 , \dots, t-1 \big)$. The trick might be to express the quantity $\nu_{\scriptscriptstyle T}(\lambda)$ in terms of the coordinates of the corresponding GKS-vector $\vec{n}$. Consider the case of $3$-cores: If my understanding of O. Brunat and R. Nath's pointed abacus construction is correct (see https://arxiv.org/pdf/2101.01512.pdf) a $3$-core partition $\lambda$ with GKS-vector $\vec{n}= \big(n_0, n_1, n_2 \big)$ has an arm of length $3p + r$ with residue $0 \leq r \leq 2$ if and only if $n_r$ is positive and $0 \leq p \leq n_r - 1$. Likewise $\lambda$ will have a leg of length $3p + 2 - r$ with $0 \leq r \leq 2$ if and only if $n_r$ is negative and $0 \leq p \leq | n_r | - 1 $. As mentioned in the comments \begin{equation} \begin{array}{ll} \nu_{\scriptscriptstyle T}(\lambda) &\displaystyle = \ {1 \over 2} \, \sum_{j=1}^k \, \Big(a_j + {1 \over 2} \Big)^2 - \Big(b_j + {1 \over 2} \Big)^2 \\ &\displaystyle = \ \sum_{j=1}^k \, a_j + {1 \over 2} a_j^2 \ - \ \sum_{j=1}^k b_j + {1 \over 2} b_j^2 \end{array} \end{equation} where $a_j$ and $b_j$ are the respective $j$-th arm and length lengths of the partition $\lambda$. So it should be possible to write $H_3(x \, ; q)$ as a piecewise polynomial function of the GKS-coordinates $n_0$, $n_1$, $n_2$. As illustration consider the situation where $n_0 < 0$ and $n_1 \geq -n_0$ and $n_2 = -n_0 - n_1 \leq 0$ which is one of the of six possible sign configurations of the three GKS-coordinates $n_0$, $n_1$, and $n_2$. By the Brunat-Nath recipe only $n_1$ will contribute arm lengths while $n_0$ and $n_2$ will contribute leg lengths. The arm contribution to $\nu_{\scriptscriptstyle T}(\lambda)$ will be \begin{equation} \begin{array}{ll} \displaystyle \sum_{j=1}^k a_j + {1 \over 2}a_j^2 &\displaystyle = \ \sum_{p=0}^{n_1 - 1} \, (3p+1) + {1 \over 2}(3p+1)^2 \\ &\displaystyle = \ {1 \over 2} 3n_1 \, + \, 3n_1(n_1-1) \, + \, {3 \over 4}n_1(n_1-1)(2n_1-1) \end{array} \end{equation} while the leg contribution to $\nu_{\scriptscriptstyle T}(\lambda)$ will be \begin{equation} \begin{array}{l} \displaystyle \sum_{j=1}^k b_j + {1 \over 2}b_j^2 \ \displaystyle = \ \left\{ \begin{array}{c} \displaystyle \sum_{p=0}^{|n_0| -1} \, (3p + 2) + {1 \over 2}(3p+ 2)^2 \\ + \\ \displaystyle \sum_{p=0}^{|n_2| -1} \, (3p) + {1 \over 2}(3p)^2 \end{array} \right. \\ = \, \left\{ \begin{array}{c} \displaystyle -4n_0 \, + \, {9 \over 2}n_0(n_0+1) \, - \, {3 \over 4}n_0(n_0+1)(2n_0+1) \\ + \\ \displaystyle {3 \over 2}n_2(n_2+1) \, - \, {3 \over 4}n_2(n_2+1)(2n_2+1) \end{array} \right. \\ = \, \left\{ \begin{array}{c} \displaystyle -4n_0 \, + \, {9 \over 2}n_0(n_0+1) \, - \, {3 \over 4}n_0(n_0+1)(2n_0+1) \\ + \\ \displaystyle -{3 \over 2}(n_0+n_1)(1-n_0-n_1) \, + \, {3 \over 4}(n_0+n_1)(1-n_0 -n_1)(1-2n_0 - 2n_1) \end{array} \right. \\ \end{array} \end{equation} Taking the difference of the arm and leg contributions we get the value of $\nu_{\scriptscriptstyle T}(\lambda)$ namely \begin{equation} \nu_{\scriptscriptstyle T}(\lambda) \ = \ {1 \over 2}\Big(3n_1^2 - 3n_0^2\big(1 + 3n_1\big) + n_0\big(2 + 3n_1 - 9n_1^2 \big) \Big) \end{equation} So the lattice points of the cone in $\Bbb{Z}^2$ cut out by the inequalities $n_0 < 0$ and $n_1 \geq -n_0$ and $n_2 = -n_0 - n_1 \leq 0$ make the following contribution to $H_3(x \, ; q)$ \begin{equation} \displaystyle \sum_{n_0 < 0} \sum_{n_1 \geq -n_0} \, q^{3n_0^2 + 3n_1^2 + 3n_0n_1 - 2n_0 - n_1} \exp \Big\{ {x \over 2}\Big(3n_1^2 - 3n_0^2\big(1 + 3n_1\big) + n_0\big(2 + 3n_1 - 9n_1^2 \big) \Big) \Big\} \end{equation} A similiar calculation can be undertaken for the remaining five cones in $\Bbb{Z}^2$. Does anyone recognize this kind of sum? REPLY [2 votes]: This posting shows how to handle the case of $3$-cores and compute $H_3(x \, ; q)$. Let's begin with the straight forward observation that $\lambda$ is a $t$-core if and only if its conjugate partition $\lambda'$ is a $t$-core. Furthermore, taking advantage of R. Stanley's comment \begin{equation} \begin{array}{ll} \nu_{\scriptscriptstyle T}(\lambda) &\displaystyle = \ \eta( \lambda) \ - \ \eta(\lambda') \\ &\displaystyle = \ \sum_{i \geq 1} \, \binom{\lambda_i}{2} \ - \ \sum_{i \geq 1} \, \binom{\lambda_i'}{2} \end{array} \end{equation} we see that $\nu_{\scriptscriptstyle T}(\lambda') = -\nu_{\scriptscriptstyle T}(\lambda)$. In particular $\nu_{\scriptscriptstyle T}(\lambda)$ vanishes when $\lambda$ is a self-conjugate partition. This means that \begin{equation} H_t( x \, ; q) \ = \ 1 + \displaystyle \sum_{\ \lambda \, = \, \lambda'} \, q^{|\lambda|} \ + \displaystyle \sum_{\nu_{\scriptscriptstyle T}(\lambda) \, > \, 0} 2 \cosh \big\{ \nu_{\scriptscriptstyle T}(\lambda) \, x \big\} \, q^{|\lambda|} \end{equation} where the sums are taken over non-empty $t$-core partitions. The self-conjugate $3$-cores are precisely those $3$-cores whose GKS-vectors are of the form $(a,0,-a)$ with $a \in \Bbb{Z}$. For GKS-vectors $(a,0,-a)$ with $a > 0$ the corresponding $3$-core partition $\lambda$ will have size $|\lambda| = a(3a-2)$ and first part $\lambda_1 = 3a-2$. For GKS-vectors $(-a,0,a)$ with $a \geq 0$ the corresponding $3$-core partition $\lambda$ will have size $|\lambda| = a(3a+2)$ and first part $\lambda_1 = 3a$. These calculation are made using the Brunat-Nath set-up mentioned in second post-script of my original post. The set of $3$-cores can be arranged into the following triangular hierarchy as depicted on page 142 of these notes (https://qcpages.qc.cuny.edu/~chanusa/courses/636/14/notes/636fa14ch50.pdf) by Christopher Hanusa. After staring at the alcove pattern a bit, I'll guess that (1) a $3$-core partition is self-conjugate if and only if $\nu_{\scriptscriptstyle T}(\lambda) = 0$ and (2) any $3$-core partition with with positive $\nu_{\scriptscriptstyle T}$-value can be uniquely expressed as $\rho^k \cdot \lambda$ where $\lambda = \big(\lambda_1, \dots, \lambda_\ell \big)$ is a self-conjugate $3$-core partition, $k \geq 1$ is an integer, and $\rho \cdot \lambda := \big(\lambda_1 + 2, \lambda_1, \dots, \lambda_\ell \big)$. Here $\rho^k \cdot \lambda$ denotes the $k$-fold iteration of $\rho$. Note that for a general partition $\lambda$ we have \begin{equation} \begin{array}{rl} \displaystyle \big| \rho \cdot \lambda \big| &\displaystyle = \ |\lambda| \, + \, \lambda_1 +2 \\ \displaystyle \nu_{\scriptscriptstyle T}\big( \rho \cdot \lambda \big) &\displaystyle = \ \nu_{\scriptscriptstyle T}(\lambda) \, - \, |\lambda| \, + \, \binom{\lambda_1 +2}{2} \end{array} \end{equation} and using the Faulhaber formulae it's not too hard to see that \begin{equation} \begin{array}{rl} \big| \rho^k \cdot \lambda \big| &\displaystyle = \ |\lambda| \ + \ k\lambda_1 \ + \ k(k+1) \\ \displaystyle \nu_{\scriptscriptstyle T}\big(\rho^k \cdot \lambda \big) &\displaystyle = \ \left\{ \begin{array}{l} \displaystyle \ \ \ \, \nu_{\scriptscriptstyle T}(\lambda) \, - \, k|\lambda| \\ \displaystyle + \ {1 \over 2} \, k\lambda_1^2 \, + \, {1 \over 2} \, k(k+2)\lambda_1 \\ \displaystyle + \ {1 \over 6} \, k(k+1)(2k+1) \end{array} \right. \end{array} \end{equation} If we iteratively apply $\rho$ to a self-conjugate $3$-core $\lambda$ with GKS-vector $(a,0,-a)$ with $a > 0$ and tally the total contribution to $H_3(x \, ; q)$ made by $\rho^k \cdot \lambda$ and its conjugate partition as $k \geq 1$ varies we get: \begin{equation} \begin{array}{ll} G^+_a(x \, ; q) &\displaystyle := \ \sum_{k \geq 1} 2 \cosh \big\{ U_+(a,k) \, x \big\} \, q^{D_+(a,k)} \ \ \text{where} \\ U_+(a,k) &\displaystyle := \ \left\{ \begin{array}{l} \displaystyle \ \ \ \, {1 \over 2} \, (a-2)(3a-2)k \\ \displaystyle + \ {1 \over 2} \, (3a-2)k(k+2) \\ \displaystyle + \ {1 \over 6} \, k(k+1)(2k+1) \end{array} \right. \\ D_+(a,k) &\displaystyle = \ a(3a-2) \ + \ (3a-1)k \ + \ k^2 \end{array} \end{equation} Similarly the total contribution to $H_3(x \, ; q)$ made by $\rho^k \cdot \lambda$ and its conjugate partition as $k \geq 1$ varies and where $\lambda$ is a self-conjugate $3$-core with GKS-vector $(-a,0,a)$ and $a \geq 0$ is: \begin{equation} \begin{array}{ll} G^{-}_a(x \, ; q) &\displaystyle := \ \sum_{k \geq 1} 2 \cosh \big\{ U_{-}(a,k) \, x \big\} \, q^{D_{-}(a,k)} \ \ \text{where} \\ U_{-}(a,k) &\displaystyle := \ \left\{ \begin{array}{l} \displaystyle \ \ \ \, {1 \over 2} \, a(3a-4)k \\ \displaystyle + \ {3 \over 2} \, ak(k+2) \\ \displaystyle + \ {1 \over 6} \, k(k+1)(2k+1) \end{array} \right. \\ D_{-}(a,k) &\displaystyle = \ a(3a+2) \ + \ (3a+1)k \ + \ k^2 \end{array} \end{equation} The self-conjugate $3$-cores on the own make a contribution of \begin{equation} G(q) \ = \ 1 + \ \sum_{a \, > \, 0} \ q^{a(3a+2)} \ + \ q^{a(3a-2)} \end{equation} and thus, when taken altogether, we get \begin{equation} H_3(x \, ; q) \ = \ G(q) \ + \ \sum_{a > 0} G^+_a(x \, ; q ) \ + \sum_{a \geq 0} G^{-}_a(x \, ; q) \end{equation} Perhaps the right-hand sum will be familiar to the readership of this posting.<|endoftext|> TITLE: A generating function for non-trivial zeros of Riemann zeta function QUESTION [5 upvotes]: Suppose $0^+_\zeta$ is the set of non-trivial zeros of the Riemann zeta function $\zeta(s)$ which lie on or to the right of the critical line and above the $x$-axis, i.e, $$0^+_\zeta = \{s \in \mathbb{C} : \zeta (s) = 0,\; \operatorname{Re}(s) \geq 1/2 \text{ and } \operatorname{Im}(s) > 0\}.$$ Consider the function $$ \Sigma(\nu) = \sum\limits_{s_0 \in 0^+_\zeta} \nu^{i(s_0 - 1/2)} = \sum\limits_{1/2 +it \in 0^+_\zeta} \nu^{-t}. $$ Is this well defined for any $\nu \in (0,1)$? Has this a priori complex-valued function been studied somewhere? REPLY [3 votes]: There is no good reason to take the imaginary part of the zeros or to keep only those in $\Re(s)\ge 1/2$. From the residue theorem we get for $\Re(x) >0$ the holomorphic function $$F(x)=\sum_{\Im(\rho)>0} e^{i\rho x}=\frac1{2i\pi} \int_{-1+i\infty}^{-1}+\int_{-1}^2+\int_2^{2+i\infty} (\frac{\zeta'(s)}{\zeta(s)}+\frac1{s-1})e^{isx}ds$$ $$=A(x)+B(x)+C(x)$$ Where the first term $$A(x)= \frac1{2i\pi}\int_{-1}^2(\frac{\zeta'(s)}{\zeta(s)}+\frac1{s-1})e^{isx}ds$$ is entire, The second term depends only on the functional equation $$B(x)=\frac1{2i\pi}\int_2^{2+i\infty} \frac{e^{isx}}{s-1}ds$$ $$+\frac1{2i\pi} \int_{-1+i\infty}^{-1}(\log(2\pi)+\frac{\pi \cos(\pi s/2)}{2\sin(\pi s/2)}-\frac{\Gamma'(1-s)}{\Gamma(1-s)}+\frac1{s-1})e^{isx}ds$$ And $$C(x)=\frac1{2i\pi}\int_0^{i\infty} (\frac{\zeta'(2+s)}{\zeta(2+s)}e^{i(2+s)x}+\frac{\zeta'(2-s)}{\zeta(2-s)} e^{i(s-1)x})ds$$ $$ = \frac{1}{2i\pi} \sum_{n\ge 1}\frac{\Lambda(n)}{n^2}(\frac{e^{2ix} }{ix-\log n}+\frac{e^{-ix}}{ix+\log n})$$ $-2\pi e^{-ix/2}C(x)$ is meromorphic on the whole complex plane with simple poles at the $\pm i\log n$ of residue $\frac{\Lambda(n)}{n^{1/2}}$ In other words $F(x)$ is much simpler than it seems.<|endoftext|> TITLE: Are the “generalized Catalan numbers” of Dumitrescu–Mulase the "moments" of some "multivariate Wigner semicircle distribution"? QUESTION [6 upvotes]: The classical Catalan numbers $$ C_n = \frac{1}{n+1} \binom{2n}{n}, $$ well-known for their numerous combinatorial interpretations (the second volume of Stanley's Enumerative Combinatorics famously lists a total of 66), essentially arise as the even moments $E[X^{2n}] = \int x^{2n} d\mu$ of the Wigner semicircle distribution $d\mu$: this distribution depends on a parameter $R$, the radius of the semicircle, and in general one has $E[X^{2n}] = (R/2)^{2n} C_n$, so choosing $R=2$ gives precisely the Catalan numbers. (The odd moments are all zero.) Now, associated to this distribution are a family of very classical orthogonal polynomials: the Chebyshev polynomials of the second kind, $U_n(x)$. (Up to a simple change of variables, $U_n(x)$ coincides with the matching polynomial of the path graph on $n$ vertices; for other interpretations and discussion see this blog post and this MO question.) In their paper Lectures on the topological recursion for Higgs bundles and quantum curves, O. Dumitrescu and M. Mulase define a sequence of numbers $C_{g,n}(\mu_1,\ldots,\mu_n)$ which they call generalized Catalan numbers. The definition (found on p. 26) is as follows: Definition: Let $\vec{\Gamma}_{g,n}(\mu_1,\ldots,\mu_n)$ denote the set of arrowed cell graphs drawn on a closed, connected, oriented surface of genus $g$ with $n$ labeled vertices of degrees $\mu_1,\ldots,\mu_n$. Then $$ C_{g,n}(\mu_1,\ldots,\mu_n) := |\vec{\Gamma}_{g,n}(\mu_1,\ldots,\mu_n)|. $$ Below this, the authors justify the terminology by pointing out that $C_{0,1}(2m) = C_m = \frac{1}{m+1} \binom{2m}{m}$. I don't know much about the story of multivariate orthogonal polynomials, and there are quite a few parameters here, but I can't help but wonder: might there perhaps be some multivariate analogue of the Wigner semicircle distribution, having these quantities $C_{g,n}(\mu_1,\ldots,\mu_n)$ as its (appropriately defined) moments – and with it, a corresponding family of orthogonal polynomials? I also find it curious that the above paper discusses Laplace transforms, since the (bilateral) Laplace transform is exactly how one passes from a probability density function to its moment-generating function (as one learns in a first course on probability theory.) Admittedly, this is probably just a coincidence. REPLY [4 votes]: I'm new here and not sure if this constitutes an answer, but I saw this was open for a while and figure I can shed some light (please remove, edit, etc. if needed). You say that the $k$-th Catalan number is equal to $\int x^{2k}d\mu$ where $\mu$ is the Wigner SC distribution. Let me rephrase this in terms of the density $\rho(x)$ so that $C_k=\int x^{2k}\rho(x)dx$ (i.e., $\rho(x)$ is just the Radon-Nikodym derivative of $\mu$). The underlying story here is that we have an $n\times n$ matrix, belonging to the Gaussian unitary ensemble, whose eigenvalues have density $\rho_1(x)$. Moreover, the full joint p.d.f. for these eigenvalues is actually given by the $n$-point correlation function $\rho_n(x_1,\ldots,x_n)$; note that the $m$-point correlation function is defined by $$ \rho_m(x_1,\ldots,x_m):=\frac{(n-1)!}{(n-m)!}\int\rho_n(x_1,\ldots,x_n)dx_{m+1}\cdots dx_n. $$ If you haven't encountered this before, an example to keep in mind is that $\int_a^b\int_c^d\rho_2(x,y)dxdy$ is the joint probability that there is an eigenvalue in $(a,b)$ while there is another eigenvalue in $(c,d)$. Notice that I refer to the Wigner SC law as $\rho(x)$ and the spectral density of the $n\times n$ GUE matrix as $\rho_1(x)$. The distinction here is that $$ \rho(x):=\lim_{n\to\infty}\rho_1(x), $$ which is an important part of the story (this is where the genus parameter $g$ comes into play). You mention that the Wigner SC $\rho(x)$ is related to the Chebyshev orthogonal polynomials of the second kind; one should instead consider the Hermite orthogonal polynomials, which relate to $\rho_1(x)$ in the same way. Then, it is possible to express the $m$-point correlation functions as determinantal processes with kernel given in terms of the Hermite orthogonal polynomials. You'll want to look up chapter 5 of Mehta's book, 'Random Matrices' and maybe the Christoffel-Darboux formula. I hope this partially answers your question on orthogonal polynomials (there are of course studies on multivariate orthogonal polynomials, but I'm not sure how relevant they are to this question). Moving on, note that it is important to distinguish the eigenvalue density $\rho_1(x)$ from its large $n$ limit $\rho(x)$. Now introduce the smoothed density, $$ \tilde{\rho}_1(x):=\rho(x)+\sum_{g=1}^{\infty}\frac{\rho_{1,g}(x)}{n^{2g}}, $$ which is defined such that its moments coincide with those of $\rho_1(x)$. However, the formal series defining $\tilde{\rho}_1(x)$ and the true density $\rho_1(x)$ differ by oscillatory contributions which aren't reflected in the moments (thus, the two densities agree only in the large $n$ limit). The point here is that the generalised Catalan numbers of Dumitrescu and Mulase are such that $$ C_{g,1}(\mu)=\int x^\mu\rho_{1,g}(x)dx, $$ with $\rho_{1,0}(x):=\rho(x)$ given by the Wigner SC law. This can be expressed in a multivariate way by defining $$ \tilde{\rho}_n(x_1,\ldots,x_n):=\sum_{g=0}^{\infty}\frac{\rho_{n,g}(x_1,\ldots,x_n)}{n^{n-1+2g}} $$ so that the moments of $\tilde{\rho}_n$ and the cumulants of $\rho_n$ agree (the moment-cumulant relation is why $\tilde{\rho}_n$ is O$(n^{1-n}$). Then we have that $$ C_{g,1}(\mu)=\int\left(\frac{1}{n}\sum_{i=1}^nx_i^\mu\right)\rho_{n,g}(x_1,\ldots,x_n)dx_1\cdots dx_n $$ (use symmetry arguments to compare with the earlier equation if you doubt this), and more generally $$ C_{g,m}(\mu_1,\ldots,\mu_m)=\int\prod_{i=1}^m\left(\frac{1}{n}\sum_{j=1}^nx_j^{\mu_i}\right)\rho_{n,g}(x_1,\ldots,x_n)dx_1\cdots dx_n. $$ I hope this is helpful, though I warn that I may have some proportionality constants missing, and I've left out many details as this is quite a vast topic. I'll end with pointing out that the presence of the Laplace transform is not coincidental. Transitioning between p.d.f.s and moment generating functions is key to making things rigorous.<|endoftext|> TITLE: Reference request: Who first proved that right adjoints preserve limits? QUESTION [7 upvotes]: One of the most famous and unifying theorems in category theory is that right adjoints preserve limits. I wonder: Who was the first one to prove this fact? The notion of adjoint functors is, of course, due to Daniel Kan. But I couldn't find the mentioned fact in his paper Adjoint Functors. REPLY [19 votes]: Daniel M. Kan defined adjoint functors in his paper Adjoint functors (written in 1956). In Chapter II he defines limits and colimits of arbitrary small diagrams and proves that the limit and colimit functors are right and left adjoints to the diagonal functor in Theorems 7.8 and 8.6. In Chapter III, he defines the notion of a limit-preserving functor in Definition 13.1. In Theorem 13.8 and 13.8* he proves that left/right adjoints preserve (co)limits.<|endoftext|> TITLE: Is material set theory conservative over structural set theory? QUESTION [10 upvotes]: Suppose a statement $\phi$ that doesn't use the global $\in$-relation or the global $=$-relation in an essential way is provable in some material set theory, say bounded Zermelo with choice. (So that the statement $\phi$ itself essentially is structural while, in theory, the intermediate steps in the proof of $\phi$ might be material, thus containing the global $\in$-relation or the global $=$-relation in an essential way!) Can it be proved that $\phi$ (or, rather, the translation of $\phi$ into the appropriate structural set theory) is nevertheless provable in some structural set theory, say ETCS, meaning that all the material intermediate steps can be eliminated? (Note that since $\phi$ doesn't use the global $\in$-relation or the global $=$-relation in an essential way, such a translation should exist.) Of course, in practice, most proofs are already structural. But who knows what crazy things could be done in theory by exploiting the global $\in$-relation? I already know that ETCS and bounded Zermelo with choice are equiconsistent. Though related, it seems I'm asking for something stronger. REPLY [11 votes]: This will obviously be highly dependent on the concrete theory you are considering. But overall the answer is yes. The most general version of these results I'm aware of are in Mike Shulman's Comparing material and structural set theories. You'll find all the answer you want in this paper. But let me try to give you some pointers: I think the most relevant result here is lemma 9.1 which (roughly, see the paper for the details) says: "Given $\mathbf{Set}$ ( a suitable pretopos, for eg. a model of ETCS) then the inclusion $\mathbb{S}\mathbb{V}(\mathbf{Set}) \to \mathbf{Set}$ is an equivalence if and only if every object of $\mathbf{Set}$ can be embedded into an extentional well founded graph" where $\mathbb{V}$ denotes the construction described in Mike's paper that attach a material set theory to a structural theory and $\mathbb{S}$ takes a material set theory to its category of sets. Any formula in $\phi$ in the language of a structural set theory is in particular a forumla in the 'language of categories' so is invariant by equivalence of categories. So, if the lemma applies, a formula $\phi$ is valid in $\mathbf{Set}$ if and only if it is valid in $\mathbb{S}\mathbb{V}(\mathbf{Set})$ that is exactly if the material translation of $\phi$ is valid in $\mathbb{V}(\mathbf{Set})$. So if the material translation of $\phi$ is provable in the theory of $\mathbb{V}(\mathbf{Set})$ then it holds in $\mathbf{Set}$. If $\mathbf{Set}$ is a model of ETCS, then $\mathbb{V}(\mathbf{Set})$ satisfies Bounded Zermelo with choice ( and Mostowski principle and transitive closure, again, see mike's paper for the details). So that answer your question. Note that, In the general case, the key condition is whether in $\mathbf{Set}$ every object can be embedded into an extentional well founded graph. Working with ETCS, or more generally if you have choice then this is really not a problem (e.g. put a well order on your set), but in more generality it might be problematic: By construction material set theory (well founded and with transitive closure) always construct a set as a subset of an extentional well founded graph (its transitive closure with the $\in$ relation), so if your structural set theory cannot do this as well, this gives a counter-example to the question you are asking, as the existence of such an embedding is a meaningful structural statement. Maybe there is a generalization of this construction that allows to get non well founded set theories that would go arround this problem, but I am not aware of this having been studied (?).<|endoftext|> TITLE: Reference request: Kleiman's proof of Snapper's Lemma QUESTION [5 upvotes]: On page 4 of Nitin Nitsure's paper Construction of Hilbert and Quot Schemes, the author refers to the fact that Hilbert polynomials are indeed polynomials as a special case of Snapper's Lemma, see "An Intersection Theory for Divisors (preprint 1994)" by Steven Kleiman for a proof. Kleiman's paper (or book?) mentioned by Nitsure must have either changed its title, never gone into publication, or elsehow disappeared, since I am unable to find it. Does anybody have a link, or an alternative source for the proof? Thanks in advance. REPLY [8 votes]: Note that Nitsure's paper is part of the book FGA Explained. There is a proof of Snapper's lemma in Theorem B.7 of Appendix B ("Basic intersection theory" by Kleiman) in the same book. Kleiman's part of the book can also be found independently on arXiv: 0504020 and contains the relevant appendix.<|endoftext|> TITLE: Chevalley-Warning-Ax for double covers QUESTION [9 upvotes]: Let $f(x_1,\ldots,x_n)$ be a polynomial of degree $d$ with coefficients in the finite field $\mathbb F_q$ and let $V(f)\subseteq\mathbb F_q^n$ be its set of zeroes. Assume $d TITLE: Nondifferentiable convex function whose subdifferential admits a continuous selection QUESTION [7 upvotes]: Is there a convex function $F$ that is not differentiable, but whose subdifferential admits a continuous selection, i.e. a continuous $g$ with $g(x) \in \partial F(x)$ for all $x$ in the domain? In one dimension I think I can prove there is not: if $|\partial F(x)| > 1$ then it contains an open set, whose inverse image under $g$ is an open set, so $g$ is not a selection after all - I think it maps some $y \neq x$ to a point in the interior of $\partial F(x)$, which cannot be a subgradient at $y$. REPLY [7 votes]: The answer is no. See Rockafellar's Convex Analysis, part V. First, let $D$ be the set of points where $F$ is differentiable. Theorem 25.5 proves that $D$ is dense in the interior of the domain of $F$, with measure zero complement. And that $\nabla F$ is a continuous mapping on $D$. Next, if the domain of $F$ has non-empty interior, then at every interior point the subdifferential has the property (Theorem 25.6): $$ \partial F(x) = \mathrm{cl} ( \mathrm{conv}(S(x))) $$ where $S(x)$ is the set of limits of all sequences of the form $\nabla F(x_i)$ where $x_i \to x$ and $x_i \in D$. Now if $\partial F$ admits a continuous section $g$, then for every $x$ in the interior of the domain of $F$, we have that for every sequence $x_i \in D$ such that $x_i \to x$ it holds that $g(x_i) \to g(x)$ (continuity of $g$); but since $\partial F(x_i) = g(x_i)$ is unique, this implies that $S(x) = \{g(x)\}$. And hence $F$ is differentiable at $x$.<|endoftext|> TITLE: Are finite $G$-spectra idempotent complete? QUESTION [10 upvotes]: Question: Let $G$ be a compact Lie group (you can assume that $G$ is finite if you like). Is the category of finite $G$-spectra idempotent complete? Here, by "finite $G$-spectra", I mean those objects in the category of genuine $G$-spectra which can be constructed via a finite number of sums and cofiber sequences from orbit cells $S^n \wedge (G/H)_+$ where $H \subseteq G$ is a closed subgroup. This includes all representation spheres. By "category", I mean either the stable $\infty$-category or the triangulated category, whichever you prefer (in the stable case, a category is idempotent complete iff its homotopy category is, so it doesn't affect the question.) When $G$ is trivial, the answer is yes, but I suspect the answer in general is no, -- maybe already for $G = C_2$? By a theorem of Thomason, triangulated subcategories of a triangulated category $\mathcal T$ which generate $\mathcal T$ under splitting of idempotents are in bijection with subgroups of $K_0(\mathcal T)$. So another way to phrase the question is the following: let $\mathcal T$ be the category of compact $G$-spectra. Then is t$K_0(\mathcal T)$ generated as a group by the classes of $G$-orbits $\Sigma^n(G/H)_+$? REPLY [5 votes]: To complement Oscar's more systematic answer, let me expand my comment about the case $G = \mathbf{Z}/p\mathbf{Z}$ for a prime number $p$, where the answer is no when $\tilde{K}_0(\mathbf{Z}[G]) \neq 0$. Candidate non-finite retracts of finite $G$-spectra have already been presented in the comments, to see that they are not equivalent to finite genuine spectra we can use that the natural homomorphisms $$\tilde{K}_0(\mathbf{Z}[G]) \to \tilde{K}_0(\mathbf{Z}[\zeta_p]) \to \tilde{K}_0(\mathbf{Z}[\zeta_p,p^{-1}])$$ are both isomorphisms. The first of these is an isomorphism by a theorem of Rim. To see that the second map is an isomorphism we use that $\mathbf{Z}[\zeta_p,p^{-1}] = \mathbf{Z}[\zeta_p,(\zeta_p-1)^{-1}]$ and that $(\zeta_p - 1) \subset \mathbf{Z}[\zeta_p]$ is a prime ideal (see e.g., the proofs of Lemma 1.3 and 1.4 on page 2 of Washington's book, in fact $p = (\zeta_p - 1)^{p-1} u$ for $u \in \mathbf{Z}[\zeta_p]^\times$), so exercise 3.8(b) in chapter I of Weibel's K-book applies. For any genuine $G$-spectrum $X$, the fixed points $X^e$ for the trivial subgroup come with an action of $G$, so we can form $$C_* (X;\mathbf{Z}[\zeta_p,p^{-1}]) := C_* (X^e;\mathbf{Z}) \otimes_{\mathbf{Z}[G]} \mathbf{Z}[\zeta_p,p^{-1}].$$ This defines an exact functor $C_* (-;\mathbf{Z}[\zeta_p,p^{-1}])$ from the stable $\infty$-category of genuine $G$-spectra to $\mathcal{D} (\mathbf{Z}[\zeta_p,p^{-1}])$. This functor sends $\Sigma^\infty_+ (G/e) \mapsto \mathbf{Z}[\zeta_p,p^{-1}]$ and an easy exercise shows that it sends $\Sigma^\infty_+ (G/G) \mapsto 0$. Therefore a genuine $G$-spectrum which is finite in your sense will be sent to a compact object in $\mathcal{D} (\mathbf{Z}[\zeta_p,p^{-1}])$ representing $0 \in \tilde{K}_0(\mathbf{Z}[\zeta_p,p^{-1}])$. Any element of $\tilde{K}_0(\mathbf{Z}[\zeta_p,p^{-1}]) $ may be represented by the image under this functor of a retract of $ ( \Sigma^\infty_+ G ) ^{\oplus n} $ for some $n$. Some such retract then won't be finite when $\tilde{K}_0 (\mathbf{Z}[\zeta_p,p^{-1}]) \neq 0$.<|endoftext|> TITLE: An explicit isomorphism $K^\times/K^{\times p} \cong K^\flat/\wp K^\flat$ where $K \supset \mu_p$ is perfectoid field of mixed characteristic $(0, p)$ QUESTION [12 upvotes]: Let $K$ be a perfectoid field of mixed characteristic $(0, p)$, i.e. $K$ has characteristic $0$ but its residue field has characteristic $p$. Further suppose that $K$ contains all the $p$th roots of unity. A subgroup of $K^\times/K^{\times p}$ corresponds to an abelian extension $L/K$ of exponent $p$ by Kummer theory, which corresponds to an abelian extension $L^\flat/K^\flat$ of exponent $p$ by the tilting correspondence, which in turn corresponds to a subgroup of $K^\flat/\wp K^\flat$ by Artin–Schreier theory. Combining these correspondences should give us an isomorphism $K^\times/K^{\times p} \cong K^\flat/\wp K^\flat$, up to a choice of isomorphism $\mu_p \cong \Bbb Z/p\Bbb Z$. My question is: Can we make any direction of this isomorphism explicit? I have tried a few examples and computations to no avail. REPLY [9 votes]: Here is an explicit description of the isomorphism: It takes $a\in K^\flat$ to the class of $1+(\zeta_p-1)^p a^{1/p^n}\in K^\times$ for any large enough $n$ (the image modulo $p$-th powers is independent of the choice). Here's an explanation. Let's first analyze the quotient $K^\times/(K^\times)^p$. As $K$ is perfectoid, any class in here is represented by an element of $1+p\mathcal O_K$. In fact, thinking a bit more about it, by an element of $1+p^{p/(p-1)-\epsilon} \mathcal O_K$ for any $\epsilon>0$ (where this statement only depends on valuations, so we do not need a literal element $p^{p/(p-1)-\epsilon}$ to make sense of this). On the other hand, any element of $1+p^{p/(p-1)+\epsilon}\mathcal O_K$ is a $p$-th power. In other words, the quotient $K^\times/(K^\times)^p$ is a quotient of $(1+p^{p/(p-1)-\epsilon}\mathcal O_K)/(1+p^{p/(p-1)+\epsilon}\mathcal O_K)$, where the latter group is isomorphic via $x\mapsto 1+(\zeta_p-1)^p x$ (where $(\zeta_p-1)^p$ has the same valuation as $p^{p/(p-1)}$) to a quotient of the additive group $p^{-\epsilon}\mathcal O_K/p^\epsilon \mathcal O_K$. To figure out the quotient, we have to look at $p$-th powers of elements of the form $1+(\zeta_p-1)y$, with $y\in p^{-\epsilon}\mathcal O_K$, which gives modulo $1+p^{p/(p-1)+\epsilon}$: $$1+p(\zeta_p-1)y+(\zeta_p-1)^py^p\equiv 1+(\zeta_p-1)^p (y^p-y).$$ Using $p^{-\epsilon} \mathcal O_K/p^\epsilon \mathcal O_K\cong t^{-\epsilon} \mathcal O_{K^\flat}/t^\epsilon \mathcal O_{K^\flat}$, this quickly gives the desired isomorphism to $\mathrm{coker}(x^p-x|K^\flat)$ (using a similar analysis of the latter, to see that all elements can be represented by elements of $t^{-\epsilon}\mathcal O_{K^\flat}$ -- concretely, replace any $a$ by $a^{1/p^n}$ for large $n$ -- and the elements of $t^\epsilon \mathcal O_{K^\flat}$ are trivial). Why is this the right isomorphism? Look at a Kummer extension $T^p=1+(\zeta_p-1)^px$ with $x\in p^{-\epsilon}\mathcal O_K$. Changing coordinates as $T=1+(\zeta_p-1)U$, this translates into an equation $$1+(\zeta_p-1)^p(U^p + (\zeta_p-1)[\ldots] - U) = 1 + (\zeta_p-1)^p x,$$ i.e. $$ U^p-U = x + (\zeta_p-1)[...]. $$ Removing the terms divisible by $\zeta_p-1$ does not change the extension for $\epsilon$ small enough, and the tilt is given by the similar Artin--Schreier equation (by the recipee explained in my paper: once the discriminant of the equation is small enough, one can find the tilt by the naive operation on defining equations).<|endoftext|> TITLE: What is the theory of the random poset? QUESTION [8 upvotes]: $\DeclareMathOperator\Th{Th}$The random poset is the Fraisse limit of the class of finite posets, just like the random graph is the Fraisse limit of the class of finite graphs? That is, the random poset is the unique (upto isomorphism) countable poset $P$ such that every finite poset embeds into $P$ and every order isomorphism between finite subsets of $P$ extends to an order automorphism of $P$. Now it has been proven that $\Th(P)$ is countably categorical, with $P$ being its unique model upto isomorphism. But my question is, do we know exactly what $\Th(P)$ is, or at least the complexity of the axioms of $\Th(P)$? I’m hoping for a result similar to the $\Pi^0_2$ axiomatization of the theory of the random graph. REPLY [10 votes]: It is confusing to call this the “random poset”, as it is very different from what is usually called random posets in the literature (in accordance with random graphs): e.g., random posets have height 3 with high probability. Unambiguous descriptions of this structure found in the literature are the countable universal homogeneous poset, the countable generic poset, or the countable existentially closed poset. (I will use the latter, as the axiomatization below rather directly corresponds to existential closedness.) There is a perfectly general description. If $T$ is any universal theory in a finite relational language with the amalgamation property (and the joint embedding property, which however follows from AP in this case as long as you allow the empty structure), the Fraïssé limit of finite models of $T$ exists, and it is the unique countable existentially closed model of $T$. Its theory $T^*$ is $\omega$-categorical, and it is the theory of existentially closed models of $T$. $T^*$ can be explicitly axiomatized by $T$ + all axioms of the form $$\forall\vec x\:(\mathrm{Diag}_A(\vec x)\to\exists z\:\mathrm{Diag}_B(\vec x,z)),$$ where $A\models T$ is finite (possibly empty), $\mathrm{Diag}_A$ denotes (the conjunction of) the diagram of $A$, and $B\models T$ is an extension of $A$ of size $|A|+1$. [Proof sketch: on the one hand, the Fraïssé limit satisfies the given axioms; this amounts to the defining property that any embedding of $A$ into the structure extends to an embedding of $B$. On the other hand, a straightforward zig-zag argument shows that the theory with the given axioms is $\omega$-categorical. Thus, it is the complete theory of the Fraïssé limit.] This is essentially a special case of the (possibly infinitary) axiomatization of existentially closed models of a given $\forall_2$ theory in terms of resultants; see §8.5 in Hodges, Model theory, CUP, 1993. In the case of $T$ being the theory of posets, the axiomatization of $T^*$ above simplifies to: The axioms of partial order. The axioms $$\begin{multline} \forall x_1,\dots,x_n,y_1,\dots,y_m,u_1,\dots,u_p\:\Bigl[\bigwedge_{i,j}(x_i TITLE: Strategies for bounding the spectral norm of a tensor? QUESTION [7 upvotes]: Let $A$ be a symmetric $k$-tensor over a real or complex vector field $W$. We may define its spectral norm $|A|$ by $$|A| = \sup_{v\in W} \frac{|\langle A,x^{\otimes k}\rangle|}{|x|_2^k}.$$ (Alternatively, we may define it in a different, standard way and then show that the above inequality holds - that is a theorem of Banach's (see, e.g., https://www.stat.uchicago.edu/~lekheng/work/nuclear.pdf , beginning of section 5).) What would be some strategies -- combinatorial of preference -- to bound the norm of $|A|$? By "combinatorial" I mean something analogous to "counting paths". Let me give an example of a common strategy for $k=2$. The trace of any power $A^m$ of a real symmetric matrix can be expressed in two ways: combinatorially, that is, as a sum over closed paths (interpreting $A$ as the adjacency matrix of a graph with weights) and spectrally, as a sum $\sum_i \lambda_i^m$ over all eigenvalues $\lambda_i$. In fact, we do not need the full spectral decomposition - it is enough to note that $\langle v,A^2 v\rangle = \langle A v,A v\rangle \geq 0$ for any $v$. Then it follows that, for $m$ even, $$|A|^m\leq \textrm{Tr}\,A^m = (\text{sum over closed paths}).$$ There are various ways to amplify this bound (e.g., high multiplicity). Is an inequality even vaguely resembling this one possible for higher $k$? REPLY [6 votes]: I will show, with some non-rigorous steps, that a bound of this form that is valid for arbitrary tensors and useful for sparse tensors (fewer than $n^{k/2}$ nonvanishing entries) does not exist. First note that there is a problem with using the $\ell^2$ norm to define the spectral norm for very sparse tensors $A$, regardless of how you try to bound it. The reason is that, assuming the nonzero entries of $A$ are normalized to have size $\approx 1$, the minimum value you could possibly hope to bound the spectral norm to is $\approx 1$, by choosing some nonzero entry $A_{n_1n_2 \dots n_k}$ of $A$ and choosing $x$ a sparse vector supported on $n_1,\dots, n_k$. This means if $x$ is a vector of support of size $m$ and average size $1$, then the $\ell^2$ norm of $x$ is $m^{1/2}$, so the best bound you could hope for on $\langle A, x^{\otimes k} \rangle$ from the spectral norm is $m^{k/2}$. But the trivial bound for $\langle A, x^{\otimes k} \rangle$ is $md$ where $d$ is the maximum degree of $A$ (the maximum number of nonzero entries in a hyperplane slice of $A$). So to get a nontrivial bound you want $m < d^{ \frac{2}{k-2}}$, which is pretty small if $d$ is large! So I'll use the $\ell^p$ norm instead for the remainder, for some $p\geq k$. Let $W = \mathbb R^n$ be a real vector space of dimension $n$. Suppose that $f$ is a homogeneous polynomial in the entries of a tensor $A$ in $W^{ \otimes k}$ where we have an inequality of the form $$ \left(\sup \frac{ \langle A, x^{\otimes k} \rangle}{ |x|_p^k} \right)^{d} \leq f(A) $$ (One could think we should allow absolute values on the right side but we can just square both sides to remove them.) I will argue rigorously that the coefficient of some monomial in $f$ has size at least $n^{ dk (1/2-1/p)}/ O_{d,k}(1)$ and heuristically that no $f$ with such a large coefficient gives a good bound for highly sparse tensors. Translating $f$ by any matrix $g$ in the group $ (\mathbb Z/2) \rtimes S_n$ produces a tensor with the same spectral norm. So the minimum of $f( g\cdot A)$ for $g \in (\mathbb Z/2) \rtimes S_n$ for also bounds the spectral norm. It follows that the average of $f(g\cdot A)$ over $g \in (\mathbb Z/2) \rtimes S_n$ also bounds the spectral norm. This average cannot increase the largest coefficient of a monomial in $f$, so we may as well assume $f$ is $(\mathbb Z/2) \rtimes S_n$-invariant. By scaling, certainly $f$ must be homogeneous of degree $d$. We can write $f$ as a sum over monomials of the form $\prod_{i=1}^d A_{a_{i,1} \dots a_{i,k}}$ for some tuple $a_{i,j}$ of numbers from $1$ to $n$ indexed by $i$ from $1$ to $d$ and $j$ from $1$ to $k$. The $(\mathbb Z/2)^n$-invariance implies the coefficient of a monomial vanishes if any number appears an odd number of times among the $a_{i,j}$. The $S_n$-invariance says the coefficients of a monomial depend only on the choice of $a_{i,j}$ up to permutation. We can express this as the claim that $f$ is a linear combination of the polynomials of the form $$ \sum_{ a \colon [m] \to [n] \textrm{ injective}} \prod_{i=1}^d A_{ a (l_{i,1}) a(l_{i,2}) \dots a(l_{i,j})}$$ for some fixed tuple $l_{i,j}$ of integers from $1$ to $m$ indexed by $i$ from $1$ to $d$ and $j$ from $1$ to $k$, where each number appears an even number of times among the $l_{i,j}$ (which we may assume is at least twice). The sum is over injective maps from the numbers $1$ to $m$ to the numbers $1$ to $n$. We could, if we chose, drop the "injective" condition. Now let's consider what happens when we apply this inequality to the all-$1$s tensor. This tensor has spectral norm $\frac{n^k}{ n^{ k/p}}$, so the left side is $n^{d k (1-1/p)}$. The value of the invariant polynomial $\sum_{ a \colon [m] \to [n] \textrm{ injective}} \prod_{i=1}^d A_{ a (l_{i,1}) a(l_{i,2}) \dots a(l_{i,j})} $ is just the number of injective maps from $[m]$ to $n$, which is at most $n^m$. Because each $\ell$ occurs at least twice among the $\ell_{i,j}$, we have $m \leq d k / 2$. Thus the sum of the coefficients of these various polynomials in the linear expression for $f$ must be at least $n^{ d k (1/2-1/p)}$. Since the number of different polynomials that can appear depends only on $d$ and $k$, the coefficient of one of the monomials must be at least $n^{ d k (1/2-1/p)}/ (dk)^{dk}$. Now consider a sparse tensor $A$, say with $N$ nonvanishing entries, each of size $1$. Assuming $A$ comes from some tricky mathematical construction, we are unlikely to get an estimate for $f(A)$ with error term smaller than the largest coefficient of a monomial in $f$ as that would require getting an estimate for a difficult sum with error term less than a single term of the sum. Thus, we won't expect to get a bound for the spectral norm of $A$ better than $n^{ k (1/2-1/p)}/ (dk)^k \approx n^{ k(1/2-1/p)}$ (assuming $d,k$ grow slowly, which is reasonable). If we plug a "typical" vector $x$ into $A$, whose entries have size around $1$, we will therefore not get a bound for $\langle A, x^{\otimes k} \rangle$ better than $n^{ k /2}$. But if $N < n^{k/2}$, this is worse than the trivial bound for the sum, so our spectral norm estimate will be useless for such vectors. One can actually construct an explicit polynomial bound for the spectral norm where the coefficients of the monomials have size $\approx n^{1/2 - 1/p}$ by proving a bound for the $\ell^2$-spectral norm using Cauchy-Schwarz a bunch and then bounding the $\ell^2$ norm trivially via the $\ell^p$ norm. For $p<1/2$, the lower bound is instead $1$, which I think you can also achieve by using Holder's inequality repeatedly. An example of the Cauchy-Schwarz argument, for an $n \times n \times n$ tensor: We have $$ \left(\sum_{i_1,i_2,i_3 \in [n]} A_{i_1i_2i_3} x_{i_1} x_{i_2} x_{i_3}\right)^8 \leq |x|_2^8 \left( \sum_{i_1 \in [n]} \left( \sum_{i_2, i_3 \in [n]} A_{i_1 i_2 i_3} x_{i_2} x_{i_3} \right)^2 \right)^4 = |x|_2^8 \left( \sum_{i_1 ,i_2 ,i_3,i_4,i_5\in [n]} A_{i_1 i_2 i_3} A_{i_1 i_4 i_5} x_{i_2} x_{i_3} x_{i_4} x_{i_5} \right)^4 \leq |x|_2^{16} \left( \sum_{i_2, i_4\in [n]} \left( \sum_{i_1,i_3,i_5\in [n]} A_{i_1 i_2 i_3} A_{i_1 i_4 i_5} x_{i_3} x_{i_5} \right)^2 \right)^2 =|x|_2^{16} \left(\sum_{ i_1, i_2, i_3,i_4, i_5, i_6, i_7, i_8 \in [n]} A_{i_1 i_2 i_3} A_{i_1 i_4 i_5} A_{i_6 i_2 i_7} A_{i_6 i_4 i_8} x_{i_3} x_{i_5} x_{i_7} x_{i_8} \right)^2 \leq |x|_2^{24} \sum_{i_3, i_5 ,i_7, i_8 \in [n ] } \left( \sum_{i_1, i_2, i_4, i_6 \in [n]} A_{i_1 i_2 i_3} A_{i_1 i_4 i_5} A_{i_6 i_2 i_7} A_{i_6 i_4 i_8} \right)^2$$ which, when you expand it out, should have an elegant symmetric structure where the eight copies of $A$ correspond to the vertices of the cube and the various $i$ indices to edges.<|endoftext|> TITLE: Proofs of a character identity? QUESTION [7 upvotes]: Let $G$ be a finite group, $g \geq 0, k\geq 1 $ integers, and $(C_1,...,C_k)$ a tuple of conjugacy classes of $G$. I am interested in proofs of the following identity $$ \sum_{(c_1,...,c_k) \in C_1 \times \cdots \times C_k} \sum_{V_\lambda} d(V_\lambda)^{1-2g} \chi_\lambda(c_1^{-1}\cdots c_k^{-1}) = |C_1|\cdots|C_k|\sum_{V_\lambda} d(V_\lambda)^{2-2g-k}\chi_\lambda(c_1)\cdots \chi_\lambda(c_k) $$ where the sums with $V_\lambda$ are over all irreducible characters of $G$ and $d(V_\lambda) = \chi_\lambda(1)$ is the dimension of the representation. I obtained this from messing around with a theorem due to Frobenius (for $g=0$ and later generalized originally using a quantum topology approach - hence the $g$ for genus). Is there some nice (ideally purely character-theoretic) proof of this identity? REPLY [8 votes]: Here is another way to view things: it uses the fact that for each complex irreducible character $\chi$ of $G$, there is an algebra homomorphism $\omega_{\chi} : Z(\mathbb{C}G) \to \mathbb{C}$ defined by $\omega_{\chi}(X) = \frac{\chi(X)}{\chi(1)} .$ This is the underlying root (from a modern viewpoint) of the formula of Frobenius quoted in the question, and the formula beloved of Paul Sally quoted in L. Spice's answer. For $S$ a subset of $G$ invariant under conjugation, let $S^{+} = \sum_{s \in S} s$ in $\mathbb{C}G.$ Hence for each $\lambda$ we find that $$\frac{\chi_{\lambda}(C_{1}^{+}C_{2}^{+} \ldots C_{k}^{+})}{\chi_{\lambda}(1)} = \prod_{i=1}^{k} |C_{i}|\frac{\chi_{\lambda} (c_{i})}{\chi_{\lambda}(1)}.$$ Expressing the element $C_{1}^{+} \ldots C_{k}^{+}$ as a sum of products of group elements explains the formula of the question.<|endoftext|> TITLE: Is every permutation group on $n$ letters the symmetry group of a set of $n$ points in some euclidean space? QUESTION [6 upvotes]: This question is superficially similar to a previous question. Suppose I am given a permutation group $G \subseteq S_n$. Is it always possible to find a set $X$ of $n$ points in $\mathbb{R}^n$, such that the isometry group of $X$ (together with its natural action on $X$) is equal to $G$ as a permutation group? REPLY [14 votes]: No. If $G$ is $2$-transitive $-$ or even transitive on pairs $-$ then $X$ must be the set of vertices of a regular simplex, which has isometry group $S_n$. But there are plenty of examples of $2$-transitive groups $G$ properly contained in $S_n$ (such as the $ax+b$ group if $n$ is a prime power and $n>3$).<|endoftext|> TITLE: Affine (or Stein) tubular neighbourhood theorem QUESTION [14 upvotes]: Fix an embedding $X\subset Y$ of smooth complex affine varieties, or Stein manifolds. I would guess that in general there is no analytic neighbourhood $X\subset U\subset Y$ with a holomorphic retraction $U\to X$. But does anyone know a counterexample? (If not, a proof of the existence of $U$?) REPLY [6 votes]: My original interest in this problem was the following question. Given a holomorphic bundle $E$ on an affine/Stein scheme $X$ inside any scheme/analytic space $Y$ of finite embedding dimension, is it the restriction of a bundle on a Stein neighbourhood $U\subset Y$ of $X$? Smooth case. If $X$ and $Y$ are both smooth, then Corollary 1 in Siu's paper now gives this, by pulling $E$ back along the holomorphic retraction. Reduced case. If $X$ and $Y$ are only reduced then Mohan Ramachandran's comment below solves this. In more detail: $Y$ admits a triangulation with $X$ as a subcomplex, so there is an open neighbourhood $U'$ of $X$ which is a retract, giving a continuous (not usually holomorphic) projection $\pi:U'\to X$. Now by Siu's theorem there is a Stein neighbourhood $X\subset U\subset U'$ on which we have the topological bundle $\pi^*E|_U$. By Grauert's Oka principle there is a holomorphic structure on this which restricts to the given one on $\pi^*E|_X=E$. $X$ nonreduced. Here we apply the above result to $E|_{X^{red}}$ to get a bundle $\widetilde E$ on $U\subset Y$ with $\widetilde E|_{X^{red}}\cong E|_{X^{red}}$. But I think this means $\widetilde E|_X\cong E$ since there is at most one extension of $E|_{X^{red}}$ from $X^{red}$ to $X$ when $X$ is Stein. $Y$ nonreduced. Finally for general $Y$ we may assume it is Stein by replacing it by Siu's Stein neighbourhood of $X^{red}$. Then we can embed in some $\mathbb C^N$, find a holomorphic extension $\widetilde E$ there, then restrict back to $Y$. So thanks to Richard Lärkäng and Mohan Ramachandran I think the result is probably true. Surely there's a reference for it somewhere?<|endoftext|> TITLE: Connection between braided tensor categories and local systems on moduli of stable marked genus zero curves QUESTION [5 upvotes]: I'm looking for references regarding an unpublished Deligne's manuscript "Une descrption de catégorie tressée (inspiré par Drinfeld)" and the subject it touches, that is described in the post title. If it's not available online maybe someone can orient me to some other references regarding the same connection or to the connection itself for me to try to develop the details. Thanks, REPLY [5 votes]: Welcome to MO. One place where this idea is somewhat explained is Bezrukavnikov-Finkelberg- Schechtman, Factorizable sheaves and quantum groups (https://arxiv.org/abs/q-alg/9712001) but maybe this is where you heard about it. I'd say nowadays this idea might be best understood as a particular case of the identifications between: braided monoidal categories, and algebras over the little disks operad in Cat algebras over the little disks operad and locally constant factorization algebras on $\mathbb{R}^2$.<|endoftext|> TITLE: Outer automorphism group of Brieskorn homology sphere? QUESTION [8 upvotes]: In this post, it is discussed how a Brieskorn homology sphere $\Sigma(a_1,a_2,a_3)$ with $\displaystyle \frac{1}{a_1}+ \frac{1}{a_2}+ \frac{1}{a_3} < 1$ is an aspherical manifold with a superperfect fundamental group with non-trivial center. Would anyone know what the outer automorphism groups of their fundamental groups are? I'm looking to do semidirect products with these groups as the kernel group with another group ($\mathbb{Z} \times \mathbb{Z}$) as the quotient group, so I need to know the outer automorphism groups of these groups. EDIT: Would SnapPy be able to help with this? https://www.math.uic.edu/t3m/SnapPy/ REPLY [7 votes]: The answer is that the outer automorphism group is $\mathbb{Z}_2$. (Compare Ian Agol's answer to What is the order of the isotopy group of the Brieskorn homology 3-sphere?). This is obtained by cobbling together several standard facts. Much of this can be found in Boileau and Otal, Scindements de Heegaard et groupe des homéotopies des petites variétés de Seifert. (French) [Heegaard splittings and homeotopy group of small Seifert manifolds] Invent. Math. 106 (1991), no. 1, 85–107. The first is that the outer automorphism group of the fundamental group coincides with the diffeomorphism group modulo isotopies. The second is that any self-diffeomorphism of a Brieskorn sphere preserves orientation; see Neumann-Raymond, Seifert manifolds, plumbing, μ-invariant and orientation reversing maps. Boileau-Otal show that up to isotopy, any orientation preserving diffeomorphism preserves the fibers (setwise) and hence gives an automorphism of the 2-orbifold quotient by the circle action. Because of the relatively prime condition (easy to verify by homology calculations) there is only one automorphism of that orbifold; it is induced from a reflection in the 2-sphere and so flips the orientation. The lift to the Brieskorn sphere also reflects the fiber so is orientation preserving.<|endoftext|> TITLE: What is the proof-theoretic ordinal of bare $\mathsf{NFU}$? QUESTION [12 upvotes]: On the Stanford Encyclopedia of Philosophy article on alternative axiomatic set theories, it is stated without reference that bare $\mathsf{NFU}$ (i.e., $\mathsf{NFU}$ without the axiom of infinity) is weaker in consistency strength than $\mathsf{PA}$. In this Math SE answer, Randall Holmes states that $\mathsf{NFU}$ interprets Robinson arithmetic and furthermore says that it probably interprets bounded arithmetic with exponentiation (which I think means $\mathsf{I}\Delta_0 + \mathrm{exp}$, although I'm not completely familiar with the semi-formal terminology people use for these things). This would put $\mathsf{NFU}$ right in the range of theories for which ordinal analysis should be well-behaved (i.e., those which are at least as strong as $\mathsf{I}\Delta_0$) but also should be feasible to actually do (e.g., those which are as strong as or not too much stronger than $\mathsf{PA}$). So this raises a question that I have been unable to find any discussion of. Question 1. What is the proof-theoretic ordinal of bare $\mathsf{NFU}$? More generally, a lot is known about consistency strength and interpretation of theories of first- and second-order arithmetic between $\mathsf{I}\Delta_0$ and $\mathsf{PA}$. Proof-theoretic ordinals are supposed to be a rough measure of consistency strength, but in principle this is a more fine grained issue. Question 2. Where does bare $\mathsf{NFU}$ sit relative to well-known theories of arithmetic in terms of interpretation and consistency strength? REPLY [9 votes]: As in the question, I will use $\mathsf{NFU}$ for "bare" $\mathsf{NFU}$, i.e., the result of weakening the extensionality axiom in Quine's $\mathsf{NF}$ so as to allow urelements. Let $\mathsf{NFU}^{-\infty}$ be $\mathsf{NFU} \cup \{\lnot \mathsf{Infinity} \}$, where $\mathsf{Infinity}$ is the axiom of infinity. In 2002, Solovay proved the following results. The proofs were disseminated among some NFU enthusiasts, but alas, they remain unpublished. In what follows $\mathsf{Exp}$ is the statement asserting the totality of the exponential function, and $\mathsf{SupExp}$ is the statement asserting the totality of the superexponential function (also known as tetration), i.e., the "stack of twos" function, $f(n)$, where $f(0) = 1$ and $f(n+1) = 2^{f(n)}$. Theorem. (1) $\mathrm{I}\Delta_{0} + \mathsf{SupExp} \vdash\mathsf{Con}(\mathsf{NFU}^{-\infty}) \leftrightarrow\mathsf{Con}(\mathrm{I}\Delta_{0} + \mathsf{Exp})$. (2) $\mathrm{I}\Delta_{0} + \mathsf{Exp} \vdash \mathsf{Con} (\mathsf{NFU}^{-\infty})\rightarrow \mathsf{Con}(\mathrm{I}\Delta_{0} + \mathsf{Exp})$. (3) $\mathrm{I}\Delta_{0} + \mathsf{Exp} \nvdash \mathsf{Con} (\mathrm{I}\Delta_{0} + \mathsf{Exp})\rightarrow\mathsf{Con} (\mathsf{NFU}^{-\infty})$. Suspected Answer to Question 1. Based on the above theorem (part 1), the proof theoretic ordinal of $\mathsf{NFU}^{-\infty}$ is likely to be the same as the proof theoretic ordinal of $\mathrm{I}\Delta_{0} + \mathsf{Exp}$, i.e., is $\omega^3$, if one were to follow the procedure given by Taranovsky to this question. Answer to Question 2 Solovay's proof of the above theorem makes it clear that $\mathsf{NFU}^{-\infty}$ interprets $\mathrm{I}\Delta_{0} + \mathsf{Exp}$, but not vice versa; thus the interpretability relation between $\mathsf{NFU}^{-\infty}$ and $\mathrm{I}\Delta_{0} + \mathsf{Exp}$ is similar to the interpretability relation between $\mathsf{ACA}_0$ and $\mathsf{PA}$.<|endoftext|> TITLE: What's the advantage of defining Lie algebra cohomology using derived functors? QUESTION [9 upvotes]: The way I learned Lie algebra cohomology in the context of Lie groups was a direct construction: one defines the Chevalley-Eilenberg complex with coefficients in a vector space $V$ (we assume the real case) as $$C^p(\mathfrak g; V) := \operatorname{Hom}(\bigwedge^p \mathfrak g, V)$$ and explicitly defines the boundary map $\delta^p: C^p(\mathfrak g; V) \to C^{p+1}(\mathfrak g; V)$ by a formula similar to the coordinate-free definition of the de Rham differential. Finally, one takes the homology of this cochain complex. With this approach, the connection between the Chevalley-Eilenberg cohomology and the left-invariant de Rham cohomology is obvious. This was roughly the approach used by Chevalley and Eilenberg themselves However, Wikipedia and some other sources rather define $$H^n(\mathfrak g; V) = \operatorname{Ext}^n_{U(\mathfrak g)}(\mathbb R, V)$$ where one constructs so-called universal enveloping algebra $U(\mathfrak{g})$, whose motivation isn't clear to me, even though I understand the formal definition. Even when one knows what a derived functor is (which I do), this definition still requires a lot of work, such as the introduction of the universal enveloping algebra, finding the projective resolutions, etc. At first I thought that the $\operatorname{Ext}$ approach might just be abstract restatement of the same procedure that we carry out while defining the cohomology through the Chevalley-Eilenberg complex, but I don't really see why it should be that way. Well, we take homology of the $\operatorname{Hom}$ complex, but it's where the analogy seems to end because of this universal enveloping algebra, which doesn't have a clear counterpart in the explicit construction. Is there any advantage to use the second definition of the Lie algebra cohomology? The only reason I could see is the derived functor LES, but it would probably be much easier to show it directly. There's also a clear analogy with the group cohomology defined that way - one just takes the group ring $\mathbb Z[G]$ instead of the universal enveloping algebra $U(\mathfrak{g})$, but group cohomology isn't hard to construct explicitly and the derived functor definition seems so abstract that it's useless. Maybe my confusion stems from the fact that I learned homological algebra separately in a very abstract setting, roughly following Weibel's book, and while I understood the definitions I don't think I understood the motivations and the big ideas. I asked a similar question on Math.SE, but now I realized that MO is a better place to ask. REPLY [10 votes]: What is your motivation for thinking about Lie algebra homology? If all you want to do is compute it for a single fairly explicit Lie algebra, then the Chevalley-Eilenberg complex will do the job. But for actually proving non-homological theorems using Lie algebra homology, it's much better to have the more flexible derived functor definition. For instance, one of my favorite easy applications is to give very efficient proofs of things like semi-simplicity of finite-dimensional representations of semisimple Lie algebras over fields $k$ of characteristic $0$ (like $\mathbb{R}$). For this, the key fact is that extensions of $U$ by $V$ are classified by $\text{Ext}^1_{U(\mathfrak{g})}(U,V)$, and this can be identified with Lie algebra cohomology via the sequence of isomorphisms $$\text{Ext}^1_{U(\mathfrak{g})}(U,V) = \text{Ext}^1_{U(\mathfrak{g})}(k,\text{Hom}_k(U,V)) \cong H^1(\mathfrak{g};\text{Hom}_k(U,V)).$$ So what we want to do is show that $H^1(\mathfrak{g};W)=0$ for all finite-dimensional $W$ (assuming semisimplicity for $\mathfrak{g}$). You do this in two steps. First, you use the quadratic Casimir element in the universal enveloping algebra to kill $\text{Ext}^1_{U(\mathfrak{g})}(k,W)$ for nontrivial simple $W$ (this is the key use of semisimplicity!), and then you use the fact that $\mathfrak{g}$ has trivial abelianization to see that $H^1(\mathfrak{g};k)=0$, and then finally you deal with arbitrary finite-dimensional $W$ by filtering them so that the associated graded pieces are simple. If you're comfortable with homological algebra, then the above is a pretty natural proof outline, and it carefully isolates the non-formal part (the existence of the Casimir) from the formal bookkeeping. But if you insisted on defining cohomology using the Chevalley-Eilenberg complex, I would have no idea how to come up with it. Why would cohomology have something to do with extensions?<|endoftext|> TITLE: $\mathbb{R}^3$ as the union of disjoint circles QUESTION [15 upvotes]: In the question Covering the space by disjoint unit circles the following result is attributed to Sierpinski. Theorem. The Euclidean space $\mathbb{R}^3$ is a union of nondegenerate disjoint circles. But the usual reference on this result is Szulkin, Andrzej, $\mathbb{R}^ 3$ is the union of disjoint circles, Amer. Math. Monthly. 90, 640-641 (1983). ZBL0521.52011. MR0719756. What is the correct reference? I failed to find this result among Sierpinski's works. REPLY [2 votes]: You seem to be correct. I was wrong, probably because this result appears in Ciesielski's book ("Set theory for the working mathematician" CUP, 1997; Theorem 6.1.3) near a result of Sierpinski. I corrected my question accordingly.<|endoftext|> TITLE: What is the limit of $a (n + 1) / a (n)$? QUESTION [11 upvotes]: Let $a(n) = f(n,n)$ where $f(m,n) = 1$ if $m < 2 $ or $ n < 2$ and $f(m,n) = f(m-1,n-1) + f(m-1,n-2) + 2 f(m-2,n-1)$ otherwise. What is the limit of $a(n + 1) / a (n)$? $(2.71...)$ REPLY [9 votes]: Decided to do a separate answer as there is a subtle point involved which is not mentioned in my comments to the answer by @Max Starting from the generating function by Max Alexeyev $$ \sum_{m,n\geqslant0}f(m,n)x^my^n=\frac1{(1-x)(1-y)}\left(1+\frac{3x^2y^2}{1-xy(1+2x+y)}\right) $$ we need to find the generating function for $F(t):=\sum f(n,n)t^n$. It can be done, as mentioned by @robinpemantle, using the method of residues, and this gives $$ F(t)=\frac1{2(1-2t-2t^2)(1-3t-t^2)}\left(2-8t+4t^2+5t^3-3t^3\frac{3-7t-4t^2}{\sqrt{1-2t+t^2-8t^3}}\right). $$ The subtle point here is that the apparent poles at the roots of $1-2t-2t^2$ and $1-3t-t^2$ (respectively, $\approx.366025$ and $\approx.302776$) are closer to the origin than the 2-branching pole at the root of $1-2t+t^2-8t^3$ ($\approx.36816293915706916$). However it turns out that these poles are actually cancelled out by zeros. To see this, observe that $$ 2-8t+4t^2+5t^3-3t^3\frac{3-7t-4t^2}{\sqrt{1-2t+t^2-8t^3}}=0 $$ happens when $$ R(t):=1-2t+t^2-8t^3-\left(3t^3\frac{3-7t-4t^2}{2-8t+4t^2+5t^3}\right)^2 $$ is zero. But $$ R(t)=\left(\frac2{2-8t+4t^2+5t^3}\right)^2(1-2t-2t^2)(1-3t-t^2)(1-5t+9t^2-15t^3+26t^4-16t^5-18t^6)$$ so that $R(t)$ vanishes at the roots of $1-2t-2t^2$ and $1-3t-t^2$ (to be entirely rigorous, one has to check that these do not have common roots with $2-8t+4t^2+5t^3$, which can be checked e. g. by looking at resultants). Thus the singularity of $F(t)$ nearest to the origin is the root of $1-2t+t^2-8t^3$ with smallest absolute value, so that the leading asymptotics is given by the root of $x^3-2x^2+x-8$ with largest absolute value (as in the answer by Robin Pemantle), i. e. $$ \frac{1}{3} \left(\sqrt[3]{107+6 \sqrt{318}}+\sqrt[3]{107-6 \sqrt{318}}+2\right)\approx2.7161886589931057 $$ PS As, judging by upvotes (thanks!) this answer seems to attract some attention, I decided to re-check it still more carefully, and it seems that that subtle issue is not actually fully exhausted in this answer. Not to repeat lengthy expressions, let me abbreviate $F(t)=(1/P)(A-B/\sqrt Q)$ (where $P$, $Q$, $A$, $B$ are polynomials in $t$). I claimed something like that if $P=0$, then $A-B/\sqrt Q=0$, and this is actually not true. What is true is that if $P=0$, then $A^2-B^2/Q=0$. Now $P$ has four roots (all real), of which only two are relevant (meaning that of the four only these two are closer to the origin than the smallest root of $Q$, which might influence the final answer). What actually happens is that in the vicinity of all four $Q$ is positive, and, denoting by $\sqrt Q$ the positive square root of $Q$ there, $A+B/\sqrt Q$ vanishes at the irrelevant roots of $P$, while $A-B/\sqrt Q$ vanishes at the relevant ones. And this requires additional analysis. I just now checked numerically that $A+B/\sqrt Q$ is away from zero at both relevant roots, which suffices after one knows about $A^2-B^2/Q$ rigorously. I can supply more detailed explanation if anybody requests it.<|endoftext|> TITLE: Coding the universe into a real over better core models QUESTION [11 upvotes]: One of the most incredible results in modern set theory, due to Jensen, is that given any model of $\sf ZFC$, there is a class forcing which adds a real number $r$ and in the extension $V=L[r]$. Moreover, if we assume some basic things like $\sf GCH$, then cardinals are not collapsed, so in particular things like inaccessible cardinals are preserved. So in Jensen's result the "coding model" is $L$. But $L[x]$ is somewhat of a dull model. It doesn't have any sharps, measurable cardinals, or larger cardinals. Even those with reasonably canonical inner models. Question. Given any reasonably canonical core model $K$, can we code the universe into a real over $K$ while preserving large cardinals that are captured by $K$? In other words, can we code the universe into a real while preserving measurable cardinals? Can we code the universe into a real while preserving strong, Woodin, etc.? And the obvious follow-up question, what is the bare minimum needed from an inner model $M$ to be a coding model? REPLY [10 votes]: For measurable cardinals, the answer is yes and is due to Sy Friedman. See Coding Over a Measurable Cardinal. There is some difficulty to extend the result to the context of Woodin cardinals, see Genericiy and large cardinals. For strong cardinals, this is open, though for some simple sets, say the collection of Prikry sequences for the Prikry product, the answer is yes, see Killing the GCH everywhere with a single real. See also Coding over core models<|endoftext|> TITLE: Principal ideal domains with finitely many units QUESTION [16 upvotes]: Question: What are the (in characteristic 0 if needed) principal ideal domains that have finitely many units? Can such rings be classified? (This is a more specialised version of the question in Integral domains with finitely many units and got split off of this thread) The well known examples are the imaginary quadratic integer rings of $\mathbb{Q} (\sqrt{d})$ for $d \in \{−1, −2, −3, −7, −11, −19, −43, −67, −163 \}$. Is there a nice infinite family in characteristic 0? REPLY [16 votes]: There is not likely to be a good answer to this question, because of a very annoying Theorem due to Heinzer and Roitman: If $D$ is any UFD, then there is a PID $R$ containing $D$ which has the same unit group as $D$ and such that every prime of $D$ remains prime in $R$. Since UFD's with finite unit group are common, this will give lots of examples. Here is a sketch of the proof of Heinzer and Roitman's result: If $D$ is a UFD, and $a$ and $b$ are relatively prime elements of $D$, set $D_{a,b} = D[x,y]/(ax+by-1)$. One can check that $D_{a,b}$ is a UFD and that the ideals $(a)$ and $(b)$ are comaximal in $D_{a,b}$; one can also check that $D^{\times} = D^{\times}_{a,b}$ and that prime elements of $D$ stay prime in $D_{a,b}$. Using the $D_{a,b}$ construction and a transfinite induction procedure, we embed $D$ into a ring $R$ such that $R$ is a UFD and any two relatively prime elements of $R$ generate comaximal ideals. Such a UFD is necessarily a PID. One also has $R^{\times} = D^{\times}$, and every prime of $D$ stays prime in $R$. See Section 4 of Heinzer, William; Roitman, Moshe, Principal ideal domains and Euclidean domains having 1 as the only unit, Commun. Algebra 29, No. 11, 5197-5208 (2001). ZBL1094.13532. for the details. As observed in the comments, if such a ring $A$ is finitely generated over $\mathbb{Z}$, it must be either be an order in the ring of integers in a number field, or else it must be the coordinate ring of an affine curve over $\mathbb{F}_q$. In the first case, by Dirichlet's unit theorem, the number field $K$ must be an imaginary complkex field. Since PID's are integrally closed, $A$ must be the full ring of integers. The imaginary quadratic number fields with class number $1$ are well known, they are $\mathbb{Z}[\sqrt{-1}]$, $\mathbb{Z}[\sqrt{-2}]$ and $\mathbb{Z}[\tfrac{1+\sqrt{-D}}{2}]$ for $D \in \{ 3, 7, 11, 19, 43, 67, 163 \}$. In the second case, let $A$ be the coordinate ring of an affine curve $U$ over $\mathbb{F}_q$, and let $X$ be the smooth projective curve completing $U$. By Dirichlet's unit theorem for function fields, $U \setminus X$ is a single closed point $x$; let that point $x$ have residue field $q^f$. Then $\mathrm{Pic}(U) = \mathrm{Pic}(X)/[x]$. We have a short exact sequence $$0 \to \mathrm{Pic}^0(X) \to \mathrm{Pic}(X) \to \mathbb{Z} \to 0$$ and the image of $[x]$ in $\mathbb{Z}$ is $f$ times the generator, so $\mathrm{Pic}(U)$ has a $\mathbb{Z}/f \mathbb{Z}$ quotient. Our assumption that $A$ is a PID means that $\mathrm{Pic}(U)$ is trivial, so we deduce that $f=1$ and $x$ is an $\mathbb{F}_q$ point. In that case, we have $\mathrm{Pic}(U) = \mathrm{Pic}^0(X)$, so we must additionally have $\mathrm{Pic}^0(X)$ trivial. In other words, if $J$ is the Jacobian of $X$, we must have $J(\mathbb{F}_q)$ trivial. Let's assume from now on that $X$ has genus $>0$, since genus $0$ gives us the case $\mathbb{F}_q[x]$, which we already know. Note that, with this assumption, requiring that $J(\mathbb{F}_q)$ be trivial imposes in particular that $X(\mathbb{F}_q)$ has at most one point. Let $\lambda_1$, $\lambda_2$, ..., $\lambda_{2g}$ be the eigenvalues of $q$-power Frobenius on $X$. Then the size of $J(\mathbb{F}_q)$ is $\prod (\lambda_i-1)$. Each $\lambda_i$ is a complex number of norm $\sqrt{q}$, so $\lambda_i$ is a complex number of norm at least $\sqrt{q}-1$. Thus, all examples have $\sqrt{q}-1 \leq 1$, so $q \in \{ 2,3,4 \}$. Heinzer and Roitman look at the $q=2$ case and give the examples of $y^2+y = x^3+x^2+1$ (genus one) and $y^2+y = x^5+x^3+1$ (genus two). For $q=3$, I noticed $y^2 = x^3-x+1$ (genus one) and for $q=4$ I noticed $y^2+y = x^3+\omega$ (genus one), where $\omega$ is a root of $\omega^2+\omega+1=0$ in $\mathbb{F}_4$. The eigenvalues of Frobenius in the genus $1$ cases are $\sqrt{2} e^{\pm(2 \pi i)/8} =1\pm i$, $\sqrt{3} e^{\pm(2 \pi i)/12} = \tfrac{3}{2} \pm \tfrac{\sqrt{-3}}{2}$ and $2$ (with multiplicity two). The eigenvalues of Frobenius in the genus $2$ case are $\sqrt{2} e^{\pm(2 \pi i)/24}$ and $\sqrt{2} e^{\pm 7(2 \pi i)/24}$, also known as $(1\pm i)\left( \tfrac{1\pm \sqrt{-3}}{2} \right)$. I attempted to prove that the curves above were the only examples, but my proof had a gap. Fortunately, pregunton discovered a paper of Mercuri and Stirpe which lists all curves with $J(\mathbb{F}_q)$ trivial (both with $X(\mathbb{F}_q)$ singleton, like we want, and with $X(\mathbb{F}_q)$ empty). The curves listed above are cases (i), (ii), (vi) and (vii) on their list; I believe all the other curves on their list have $X(\mathbb{F}_q)$ empty. Here is what I can salvage from my computation, which I still like: There are no examples with $q=4$ and genus $>1$. If there were, then we would have $\lambda_1 = \lambda_2 = \cdots = \lambda_{2g}=2$. But then the number of points in $X(\mathbb{F}_4)$ would be $4-\sum \lambda_i + 1 = 5-4g<0$, a contradiction. Now let $q$ be $2$ or $3$. We observe that $\sqrt{q}$ must have even multiplicity as an eigenvalue of Frobenius. Proof: We have $\#J(\mathbb{F}_q) = \prod (1-\lambda_j)$. Group together the terms with complex conjguate eigenvalues; then every factor except the one coming from $1-\sqrt{q}$ is a positive real, so $(1-\sqrt{q})$ must be raised to an even power as well. Using $\#J(\mathbb{F}_q) = \prod (\lambda_j-1)$, we see that $-\sqrt{q}$ likewise has even multiplicity. Therefore, we can write the eigenvalues of Frobenius as $\sqrt{q} e^{\pm i \theta_j}$ for $1\leq \theta_j \leq g$, remembering to take both signs in the exponent even when $\theta_j$ is $0$ or $\pi$. Let $c_j = \cos \theta_j$. The equation $X(\mathbb{F}_q)=1$ translates to $$q-2 \sqrt{q} \sum \cos \theta_j +1 =1 \ \implies \ \sum c_j = \tfrac{\sqrt{q}}{2} \quad (\ast)$$ The equation $J(\mathbb{F}_q)=1$ translates to $$\prod (q+1-2 \sqrt{q} c_j) = 1 \quad (\dagger).$$ We note that $(\ast)$ is a convex polytope whose vertices are $(1,1,\ldots,1, -1, -1, \ldots, -1, \tfrac{\sqrt{q}}{2})$ for $g$ odd and $(1,1,\ldots,1, -1, -1, \ldots, -1, \tfrac{\sqrt{q}}{2}-1)$ for $g$ even. We will try to show that $\prod (q+1-2 \sqrt{q} c_j) \geq 1$ everywhere on this polytope. Note that $\log \prod (q+1-2 \sqrt{q} c_j)$ is concave, so it is enough to check the inequality at the vertices. We compute that the value at the vertices is $$\begin{array}{c|cc} & g=2k+1 & g=2k+2 \\ \hline q=2 & 1 & (3-2 \sqrt{2}) (3 - 2 \sqrt{2} (\tfrac{\sqrt{2}}{2}-1)) \approx 0.657 \\ q=3 & 4^k & 4^k (4-2 \sqrt{3}) (4 - 2 \sqrt{3} (\tfrac{\sqrt{3}}{2}-1)) \approx 2.39 \times 4^k \\ \end{array}.$$ We see that the only solution with $q=3$ is $g=1$ with $c_1 = \tfrac{\sqrt{3}}{2}$. If $q=2$ and $g=2k+1$, then $(\dagger)$ only occurs at the vertices of the polytope, corresponding to the eigenvalue sequence $\sqrt{2}$ (multiplicity $2k$), $-\sqrt{2}$ (multiplicity $2k$) and $1 \pm i$. However, this would lead to a curve with $1-4k$ points over $\mathbb{F}_4$, so this can only happen for $k=0$ (and thus $g=1$.) However, I am unclear as to how to deal with the case of $q=2$, $g$ even. When $g=2$, the polytope is just a line segment and $(\dagger)$ occurs at two symmetrically placed points of the line segment, corresponding to $(c_1, c_2) = ( \tfrac{1 \pm \sqrt{3}}{\sqrt{2}}, \tfrac{1 \mp \sqrt{3}}{\sqrt{2}})$. However, for larger $k$, $(\dagger)$ occurs on a little $2k$-dimensional manifold in the corners of the polytope, and it isn't clear to me how to rule out more solutions here. A fun fact is to note that none of these affine curve examples (except genus zero) will be Euclidean. This is due to the following nice exercise: If $A$ is a Euclidean domain and not a field, then there is some prime $\mathfrak{p}$ of $A$ such that $A^{\times} \to (A/\mathfrak{p})^{\times}$ is surjective. (Hint: Take a non-unit of minimal norm.) Since $U$ has no $\mathbb{F}_q$ points in any of these examples, all of the residue fields $A/\mathfrak{p}$ are proper extensions of $\mathbb{F}_q$, whereas $U^{\times} = \mathbb{F}_q^{\times}$.<|endoftext|> TITLE: How does the bound in the large sieve depend on the norm on the lattice? QUESTION [8 upvotes]: I've been reading about the large sieve inequality in Serre's "Lectures on the Mordell-Weil theorem", which states it in the following setting, which I've simplified a bit here: Suppose $\Lambda \cong \Bbb Z^n$ is a lattice and we have a norm $\| \cdot \|$ on $\Lambda_{\Bbb R} = \Lambda \otimes \Bbb R \cong \Bbb R^n$. Then there is a constant $c = c(\Lambda, \|\cdot \|)$ depending only on $\Lambda$ and the norm $\| \cdot \|$, such that for any set $X$ with the following properties: $X$ is contained in a ball of radius $N$ for the norm $\| \cdot \|$ some "local sieving" conditions at primes $\le Q$, we have $|X| \le c \sup(N^n, Q^{2n})/L(Q)$, where $L(Q)$ is a function of the local densities used in the sieve. Is there a treatment of the large sieve that spells out how the upper bound in this sieve depends on the choice of norm $\| \cdot \|$, or equiavlently on the shape of the ball bounding $X$? I don't want explicit constants, just a qualitative sense of the dependence. I've looked at the example of Chapter 4 of Kowalski's book on the large sieve, which gives a bound in the case where the ball is a paralellepiped aligned with the coordinate axes, but I'm interested in a more general situation (e.g. a general parallelepiped or ellipsoid). REPLY [4 votes]: One approach is to reduce to the parallelepiped case. Let $\Lambda'$ be the dual lattice of $\Lambda$ with the dual norm $$|| y|| = sup_{x \in \Lambda} \frac{ x \cdot y}{ || x||}.$$ Let $\lambda_1,\dots, \lambda_n$ be the successive minima of $\Lambda'$. Let $w_1,\dots, w_n$ be the associated shortest vectors, forming a basis of $\Lambda'$. Let $v_1,\dots, v_n$ be the dual basis of $\Lambda$. Then we have $$ \left|\left|\sum_i a_i v_i \right|\right| \leq \sup_{j\in [n]} \frac{|\sum_i a_i v_i \cdot w_j|} { \lambda_j} = \sup_{j\in [n]} \frac{ |a_j| }{\lambda_j}$$ so the ball of radius $N$ is contained in the parallelepiped $$\left \{ \sum_i a_i v_i , |a_i| \leq \lambda_i N \right\}$$ and then you can apply the version from Kowalski's book.<|endoftext|> TITLE: Reference for behavior of Artin $L$-functions at $\Re(s) = 1$ QUESTION [6 upvotes]: Would anyone know a reference that proves the basic facts about Artin $L$-functions at $\Re(s) = 1$? Namely, the non-vanishing and holomorphicity for non-trivial characters. I assume this was done in Artin's original paper, but a modern source would be most welcome. REPLY [5 votes]: I also had a hard time finding a reference for this a while back. See page 225 of Chapter VIII "Zeta-Functions and L-functions" by Heilbronn in "Algebraic number theory" edited by Cassels and Fröhlich, MR0218327 Also see $\S$12 of Chapter XIII of Weil's book "Basic Number Theory", MR0234930<|endoftext|> TITLE: What is the category of covariant and contravariant functors? QUESTION [11 upvotes]: Let $\bf Cat'$ be the category that has as objects small categories $A, B...$, and as arrows functors $F:A\to B$ that are either covariant or contravariant. The identity on $A\in\bf Cat'$ is the usual identity functor; the composition of a covariant and a contravariant functor is contravariant and the composition of two contravariant functors is covariant. (Here, by a contravariant functor $F:A\to B$, I mean a mapping $F: obj A \to obj B$ and for any arrow $f:x\to y$ an arrow $Ff:Fy\to Fx$ with the usual axioms on identities and composition.) I wonder which are the abstract properties of $\bf Cat'$. Note for instance that in $\bf Cat'$ any category is isomorphic to its dual; so I suppose that it is not possible to recover the covariant arrows in an abstract way. Maybe the obvious functor $\bf Cat \to \bf Cat'$ has some universal property? REPLY [5 votes]: You may also be interested in my paper Contravariance through enrichment, which shows that $\bf Cat'$ is enriched over a certain non-symmetric monoidal structure on $\bf Cat\times Cat$. This allows distinguishing the covariant and contravariant functors while still keeping them both present in the structure. Thus for instance we can distinguish the "contravariant isomorphism" $A\cong A^{\rm op}$ from a normal covariant isomorphism, and indeed use the former to characterize $A^{\rm op}$ as "the unique (up to covariant isomorphism) object contravariantly isomorphic to $A$", which is also a copower of $A$ in the sense of enriched category theory. This structure is actually closely related to the construction in Simon's answer; the monoidal structure on $\bf Cat\times Cat$ is constructed using the $\mathbb{Z}/2\mathbb{Z}$ action on $\bf Cat$, and can be generalized to other group actions. Also I should note that when I say $\bf Cat'$ "is" enriched over $\bf Cat\times Cat$, I don't mean in the precise sense that $\bf Cat'$ is the "underlying ordinary category" of an enriched category in the standard sense of enriched category theory; the underlying ordinary category of the enriched form of $\bf Cat'$ is actually just $\bf Cat$. This is similar to the situation of group actions, where a category enriched over $G$-sets is a category with $G$-actions on its hom-sets, but the "underlying ordinary category" takes the fixed-point sets of those hom-sets (rather than, for instance, forgetting the $G$-actions).<|endoftext|> TITLE: Historical question about the dimension subgroup conjecture QUESTION [6 upvotes]: Let $G$ be a group and let $I$ be the augmentation ideal in the group ring $\mathbb{Z}[G]$. The k-th dimension subgroup of $G$ is $$D_k(G) = \{\text{$g \in G$ $|$ $g-1 \in I^k$}\}.$$ It is not hard to see that this is a central series, so if $\gamma_k(G)$ denotes the k-th term of the lower central series of $G$ then $\gamma_k(G) \subset D_k(G)$. The dimension subgroup conjecture asserts that these are always equal. Unfortunately, it is false (a counterexample was found by Rips). However, it is often true. In particular, it is known that if the lower central series quotients $\gamma_k(G)/\gamma_{k+1}(G)$ are all torsion-free, then $\gamma_k(G) = D_k(G)$ for all $k$. I first learned about this from Quillen's paper Quillen, Daniel G., On the associated graded ring of a group ring., J. Algebra 10 (1968), 411-418. where it is Corollary 4.2. However, there is a vast literature on the dimension subgroup problem that never mentions Quillen's paper at all (despite it being heavily cited elsewhere!). In this literature, this result seems to be attributed to Hall and Jennings. This is done for instance in Hurley, Ted, Dimension and Fox subgroups, Around group rings (Jasper, AB, 2001). Resenhas 5 (2002), no. 4, 293–304. which can be downloaded here. However, I have been unable to located any paper of Hall or Jennings that proves it. What's the story here? Whom should I attribute this result to? REPLY [4 votes]: I'm going to write an answer based on information that I figured out while consulting the references that Benjamin Steinberg suggested in the comments. It will involve a certain amount of speculation, so I invite clarifications from anyone who actually knows the historical record. In his paper Jennings, S. A., The group ring of a class of infinite nilpotent groups, Canadian J. Math. 7 (1955), 169-187. Jennings claims the result in question at the end of Part I. He says that the proof will appear in a forthcoming paper. I gather that this paper never appeared. However, Philip Hall lectured about it at the Summer Seminar of the Canadian Mathematical Congress held at the University of Alberta in 1957. These lectures were later published in 1969 as part of the Queen Mary College Mathematics Notes. The reference is here: Hall, Philip, The Edmonton notes on nilpotent groups, Queen Mary College Mathematics Notes. Mathematics Department, Queen Mary College, London 1969 iii+76 pp. The theorem in question is Theorem 7.1, which Hall proves and attributes to Jennings. So what's the deal with Quillen's paper? I'm going to have to speculate here. Quillen's paper appeared in 1968, which is between Jennings's paper and the formal publication of Hall's lecture notes. My guess is that he wasn't aware of Jennings's claim since it is buried 2/3 of the way through a paper on something else and not proved there. Hall's notes were apparently circulated privately among people working in group theory long before they were published, but since Quillen was a topologist it stands to reason that he would not know about them. Quillen's proof is quite different from the ones given by the group-theorists, and depends in part on Milnor-Moore's work on Hopf algebras. So this is really just a matter of the imperfect circulation of informal word-of-mouth mathematical knowledge, especially before the internet. As for why Quillen's paper never seems to be mentioned by group theorists working on dimension subgroups, they certainly would not be obligated to cite it since he was not the first to prove the result. Also, I suspect that the methods/language used were not very natural to hard-core group theorists, so I guess it didn't influence them (though Quillen's paper has 77 citations in mathscinet, so certainly other people found it useful!).<|endoftext|> TITLE: Packing equal-size disks in a unit disk QUESTION [7 upvotes]: Inspired by the delicious buns and Siu Mai in bamboo steamers I saw tonight in a food show about Cantonese Dim Sum, here is a natural question. It probably has been well studied in the literature, but I cannot find related reference. Given a real number $r \le 1$, let $f(r)$ be the maximum number of radius-$r$ disks that can be packed into a unit disk. For example, $f(1)=1$ for $r \in (1/2, 1]$, $f(r)=2$ for $r \in (2\sqrt{3}-3, 1/2]$, etc. Question: Is it true that $\{f(r): r \in (0, 1]\}=\mathbb{N}$? Noam Elkies immediately pointed out that the maximum radius allowed for packing $6$ and $7$ balls are the same (both $1/3$). Now let me change the problem to just make it harder :) Modified question: Is $\mathbb{N} \setminus \{f(r): r \in (0, 1]\}$ a finite set? REPLY [13 votes]: No: according to the pictures in https://en.wikipedia.org/wiki/Circle_packing_in_a_circle, $f(r)$ is never $6$, with $f(\frac13) = 7$ but $f(\frac13 + \epsilon) = 5$. (This result is attributed to the late R.L.Graham's solution in 1968 of a problem in the American Math. Monthly.) It is also known that $f(r_0) = 19$ for $r_0 = 1 / (1+\sqrt2+\sqrt6)$, and conjectured that $f(r_0 + \epsilon) = 17$, which would mean that $f(r)$ is never 18 either. The conjecture is in a 1998 article co-authored by the same R.L.Graham: Graham RL, Lubachevsky BD, Nurmela KJ, Ostergard PRJ. Dense packings of congruent circles in a circle. Discrete Math 1998;181:139–154.<|endoftext|> TITLE: Iterated antiderivatives of polynomials having many real roots QUESTION [18 upvotes]: Question For which polynomials $p_n:\mathbb{R} \rightarrow \mathbb{R}$ having $n$ distinct real roots can we find an infinite sequence of polynomials $$ p_n, p_{n+1}, p_{n+2} , p_{n+3}, \dots, $$ such that $p_i$ is a polynomial of degree $i$ with $i$ distinct real roots and $p_{i+1}$ is an anti-derivative of $p_i$ for all $i \geq n$? Hermite polynomials have this property. The $n-$th Hermite polynomial $H_n$ has $n$ distinct roots and satisfies the derivative relation $$ \frac{d}{dx} H_n = n H_{n-1}.$$ This means that when computing the antiderivative of $H_{n}$, one can always find the correct constant to turn the anti-derivative into a multiple of $H_{n+1}$ and multiples of $H_{n+1}$ have exactly $n+1$ roots. My gut feeling would be that this is a very special property that not many polynomials have and I am wondering whether there is any sort of result in that direction. Motivation This question is motivated by the asymptotic behavior of roots of polynomials when polynomials are differentiated many times. A while back I proposed a PDE to describe this and the PDE seems to be smoothing, so there should be a loss of information as one differentiates. More recently, Jeremy Hoskins and I proved a result that also hints towards loss of information. This sort of suggests that having these infinite chains should be somehow `rare' in some sense but it's more a vague connection than anything theorem-based. REPLY [5 votes]: The question seems to be answered at the "Untitled" link from canvas.wisc.edu when you do a google search on "appell sequence real zeros".<|endoftext|> TITLE: Hakim's definition of a locally ringed topos QUESTION [5 upvotes]: In Hakim's book "Topos annelés et schémas relatifs", Chap. III, Def. 2.3 states that a ringed topos $(X,A)$ is a locally ringed topos when two equivalent conditions are satisfied: (i) For each $U \in X$ and each section $s \in A(U)$ one has $U = U_s \cup U_{1-s}$. (ii) For each $U \in X$ and each family $(s_i)_{i \in I}$ in $A(U)$ generating the unit ideal one has $U = \bigcup_{i \in I} U_{s_i}$. Here $ U_s \subseteq U$ is the largest subobject on which $s$ is invertible. But I don't think that (i) is equivalent to (ii), since (i) is satisfied for $A=0$, right? Notice that (ii) implies that $A(U)=0 \implies U=0$ (take $I=\emptyset$, cf. MO/45951), which I would expect from a local ring object (see also here). Am I missing something? REPLY [4 votes]: Agreed. $(ii)$ is equivalent to "$(i)$ and $( 0 = 1$ in $A(U) ) \Rightarrow U= \emptyset$". Remark: I haven't looked at Hakim's convention, by $U= \emptyset$ I mean "as a sheaf", that is, if we are talking about an object of a site and not an object of a topos it means that the empty sieve is a covering of $U$. This corresponds to the fact that (constructively) a Ring is local if either: $(i)$ $0 \neq 1$ and $\forall s \in A(U)$, either $s$ or $1-s$ is invertible. $(ii)$ If $\sum s_i =1$ then $\exists i$ such that $s_i$ is invertible. The interpretration of these claim ni the sheaf semantics corresponds exactly to the proposition given in Hakim's these, except the missing condition $0 \neq 1$ which interprets as $( 0 = 1$ in $A(U) ) \Rightarrow U= \emptyset$. One can also say (but I don't think that it is the intended meaning) that $(i)$ (without $0 \neq 1$) is equivalent to $(ii)$ restricted to familly with at least one elements.<|endoftext|> TITLE: Can I glue the X axis to the Y axis? QUESTION [20 upvotes]: Consider the following diagram of algebraic varieties: $$\mathbb{A}^0 \to \mathbb{A}^1 \rightrightarrows \mathbb{A}^2$$ Here the first arrow is the inclusion of the origin into the line, and the next two are the inclusion of the line into the plane as the X and the Y axes. Does this diagram have a colimit in the category of schemes? (The first arrow giving inclusion of the point is not relevant; it just so happens that it was there when I met this question.) REPLY [6 votes]: Dan Petersen has already answered the hard part -- how to show that the affine pushout $\{ f \in k[x,y] : f(t,0) = f(0,t) \}$ is also the scheme push out. I write to record a discussion in the comments about how to see that this ring is finitely generated and obtain an explicit list of generators. Let $R = k[x,y]$ and let $S = k[x,y]^{S_2} = k[b,c]$ where $b=x+y$ and $c = xy$. Let $A = \{ f \in k[x,y] : f(t,0) = f(0,t) \}$. Then clearly $A$ is an $S$-submodule of $R$. Since $S$ is noetherian and $R$ is generated as an $S$-module by $1$ and $x$, this shows that $A$ is finitely generated as an $S$-module. If the characteristic of $k$ is not $2$, one can get explicit generators easily. $A$ is invariant under switching the generators $x$ and $y$, so $A$ splits into positive and negative eigenspaces, call them $A_+$ and $A_-$, for this switch. The positive eigenspace is just $S$. The negative eigenspace $A_-$ is $xy(x-y) S$, since it is easy to see that anything in $A_-$ is divisibly by $xy(x-y)$, and the quotient is in $A_+$. So $A = S \oplus xy(x-y) S$ and the ring is generated by $b = x+y$, $c = xy$ and $f=xy(x-y)$, with the defining relation $f^2 = c^2 (b^2-4c)$.<|endoftext|> TITLE: Anti Arzela-Ascoli QUESTION [12 upvotes]: Notation: We say a sequence of real numbers diverges if it does not converge to a finite limit. We say a sequence $f_n$ of real valued functions on $[0, 1] $ are equibounded if $\sup_{n \in \mathbb N}\sup_{x \in [0, 1]} |f_n (x)| < \infty$. Some motivation: The Arzela Ascoli theorem for $C[0, 1]$ says that if we have an equibounded, equicontinuous sequence of functions, we have uniform convergence along a subsequence. What happens if we drop equicontinuity (but retain continuity of the functions)? What kind of convergence can we expect? For any countable subset $S$ of $[0, 1]$, we can still diagonalize to get pointwise convergence on $S$ along a subsequence $f_{n_k}$ that depends on $S$. But conjecturally this is the best we can do in general. Indeed as far as pointwise convergence is concerned, we have: Theorem 1: There exists an equibounded sequence of continuous functions $f_n: [0, 1] \to \mathbb R$ such that for every increasing sequence $n_k$ of naturals, $f_{n_k} (x)$ diverges for almost every $x \in [0, 1]$. However, the examples that I am familiar with all rely on some sort of “independence” or equidistribution type argument. For such examples, one intuitively expects $f_n$ to converge in Cesaro sense. To illustrate, we consider the following two examples. The first of these was proposed by Yuval Peres in discussion on a seperate forum. Example 1: Take $f_n (x) = \sin(nx)$.That $f_n$ satisfy the conditions in Theorem 1 can be seen by noting that by Weyl’s criterion for equidistribution, the sequence $n_{k}x \ \text{mod} \ 1$ is equidistributed for a.e. $x \in [0, 1]$. However by the same coin, we have that for any subsequence, $f_{n_k} (x)$ converges in Cesaro sense for almost every $x$. Example 2: Consider the domain $[0, 1]$ as a probability space, and take $g_n$ to be the indicator function of independent events with probability $1/2$ each. Then an argument based on the second Borel Cantelli lemma gives us that the $g_n$ satisfy all conditions in Theorem 1 except continuity. We can then approximate the $g_n$ by a sequence $f_n$ of continuous functions, whence $f_n$ satisfy the conditions in Theorem 1. But again it can be shown that for any subsequence, $f_{n_k} (x)$ converges in Cesaro sense almost everywhere. This suggests the following question: Question: Does there exist an equibounded sequence of continuous functions $f_n: [0, 1] \to \mathbb R$ such that for every increasing sequence $n_k$ of naturals, $\lim_{N \to \infty} \frac{1}{N} \sum_{k = 0}^{N-1} f_{n_k}(x)$ almost everywhere fails to exist? REPLY [21 votes]: Under the stated conditions, there always exists a subsequence that Cesaro converges almost everywhere. This was a question of Steinhaus, solved by Revesz [1]. More generally, it suffices that the sequence $f_n$ be uniformly bounded in $L^1$; This is a striking Theorem of Komlos [2] which in particular implies the Kolmogorov strong law of large numbers. [1] P. Revesz, On a problem of Steinhaus, Acta Mathematica Academiae Scientiarum Hungaricae, 16 (1965), pp. 310–318. [2] Janos Komlos, A generalization of a problem of Steinhaus, Acta Mathematica Academiae Scientiarum Hungaricae, 18 (1967), pp. 217–229. https://scholarship.libraries.rutgers.edu/discovery/delivery?vid=01RUT_INST:ResearchRepository&repId=12643427010004646#13643523800004646<|endoftext|> TITLE: Comparing the existing formulations of universal algebra and their levels of generality QUESTION [11 upvotes]: I am a newcomer to universal algebra and I just read this (very good, IMO) book on the topic: Adámek, J., Rosický, J., & Vitale, E. M. (2010). Algebraic theories: a categorical introduction to general algebra (Vol. 184). Cambridge University Press. The authors most notably treat the case of 1-sorted first-order finitary Lawvere theories (equivalently: finitary monads). This is, as I understand it, the most common level of generality appearing in the literature to deal with (categorical) universal algebra. Additionally the authors explain the relation between "old-fashioned" universal algebra -- which we'll call equational theories here -- and Lawvere theories, as well as the relation between Lawvere theories and categories of models of Lawvere theories. The quick and informal upshot is that there is a pair of such adjunctions: old universal algebra, lawvere thies, models Where $\mathrm{EqTh}$ is the category of equational theories (that is: signatures + equations), $\mathrm{AlgTh}$ is the category of Lawvere theories, $\mathrm{AlgCat}$ is the category of algebraic categories. Algebraic categories are basically defined to be categories of models of Lawvere theories. The first adjunction connects particular presentations of theories to Lawvere theories. The second adjunction is some kind of Gabriel-Ulmer duality and links theories to their models. My questions is: what are the existing ways we can generalize this picture? And is there an ultimate framework encompassing all those generalizations? For instance I would be interested to hear about what can be done to obtain type theories as some kind algebraic theories. Some examples of generalisations (either on the presentation side or on the Lawvere theory side or on both sides): higher-order logic, enriched Lawvere theories, sketches, Makkai's first order logic dependent sorts (FOLDS), infinitary theories. I'm sure there are a lot more! EDIT: Here is a paper about higher order algebraic theories REPLY [4 votes]: @focso’s answer and @varkor’s answer give excellent answers covering most aspects of your question. I’ll just add a little more on extending these frameworks to cover “type theories” — i.e. to include both sort-dependence and variable binding. This is a quite new and open topic, but a good bit of groundwork has been laid. The well-established part dates back to the 80’s, starting with Cartmell’s work. This defines contextual categories, and later various slight variants (categories with attributes (CwA’s), categories with families (CwF’s)…) corresponding to a so-called “generalised algebraic theory” or “dependent algebraic theory”. This gives a nice treatment of dependent sorts, but not yet of variable binding. Specific type theories with binders then correspond to extra algebraic structure on contextual categories (or CwA’s, CwF’s, etc). A good introduction is Martin Hofmann’s chapter Syntax and semantics of dependent types, doi:10.1007/978-1-4471-0963-1_2, un-paywalled pdf here. For individual type theories, this correspondence works satisfactorily (and has been developed by many authors — I won’t try to give a bibliography here); but until quite recently, it wasn’t generalised to a general notion of “type theory”. Approaches to that so far include: Isaev, Algebraic Presentations of Dependent Type Theories, arXiv:1602.08504. This defines type theories directly as certain essentially-algebraic extensions of CwA’s. Uemura, A General Framework for the Semantics of Type Theory, arXiv:1904.04097. This in my opinion is the most powerful and best developed approach so far. It gives a categorical abstract definition of type theories as representable map categories (an imaginative and far-reaching generalisation of Fiore/Awodey’s re-analysis of CwF’s), and a corresponding syntactic treatment (based on the “logical framework” approach). It’s much more general than either of the other approaches I mention here, and very cleanly set up. Bauer–Lumsdaine–Haselwarter, A general definition of dependent type theories, arXiv:2009.05539. A syntactic definition of general type theories, taking variable binding seriously as primitive, and written with a categorical analysis strongly in mind (though not given in the paper). This should correspond to a slight variant of Isaev’s algebraic definition, and to a subclass of Uemura’s. I very much hope that these and the connections between them will be developed further in the near future, and that this part of your question will have a better answer in five years than it does now!<|endoftext|> TITLE: Branched covers of the sphere branched over few points QUESTION [5 upvotes]: Let $X$ be a compact Riemann surface of genus $g\geq 2$. By the Riemann-Roch theorem, $X$ is a branched cover of the sphere, branched over finitely many branched values. What is the smallest number of such values, for general $X$ of genus $g$? Since moduli space has complex dimension $3g-3$, and branched covers branched over $B$ points are contained in the union of countably many varieties of dimension at most $B-3$, for general $X$ we need at least $3g$ branched values. Riemann-Roch shows that there is a holomorphic $X\to\hat{\mathbb{C}}$ of degree $g+1$ having a single pole, of degree $g+1$. By Riemann-Hurwitz, this function has $4g$ critical points, of which $g$ are over infinity. So in total there are at most $3g+1$ critical values. So my question becomes: Question. Let $X$ be a compact Riemann surface of genus $g\geq 2$. Is there a holomorphic function $f\colon X\to \hat{\mathbb{C}}$ which has at most $3g$ critical values? It is plausible that the answer is positive, but in either case the answer is surely known. Does anyone know a reference? I have been told that Brill-Noether theory shows the existence of a meromorphic function of degree $\lfloor(g+3)/2\rfloor$ on $X$. For even $g$, applying Riemann-Hurwitz then shows that there are $3g$ critical points for this function. That answers the question in the positive for even $g$. But for odd $g$, we get the same number $3g+1$ as via Riemann-Roch. REPLY [7 votes]: Let me post my comment as an answer. Take a Weierstrass point on $X$, that is, a point $P$ for which there exists a meromorphic function $f$ with a pole of order $k\leq g$ at $P$ (there always exists such a point). Then apply the argument in the post: by Riemann-Hurwitz the number of critical points of $f$, counted with multiplicity, is $2g-2+2k$. But $P$ appears with multiplicity $k-1$ in this count, so the number of critical values outside $\infty$ is $\leq 2g-2+2k-(k-1)=2g+k-1$, and the total number of critical values is $\leq 2g+k\leq 3g$.<|endoftext|> TITLE: Elementary proof that an open subset of $\Bbb{R}^n$ does not have measure zero? QUESTION [7 upvotes]: There is an elementary theory of subsets of $\Bbb{R}^n$ of measure zero, namely one defines the volume of a cube in the obvious way and one says that a subset $A$ has measure zero if given any $\epsilon>0$ there exists a countable number of cubes that cover $A$ and such that the sum of the volumes of the cubes is $\leq \epsilon$. One can show, with modest effort, that this notion is invariant under diffeomorphisms and thus leads to the notion of subsets of measure zero on a smooth manifold. This notion shows up in Sard's Theorem which says that the set of critical values has measure zero. Is there an elementary argument why non-empty open subsets do not have measure zero? Evidently it follows from standard measure theory, but for my topology course I would appreciate it if there was an elementary argument, but I can't think of one and I can't find one. This is stated as an exercise in Lee's book on smooth manifolds, but it's not obvious to me. Note that even $n=1$ seems tricky. REPLY [2 votes]: It should also be mentioned that $[0,1]$ being not a null Lebesgue set can also be derived by the Lebesgue-Vitali characterization theorem for Riemann integrable function. If $[0,1]$ were negligible, any bounded function on it would be Riemann integrable, but we know that for instance the Dirichlet function $\chi_{[0,1]\cap\mathbb{Q}}$ is not.<|endoftext|> TITLE: Alternating sum of hook lengths: Part I QUESTION [12 upvotes]: Given $\lambda$ an integer partition of $n$, let $h_{ij}(\lambda)$ denote the hook length of cell $(i,j)$ in the Young diagram of $\lambda$. Is there a closed formula or a generating function for the following sequence? $$f_n=\sum_{\lambda\vdash n}\sum_{(i,j)\in\lambda}(-1)^{i+j}\cdot h_{ij}(\lambda)$$ REPLY [3 votes]: This is a follow up note on Gjergji Zaimi's fine response. It seems that $$\sum_nA_nq^n=\sum_k\frac{q^{2k-1}}{1-q^{2k-1}}=\sum_n\frac{d(n)}{\nu_2(2n)}q^n,$$ where $d(n)$ is the sum of (positive) divisors of $n$ and $\nu_2(m)$ is the $2$-adic valuation of $m$. Therefore, we obtain the convolution sum $$f_n=\sum_{k=1}^n\frac{d(k)\cdot p(n-k)}{\nu_2(2k)}$$ where $p(m)$ is the number of integer partition of $m$.<|endoftext|> TITLE: Vector bundles on adic spaces QUESTION [6 upvotes]: Let $X = \mathrm{Spa}(A,A^+)$ be an analytic sheafy adic space. Let $\mathcal{E}$ be a locally finite free $\mathcal{O}_X$ sheaf. Does $\mathcal{E}$ correspond to a geometric vector bundle over $X$? In other words, does there exist an analytic adic space $E$ with a morphism to $X$, unique up to isomorphism as an analytic adic space over $X$, such that for every open immersion of analytic adic spaces $S\rightarrow X$ there is a natural isomorphism $\mathcal{E}(S) \cong \mathrm{Hom}_X(S,E)$? By a theorem of Kedlaya-Liu, we do know that there is a correspondence between $\mathcal{E}$'s and finite projective $A$-modules, which is achieved by mapping $\mathcal{E}$ to the global sections $M=\mathrm{H}^0(X,\mathcal{E})$. So a natural guess for $E$ would be to take $\mathrm{Spa}(B,B^+)$ with $B=\mathrm{Sym}_A(M^\vee)$; however, I do not know what to take for $B^+$. (Is there some sort of $M^+$?) Thanks! REPLY [8 votes]: The question is local on $X$, so we may assume that $\mathcal E$ is finite free, of rank $n$, say. In that case, as also SashaP points out, the question amounts to the question whether $\mathbb A^n_X$ is an adic space. As this is covered by an increasing union of balls $\mathbb B^n_X=\mathrm{Spa}(A\langle T_1,\ldots,T_n\rangle,A^+\langle T_1,\ldots,T_n\rangle)$, it is also equivalent to ask whether $A\langle T_1,\ldots,T_n\rangle$ is sheafy. (Note that your choice of $B$ is not correct, as your $B$ is not a Huber ring. You have to complete $B$ in a certain way -- this is like the difference between $A[T_1,\ldots,T_n]$ and $A\langle T_1,\ldots,T_n\rangle$ -- and then take a union over all possible such choices of completions.) Now I'm not up to speed about the known counterexamples to sheafyness, but I think there ought to be examples where $A$ is sheafy but $A\langle T\rangle$ is not, so in this sense the answer to your question would be no. On the other hand, any practical condition guaranteeing sheafyness of $A$ usually also implies sheafyness of $A\langle T\rangle$ -- e.g., strongly noetherian, or sousperfectoid, or ... . Finally, let me also add again the advertisement that in the setting of Analytic Geometry, it is possible to generalize Huber's theory of adic spaces to incorporate non-sheafy Huber rings (in particular, by allowing the structure sheaf to be a sheaf of animated condensed rings). So in this generalized setting, the answer to the question would also be Yes.<|endoftext|> TITLE: Mindset to understand category theory QUESTION [17 upvotes]: I am a 17 years old student and I am really interested in category theory due to its abstraction and beauty. I wanted to know if you'd have any advices to approach this theory and if you have papers to begin with this. Thank you in advance for your answer. REPLY [2 votes]: Not trying to self-promote, but I have about 100 pages of basic category theory notes I was planning on eventually pushing towards internal/two-dimensional category theory and putting on the arxiv as a survey article. They've been sitting around for two years now, though, so I uploaded them in their current (very unfinished) form in case they might help you. All the content on $1$-category theory is solid (sections 1-7), and sections 8-9 are serviceable, but please ignore the later sections for now. They assume no familiarity with anything besides what a set and function are, although being familiar with the naturals/integers/rationals as distinct entities will help understand an early example of a non-surjective epimorphism. If you need to develop familiarity with sets please feel free to grab any standard text on set theory and read it, or if you like my notes you can try this arxiv paper I uploaded a few years back which builds up the set theory you'll need from scratch. Again, the research at the end is shaky and shouldn't be used, but pages 1-6 give a 'from-scratch' development of the axioms of MK class theory and the stuff through page 19 offers some basic lemmas to try and prove (don't read my proofs unless you need to!) Best of luck, and remember not to beat yourself up if things get challenging! The level of abstraction in either of the above note sets is likely to be a signifigant step up from the kinds of mathematics you've encountered thus far, so be sure to come back to MSE and ask further questions! (MO is a bit too high level for this stuff generally speaking, as evidenced by the users trying to close the question, but these are completely reasonable questions over at MSE)<|endoftext|> TITLE: Class group of hypersurfaces of finite representation type QUESTION [8 upvotes]: Let $k$ be an algebraically closed field of characteristic different from $2,3$ and $5$, and let $R=k[[x,y,x_2,\dots,x_d]]/(f)$, where $f\in(x,y,x_2,\dots,x_d)^2$, $f\neq0$. By results of Buchweitz-Greuel-Schreyer and Knörrer, $R$ has finite Cohen-Macaulay representation type if and only if $R\cong k[[x,y,x_2,\dots,x_d]]/(g+x_2^2+\cdots+x_d^2)$, where $g\in k[x,y]$ is one of the following polynomials (one-dimensional ADE singularities): ($A_n$) $x^2+y^{n+1}$, $n\geq1$ ($D_n$) $x^2y+y^{n-1}$, $n\geq4$ ($E_6$) $x^3+y^4$ ($E_7$) $x^3+xy^3$ ($E_8$) $x^3+y^5$ I am interested in the class group $\mathrm{Cl(R)}$ of these rings, for any $d\geq1$. Has anyone already computed it? For $d=2$, i.e., two-dimensional ADE singularities, the class group is known and can be computed using the Auslander-Reiten quiver of $R$ (see e.g. [Yos90, $\S13$]). However, I could not find similar (even partial) computations for $d>2$ in the literature. Could you point me to some references in this direction? Thank you in advance! References: [Yos90] Yoshino, Yuji, Cohen-Macaulay modules over Cohen-Macaulay rings. London Mathematical Society Lecture Note Series, 146. Cambridge University Press, Cambridge, 1990. REPLY [2 votes]: When $d\geq 3$, these are isolated hypersurface singularities of dimension at least $4$, so are UFD by the Grothendieck's local Lefschetz Theorem. When $d=2$ and the field has characteristic $0$, the class group is $\mathbb Z^{r-1}$ where $r$ is the number of branches of $g$. See 2.2.6 of Kollár's paper "Flip, flops, minimal models, etc". That result was for the Picard group of the punctured spectrum of the affine hypersurface, but it should agree with the class group of the completed local rings (you might need to use old results by Danilov here). I do not know if anyone has worked out the case of positive char. But at least we know that the class group is torsion-free, see the references in this MO question and answer. A good reference for local Picard group (which for a normal local point of dimension $2$ or higher is just the class group at that point) is Kollár's paper "Maps between local Picard groups", where you can also find reference to the Grothendieck-Lefchetz theorem mentioned above.<|endoftext|> TITLE: Embedding of $C(X)$ into $B(H)$ where $H$ is separable QUESTION [6 upvotes]: I would like to ask a question which may look strange at the first sight nevertheless I find it interesting. Let $H$ be a separable Hilbert space: for any separable $C^*$-algebra $A$ one can embed $A$ into $B(H)$. Very often one can do even better and many nonseparable $C^*$-algebras still embed into $B(H)$, for example $\ell^{\infty}, L^{\infty}[0,1]$ or $B(H)$ itself. However it is not always possible even for quotients of such algebras, $\ell^{\infty}/c_0$ being an example. One can view $\ell^{\infty}$ as $C(\beta\mathbb{N})$ where $\beta\mathbb{N}$ is the Stone-Cech compactification of $\mathbb{N}$ (which is of cardinality $2^{\mathfrak{c}}$) and $\ell^{\infty}/c_0$ as $C(\beta\mathbb{N} \setminus \mathbb{N})$ (still of cardinality $2^{\mathfrak{c}}$). So here comes my question: Let $X$ be a compact nonmetrizable Hausdorff topological space of cardinality $\mathfrak{c}$. Is it true that $C(X)$ embeds into $B(H)$ for separable $H$? REPLY [7 votes]: No, not necessarily. If $C(X)$ embeds into $B(H)$ then $X$ must be ccc, but there exist compact Hausdorff spaces of cardinality $\mathfrak{c}$ which are not ccc. (One example is to take $\mathfrak{c}$ with its discrete topology and form the one-point compactification $\mathfrak{c} \cup \{\infty\}$; then all the singletons except $\infty$ are disjoint open sets. Another is the ordinal $\mathfrak{c}+1$ with its order topology; all the successor singletons $\{\alpha+1\}$, $\alpha < \mathfrak{c}$, are disjoint open sets.) To see this, suppose that $X$ is not ccc, so that there is an uncountable family $\{U_\alpha\}$ of disjoint nonempty open sets. By the appropriate version of Urysohn's lemma, for each $\alpha$ there is a nonzero real-valued $f_\alpha \in C(X)$ supported inside $U_\alpha$; in particular, $f_\alpha f_\beta = 0$ for $\alpha \ne \beta$. Now if there is an injective *-homomorphism $\Phi : C(X) \to B(H)$, then each $\Phi(f_\alpha)$ is nonzero, so we may find $h_\alpha \in H$ so that $\|\Phi(f_\alpha) h_\alpha\|=1$. Also, $\Phi(f_\alpha)$ is self-adjoint, so for $\alpha \ne \beta$ we have $$\langle \Phi(f_\alpha) h_\alpha, \Phi(f_\beta) h_\beta\rangle = \langle h_\alpha, \Phi(f_\alpha)\Phi(f_\beta) h_\beta\rangle = \langle h_\alpha, \Phi(f_\alpha f_\beta) h_\beta\rangle = 0$$ so that $\{\Phi(f_\alpha) h_\alpha\} \subset H$ is an uncountable orthonormal set, which is impossible if $H$ is separable. This is fairly close to optimal because every separable compact Hausdorff $X$ does have $C(X)$ embeddable into $B(H)$ (recall that separable implies ccc but not conversely). Indeed, if $\{x_n\}$ is a countable dense subset of $X$, then identifying $f \in C(X)$ with the sequence $(f(x_1), f(x_2), \dots)$ gives an isometric embedding of $C(X)$ into $\ell^\infty$, which as we know embeds in $B(\ell^2)$. Presumably there is a necessary and sufficient condition somewhere in between, but I don't know what it is.<|endoftext|> TITLE: Jumping conics in Grassmannians QUESTION [5 upvotes]: Let $Gr(1,n)$ be the Grassmannian of lines in $\mathbb{P}^n$, and $f:\mathbb{P}^1\rightarrow Gr(1,n)$ a morphism of degree two. The pull-back $f^{*}S$ of the tautological bundle $S$ on $Gr(1,n)$ splits as $f^{*}S = \mathcal{O}_{\mathbb{P}^1}(a)\oplus \mathcal{O}_{\mathbb{P}^1}(b)$. If $f$ is general in the moduli space of degree two morphisms $\mathbb{P}^1\rightarrow Gr(1,n)$ then $a = b = -1$. However, for some morphisms $f$ we may have $a = 0, b = -2$. Is there any geometric characterization of the morphisms $f:\mathbb{P}^1\rightarrow Gr(1,n)$ yielding $a = 0,b = -2$? REPLY [8 votes]: These conics are exactly conics contained in some $$ \mathbb{P}^{n-1} \subset Gr(1,n), $$ that parameterizes all lines in $\mathbb{P}^n$ passing through a fixed point.<|endoftext|> TITLE: Finite simple groups all of whose Sylow subgroups of odd order are cyclic QUESTION [10 upvotes]: Let $G$ be a nonabelian finite simple group all of whose Sylow subgroups of odd order are cyclic. If we further assume that its Sylow $2$-subgroup is dihedral, then due to Suzuki, we know that $G\cong \operatorname{PSL}(2,p)$ for a prime $p>3$. Without any further assumption, what is the list of finite nonabelian simple groups all of whose Sylow subgroups of odd order are cyclic? Secondly, is there any reference for that not appealing to the full classification of finite simple groups? REPLY [17 votes]: Such simple groups are a subset of the finite simple "thin" groups, and the latter have been classified (by Michael Aschbacher in 1976 and 1978). A $2$-local subgroup of a finite group is the normalizer of some non-trivial $2$-subgroup. A finite simple group $G$ is said to be "thin" if all its $2$-local subgroups have cyclic Sylow subgroups for all odd primes. We may note that Feit and Thompson proved early in their odd order paper (using transfer theorems) that if $G$ is a finite group of odd order which has no elementary Abelian subgroup $p$-subgroup of rank $3$ or more for any prime $p$, then $G$ has a normal Sylow $q$-subgroup where $q$ is the largest prime divisor of $|G|$. Hence (without using the full force of the odd order theorem), it is indeed the case that if a finite non-Abelian simple group $G$ has cyclic Sylow subgroups for all odd primes $p$, then $G$ has even order, and is, in particular, a thin group. However, some thin groups do not satisfy the condition of the question. According to Aschbacher's classification, the list of thin finite simple groups is ${\rm SL}(2,2^{n})$ and ${\rm PSL}(2,q)$ for $q$ an odd prime power (but we can remove the cases when the odd $q$ is not prime in answer to the present question since the Sylow subgroup of order $q$ is not cyclic), ${\rm PSL}(3,4)$ (but we can remove this in answer to the present question since it has a non-cyclic subgroup of order $9$), ${\rm PSL}(3,p)$ for $p$ of the form $1+2^{a}3^{b}$ (which can be removed in answer to the present question when $p$ is odd, since there are non-cyclic $p$-subgroups in that case), ${\rm PSU}(3,p)$ ($p = 2^{a}3^{b}-1$, $b \in \{0,1\}$ (but these can be omitted in answer to the present question as there are non-cyclic $p$-subgroups), ${\rm PSU}(3,2^{n})$ (but these can be omitted in answer to the present question, since ${\rm PSU}(3,4)$ has an elementary Abelian subgroup of order $25$), ${\rm Sz}(2^{n})$ , the Tits group $^{2}F_{4}(2)^{\prime}$ (which can be omitted in answer to the present question, as it contains an elementary Abelian subgroup of order $9$), the Steinberg triality group $^{3}D_{4}(2)$ (which can be omitted in answer to the present question since it has elementary Abelian subgroups of order $9$), the Mathieu group $M_{11}$ (which can be omitted in answer to the present question as it has elementary Abelian subgroups of order $9$), and the first Janko group $J_{1}.$ Later edit: Notice that Aschbacher's classification of the thin finite simple groups predates the full classification of the finite simple groups. As a matter of historical interest, the classification of finite simple groups in which all $2$-local subgroups have low $p$-rank for all odd primes $p$ is a difficult part of the current proofs (and of ongoing revisions) of the classification of finite simple groups.<|endoftext|> TITLE: Boundary Maslov index of holomorphic disks in Calabi-Yau manifolds QUESTION [5 upvotes]: Let $u \colon \Sigma^2 \to M^{2n}$ be a holomorphic disk (so $\Sigma = \{z \in \mathbb{C} \colon |z| \leq 1\}$) in a compact Calabi-Yau manifold $M$ of real dimension $2n$ with boundary on a Lagrangian submanifold $L^n \subset M^{2n}$. Note that the normal bundle $N\Sigma \to \Sigma$ is a complex vector bundle of complex rank $n-1$. At points of $\partial \Sigma$, let's orthogonally decompose $TL|_{\partial \Sigma} = T(\partial \Sigma)|_{\partial \Sigma} \oplus F$, so that $F \subset N\Sigma|_{\partial \Sigma}$ is a totally real subbundle of real rank $n-1$. I am interested in the boundary Maslov index $\mu(N\Sigma, F) \in \mathbb{Z}$. My question is: Among all possible pairs $(u, L)$ of holomorphic curves with Lagrangian boundary, can $\mu(N\Sigma, F)$ attain every possible integer value, or are there restrictions (coming from, say, the Calabi-Yau assumption on $M$)? I am not a symplectic geometer, so I don't know what to expect. In particular, I don't know whether there are standard examples of holomorphic curves with Lagrangian boundary for which $\mu(N\Sigma, F)$ is easily computable. REPLY [2 votes]: I assume you're asking about embedded or at least immersed discs, in order to make sense of the normal bundle. If so, let's fix an immersion $\iota\colon\Sigma\to M$ and observe that the pullback bundle-pair $(\iota^*TM,\iota^*TL)$ splits as $(T\Sigma\oplus N\Sigma,TS^1\oplus F)$ for some totally real subbundle $F\subset\iota^*TL$. The total Maslov number $\mu$ of the disc splits as $2+n$ where $n$ is the Maslov index of the normal component (which is what you're interested in). So we might as well ask about the ordinary Maslov number and then subtract 2. First thing to notice is that the Maslov number needs to be even if $L$ is orientable (this is because the orientation allows you to lift the classifying map for the bundle pair to the double cover which classifies orientable bundle pairs). Now it's easy to get any positive even number as the Maslov index of a disc. To see this, note that any symplectic manifold (in particular a compact Calabi-Yau) contains a Darboux ball (symplectomorphic to an open ball in $\mathbb{C}^n$, and inside a Darboux ball you can find a Lagrangian product torus which bounds discs of Maslov index 2, 4, 6, etc. This gives you normal Maslov indices 0, 2, 4, etc. I imagine it's possible to cook up examples with any negative Maslov index. If you're interested in things like this, you might be interested in the following papers by Globevnik and Oh: Josip Globevnik. Perturbation by analytic discs along maximal real submanifolds of C N . Math. Z., 217(2):287–316, 1994. Yong-Geun Oh. Riemann-Hilbert problem and application to the perturbation theory of analytic discs. Kyungpook Math. J., 35(1):39–75, 1995 They study holomorphic discs in complex manifolds with boundary on a totally real submanifold, and show that there is a splitting theorem like the Birkhoff-Grothendieck splitting for holomorphic vector bundles over $\mathbb{CP}^1$. Namely, every holomorphic bundle pair (holomorphic bundle over the disc together with a totally real subbundle around the boundary) splits as a direct sum of line bundles.<|endoftext|> TITLE: Calculation of a series QUESTION [14 upvotes]: It seems that we have: $$\sum_{n\geq 1} \frac{2^n}{3^{2^{n-1}}+1}=1.$$ Please, how can one prove it? REPLY [49 votes]: This is the special case $q=3$ of a formula $$ \qquad\qquad \sum_{n=1}^\infty \frac{2^n}{q^{2^{n-1}}+1} = \frac{2}{q-1} \qquad\qquad(*) $$ which holds for all $q$ such that the sum converges, i.e. such that $|q|>1$. This follows from the identity $$ \frac{1}{x-1} - \frac{2}{x^2-1} = \frac{1}{x+1}. $$ Substitute $q^{2^{n-1}}$ for $x$, multiply by $2^n$, and sum from $n=1$ to $n=N$ to obtain the telescoping series $$ \frac{2}{q-1} - \frac{2^{N+1}}{q^{2^N}-1} = \sum_{n=1}^N \left( \frac{2^n}{q^{2^{n-1}}-1} - \frac{2^{n+1}}{q^{2^n}-1} \right) = \sum_{n=1}^N \frac{2^{n-1}} {q^{2^n} + 1}. $$ Taking the limit as $N \to \infty$ yields the claimed formula $(*)$.<|endoftext|> TITLE: Chip-firing clocks QUESTION [6 upvotes]: Let $G$ be some outdegree-regular directed graph with $n$ vertices and let $H$ be the Laplacian of $G$, so that the rows of $H$ correspond to chip-firing moves. I’m interested in linear functions $f$ from $\mathbb{Z}^n$ to $\mathbb{Z}/k\mathbb{Z}$ with the property that performing a chip-firing move on a vector in $\mathbb{Z}^n$ increases the value of $f$ by 1 mod $k$ (I call such a function $f$ a chip-firing “clock”). Such an $f$ will exist only for certain values of $k$. This feels like it’s related to the structure of the cokernel of the Laplacian but I’m having trouble seeing exactly what's going on. I’m interested in knowing about all such functions $f$ for a given $G$. REPLY [2 votes]: This is just to record the observation of lambda from the comments. I'll keep your convention that the rows of $H$ determine the chip-firing moves, so that for a sequence of firings $v\in\mathbb{Z}^n$ the result of caring out these firings (starting from the zero configuration) is $vH$. (But this means that we should really be talking about the kernel and cokernel of $H^t$ everywhere...) For $x\in \mathbb{Z}^n$, the linear function $f(v)=\langle v, x\rangle \mod k$ is a clock if and only if $Hx=\mathbf{1} \mod k$. (Here $\langle \cdot , \cdot \rangle$ is the usual inner product, and $\mathbf{1}$ is the all ones vector.) This is because the requirement we have to satisfy to be a clock is that $f(vH)=\langle v, \mathbf{1}\rangle \mod k$ for all $v\in\mathbb{Z}^n$, so we need $f(vH)=\langle vH, x\rangle =\langle v,Hx\rangle$ to be equal moulo $k$ to $\langle v, \mathbf{1}\rangle$ for all $v\in \mathbb{Z}^n$, which happens if and only if $Hx=\mathbf{1} \mod k$. It still is not totally clear how given a graph $G$ to find the (finite!) list of $k$ for which a clock exists, but for a fixed $k$ at least this makes it clear that the question of whether such a $k$ exists is a "linear algebra" problem (if $k$ is prime then indeed we're talking about linear algebra over a field). [ By the way, here is the argument that the set of such $k$ is finite. $H^t$ is a singular M-matrix, so there is some vector $v_{*}\in\mathrm{ker}(H^t)$ with $v_{*}\neq 0$ but all entries of $v_{*}$ nonnegative. Hence in particular $\langle v_{*}, \mathbf{1} \rangle > 0$. But, as mentioned, if there is a $k$-clock we need that $\langle v, \mathbf{1} \rangle = 0 \mod k$ for any $v \in \mathrm{ker}(H^t)$, so as long as $k > \langle v_{*}, \mathbf{1}\rangle$ then there cannot be a $k$-clock. ]<|endoftext|> TITLE: Expectation of period length of functions $f:\{1,\ldots,n\}\to \{1,\ldots,n\}$ QUESTION [5 upvotes]: For $n\in\mathbb{N}$, let $[n]:= \{1,\ldots,n\}$. Let $\text{Fun}(n)$ denote the set of all functions $f:[n]\to[n]$. To $f\in\text{Fun}(n)$ associate a sequence $\text{seq}(f))$ defined recursively by $\text{seq}(f)_1 = f(1)$, and $\text{seq}(f)_{k+1} = f(\text{seq}(f)_k)$ for all $k\in\mathbb{N}$. Eventually $\text{seq}(f)$ will be periodic, and with $\text{per}(f)$ we denote the length of the period of $\text{seq}(f)$. By $E_n$ we denote the expected value of $\text{per}(f)$ for any $f\in\text{Fun}(n)$. Explicitly, we have $$E_n = \frac{1}{n^n}\sum_{f\in\text{Fun}(n)}\text{per}(f).$$ Questions. Do we have $\lim\sup_{n\to\infty} E_n/n > 0$, and if yes, what is that value? If no, do we have $\lim\sup_{n\to\infty} E_n/\log(n) > 0$? REPLY [5 votes]: A standard reference for all sorts of average statistics of random maps of $\{1,\ldots,,N\}$ is Probability distributions related to random mappings, Bernard Harris, Ann. Math. Statist. 31 (1960), 1045-1062.<|endoftext|> TITLE: How professional mathematicians deal with discouragement? QUESTION [10 upvotes]: All professional mathematicians feel discouraged occasionally due to some issue. My question is: How do professional mathematicians deal with discouragement? In this link , Andrew Wiles say that I would go out for a walk. I'd often walk down by the lake. Walking has a very good effect in that you're in this state of relaxation. REPLY [28 votes]: Pour yourself a beer and reflect on how crazy it is that 1) modern life actually needs knowledge about stuff like "elliptic curves over finite fields" and 2) you are lucky enough to make a living thinking and teaching about them. Works for me every time. REPLY [12 votes]: It helps me to remember that none of this stuff really matters. That might sound discouraging at first; why do something if it doesn't matter? Because nothing really matters, and we have to occupy our time! I was intentionally provocative in the presentation of the above idea, but in a more moderate tone it would probably read something like 'if mathematics makes you happy, do it because it makes you happy and forget the rest', or 'do it because it makes you happy and remember that we are all very small in a very big universe'. It is somewhat Absurd to pursue challenging tasks in an indifferent universe, but while we're here we should occupy our time with things we find interesting and which bring joy to us and others -- we must imagine Sisyphus happy. Mathematical research is somewhat intrinsically difficult, but for most people who ultimately pursue it the labor is one of love, and a source of limitless wonderment and joy (or wonder and joy with limits if you work in category theory/calculus ;). All of this is to say, don't sweat the small stuff. Pursue the things you find fascinating, and what will be will be -- que sera sera. (I find that good music helps too.) After reading this answer before posting, it occurrs to me that none of the content of this message is specific to mathematics in the sense that we could swap out the word 'mathematics' for any other challenging pursuit and the same point would stand. I've flagged the question to be converted to CW (community wiki) by a moderator and would like to wait to post this answer until that happens, but I'm posting it now because I've waited an hour after flagging and would like to post it and go to sleep.<|endoftext|> TITLE: Is Koszul homology of a monomial ideal always generated by the "obvious" things? QUESTION [5 upvotes]: Let $R = k[x_1 , \dots , x_n]$ be a polynomial ring over a field and $I$ a monomial ideal in $R$. Then, is it true that the Koszul homology of $R/I$ is always generated by elements of the form $$r e_{i_1} \wedge \cdots \wedge e_{i_k} \quad \textrm{where} \ x_{i_\ell} r \in I \ \textrm{for all} \ 1 \leq \ell \leq k ?$$ These elements are certainly contained in the Koszul homology. Moreover, this does constitute a generating set, for example, for stable ideals, since one can show that the Koszul homology is actually minimally generated by a subset of elements of the above form. I have computed a fair amount of examples and it seems true more generally that this is a generating set. I'm not sure if this is well-known or perhaps false, and any help or references for this would be greatly appreciated. REPLY [7 votes]: This holds for $n\leq 3$ but may fail for $n=4$ and higher. See Proposition 2.6 and Example 2.9 in the paper "On monomial Golod ideals" (but probably known to experts before).<|endoftext|> TITLE: Perron-Frobenius and Markov chains on countable state space QUESTION [6 upvotes]: The following question naturally arises in the theory of Markov chains with countable state space to which I would be curious to know the answer: Let $A:\ell^1 \rightarrow \ell^1$ be a contraction, i.e. $\Vert A \Vert \le1 ,$ such that $A$ is positivity preserving. Moreover, let $A$ have the property that it preserves probabilities, i.e. let $x=(x_i) \in \ell^1$ such that $x_i \ge 0$ and $\sum_i x_i =1$, then also $\sum_i (Ax)_i=1.$ If we then know that the spectrum of $A$ on the unit circle consists of isolated point spectrum and let $\lambda \in \sigma(A)$ be one of them, i.e. $\vert \lambda \vert=1$. We can then study the spectral projection $\text{Proj}_{\lambda}$ associated with $\lambda$. Does it follow that $$(A-\lambda) \text{Proj}_{\lambda}=0?$$-Maybe at least for $\lambda=1$? REPLY [3 votes]: What you are looking for is actually true for every power-bounded operator, without any appeal to positivity: Theorem. Let $E$ be a Banach space and let $A: E \to E$ be a bounded linear operator such that $\sup_{n \in \mathbb{N}_0} \|A^n\| < \infty$. If $\lambda$ is an isolated spectral value of $A$ and a pole of the resolvent, and has modulus $1$, then the corresponding pole order is $1$ and hence, $\lambda$ is a semi-simple eigenvalue. In particular the range of the corresponding spectral projection $P$ coincides with the eigenspace $\ker(\lambda-A)$. Sketch of proof. This is all classical spectral theory. The essence of the proof is as follows: if the eigenvalue was not semi-simple, then we could find a generalized eigenvector $x$ of rank $2$. For the non-zero vector $y := (A-\lambda)x$ we would then obtain $$ A^nx = n\lambda^{n-1}y + \lambda^n x \qquad \text{for each integer } n \ge 0, $$ which contradicts the power-boundedness of $A$. $\square$ I summed up several such results in Appendix A here: DOI: 10.18725/OPARU-4238 (without many proofs, bit with detailed references to the literature).<|endoftext|> TITLE: Planar flow with bounded orbits and a single equilibrium point QUESTION [7 upvotes]: Is there a $C^1$ flow $\varphi_t$ defined on ${\bf R}^2$ with a single fixed point $0$ and such that for all x, $$\lim_{t\rightarrow +\infty}\varphi_t(x) = 0,$$ $$\lim_{t\rightarrow -\infty}\varphi_t(x) = 0?$$ I think that this should not exist but I can't find a simple argument to rule out the existence of such a flow. Same question for a $C^0$ flow. REPLY [3 votes]: (edited to include Willie Wong's idea for $C^0$ case.) This kind of flow can't exist in any dimension. Let $S$ be the unit sphere and $B$ be the open unit ball. If the origin is a global attractor for $\varphi$, then $S \subset \bigcup\limits_{t>0}{\varphi_{-t}(B)}$. By compactness, $S$ is covered by a union of a finite subset of the $\varphi_{-t}(B)$. That implies that there is a constant $T$ such that no point on $S$ (or $\overline{B}$) flows for more than time $T$ outside $\overline{B}$. Since the image of $[0, T] \times \overline{B}$ under $\varphi$ is compact, it can't cover the whole space, so the origin is not a global attractor for $\varphi^{-1}$.<|endoftext|> TITLE: Is there a good general definition of "sheaves with values in a category"? QUESTION [24 upvotes]: Let $\mathcal{A}$ be a category. There is a common definition of "sheaves with values in $\mathcal{A}$", which is what one obtains by taking the Grothendieck-style definition of "sheaf of sets" (i.e. in terms of presheaves satisfying a certain limit condition with respect to all covering sieves) and blithely replacing $\textbf{Set}$ with $\mathcal{A}$. In my view, this is a bad definition if we do not assume $\mathcal{A}$ is sufficiently nice – say, locally finitely presentable. When $\mathcal{A}$ is locally finitely presentable, we obtain various properties I consider to be desiderata for a "good" definition of "sheaves with values in $\mathcal{A}$", namely: The properties of limits and colimits in the category of sheaves on a general site with values in $\mathcal{A}$ are "similar" to those of $\mathcal{A}$ itself. (I am being vague here because even when $\mathcal{A}$ is locally finitely presentable, the category of sheaves with values in $\mathcal{A}$ may not be locally finitely presentable – this already happens for $\mathcal{A} = \textbf{Set}$.) The category of sheaves on a site $(\mathcal{C}, J)$ with values in $\mathcal{A}$ is (pseudo)functorial in $(\mathcal{C}, J)$ with respect to morphisms of sites. (By "morphism of sites" I mean the notion that contravariantly induces geometric morphisms.) The construction respects Morita equivalence of sites, i.e. factors through the (bi)category of Grothendieck toposes. The construction respects "good" (bi)colimits in the (bi)category of Grothendieck toposes, i.e. sends them to (bi)limits of categories. (I don't know what "good" should mean here, but at minimum it should include coproducts. When $\mathcal{A}$ is locally finitely presentable, there is a classifying topos, so in fact the construction respects all (bi)colimits.) The category of sheaves on the point with values in $\mathcal{A}$ is canonically equivalent to $\mathcal{A}$. The category of sheaves on the Sierpiński space with values in $\mathcal{A}$ is canonically equivalent to the arrow category of $\mathcal{A}$. Question. What is a (the?) "good" definition of "sheaves with values in $\mathcal{A}$"? ... when $\mathcal{A}$ is finitely accessible, not necessarily cocomplete, e.g. the category of Kan complexes, or the category of divisible abelian groups? ... when $\mathcal{A}$ is an abelian category, not necessarily accessible, e.g. the category of finite abelian groups, or the category of finitely generated abelian groups? ... when $\mathcal{A}$ is a Grothendieck abelian category, not necessarily locally finitely presentable? Perhaps something like right Kan extension along the inclusion of the (bi)category of presheaf toposes into the (bi)category of Grothendieck toposes might work – but the existence of toposes with no points suggests it may not – but it would be nice to have a somewhat more concrete description. There is a temptation to strengthen desideratum 6 to require that the category of presheaves on a (small) category $\mathcal{C}$ be equivalent to the category of functors $\mathcal{C}^\textrm{op} \to \mathcal{A}$, but this does not appear to be a good idea. As Simon Henry remarks, if $\mathcal{C}^\textrm{op}$ is filtered, then the unique functor $\mathcal{C} \to \mathbf{1}$ is a morphism of sites corresponding to a geometric morphism that has a right adjoint (i.e. the inverse image functor itself has a left adjoint that preserves finite limits), so contravariant (bi)functoriality with respect to geometric morphisms forces the induced $\mathcal{A} \to [\mathcal{C}^\textrm{op}, \mathcal{A}]$ to have a left adjoint, i.e. $\mathcal{A}$ must have colimits of shape $\mathcal{C}^\textrm{op}$. This is the same argument that shows that the category of points of a topos must have filtered colimits. Since I am looking for a construction where $\mathcal{A}$ is not necessarily the category of models of a geometric theory, I conclude that I cannot require the category of presheaves on $\mathcal{C}$ to be $[\mathcal{C}^\textrm{op}, \mathcal{A}]$. REPLY [16 votes]: In my view, the correct notion of "sheaf of Xs" is "internal X in the topos (or $\infty$-topos) of sheaves of sets (or spaces)". (I mentioned this previously on MO here.) Since sheaves of sets are a limit theory, if X is also defined by a limit theory (i.e. the category of Xs is locally presentable), then by commutation of limits this is the same as a sheaf of Xs in the naive sense. But for other values of X it gives different answers. In fact, the answer it gives may depend on exactly how the theory of X is presented; but that's reasonable becaues sometimes there is more than one correct notion of "sheaf of Xs" (equivalently, there is more than one version of X in the internal constructive logic of a topos). For instance: If X = fields, there are discrete fields, Heyting fields, and residue fields. I think discrete fields are the one that corresponds to viewing fields as models of a limit-colimit sketch (i.e. as an accessible category), but the others are often more useful (e.g. Heyting fields include the sheaf of continuous real-valued functions on a topological space). The case of X = Kan complexes has already been mentioned in other answers. Although in general once you're talking about homotopy theory, it's better to incorporate the homotopy theory into the ambient $\infty$-topos and work with stacks. If X = finite abelian groups, there are different notions of finite object in a topos. If X = topological spaces, you can internalize that directly, but often more useful is to internalize the notion of locale -- for instance, a "sheaf of locales" on a sufficiently nice topological space $Y$ is equivalent to a space over $Y$. If X = local rings, written as a geometric theory, this definition gives you the generally accepted definition of "sheaf of local rings", i.e. a sheaf of rings whose stalks are local. This definition of "sheaf of Xs" satisfies your criteria (3) and (5). It also satisfies your criterion (1) in as strong a way as I think could be expected: the category of internal Xs in a topos behaves exactly like the ordinary category of Xs, as long as the latter is interpreted using constructive logic. And it satisfies your criteria (2), (4), and (6) if the theory of Xs is geometric, hence has a classifying topos -- which I think is the most general situation in which one can expect these properties to hold. (Note, by the way, that your criterion (6), as well as the stronger version referring to all presheaf toposes, is a special case of your (4), since presheaves on $C$ are the Cat-enriched copower of the terminal topos by $C$ in the bicategory of toposes.)<|endoftext|> TITLE: Simply connected threefold with nef $K_X$ and $b_3=0$ QUESTION [5 upvotes]: Is there a simply connected smooth projective threefold $X$ with nef canonical divisor and vanishing third Betti number? $X$ cannot have the same integral cohomology ring as the projective space (Fujita). $X$ cannot be of general type. The discussion in the introduction to this paper seems relevant. REPLY [6 votes]: Such a threefold does not exist, even assuming only $b_1(X)=0$ instead of $\pi_1(X)=0$. Indeed the Miyaoka-Yau inequality holds for projective manifolds with $K$ nef -- see this paper. For threefolds, this reads $K_X^3\leq -64\chi (\mathscr{O}_X)$, hence $\chi (\mathscr{O}_X)\leq 0$ because $K_X$ is nef. Since $h^1(\mathscr{O}_X)=0$, this implies $h^3(\mathscr{O}_X)\geq 1$, hence $b_3>0$.<|endoftext|> TITLE: Constructing permutations avoiding a pattern QUESTION [7 upvotes]: See here for some theory. It is fairly easy to explicitly generate all permutations of $n$ elements that have a pattern (just begin with the pattern and add the rest in all possible positions), but can I do the same for pattern-avoiding? E.g. consider all permutations that avoid both $[1,2,3]$ and $[3,2,1]$, then the set even for any $n$ is finite, as already for $n=5$ none exists. But it's probably not that easy in general... Note that I don't care for an enumeration for arbitrarily large $n$ (haaaard, see link) but for an effective algorithm for medium $n$ (say, directly creating $<10^9$ permutations that avoid the pattern[s] for $n=20$, instead of checking $20!$ permutations for validity). A partial result (like "combining pattern foo and bar makes the set empty for large $n$") would be nice too. REPLY [14 votes]: As far as "combining pattern foo and bar makes the set empty for large $n$", there is an answer and it is fairly trivial. There are no permutations of length longer than $(k-1)(\ell-1)+1$ that avoid both $12\cdots k$ and $\ell\cdots 21$ by the Erdős–Szekeres theorem. On the other hand, given any set $B$ of patterns, if $B$ does not contain an increasing pattern, then all decreasing permutations will avoid every pattern in $B$. Similarly, if $B$ does not include a decreasing pattern, then all increasing permutations will avoid every pattern in it. Thus there are arbitrarily long $B$-avoiding permutations if and only if $B$ doesn't contain both an increasing pattern and a decreasing pattern. As for constructing all pattern-avoiding permutations of "reasonable" length, there has been some thought spent on this. Alex points out that there is an approach called "generating trees" that was used in the 90s, both for constructing these permutations and for counting them. The very basic idea is that if you have a list of all pattern-avoiding permutations of length $n-1$, then you can get the pattern-avoiding permutations of length $n$ by inserting $n$ somewhere in those. The less basic version is that you can keep track of which insertions "failed" earlier on, and not try them again. The advanced version is that for certain sets of patterns, there is a pattern to the allowable places for insertions, and this can let you construct those permutations quickly, or even enumerate them without constructing them. But that's only if you get lucky and there is a nice pattern. Since you asked about constructing the pattern-avoiding permutations, I'll focus on that. PermLab by Michael Albert is a Java program that constructs pattern-avoiding permutations of reasonable length quickly in practice (at least as quickly as a generating tree approach would do, possibly more quickly, and certainly quickly enough for my demands). I'm not sure if Michael has ever written about how it actually does that (maybe he could add a comment if he has?). PermPy by Michael Engen, Cheyne Homberger, and Jay Pantone is a Python library that uses many of the same tricks as PermLab. All that said, if you want to think more about this, I would recommend starting off with William Kuszmaul's paper "Fast algorithms for finding pattern avoiders and counting pattern occurrences in permutations". There are a lot of neat ideas in there and references to other work. One of his ideas that I find very cute, just as a fan of bit-level programming, is that he focuses on permutations of length 16, so that he can represent each one as a single int (on a 64-bit processor), and use various bit operations to manipulate them.<|endoftext|> TITLE: The period map and the Kodaira--Spencer map QUESTION [6 upvotes]: Let $f : X \to B$ be a non-isotrivial holomorphic submersion with connected fibres between compact Kähler manifolds of positive relative dimension. Suppose the fibres of $f$ are Calabi--Yau $(c_{1,\mathbb{R}}=0$) or canonically polarised ($c_1<0$). The differential of the moduli map $\mu : B^{\circ} \to \mathcal{M}$ is the Kodaira--Spencer map $\tau = d\mu$, measuring in the complex structure of the smooth fibres of the family. The period map $p : B^{\circ} \to D$ is a holomorphic map which measures the variation of the Hodge decomposition of the fibres. Is there a relationship between the Kodaira--Spencer map and the period map? REPLY [10 votes]: Differential of period map $d P^{p+q,p}$ is composition of KS-map $T_{B,0} \to H^1(X_0,T_{X_0})$ with natural map $H^1(X_0,T_{X_0}) \to Hom(H^{p,q}(X_0),H^{p-1,q+1}(X_0))$ (given by the cup product and the interior product). See Voisin "Hodge theory and complex algebraic geometry" Theorem 10.4<|endoftext|> TITLE: Visualizing the wave operator in two dimensions QUESTION [7 upvotes]: For $n\geq 1$, let $D_n$ be the Dirac operator on the spinor bundle on the $n$-dimensional sphere $S^n$. For example, $D_1$ acts on the trivial bundle $S^1\times\mathbb{C}\to S^1$, and can be explicitly written as $-i\frac{d}{d\theta}$. Let $e^{itD_n}$ be the wave operator associated to $D_n$; that is, the operator that solves the wave equation $$\frac{du}{dt}=iD_nu,$$ where $u$ is an $L^2$-spinor. In the case of $n=1$, $e^{itD_n}$ is just the operator that "shifts" everything on the circle by $t$ in the $\theta$-direction. Question 1: Let $D_{\mathbb{R}^2}$ be the Dirac operator on $\mathbb{R}^2$. How might one visualize $e^{itD_{\mathbb{R}^2}}$? Question 2: How might one visualize $e^{itD_2}$ on $S^2$? REPLY [5 votes]: The Dirac operator on $\mathbb{R}^2$, $$ D=\begin{pmatrix}0&k_x-ik_y\\ k_x+ik_y&0\end{pmatrix},\;\;\mathbf{k}=-i\frac{\partial}{\partial \mathbf{r}},$$ acting as $e^{-iDt}$ on a spinor plane wave $e^{i\mathbf{q}\cdot\mathbf{r}}\psi$, rotates the spinor $\psi$ on the Bloch sphere, by an angle $\theta=\tfrac{1}{2}t|\mathbf{q}|$ around the axis parallel to the vector $\mathbf{q}$.<|endoftext|> TITLE: Simplicial spaces internally to simplicial sets QUESTION [9 upvotes]: I am a master’s student with interest in topos theory and its applications (motivated by Ingo Blechschmidt’s thesis, as seems to be usual). After finding out about some of the uses of simplicial spaces (as nice “resolutions” to ordinary spaces), particularly for understanding mixed Hodge theory, I was wondering if there is a way to understand (or replace) the theory of simplicial spaces with a theory of internal locales to $\mathbf{sSet}$ or something similar. Question. Can you create a constructive version of classical Hodge theory, such that internalized to simplicial sets we retrieve the theory of mixed Hodge theory? Is there some place where the internal language of simplicial sets has been developed? Perhaps the answer to the question comes from infinity topos theory or similar. I don’t know much about this subject but would still appreciate an answer following this route. REPLY [4 votes]: This is a small clarification of Jens Hemelaer's answer, but it is essentially the same point: Internal locales in a presheaf category are actually very different from "presheaf of locales". I'm denoting by $\mathcal{Loc}$ the category of locales, with morphisms in the geometric direction (i.e. if $f:X \to Y$ is a morphism in $\mathcal{Loc}$, then we have a morphism of frame $f^*: \mathcal{O}(Y) \to \mathcal{O}(X)$. In An extension of the Galois theory of Grothendieck (Memoirs of the American Mathematical Society vol 51 (1984) — Which I unfortunately cannot find an online version) Joyal and Tierney prove the following (it also appears as pointed out by Jens as Lemma C.1.6.9 in Sketches of an Elephant): Theorem Let $A$ be a category with finite limits. Then a locale in the presheaf topos $Psh(A)$ is given by a functor $X: A \to Loc$ such that (1) For each arrow $f:A \to B$, the induced map of locales $X(A) \to X(B)$ is an open map. That is: (1a) the map $f^*:\mathcal{O}(X(B)) \to \mathcal{O}(X(A))$ has a further left adjoint $f_!$ and (1b) it satisfies the Frobenius identity $f_!(f^*(x) \wedge y ) = x \wedge f_!(y)$. (2) For each pullback square you have a Beck–Chevalley condition: $g^* f_! = k_! h^*$ However, that doesn't apply to $\Delta$ immediately as it does not have finite limits. What I wanted to say is that, using the same methods their results can be improved as follows: Proposition: If one remove the assumption that $A$ has finite limits in the theorem, then the result still holds if we replace condition (2) by: (2') for every copan $X \overset{f} \to B \overset{g}{\leftarrow} Y$ we have $$ g^* f_! = \sup k_! h^*$$ where the supremum is over all pairs of maps $(k,h)$ in $A$ that complete the cospan $(f,g)$ into a commutative square. Note that it is easy to recover condition (2) from condition (2') when the pullback exists, as the $k_! f^*$ corresponding to the pullback is easily seen to be maximal amongst all these corresponding to arbitrary commutative square. I worked that out myself a long time ago and never published it, but it might very well be somewhere else in the literature, I'd be happy to add a reference if someone knows one! (to some extent, it is relatively easy to deduce it from Corollary C;1.6.10 of the Elephant, though) I suspect that condition (2) and (2') have some sort of geometric interpretation, but I haven't worked it out explicitly. For these interested one can even generalize to the situation where $A$ has a Grothendieck topology, in this case internal locales in the sheaf topos are these that further satisfy the condition: (3) For each covering family $(B_i \to B)$ in $A$, the maps $X(B_i) \to X(B)$ are jointly a covering. Here I mean that the open map $\left( \coprod X(B_i) \right) \to X(B)$ is actually an open surjection. So in any case, internal locales in simplicial sets are fairly different from simplicial spaces. They are cosimplicial spaces whose structural map are open and that further satisfies this condition (2').<|endoftext|> TITLE: Open immersion of affinoid adic spaces QUESTION [12 upvotes]: If $R$ and $S$ are complete Huber rings with $\varphi: R \to S$ a continuous map, then is it true in general that if $\mathrm{Spa}(S, S^\circ) \to \mathrm{Spa}(R, R^\circ)$ is an open immersion of adic spaces (here $S^\circ$ and $R^\circ$ are the power-bounded subrings) then $\mathrm{Spec}(S) \to \mathrm{Spec}(R)$ is injective? For example, this is true if $R$ and $S$ both have the discrete topology, because if $\frak p$ and $\frak q$ are two prime ideals in $S$ which are equal after restricting to $R$ then $(\frak p, |\cdot|_{\rm triv})$ and $(\frak q, |\cdot|_{\rm triv})$ (trivial valuations), which are both points in $\mathrm{Spa}(S,S)$, restrict to the trivial valuation on $R/\varphi^{-1}(\frak p)$. But I'm not sure how generally to expect that this is true. If it makes it easier, we can assume that $R$ and $S$ are Tate, so the adic spaces are analytic. REPLY [6 votes]: This is not correct in general. There are in fact two examples in Bosch's Lectures on Formal and Rigid Geometry p.61-63. Let me sketch the first one. While it uses rigid-analytic spaces, it can be easily transferred to adic spaces: Weierstraß subdomains can be seen as special cases of rational subdomains (in adic spaces). Take $(R,R^\circ)=(K,\mathcal{O}_K)$ a non-archimedean field and let $D=Spa(R,R^\circ )$ be the unit disk. Choosing a $c\in K$ such that $0 < |c| < 1$, we can look at the subspace $$ U=\{x\in D \mid |T(x)(T(x)-1)|\leq |c|\}.$$ Then $U=Spa(S,S^\circ)$ and $S\cong K\langle \frac{T(T-1)}{c}\rangle$, which in turn is isomorphic to $$ S_0\times S_1:=K\langle\frac{T}{c}\rangle\times K\langle\frac{T-1}{c}\rangle.$$ One way to see this is that as rigid-analytic spaces we have $$\{x\in D \mid |T(x)(T(x)-1)|\leq |c|\}=\{x\in D \mid |T(x)|\leq |c|\}\coprod \{x\in D \mid |(T(x)-1)|\leq |c|\}. $$ But now $$ Spec(S)\cong Spec(S_1)\coprod Spec(S_2)\to Spec(R)$$ is not injective: Both the generic point of $S_1$ and $S_2$ map to the generic point of $R$.<|endoftext|> TITLE: Pairs of vertices with high degree difference QUESTION [9 upvotes]: Consider a finite, simple and undirected graph $G=(V,E)$ with $V=\{v_1,\dots, v_n\}$. Let us also fix an integer $k> n/2$. What are we able to say about the following quantity: $$\mathcal{I}_k(G) := \sum_{1\le ia\geq b.\tag1 $$ Thus \begin{align*} \mathcal{I}_k(G)&\leq s_A(b-s_B)+s_B(a-s_A)+(a-s_A)(b-s_B)\\ &\leq s_A(b-s_B)+s_B(a-s_A)+(a-s_A)s_A+(b-s_B)s_B\\ &=(s_A+s_B)(n-s_A-s_B)\\ &\leq (k+1)(n-k-1) \end{align*} where we used ($1$) in the second inequality and $\mathrm{sp}(G,k-1)\geq k+1>n/2$ in the last inequality. Note that this calculation gives an upper bound in terms of $\mathrm{sp}(G, k-1)=s_A+s_B$ and shows that even when $\mathrm{sp}(G, k-1)=s_A+s_B=k+1$ we have a strict inequality provided $s_A TITLE: Kulkarni-Nomizu square root of the Riemann tensor QUESTION [8 upvotes]: Given a Riemann tensor $Riem$, what are conditions such that $Riem=B\star B$ for some bilinear symmetric form $B$, where $\star$ is the Kulkarni-Nomizu product? It follows from the proof of Proposition 15 (Chapter 4, Section 2, page 99) in the book of Petersen "Riemannian geometry" - Second Edition, that positivity of the curvature operator ${\mathscr{R}}$ on $\Lambda^2T_pM$ is sufficient in dimension 3. Moreover, when $n=3$, $\det{\mathscr{R}}>0$ is sufficient. Anyone have a reference for $n>3$, please? This is how to know if there exists a square matrix $n\times n$ with assigned all its $2\times2$ subdeterminants (and possibly how to recover it). In the following paper of Greenhill http://web.maths.unsw.edu.au/~csg/papers/ext-matrix.pdf, an algorithm to find such a matrix is showed, but apparently it does not state a theoretical ("easy") criterium for its existence. My question is related to my other question 2x2 subdeterminants of a matrix which was related to uniqueness. REPLY [3 votes]: Note that Petersen's Prop 15 does not directly address your question. The assumption that there already exists an embedding from the $n$ dimensional $M$ to $(n+1)$ dimensional Euclidean space is crucial. Petersen's Proposition 15 concerns not solvability of $\mathrm{Riem} = S\star S$ but whether the solution $S$ is unique. What you are looking for is a theorem due to Jaak Vilms. He described it in two different places: Local isometric imbedding of Riemannian $n$-manifolds into Euclidean $(n+1)$-space, J. Differential Geom. 12(2): 197-202 (1977). DOI: 10.4310/jdg/1214433981 Factorization of Curvature Operators, Transactions of the American Mathematical Society Vol. 260, No. 2 (Aug., 1980), pp. 595-605. https://www.jstor.org/stable/1998025 Note that this discussion presumes that you have a given Riemannian metric (or I guess in your case a positive definite inner product on $T_pM$). Essentially: if one looks at the Gauss-Codazzi equations, one sees that when $M$ has an isometric immersion into Euclidean space of $p$ dimensions higher, then one has a decomposition of $$ \mathrm{Riem} = \sum_{i = 1}^p S_i \star S_i $$ And so the question you are asking for is identical the (local) isometric immersivity of $M^n$ into $E^{n+1}$. Notation Denote by $\mathscr{R}: \Lambda^2 T_pM \to \Lambda^2 T_pM$ the curvature operator acting on two forms. In indices its components are $R^{ij}_{kl}$ antisymmetric in $i,j$ and in $k,l$ separately. Define $$ \phi(\mathscr{R}) = R^{ij}_{kl} R^{kp}_{iq} R^{lq}_{jp} $$ and $$ \psi(\mathscr{R}) = R^{ij}_{kl} R^{kl}_{pq} R^{pq}_{ij} $$ The curvature operator $\mathscr{R}$ is assumed to satisfy Bianchi identity. The curvature operator $\mathscr{R}$ is said to "preserve decomposability" if for arbitrary 2 forms $\alpha, \beta$: $$ \mathscr{R}\alpha \wedge \mathscr{R}\beta = 0 \iff \alpha \wedge \beta = 0 $$ Note that this condition is trivial if $n = 3$. Given a curvature operator $\mathscr{R}$, its rank subspace is the smallest subspace $W$ of $T_pM$ such that the image of $\mathscr{R}$ lies in $\Lambda^2 W$; below when we refer to rank of $\mathscr{R}$ we mean the dimension of $W$. Theorem A curvature operator with rank $> 1$ decomposes as $S\star S$ if and only if $\mathcal{R}$ preserves decomposability $\phi + \frac14 \psi > 0$; in the case $n = 3 \pmod 4$ this condition can be replaced by positivity of $\det \mathscr{R}$ when restricted to the complement of its kernel. a technical "non singularity condition". The non-singularity condition is a bit annoying to state; see the second paper for details. It is trivially satisfied when $\mathscr{R}$ is positive (or negative) definite, and $n$ is either 3 or $>4$. (I think it may also be satisfied when $n = 4$ with definite curvature operator, but the computations are less obvious.)<|endoftext|> TITLE: Braid groups and Kazhdan's property (T) QUESTION [8 upvotes]: In Nica's dissertation Group actions on median spaces, we can read the following assertion: Braid groups do not contain infinite subgroups satisfying Kazhdan's property (T). This is used in order to motivate the question (which is still open up to my knowledge) of whether braid groups satisfy the Haagerup property. Does anyone have a reference and/or a quick justification of the above claim? REPLY [5 votes]: It is enough to see that the pure braid group $P_n$ does not contain subgroups which have Kazhdan property (T). There is an exact sequence $$1 \rightarrow F_{n-1} \rightarrow P_n \rightarrow P_{n-1} \rightarrow 1 $$ which reduces, by induction, to showing that $F_n$ does not contain subgroups with Kazhdan property (T). The latter is easy. REPLY [5 votes]: As suggested by Will Sawin in the comments, I took a look at Bekka, de la Harpe and Valette's book. The claim is a straightforward consequence of the following statement: Theorem. (Artin) For every $k \geq 2$, the morphism between pure braid groups $P_k \to P_{k-1}$ that forgets the last strand has a free kernel. Indeed, let $k \geq 1$ be an integer and let $H \leq B_k$ be a subgroup having (T). We want to prove that $H$ is trivial. Because $P_k$ has finite index in $B_k$, we can assume that $H$ lies in $P_k$ up to replacing $H$ with $H \cap P_k$. Now, we argue by induction over $k$. If $k=1$ then the conclusion is clear. If $k \geq 2$, then the image of $H$ under $P_k \to P_{k-1}$ must be trivial as a consequence of our induction hypothesis. In other words, $H$ must lie in the kernel of $P_k \to P_{k-1}$, which is free according to Artin's theorem. This implies that $H$ is trivial, as desired. Actually, the argument applies to any group property $(P)$ satisfying the following axioms: If $G$ has $(P)$, then so does the image of $G$ under any homomorphism. If $G$ has $(P)$, then so do the finite-index subgroups of $G$. Free groups of rank $\geq 1$ do not have $(P)$. This includes many properties weaker than (T) such that the hereditary property (FA) or the property (FW).<|endoftext|> TITLE: Using the universal property of spaces QUESTION [6 upvotes]: The $\infty$-category of spaces has the following properties: It is the $\infty$-category obtained from the (ordinary) category of finite sets by freely adding sifted colimits. (See e.g. Cesnavicius-Scholze https://arxiv.org/abs/1912.10932 §5.1 for a review of this notion and for pointers to Lurie's HTT where this is proven.) (As Tim Campion points out in a comment, another characterization, also in HTT): Spaces are obtained by freely adding arbitrary colimits to the category $\{*\}$. Can either of these characterizations be used (ideally without referring to the model of quasi-categories) to show other properties, such as: that colimits in spaces are universal (proven by Lurie in HTT Lemma 6.1.3.14)? possibly even that $Cat_\infty$ is generated by (the compact objects) $*$ and $\Delta^1$? REPLY [3 votes]: Here is a "model-independent" proof that colimits in $Spaces$ are universal. Of course, the ingredients going into the proof may have model-dependent proofs. Fact: Colimits in $Spaces$ are universal. Proof: We want to show that for any map of spaces $f: Y \to X$, the pullback functor $$f^\ast: Spaces_{/X} \to Spaces_{/Y}$$ preserves colimits. We may view $X,Y$ as $\infty$-categories which happen to be $\infty$-groupoids and $f: Y \to X$ as a functor between them. By straightening / unstraightening, the functor $f^\ast :Spaces_{/X} \to Spaces_{/Y}$ is identitified with the "precompose $f$" functor $$f^\ast: Psh(X) \to Psh(Y)$$ Now, $Psh(X),Psh(Y)$ are functor categories with values in the cocomplete category $Spaces$. So colimits are computed "objectwise" in these categories. That is, for $F: I \to Psh(X)$, we have $(\varinjlim_{i \in I} F(i))(x) = \varinjlim_{i \in I} (F(i)(x))$, and similarly in $Y$. From these formulas, it is immediate that precomposing $f$ preserves colimits. That is, we have $$(\varinjlim_{i \in I} f^\ast F(i))(y) = \varinjlim_{i \in I} F(i)(f(y)) = \varinjlim_{i \in I} (f^\ast F(i))(y)$$ as desired. This proof doesn't precisely use the universal property of $Psh(X),Psh(Y)$ of freely adding colimits, but perhaps it could be tweaked to do so.<|endoftext|> TITLE: Hartogs' theorem for real-analytic subvarieties QUESTION [10 upvotes]: One version of Hartogs' extension theorem is the following (see, e.g. [1], Theorem 5B, p. 50). Theorem. Let $U \subset \mathbb{C}^n$ be open and let $X \subset U$ be a complex-analytic subvariety of codimension $>1$. Then, any holomorphic function $U \setminus X \to \mathbb{C}$ has a unique holomorphic extension $U \to \mathbb{C}$. Question. Is this theorem still true if we replace $X$ by a real-analytic subvariety of codimension $>2$? Note that I am still considering holomorphic functions. (The real-analytic version of Hartogs' theorem is false.) References. [1] Whitney, Hassler. Complex analytic varieties. Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills, Ont., 1972. xii+399 pp. REPLY [17 votes]: The answer to your question is yes. It is enough to have the $2n-2$ dimensional Hausdorff measure of $X$ be zero and $X$ is closed. See the book of E. M. Chirka Complex Analytic Sets, page 298 proposition 3.<|endoftext|> TITLE: What can one say about $\sum\limits_{i=1}^\infty \frac{1}{p_{i+1}^2-p_i^2}$? QUESTION [10 upvotes]: Denoting by $p_i$ the $i$-th prime, is it known that $\displaystyle \sum_{i=1}^\infty \frac{1}{p_{i+1}^2-p_i^2}$ converges? Can one compute a few digits based on euristic considerations or plausible conjectures about distributions of primes and prime gaps? I think it may be a bit less that 0.63, but I'm not at all confident. REPLY [16 votes]: In the paper [On the sum of the reciprocals of the differences between consecutive primes, Ramanujan J., 47,427–433(2018)] by me, I proved under the Hardy–Littlewood prime-pair conjecture that $$\sum_{n\le X}\frac{1}{p_{n+1}-p_n}\sim \frac{X\log\log X}{\log X},$$ and without the Hardy–Littlewood prime-pair conjecture, one has $$ \sum_{n\le X}\frac{1}{p_{n+1}-p_n}\ll \frac{X\log\log X}{\log X}. $$ Therefore, by using Abel’s summation formula, one can prove that the conjecture is true unconditionally. In fact, this problem has been investigated by Erdős and Nathanson [On the sum of the reciprocals of the differences between consecutive primes. In: Chudnovsky, D.V., Chudnovsky, G.V., Nathanson, M.B. (eds.) Number theory: New York Seminar 1991–1995, pp. 97–101. Springer, New York (1996)]. They proved $$ \sum_{n\ge 2}\frac{1}{(p_{n+1}-p_n)n(\log\log n)^c}<+\infty, $$ for all $c>2$. Then by noting that $p_n\sim n\log n$, one can give an alternative proof.<|endoftext|> TITLE: Is Tarskian hyperbolic geometry consistent, complete & decidable? QUESTION [8 upvotes]: Tarski developed an axiomatic description of Euclidean geometry in first order logic. Its primitive notions are points and its primitive relations are betweeness and congruence of points. The Parallel axiom is stated using betweeness. He proved it Consistent. That is, it does not prove both any sentence and its negation. Complete. That is, it proves any sentence or its negation to be true. Decidable. There is an algorithm assigning a truth value to every sentence. Obviously, we can negate the Parallel axiom to obtain Tarskian hyperbolic geometry. Q. Is this geometry also consistent, complete and decidable? REPLY [13 votes]: The canonical reference for Tarski-style elementary geometry is the monograph Schwabhäuser, Szmielew, Tarski [1]. This includes a treatment of hyperbolic geometry in parallel with Euclidean geometry; in particular, the consistency, completeness, and decidability of $n$-dimensional hyperbolic geometry for any $n\ge2$ is proved in Satz II.3.65. Reference: [1] Wolfram Schwabhäuser, Wanda Szmielew, Alfred Tarski: Metamathematische Methoden in der Geometrie, Springer-Verlag, 1983.<|endoftext|> TITLE: Modern proofs of the Verlinde formula? QUESTION [6 upvotes]: Let $G$ be a semisimple algebraic group and $\Sigma$ a smooth proper curve. Then $\text{Bun}_G(\Sigma)$ comes equipped with a line bundle $\mathcal{L}$ which generates the torsion free part of $\text{Pic}\text{Bun}_G(\Sigma)$ (e.g. in type A or C it's the determinant bundle or in type B or D the Pfaffian bundle). Then the Verlinde formula is an explicit formula for $$H^0(\text{Bun}_G(\Sigma),\mathcal{L}^{\otimes k}),$$ intimately related to fusion products. There are lots of nice proofs of the Verlinde formula published in the 1990s, e.g. Beauville's Conformal blocks, fusion rules and the Verlinde formula. However, they are all quite algebraic which makes it hard (at least for me) to understand what's going on. Given how much better the geometric side has been understood in recent decades (e.g. fusion and the BD Grassmannian), is there written up anywhere a slightly cleaner/more geometric proof of the Verlinde formula? Edit: I'm mainly curious about whether there is a "more geometric" version of Beaville's proof, but am very happy to see other methods also. REPLY [3 votes]: This proof is geometric and from 2020, so it might satisfy the two requirements of the OP: A finite dimensional proof of Verlinde Formula.<|endoftext|> TITLE: Descriptive set theory for computer scientists? QUESTION [12 upvotes]: It seems to me that there are scattered references of deep relationships between descriptive set theory and computability theory. For one, the relationship between the Borel hierarchy and the Polynomial hierarchy for one, and the relationship between topology and computability. I would like to understand things such as: What is forcing, and how does one use it to prove the independence of the Continuum Hypothesis? What precisely is an inner model? Why are Polish spaces an important setting for descriptive set theory? It would be fantastic if these could be answered from the perspective of computability. For example, I know that the Cantor set has good computability properties, and I (possibly very incorrect) believe that this is related to descriptive set theory. I'd greatly appreciate references to lecture notes, talks, video lectures, textbooks. Pretty much anything where I can study descriptive set theory with a computational bent would be great. REPLY [12 votes]: For the last point, besides texts explicitly on computability-theoretic descriptive set theory (e.g. the hard-to-find Mansfield-Weitkampf, the freely-accessible section $3$ of Moschovakis' book, or these brief notes of Hjorth) I think the literature on represented spaces will be useful. There is a lot written about this; I recommend this recent survey of Schroeder (and the sources in its bibliography). Granted, this is somewhat more geared towards computable analysis than (effective) descriptive set theory, but there is still strong overlap. Roughly speaking, suppose we have a topological space $\mathcal{X}$ equipped with a canonical basis. Each point in $\mathcal{X}$ can be described by a set of basis elements, and examples like $\mathbb{R}, 2^\mathbb{N}$, and $\mathbb{N}^\mathbb{N}$ suggest that with "naturally-occuring" spaces it is often the case that such representations yield a good computability theory (we have notions of "computable elements" of each of those spaces). Polish spaces arise as a natural generalization of this setting; note that completeness can be thought of as the analogue of "every infinite sequence of decimal digits corresponds to a real number." Of course the situation is a bit more complicated than this, as the theory of represented spaces indicates, but this is still a good starting point. Moreover, since all "nontrivial" Polish spaces are Borel-isomorphic, we get a general independence principle for "high-level" descriptive set theoretic questions. This helps motivate them on the classical side. However, I don't really think that (set-theoretic) forcing and inner models are best approached via computability theory, even for a computability theorist; while they do have computability-theoretic aspects, focusing too much on this will obscure the fundamentally set-theoretic nature of the topic which is necessary to any deep understanding. In particular, while there is a notion of forcing in classical computability theory, in my opinion it's far too weak to yield a good bridge to set-theoretic forcing. I think to a large extent there's nothing to do but grab the bull by the horns. Certainly not all expositions of forcing are equally clear, but in my experience ones which attempt a significantly simplified presentation (e.g. this "analytic" approach of Scott) don't really serve well as introductions to the fully general theory. The two standard approaches to forcing are via posets and Boolean-valued models, and I would just pick one and study it directly. In my experience the poset-based approach is far simpler, and it's what I learned; however, the Boolean-valued approach avoids any need to talk about generic objects directly, which may be a significant intuitive benefit if one is uncomfortable with model theory. For general set-theoretic forcing and inner models, I think the right starting point (overlooking the concern of the paragraph above) is hyperarithmetic theory. Not only is this a "lightface" analogue of the Borel hierarchy (so it's relevant to the previous point too), it is tightly connected to the construction of $L$: in a very strong sense, $L$ is what you get when you "continue the hyperarithmetic hierarchy through the ordinals" (the relevant term here is "master codes" - see here). Hyperarithmetic theory is also useful for understanding set-theoretic forcing. Forcing in classical computability theory (e.g. Jockusch-Soare forcing) is a bit too simple to provide a good starting point for set-theoretic forcing in my opinion. One wants to shift from classical computability theoretic forcing to forcing over some "set theory flavored" structure. It turns out that forcing over levels of $L$ is a very good context for doing this. For example, consider the proof that a comeager set of reals $x$ satisfy $\omega_1^{CK}(x)=\omega_1^{CK}$; this is basically "Cohen forcing over $L_{\omega_1^{CK}}$." In general, there's a rich theory of forcing over admissible sets. My personal favorite treatment of the hyperarithmetic hierarchy and related notions is (the first few parts of) Sacks' book Higher recursion theory (it should be available via ProjectEuclid, but that link doesn't work at the moment). Now granted, in a precise sense we can make sense of forcing "computably" rather than "hyperarithmetically;" however, while it's a very nice result I don't think this is a good way to approach the subject as a beginner from computability theory. One final topic worth mentioning re: forcing, specifically connected with combinatorial topics like $\mathsf{CH}$, is the theory of effective cardinal characteristics of the continuum. This amounts to a "computable analogue" of the analysis of the interval $[\aleph_1,2^{\aleph_0}]$ via forcing, with set-theoretic forcing notions usually having corresponding computability-theoretic ones which yield analogous results. For example, we have a direct connection between the following two facts: It is (relatively) consistent with $\mathsf{ZFC}$ that there is a cardinal $\kappa$ such that $(i)$ there is a set $F$ of functions $\mathbb{N}\rightarrow\mathbb{N}$ such that every $g:\mathbb{N}\rightarrow\mathbb{N}$ is escaped by some $f\in F$, but $(ii)$ there is a set $F$ of functions $\mathbb{N}\rightarrow\mathbb{N}$ such that every $g:\mathbb{N}\rightarrow\mathbb{N}$ is dominated by some $f\in F$. There are hyperimmune degrees (= compute a function escaping all computable functions) which are not high (= compute a function dominating all computable functions). Note that a consistency result is replaced by an outright fact. Here the idea is to shift from cardinality on the set-theoretic side to lowness/highness notions on the computability-theoretic side. Moreover, while the applications generally go from the set-theoretic version to the computability-theoretic version, there is at least one case where we get a result on the set-theoretic side building on an initial result on the computability-theoretic side; the set-theoretic result (which was much harder) was gotten by Ivan Ongay-Valverde, and the computability-theoretic result is joint between Ivan and myself. However, I think that this is a bit of a double-edged sword: while much more accessible to the classical computability theorist than hyperarithmetic theory, the conceptual shift in the previous sentence does lose a lot of the set-theoretic "spirit." I'm not sure how much I recommend it as a pedagogical starting point.<|endoftext|> TITLE: Decidability of knot equivalence in general 3-manifolds? Surface equivalence? QUESTION [6 upvotes]: Given a closed orientable 3-manifold $M^3$ and two knots $K_1$ and $K_2$ in $M$, is there an algorithm to decide if $K_1$ and $K_2$ are isotopic? Is there an algorithm to decide if there is a homeomorphism $f: M \to M$ with $f(K_1) = K_2$? I would expect these questions to be involved (at least as difficult as the recognition problem for 3-manifolds) but I thought that the answer is possibly a folklore consequence of geometrization. The special case of the first question for $K_1$ unknotted follows from a result of Hass and Lagarias (see Theorem 1.2). The case where $M = S^3$ is a result of Waldhausen using normal surface theory. As an aside, I don't know anything about the corresponding questions where instead of considering knots we consider closed embedded surfaces $F_1$ and $F_2$. Even in the case where $M = S^3$ (so only the isotopy question is relevant) the genus of $F_1$ and $F_2$ is greater than one, this is a total mystery to me. By looking at tori bounding regular neighborhoods of knots, this is strictly more difficult than the corresponding questions for knots. Are there results in this direction? REPLY [4 votes]: Regarding the second question. This reduces to the homeomorphism problem for three manifolds, which is known due to geometrisation. Here is a sketch. Fix $M$ as well as the knots $K$ and $L$. Let $X_K$ and $X_L$ be the knot complements. We mark the boundary of each with the meridional slope of $K$ and $L$, respectively. We ask our solution to the homeomorphism problem if $X_K$ and $X_L$ are homeomorphic. If the answer is "no" then we are done. If the answer is "yes" we ask for all homeomorphisms from $X_K$ to $X_L$. For each of these we restrict to the boundary and see if the resulting map takes the meridian of $K$ to that of $L$. If any do, the answer is "yes". If none do, the answer is "no". [Edit - as pointed out in the comments, there may be infinitely many homoeomorphisms between $X_K$ and $X_L$. This happens, for example, when $K$ and $L$ are both unknots in the three-sphere. In any case, the self-homeomorphisms of $X_K$ restrict to give a virtually cyclic group of homeomorphisms of $\partial X_K$. When this group is infinite, the cyclic subgroup is generated by a Dehn twist about the slope that dies in homology.] Regarding the first question. If $M = S^3$ then again this reduces to the homeomorphism problem for three-manifolds, by appealing to Property P (due to Gordon-Luecke). Some fiddling about is needed to deal with mirror images. I don't know the status of the first question for other manifolds off the top of my head. I'll report back if anything occurs to me. :)<|endoftext|> TITLE: Does inclusion from n-stacks into (n+1)-stacks preserve the sheaf condition? QUESTION [8 upvotes]: I'm going to describe two situations that seem to contradict each other, and I'm interested to know precisely what's wrong with this reasoning. Let $M$ be a manifold, and consider the presheaf $C^*(-,\mathbb{Z})$ on $M$ sending $U$ to $C^*(U;\mathbb{Z})$, the complex of (singular) cochains on $U$. Then this association is a homotopy sheaf in the sense that for any $U$ and a covering $\{U_i\}_i$ of $U$, the map $$C^*(U;\mathbb{Z}) \to \mathrm{holim} \, C^*(N(\{U_i\});\mathbb{Z})$$ is an equivalence (e.g., in Lurie's formalism by considering the quasicategory $\operatorname{Mod}_{\mathbb{Z}}$ and ordinary limits in that sense, or homotopy limits in the model category of chain complexes). For a covering $U = U_1 \cup U_2$, the fact that it's a homotopy sheaf is essentially equivalent to Mayer-Vietoris (and more generally, it corresponds to the Mayer-Vietoris spectral sequence for a general covering). Here's the crux: when viewed as taking values in chain complexes, the constant sheaf $\underline{\mathbb{Z}}$ is NOT a homotopy sheaf (even though it is an ordinary sheaf), and the map $$ \underline{\mathbb{Z}} \to C^*(-,\mathbb{Z}) $$ is a homotopy sheafification. What this tells us is: even something that's a sheaf in the ordinary $1$-categorical sense is NOT necessarily a homotopy sheaf (e.g. assuming $M$ has nontrivial cohomology). Since the sheaf condition is a condition on limits, this tells us that the inclusion of quasicategories from $\operatorname{Ab}$ into $\operatorname{Mod}_{\mathbb{Z}}$ does not preserve limits. Something similar happens with etale cohomology on the etale site. It's well-known that schemes are algebraic stacks. In particular, if I have a representable functor on the category of schemes, then if we view it as a functor from schemes to groupoids (via the inclusion of sets into groupoids), then it's a sheaf (aka stack) for the fppf topology. I assume this means that if I have ANY set-valued sheaf for the fppf topology, then if I consider it as a presheaf valued in groupoids, it remains a sheaf (in the appropriate $2$-categorical sense). More generally, a scheme is an algebraic $n$-stack, and an $n$-stack is an $m$-stack for $m \ge n$ (c.f. https://arxiv.org/pdf/alg-geom/9609014.pdf, p.3). There's no discussion of needing to sheafify when you do this. So what's the difference? Am I wrong about one of these two situations? (and if so, which one and how?) Is it some fundamental difference between the abelian and non-abelian settings? Or something else? I talked to my advisor, who has some expertise in this area, but they were confused by this as well. REPLY [16 votes]: You said it yourself in the question! The reason that sheaves of abelian groups are not $\infty$-sheaves in general, when considered as presheaves taking values in the $\infty$-category $\mathsf{Mod}_{\mathbb Z}$, is that the $\infty$-functor $\mathsf {Ab} \longrightarrow \mathsf{Mod}_{\mathbb Z}$ does not preserve limits. And the reason that sheaves of sets are $\infty$-sheaves is that the $\infty$-functor $\mathsf{Set} \longrightarrow \mathsf{Spaces}$ does preserve limits: it has a left adjoint, given by $\pi_0$. The same holds for the inclusion of $k$-truncated spaces into $(k+1)$-truncated spaces, in which case the adjoint is Postnikov truncation, and this answers your question about higher stacks. More generally the full subcategory of $k$-truncated objects in any $\infty$-category is closed under all limits. You wrote: "Is it some fundamental difference between the abelian and non-abelian settings?". Perhaps not. The inclusion of the 1-category of abelian groups into the $\infty$-category of nonnegatively graded chain complexes preserves all limits, and this is arguably the "true" abelian analogue of the inclusion of sets into spaces via Dold-Kan. REPLY [9 votes]: the constant sheaf Z is NOT a homotopy sheaf (even though it is an ordinary sheaf) This type of phrasing is ambiguous and is probably responsible for the confusion. In this sentence, Z is used to refer to two completely different presheaves: the presheaf Z of abelian groups, which sends U to the set of locally constant Z-valued functions on U; the presheaf Z[0] of unbounded chain complexes, which sends U to the unbounded chain complex concentrated in degree 0, where it is given by the abelian group of locally constant Z-valued functions on U. The presheaf Z is indeed a 1-sheaf and an ∞-sheaf of abelian groups. The presheaf Z[0] is a 1-sheaf of unbounded chain complexes. It is not an ∞-sheaf of unbounded chain complexes and its ∞-sheafification can be computed as the ∞-sheaf of integral singular cochains. To summarize, regardless of the site (manifolds or schemes), presheaves of connective objects (i.e., with homotopy groups concentrated in nonnegative degrees), whether sets, abelian groups, simplicial sets, nonnegatively graded chain complexes, or connective spectra, are automatically ∞-sheaves, provided their individual values have vanishing homotopy groups in degree 1 and above, and their π_0 is a sheaf in the ordinary sense. On the other hand, passing to the nonconnective setting, whether unbounded chain complexes or spectra, almost always destroys the ∞-sheaf property, precisely because higher sheaf cohomology groups can be nonvanishing. (Exceptions exist in special cases, e.g., quasicoherent sheaves on affine schemes or Stein spaces, or representable sheaves of abelian Lie groups on cartesian spaces.) Indeed, the inclusion of connective objects into nonconnective objects already fails to preserve (say) pullbacks. For example, consider an object A with trivial π_k for all k≥1. (Assume all objects are pointed, for simplicity.) The loop space of A is the homotopy pullback of 1→A←1. In the nonconnective setting, π_{−1} of this homotopy pullback will be nontrivial (namely, A in the case of unbounded chain complexes or spectra), whereas in the connective setting the homotopy pullback is 1, with trivial homotopy groups in all degrees.<|endoftext|> TITLE: $1$-cocycle associated to universal $G$-bundle $EG \to BG$ QUESTION [5 upvotes]: Let $G$ be a (topological) group whose identity element $e_G$ is a nondegenerated basepoint (e.g. if $G$ is a Lie group). Then that's a known fact that there is for every 'nice' enough topological space $X$ (eg paracompact sp, or more elementary a CW complex) a natural bijection $$ [X, BG] \cong \mathcal{P}_G(X) $$ between homotopy classes of cont maps $X \to BG$ and isomorphism classes of of principal $G$-bundles over base $X$. Explicitely this is given by associating to a homotopy class of a map $f: X \to BG$ the pullback bundle $f^*EG$ of the universal principal $G$-bundle $u:EG \to BG$. Now like the trivial bundle the universal bundle $EG$ is the second 'canonical' (up to isomorphism) $G$-bundle and besides it's universal property I'm trying to understand if it's possible to deduce it's universal property only from study of it's $1$-cocycle represenation. Let's recall that we can over every nice enough base space $X$ encode the complete information about a $G$-bundle $E \to X$ with fiber $F$ (therefore $G \subset Aut(F)$; in our case later it will be $G$ itself since we are interested on principal bundles) in terms of a $1$-cocycle. Let $\mathcal{U}= \{U_i \}_i$ covering of $X$ over which $E$ trivializes, then (1) a colection of homeomorphisms $\psi_i: U_i \times F \to p^{-1}(U_i)$ compatible with projection $p$ (2) the $1$-cycle (ie the set of transition functions) $g_{ij}: U_j \cap U_i \to G$ such that for every $(x, f) \in (U_j \cap U_i) \times F$ we have $$ \psi_j^{-1} \circ \psi_i (x,f) = (x, (g_{ji}(x))(f))$$ (3) the transition functions satisfy $g_{ii} = e_G$ (constant function), $g_{ji}= g_{ij}^{-1}$ and $g_{kj} \cdot g_{ji}= g_{ki}$ Moreover two $G$-bundles $E_1, E_2$ over same base $X$ are isomorphic if there exist fine enough covering $\{U_i \}_i$ of $X$ such that there eist a family $g_i: U_i \to G$ with $$ g^2_{ij} = g_i \cdot g^1_{ij} \cdot g_j^{-1} $$ where $\{g^1_{ij}\}$ and $\{g^2_{ij}\}$ are $1$-cycles of $E_1$ and $E_2$. Question: Is it known which $1$-cocycle $\{g^{EG}_{ij}\}$ is naturally associated to the universal bundle $u:EG \to BG$? Is there any canonical choice of the covering $\{U_i \}_i$ of $BG$ known over which this $1$-cocycle can be in most transparent way be declared? Is it known at least if we work with a finite group? The model of $EG$ and $BG$ I have here in mind is that one by Milnor explaned in detail for example here. If we work with this model for $EG \to BG$ can it's $1$-cocycle be writen down? It must be something canonical but I don't know how to determine it. My motivation is to understand if it's possible to understand only by study the structure of it's $1$-cocycle that this $G$-bundle is in certain way the most 'flexible' one in the sense such that every other can be derived up to iso by the pullback of it, as the universal property says. So speaking informally every 'twist' of $G$-bundle should be already 'known' in some way already to the universal bundle, such it 'knows' all possible twists can happen only by taking pullbacks. (Ok, I admit, that the last formulation is a bit vague, but that's the core of my motivation: I understand 'formally' the universal property of the universal bundle, but have no intuition on it's intrinsical geometry why and why exacly this geometry is the most 'flexible' to allow to recover all over $G$-bundles by taking pullbacks of it. REPLY [5 votes]: For any $x \in (0,1)$, I will define an open set $U_x$ of $BG$. The way we can define it is by defining an open set $\tilde{U}_x$ of $EG$, stable under the action of $G$, with $|G|$ connected components that are permuted by the action of $G$. Then any one of these components will project to an open subset of $BG$. The set $\tilde{U}_x$ in each simplex of $E_G$ by the condition that there exists some $i$ such that $\sum_{j< i} t_j < x$ and $\sum_{j \leq i } t_j > x$. Because these are strict inequalities, this is an open subset of each simplex. Because they are unaffected by inserting a new $i$ with $t_i=0$, this remains open in $EG$. Because the tuple $g_1,\dots, g_n$ is not used in the definition, it is certainly stable under the action of $G$. The connected components associated to $g\in G$ consists of those points where $g_i=g$ for the unique $i$ satisfying $\sum_{j< i} t_j < x$ and $\sum_{j \leq i } t_j > x$. This is a component because the function $g_i$ is locally constant, because, on each simplex, the unique value of $i$ satisfying these inequalities is locally constant - it only changes at the points where $\sum_{j < i} t_i =x$ or $\sum_{j\leq i} t_i = x$, which are not in the set. The open sets $U_x$ cover $BG$ (it suffices to take any infinite set of $x$) since at each point, such an $i$ exists for all but finitely many $x$. The cocycle is defined as follows: Given a point in $BG$, expressed as a tuple $t_1,\dots, t_n$ of nonnegative reals summing to one and a tuple $g_1,\dots, g_n \in G$ modulo the left action of $G$, that lies in $U_x \cap U_y$, let $i_x$ be the unique $i$ such that $\sum_{j < i} t_i x$, and define $i_y$ similarly, and then set $$\psi_{xy} ( ((t_1,\dots, t_n), (g_1,\dots, g_n)) = g_{i_x}^{-1} g_{i_y} .$$ This cocycle is flexible in the sense that for any distinct tuple $x_1,\dots, x_n$ of open sets $U_{x_1},\dots, U_{x_n}$, for any tuple of $\psi_{ij} \in G$ for $1\leq i,j\leq n$ such that $\psi_{ik} = \psi_{ij} \psi_{jk}$ for all $i,j,k$, there is a unique component of $U_{x_1} \cap \dots \cap U_{x_n}$ such that $\psi_{x_i x_j} $ restricted to this component is $\psi_{ij}$. Indeed, such a tuple must have the form $\psi_{ij} = c_i^{-1} c_j$ for some tuple $c_i \in G$ for $i$ from $1$ to $n$, and then the component where $g_{i_{x_j}} = c_j$ for all $j$ does the trick. But possibly this is not actually what you want to do. Probably you want to use a different cocycle presentation of $G$-bundles, the one where you fix a simplicial complex structure on the base, and present a cocycle as a tuple consisting of one element of $G$ for each edge satisfying one relation for each triangle. For this one, the key point is that given a simplicial complex with such a cocycle, you can map it to BG by mapping each edge to the corresponding edge of $BG$, and then there will be a unique way of extending to triangles and then to higher simplices.<|endoftext|> TITLE: A number characterizing the deviation of a triangle from the regular triangle QUESTION [5 upvotes]: Given a triangle $\Delta$ with sides of length $a\le b\le c$, consider the number $$q=\frac{a^4+b^4+c^4}{(a^2+b^2+c^2)^2}$$ and observe that $\frac13\le q\le\frac12$ and the extremal values of $q$ characterize some geometric properties of the triangle $\Delta$. Namely: $\bullet$ $q=\frac13$ if and only if $a=b=c$ (which means that the triangle $\Delta$ is regular); $\bullet$ $q=\frac12$ if and only if $c=a+b$ (which means that the triangle $\Delta$ is degenerated). I am writing a paper (in applications of math to Electric Engineering) where the number $q$ is applied for evaluation of the deviation of a triangle (describing the quality of 3-phase electric energy) from being regular, and need to call the number $q$ somewhow (for example, quadrofactror), but wonder if $q$ already has some standard name. This motivates my Question. Has the number $q$ some standard name in Plane Geometry? REPLY [5 votes]: Added to my comment above, this time taking care of my carelessness in not normalising: one has the formula $$\frac{16 A^2}{(a^2+b^2+c^2)^2}=1-\frac{2(a^4+b^4+c^4)}{(a^2+b^2+c^2)^2} $$ which shows, at least in my book, that a normalised version of the area $A$ (more precisely of its square) does the trick.<|endoftext|> TITLE: Simplicity of group $C^\ast$-algebra implies fullness of group-von Neumann algebra? QUESTION [8 upvotes]: Let $\Gamma$ be a discrete group whose reduced group $C^\ast$-algebra is simple. Can we conclude that the corresponding group-von Neumann algebra $\mathcal{L}(G)$ is a full $\text{II}_1$-factor, meaning that every uniformly bounded net $(x_i)_{i \in I}$ that satisfies $\lim_i \Vert x_ia-ax_i\Vert_2 =0$ for all $a \in \mathcal L(\Gamma)$ we must have $\lim_i \Vert x_i -\tau (x_i)1\Vert=0$ where $\tau$ is the canonical tracial state of $\mathcal{L}(\Gamma)$? REPLY [12 votes]: No, whenever $\Gamma$ is an infinite direct product of C$^*$-simple groups, we obtain a counterexample. For instance, taking $\Gamma = \mathbb{F}_2^{(\mathbb{N})}$ to be the direct sum of infinitely many copies of the free group $\mathbb{F}_2$, we get that $C^*_r(\Gamma)$ is simple (because this is true for every finite direct product). On the other hand, taking a sequence of group elements $g_n \in \Gamma$ in the $n$-th factor of the direct product, the corresponding sequence of unitaries in $L(\Gamma)$ is a nontrivial central sequence, so that $L(\Gamma)$ is not full.<|endoftext|> TITLE: Even, non liftable Stiefel-Whitney class QUESTION [7 upvotes]: Let $M$ be a smooth manifold and $E$ a smooth real vector bundle of even rank over $M$. If $E$ admits of a complex vector bundle structure $\mathcal E$ ($\mathcal E_\mathbb R=E$) then all odd Stiefel-Whitney classes of $E$ vanish: $$w_{2i+1}(E)=0$$ Moreover the even Stiefel-Whitney classes of $E$ are the images under the reduction morphism $\operatorname {red}^{2i}:H^{2i}(M,\mathbb Z)\to H^{2i}(M,\mathbb F_2)$ of its Chern classes, namely $$w_{2i}(E)=\operatorname {red}^{2i}(c_i(\mathcal E))$$ My question Is there a real vector bundle of even rank $E$ with all odd $w_{2i+1}(E)=0$ that nevertheless cannot be endowed with a complex structure just because some $w_{2i}(E)\in H^{2i}(M,\mathbb F_2)$ cannot be lifted to $\mathbb Z$? Explicitly, the equation $$\operatorname {red}^{2i}(c_i)=w_{2i}(E)\in H^{2i}(M,\mathbb F_2)$$ has no solution $c_i\in H^{2i}(M,\mathbb Z)$ . REPLY [3 votes]: As a complement to Bertram Arnold's excellent answer, one could add a reference to Teichner, Peter, 6-dimensional manifolds without totally algebraic homology, Proc. Am. Math. Soc. 123, No. 9, 2909-2914 (1995). ZBL0858.57033. In particular, Lemma 2 states that if $M$ is a closed $4$-manifold with $\pi_1(M)=\mathbb{Z}/4$, then there exists a rank $3$ vector bundle $E$ over $M$ with $w_1(E)=w_1(M)$ and $w_3(E)=0$, and $w_2(E)$ is not the reduction of a class in $H^2(M;\mathbb{Z}^{w_1(E)})$ (where $\mathbb{Z}^{w_1(E)}$ denotes the local system of integer coefficients twisted by $w_1(E)$). Thus over any orientable $4$-manifold with $\pi_1=\mathbb{Z}/4$ we get an example. (Take direct sum with a trivial line bundle to get an even rank example, as in Bertram's answer.)<|endoftext|> TITLE: What is a subdivision of an abstract simplicial complex? QUESTION [8 upvotes]: I am looking for the definition of the subdivision of a simplicial complex. When the complex is defined in a geometric way, then the definition is pretty simple : the complex σ(C) is a subdivision of C if each simplex of σ(C) is contained in a simplex in C, and if each simplex of C is the union of finitely many simplexes in σ(C) (definition found in Herlihy-Shavit 1999). When we deal with the abstract definition of simplicial complexes (i.e. a subset-closed family of subsets), I fail to find a similar definition. I found definitions of specific subdivisions (e.g. standard chromatic subdivision of the n-simplex) but not a general formulation like the one I gave. A difficulty is the fact that an abstract simplex cannot be the union of its subdivision, since the subdivision contains new edges that do not appear in the first simplex (they are not in the same "space" because there is no convenient notion of "space" in the abstract definition). Thanks a lot for your help ! REPLY [4 votes]: A generalization of a barycentric subdivision is the notion of a stellar subdivision, where one performs a barycentric subdivison of a simplex $s$ and subdivides all simplices accordingly that contain $s$ as a face. The inverse of a stellar subdivision is called a stellar weld. Now let $K$ and $L$ be two abstract simplicial complexes. If the topological realizations $|K|$ and $|L$ admit subdivisions that are isomorphic as simplicial complexes, then the Alexander-Newman theorem says that $K$ and $L$ are stellar equivalent, i.e. they are related by a finite sequence of stellar subdivisions and stellar welds. See Chapter 64.5 of my lecture notes for details and references https://www.uni-regensburg.de/Fakultaeten/nat_Fak_I/friedl/papers/1at-uptodate.pdf<|endoftext|> TITLE: Groups with three conjugacy classes that define an ordering QUESTION [6 upvotes]: Consider the following property for a group $(\mathcal{G},\cdot,1)$: There are exactly three conjugacy classes $\{1\}$, $\mathcal{C}_1$, $\mathcal{C}_2$ in $\mathcal{G}$, and we have $\mathcal{C}_1 \mathcal{C}_1 \subseteq \mathcal{C}_1$ and $\mathcal{C}_2=\mathcal{C}^{-1}_1$. Note that the only finite groups with exactly three conjugacy classes are the cyclic group of order $3$ and the symmetric group of order $6$. Those do not satisfy the property above. So any such group must be infnite, hence also non-abelian. In fact, given such a group $\mathcal{G}$, define $x1 \Longleftrightarrow x \cdot y \cdot z> x \cdot z$. So it is perhaps best to think of those groups as linearly bi-ordered groups where any two strictly positive elements are conjugated. I suspect not so many such groups are known. So my question is: are there known examples of such groups? Or better yet: have they been studied to some extent? One could also expect such groups to be related to the first-order theory $T$ of linearly bi-ordered non-trivial groups with trivial center. The question would be: does any algebraically closed model of $T$ satisfy the property above? Indeed I believe that László Fuchs proved that in the case of partially bi-ordered, lattice ordered groups, the algebraically closed models have the property that any two strictly positive elements are conjugated. In that case however there may still be more conjugacy classes than three because not every element must be positive or negative. REPLY [9 votes]: There are currently no known examples of bi-orderable groups where all positive elements are conjugate. The question of their existence appears as Problem 3.31 of the 2009 problem list Unsolved Problems in Ordered and Orderable Groups compiled by Bludov, Glass, Kopytov and Medvedev. It is also apparently recorded in the Black Swamp Problem Book. On the other hand, it is known that every lattice-ordered group can be embedded in one with exactly four conjugacy classes, and that every right ordered group can be embedded in one where all non-trivial elements are conjugate: see V. V. Bludov and A. M. W. Glass, Conjugacy in lattice-ordered groups and right ordered groups (2008).<|endoftext|> TITLE: Bounds on homological dimension of functor categories QUESTION [8 upvotes]: Let $A$ be a Grothendieck abelian category. I will say that $A$ is of global dimension less or equal to $n$ if $Ext^{k}_{A}(a, b) = 0$ for $k > n$ and all $a, b \in A$. This is equivalent to saying that any object of $A$ admits an injective resolution of length at most $n$. Let $I$ be a small diagram category, so that the category of functors $Fun(I, A)$ is again Grothendieck. Are there any general bounds on the dimension of this functor category in terms of the dimension of $A$ and some invariant of the diagram category? Note that I do not expect this dimension would be finite for arbitrary $I$, I am rather looking for examples of "nice" $I$ for which we have such a bound for arbitrary Grothendieck category. In the simplest possible case, I would like to know the answer even if $A$ is a category of modules over a ring. REPLY [5 votes]: Claim: If the simplicial nerve of $I$ has dimension $m$ and $A$ has global dimension $n$, then $\operatorname{Fun}(I, A)$ has global dimension (at most) $m+n$. To prove it, you can use the standard simplicial resolution of a functor $F\colon I \to A$ by representable functors. $$F(-)\leftarrow \bigoplus_{c_0\in Ob(I)} F(c_0)\times \operatorname{mor}_I(c_0,-) \Leftarrow \bigoplus_{c_0\to c_1} F(c_0)\times \operatorname{mor}_I(c_1,-)\cdots$$ Taking $\operatorname{nat}(-, G)$ we obtain a cosimplicial resolution of $\operatorname{nat}(F, G)$. The cosimplicial object has dimension $m$, and each term has homological dimension $n$. It follows that $\operatorname{Ext}^k(F, G)=0$ for $k>m+n$.<|endoftext|> TITLE: Invariant ideal generated by invariant elements QUESTION [5 upvotes]: Let $G$ be a complex reductive group acting linearly on $\mathbb{C}^n$ and let $X$ be a $G$-invariant closed subvariety of $\mathbb{C}^n$. Is $X$ the zero-set of finitely many $G$-invariant functions? In other words, is the ideal $I \subset \mathbb{C}[x_1, \ldots, x_n]$ defining $X$ generated by finitely many invariant polynomials? Since $G$ is reductive, the set $I^G$ of $G$-invariant elements of $I$ is an ideal in the Noetherian ring $\mathbb{C}[x_1, \ldots, x_n]^G$, so it suffices to show that $I^G$ generates $I$. REPLY [2 votes]: In general, the answer is no, as Friedrich Knop intimates in a comment. Consider the reductive group $G:=\mathbb{C}^\times = GL_1(\mathbb{C})$ acting on $\mathbb{C}^2$ by $\alpha\cdot (x,y)\mapsto (\alpha x, \alpha y)$. Then the invariant ring $\mathbb{C}[x,y]^G$ is just $\mathbb{C}$, and the only subvarieties of $\mathbb{C}^2$ it's able to "see" (more precisely the only subvarieties of $\mathbb{C}^2$ cut out by an ideal that is generated by invariants) are the empty set and everything. In particular, $\{0\}$ and any complex line through the origin are invariant subvarieties that are not cut out by invariants. There is a qualified positive answer in the special case that $G$ is finite, however, or more generally in the situation that the action of $G$ is closed (i.e., every orbit of $G$ is Zariski-closed in $\mathbb{C}^n$). In this case, the answer to your question about ideals is still no: for example the action of $\mathbb{Z}/2\mathbb{Z}$ on $\mathbb{C}^1$ via the sign representation has no degree $1$ invariants (the invariant ring is $\mathbb{C}[x^2]$), but the ideal $(x)\subset\mathbb{C}[x]$ is $G$-stable, so this is a $G$-stable ideal that is not generated by invariants. That said, the answer is yes up to radical; equivalently, the answer to your question about varieties is yes in this situation. This is because when the action of $G$ on $\mathbb{C}^n$ is closed (for example if $G$ is finite), the map $$\mathbb{A}_{\mathbb{C}}^n \rightarrow \operatorname{Spec}\mathbb{C}[x_1,\dots,x_n]^G$$ induced by the ring inclusion $\mathbb{C}[x_1,\dots,x_n]^G\hookrightarrow\mathbb{C}[x_1,\dots,x_n]$ is a geometric quotient (see Mumford, GIT, Chapter 1 Section 2). In particular it is a topological quotient on the underlying topological spaces; we may identify the MaxSpec of $\mathbb{C}[x_1,\dots,x_n]^G$ with the topological quotient $\mathbb{C}^n/G$, so a $G$-stable Zariski-closed set in $\mathbb{C}^n$ maps to a closed set in $\mathbb{C}^n/G$, which, because the latter is an affine variety, is cut out by regular functions on this quotient, which are precisely invariant polynomials on the original $\mathbb{C}^n$. So any $G$-stable subvariety of $\mathbb{C}^n$ is indeed cut out by invariant polynomials. As an illustration, in the situation above with $\mathbb{Z}/2\mathbb{Z}$, the ideal $(x)$ may not be generated by invariants, but the ideal generated by the invariant $x^2$ has the same radical, so it cuts out the same variety $\{0\}$. (Aside: I did not engage the "finitely many" language in the question because it does not add a new requirement. As you note, reductivity of $G$ implies the ring $\mathbb{C}[x_1,\dots,x_n]^G$ is noetherian; thus any variety cut out by invariants will be cut out by finitely many.)<|endoftext|> TITLE: Conclusion of Hurewicz for $H_3$ without vanishing fundamental group? QUESTION [7 upvotes]: Fix a space $X$, which I want to assume is a manifold. Under the assumption of simple-connectivity, Hurewicz's theorem tells us that $$ \pi_3(X)\to H_3(X,\mathbb{Z})\qquad (*) $$ is surjective, hence that every homology class is represented by a map $S^3\to X$. There is lots of hard work gone into worrying when homology classes are represented by embedded submanifolds, but that is not what I'm interested in here. What I want to know is: Are there are nontrivial examples of manifolds $X$ with $(*)$ surjective, but with infinite $\pi_1(X)$? I don't need the case when $\pi_1$ is finite, since I have a different proof that doesn't need this more subtle property. An equivalent question is asking whether $H_3(\tilde{X},\mathbb{Z})\to H_3(X,\mathbb{Z})$ is surjective, for $\tilde{X}$ the universal covering space. Since $\pi_1(X)$ is nontrivial, then questions about the Eilenberg–Moore spectral sequence become more subtle, but I'm only looking at such a low-dimensional group that maybe things are not so bad (I don't really understand the EM spectral sequence, even relative to my background knowledge of spectral sequences, so I don't quite know how to start extracting information from that). Added: I was interested to know if there are general hypothesis that allow me to conclude $(\ast)$ is onto, but given a comment below by user51223, I think all I care about is the weaker statement that $$ \pi_3(X)\to H_3(X,\mathbb{Z}) \to H_3(X,\mathbb{Z})/\text{torsion} \qquad (**) $$ is surjective. So: Are there general conditions that guarantee this weaker statement is true? REPLY [6 votes]: For closed 3-manifolds, this holds iff it is the 3-sphere. If the fundamental group of a closed 3-manifold $X$ is infinite, then $H_3({\tilde X})$ is trivial. If $\pi_1(X)$ is finite, then $\tilde{X}\cong S^3$, and $H_3(\tilde{X})\cong \pi_3(\tilde{X})$. So the map $\pi_3(X) \to H_3(X)$ will have range a subgroup of index $|\pi_1(X)|$. For a closed 4-manifold $X$, $H_3({\tilde X})$ will be torsion-free. If it’s non-trivial, then $\pi_1 X$ will have more than one end, and hence by Stallings’ theorem $\pi_1 X$ will split over a finite group. Take a maximal splitting of $\pi_1 X$ as a graph of groups with finite edge groups. Then if the vertex groups have $H^1( ;\mathbb{Q})=0$, then your condition (*) should hold with rational coefficients. In fact I think this is also necessary, but one might have to analyze Stallings’ proof to show this. Equivalently, the map from the manifold to the graph defining the graph of groups should induce a surjection on $H^1( ; \mathbb{Z})$ from the graph to the manifold. As in the 3-manifold case, I’m not sure that this can hold with integral coefficients when the edge groups of the graph of groups are non-trivial finite groups (in this case one might only hit a finite-index subgroup of $H_3(X)$). So I’m guessing that a necessary and sufficient condition is that the manifold is a connect sum of a manifold $X$ with $H_3(X)=0$ and $\#^k (S^1\times S^3)$ (or the twisted version $S^1\tilde{\times} S^3$ if non-orientable manifolds are included).<|endoftext|> TITLE: Is every additive, left exact functor isomorphic to a hom functor? QUESTION [9 upvotes]: Let $A$ be an Artin algebra, $\text{mod}\,A$ the category of finitely generated $A$-modules and $\text{Ab}$ the category of abelian groups. Is every additive, covariant, left-exact functor $F:\text{mod}\,A \rightarrow \text{Ab}$ (natural) isomorphic to $\text{Hom}(X,-)$ for some $X\in \text{mod}\,A$? How do we obtain $X$? REPLY [6 votes]: $\DeclareMathOperator{\Set}{Set} \DeclareMathOperator{\El}{El} \DeclareMathOperator*{\colim}{colim} \newcommand{\fC}{\mathcal C}$By abstract nonsense, for any category $\fC$, a functor $F:\fC\to \Set$ is a colimit of corepresented functors. Explicitly, let $\El(F)$ be the category of elements of $F$, whose objects are pairs $(c,x\in F(c))$ and morphisms $(c,x)\to (d,y)$ are morphisms $f:c\to d$ in $C$ such that $F(f)(x) = y$. There is an evident functor $p:\El(f)\to C$ which forgets the additional datum $x$, and we have $F\cong \colim_{\El(F)^{op}} y\circ p$, where $y:C^{op}\to \operatorname{Fun}(C,\Set)$ is the (opposite) Yoneda embedding. In your situation, the functor $F$ preserves finite products (by additivity) and pullbacks (by left exactness), hence all finite limits. In this case, the index category $\El(F)$ is co-filtered, i.e. any functor $G:I\to \El(F), i\mapsto (c_i,x_i)$ extends to a cone $G_\triangleleft:I_\triangleleft\to\El(F)$: Let $c = \lim_I p\circ i$ be the limit of the underlying diagram and $x\in F(c)$ be a preimage of $\{x_i\}_{i\in I}$ under the isomorphism $F(c)\cong \lim_I F\circ G\subset \prod_{i\in I} F(c_i)$, so that sending the cone point to $(c,x)$ and the cone legs to the lifts of the projections from the limit $c\to c_i$ is the required extension. (Note that if $\mathcal C$ has all products and $F$ preserves them, you can take an arbitrary small category for $I$. If you could take $\El(F)$ itself, this means that $\El(F)$ has an initial object $(c_0,x_0)$, and unraveling the definitions shows that $F$ is corepresented by $c_0$. The adjoint functor theorem in Dan Petersen's answer essentially works this way.) On the other hand, since colimits in functor categories are computed objectwise and filtered colimits commute with finite limits in Sets, any filtered colimit of corepresented functors preserves finite limits. So abstractly, the best thing you can hope for is a pro-object representing your functor, i.e. a diagram $C:I\to \fC$ from a co-filtered category such that $F(c') \cong \colim_{I^{op}} \fC(C(i),c')$. Here you can always take $I$ to be the opposite of a cardinal, considered as a poset, by choosing a cofinal subset of $\El(F)$. For example, the functor $V\mapsto V^{\oplus\infty}$ is corepresented by the pro-object with $I = \mathbb N^{op}$ and $C(n) = k^n$, with the maps $C(n+1)\to C^n$ given by projection. On the other hand, in Zhen Lin's example the category of elements $\El(F)$ has cofinality $\kappa$, which is bigger than the cofinality of any slice category, so that the functor can not be representable. Restrict now to the setting in the question, i.e. $\mathcal C$ is the category of finitely-generated modules over an Artin algebra $R$. We know that $F = \colim_{I^{op}} y(M_i)$ is a subfunctor of a corepresentable functor $y(M)$. This means that there are compatible maps $f_i:M\to M_i$. The submodules $\ker f_i$ of $M$ form a descending chain; since $R$ is Artinian and $M$ finitely generated, they stabilize at a finite level $i$. Replacing $M$ by $M/\ker f_i$ and $I$ by $I_{i/}$, we may assume that all the maps $f_i$ are injective. Similarly, the images of the maps $M_{i'}\to M_i$ form a descending chain of submodules of $M_i$, and therefore stabilize at a finite stage. Let $\{M'_i\}_{i\in I}$ be the pro-object of these images, and $F' = \colim_{I^{op}} y(M'_i)$, so that we get a factorization $F\Rightarrow F'\Rightarrow y(M)$. The first map is an isomorphism (injectivity and surjectivity can be checked by lifting elements a finite amount through the colimit). Replacing $\{M_i\}_{i\in I}$ by $\{M'_i\}_{i\in I}$, we may assume that all transition maps are surjective. Thus all transition maps in the colimit $F'(M') = \colim_{I^{op}} \fC(M_i,M')$ are injective, so that $y(M_i)$ is a subfunctor of $F$ and hence of $y(M)$, which is equivalent to the maps $M\to M_i$ being surjective. Together with injectivity, this shows that we may assume that $\{M_i\}_{i\in I} = \{M\}_{i\in I}$ is a constant pro-object, which is equivalent to corepresentability of $F$.<|endoftext|> TITLE: Is there a dense planar rational point set within which the distance of any two points is an irrational number? QUESTION [6 upvotes]: i.e. could we find a subset $X\subset \mathbb{Q}^2$ such that $\overline{X}=\mathbb{R}^2$ and that for any $x,y\in X$ the distance $|x-y|$ is an irrational number? I'm considering the following assertion of which I'm not sure : Given finite rational points $p_1,p_2,\dots,p_n$ , and an open ball $D$ on the plane, there is a rational point $x\in D$ such that $|x-p_i|\in \mathbb{R}\backslash \mathbb{Q}$ for $i=1,2,\dots,n$. But this assertion accounts to be the following seemingly number-theoretic problem: Given $n$ pairs $(a_i,b_i) (i=1,2,\dots,n)$ of positive integers such that $a_i^2+b_i^2$ is not square of any integer. Could we find an integer $N\geq 2$ such that the integral pairs $(Na_i+1,Nb_i)$ still satisfy the previous property(i.e. $(Na_i+1)^2+(Nb_i)^2$ is not square of any integer). BTW the distribution of the Pythagorean triples might help. REPLY [13 votes]: Yes. Observation: if (x,y) is half-integral, then it is at irrational distance (square root of a number that is 2 mod 4) from all integer points. Construction: Find a sequence $p_1,p_2,\dots$ of the rational points such that each point appears infinitely often, and let $G_1$ be the integer grid. For each $i$ in sequence, replace $p_i$ by the nearest half-integral point $q_i$ of grid $G_i$, and then subdivide the grid by two to produce a new grid $G_{i+1}$ in which $q_i$ is integral. Then the resulting sequence of points $q_i$ gets arbitrarily close to every rational point, so it is dense, every point is rational (more strongly dyadic rational), and by the observation every point is at irrational distance from all previous points.<|endoftext|> TITLE: Is there a topological interpretation of a module over $\Omega_{PL}(X)$? QUESTION [7 upvotes]: Associated to a DGCA (differential graded commutative algebra) $A$, we can associate to it the category of modules over $A$. Hence, for a space $X$ we can consider the category of modules over $\Omega_{PL}(X)$, modules over the PL de Rham complex of $X$. The starting point of rational homotopy theory is that $\Omega_{PL}(-)$ is an equivalence upon restricting to simply connected spaces and simply connected DGCA's in a homotopical sense. I am wondering, perhaps after replacing $\Omega_{PL}(X)$ with a Sullivan model, does this category of modules have a topological interpretation? I am also interested in the d.g. lie algebra analog. REPLY [8 votes]: TLDR: there is a contravariant adjunction between the derived category of $A_{PL}(BG)$ and the naive category of rational $G$-spectra. In some cases, this is a contravariant equivalence of categories, for example when $G$ is a connected Lie group. If I am not mistaken, the derived category of $A_{PL}(X)$ is equivalent to the homotopy category of modules over the ring spectrum $\operatorname{Map}(\Sigma^\infty X_+, H\mathbb Q)$. I think this follows from a paper of Richter and Shipley. We may as well pose the question about modules over the singular cochain complex $C^*(X)$, a.k.a modules over $\operatorname{Map}(\Sigma^\infty X_+, H\mathbb Z)$, or even more generally, about modules over the ring spectrum $D(X)=F(\Sigma^\infty X_+, S)$ (the Spanier-Whitehead dual of $X_+$). When $X$ is a disjoint union, all these rings or ring spectra split as products. So we may assume that $X$ is connected. In this case there is a Koszul duality between cochains on $X$ and chains on $\Omega X$. Or better still, there is a Koszul duality between the ring spectra $D(X)$ and $\Sigma^\infty \Omega X_+$. This implies that there is a contravariant adjunction between the homotopy categories of modules of these spectra. This adjunction restricts to an equivalence between certain subcategories of these module categories. For example, there is an equivalence between finitely generated free cellular modules over either one of the ring spectra, and the so-called nilpotent modules over the other one. Without loss of generality we may assume that $X=BG$ where $G$ is a topological groups. Then modules over $\Sigma^\infty \Omega BG_+\simeq \Sigma^\infty G_+$ are the same as spectra with an action of $G$. So there is a contravariant adjunction between the category of modules over $D(BG)$ and the naive category of $G$-spectra. Similarly, there is a Koszul duality between the ring spectra $D_{\mathbb Q}(BG)=\operatorname{Map}(\Sigma^\infty BG_+, H\mathbb Q)$ and $H\mathbb Q \wedge G_+$. If I am not mistaken, the homotopy category of modules over $\Omega_{PL}(X)$ is equivalent to modules over $D_{\mathbb Q}(BG)$. So we obtain a contravariant adjunction between this category and the naive category rational $G$-spectra. This adjunction is not an equivalence in general, but sometimes it is. For example, I think the paper "An algebraic model for free rational $G$-spectra for connected compact Lie groups", by Greenlees and Shipley, tells you that it is an equivalence when $G$ is a connected compact Lie group. They also give an explicit algebraic model for the category of modules in this case. I have the feeling that when $G$ is connected and $BG$ is finite, the adjunction is also close to being an equivalence in some sense that can be made precise. Maybe the (naive) category of rational $G$-spectra is equivalent in this case to the category of pro-nilpotent modules over $A_{PL}(BG)$, but I am not 100% sure. There is a paper about derived Koszul duality by Blumberg and Mandell that says something about it. On the other hand, if $G$ is a finite group, then $D_{\mathbb Q}(BG)\simeq H\mathbb Q$, but the category of rational $G$-spectra of course depends on $G$. (Incidentally the paper Complexes of injective $kG$-modules by Benson and Krause contains some interesting information about the derived category of $C^*(BG; \mathbb F_p)$ for finite $G$.)<|endoftext|> TITLE: Derived category of abelian sheaves on a site equivalent to sheaves on the derived category of abelian groups QUESTION [8 upvotes]: Reading Scholze's notes on Condensed Mathematics it is mentioned that when considered as $\infty$-categories, $$ D(\operatorname{Cond(Ab)}) \cong \operatorname{Cond}(D(\operatorname{Ab}))$$ and that this is not a feature of condensed abelian groups but rather of the category of sheaves on a site. I have been looking for references on this fact, but haven't been able to find anything. This thread seems to suggest that we have some additional conditions on our site. Any references to read further on this and see why this is true for this example would be greatly appreciated. REPLY [6 votes]: It's true in any $1$-topos for hypercomplete sheaves, see Theorem 2.1.2.2 in Spectral Algebraic Geometry.<|endoftext|> TITLE: Finite subgroups of $\operatorname{U}(2)$ QUESTION [7 upvotes]: Famously, the finite subgroups of $\operatorname{SU}(2)$ admit an ADE classification. Question. Is there a similar result for finite subgroups of $\operatorname{U}(2)$? Are they classified? If this is the case, what is the complete list of them? REPLY [17 votes]: Every finite subgroup of $\operatorname{GL}(2,\mathbb{C})$ is conjugate to a subgroup of $U(2)$, so you are asking first for the isomorphism types of finite subgroups of $\operatorname{GL}(2, \mathbb{C}).$ These were already known to C. Jordan. They are easy to recover. There are three types: I: Reducible subgroups: these are conjugate to groups of diagonal matrices, so (since finite), they are finite Abelian groups with at most two generators, and all of these can occur. If we restrict to groups in which all matrices are unimodular, we get all finite cyclic groups as possibilities. II: Irreducible but imprimitive subgroups: These are equivalent to monomial groups, and have a finite Abelian normal subgroup of index $2$ which (after replacing the representation by an equivalent one if necessary) is in fact cyclic (in modern terminology (for group theorists) this is a consequence of Clifford's Theorem). Each finite non-Abelian group $G$ which has an Abelian (necessarily normal) subgroup $A$ of index $2$ does occur as a finite subgroup of $\operatorname{GL}(2,\mathbb{C})$. For such a group $G$ has all irreducible characters of degree at most two, but (since $G$ is non-Abelian) there is a non-linear irreducible character of $G$. Indeed, since $A$ is cyclic, we may choose a faithful linear character of $A$ and induce it to $G$. The representing matrices for elements of $G-A$ are not even diagonal, and no non-identity element of $A$ is represented by the identity matrix in this induced representation. Hence type II gives rise to every finite non-Abelian group which has a cyclic normal subgroup of index $2$. However, if we restrict to type II groups consisting of unimodular matrices, then it is easy to check (using similar arguments) that the groups we get are just those finite non-Abelian groups $G = \langle A, u \rangle$ where $A$ is cyclic of even order and $u$ is an element of order $4$ with $u^{-1}au = a^{-1}$ for all $a \in A$. III: Irreducible primitive groups: these are the irreducible subgroups of $\operatorname{GL}(2,\mathbb{C})$ which are not conjugate to monomial groups (i.e., the $2$-dimensional representation is not equivalent to one induced from a proper subgroup). It was known to Jordan that if $G$ is such a group, then $G/Z(G)$ is isomorphic to $A_{4}$, $S_{4}$ or $A_{5}$. But we need to think explicitly how we make this identification. We take our group $G$ (in such a representation), and we multiply each element of a chosen set of generators by a (root of unity) scalar multiple so that the result has determinant one (this can be done because the original determinant is some root of unity). We get a resulting group $H \subseteq \operatorname{SL}(2,\mathbb{C})$ with $H/Z(H) \cong G/Z(G)$. There are (up to equivalence) three possible choices for $H$: we can have $H \cong \operatorname{SL}(2,3)$, $H \cong \operatorname{BO}$ or $H \cong \operatorname{SL}(2,5)$. The group I call $\operatorname{BO}$ has a generalized quaternion Sylow $2$-subgroup of order $16$ and has order $48$. It has $\operatorname{SL}(2,3)$ as a normal subgroup of index $2$. It is easy to show that each possiblity of $H$ is generated by two elements: in fact $\operatorname{SL}(2,3) = \langle a,b \rangle$ with $a$ of order $3$ and $b$ of order $4$ (with some relations), $\operatorname{BO} = \langle a,b \rangle$ with $a$ of order $3$ and $b$ of order $8$ (with relations) and $\operatorname{SL}(2,5) = \langle a,b \rangle$ with $a$ of order $3$ and $b$ of order $5$ (with relations). To produce ALL finite possibilities for $G$, we must take each choice of $H = \langle a,b \rangle$ and take all choices $G = \langle za,wb \rangle$ where we identify $a$ and $b$ with their representing matrices and where $z$, $w$ are arbitrary roots of unity. There will be repetitions of isomorphism type when we do this. The resulting $G$ has the form $C \times \langle z^{\prime}a,w^{\prime}b \rangle$ where $C$ is a finite cyclic group of scalars whose order is coprime to $30$ and $z^{\prime}$ and $w^{\prime}$ are roots of unity whose order has the form $2^{x}3^{y}5^{z}$ for non-negative integers $x$, $y$, $z$. Later edit: In fact, I think we can go further on consideration of determinant: We can always write the right-hand factor (up to isomorphism) as $D \times E{\ast}H$ where $D$ is a cyclic $\{3,5\}$-group, $E$ is a cyclic $2$-group, and $H$ is one of the three possibilities above (and ${\ast}$ denotes the central product, where $Z(H)$ (which has order $2$) is identified with the unique subgroup of order $2$ of $E$ (if $E$ > 1). It follows that the distinct possibilities for isomorphism types of $G$ of type III are of the form $ C_{1}{\ast} H$ where $H$ runs through the three possibilities above, $C_{1}$ runs over all finite cyclic groups not of twice odd order , and $\ast$ denotes the central product (which is direct when $C_{1}$ has odd order, and identifies the unique subgroup of order $2$ of each factor when $C_{1}$ has order divisible by $4$, and where $C_{1}$ and $H$ are mutually centralizing, so $C_{1}$ is represented by scalar matrices). We may omit the case that the cyclic group $C_{1}$ has twice odd order from the list because in that case, if we write $C_{1} = D \times E$ with $D$ of odd order and $E$ of order $2$, then $C_{1} \ast H \cong D \times H.$ We may note that the order of the finite cyclic group $C_{1}$ is the order of the image ${\rm det}(G)$ when $\{ {\rm det}(g): g \in G \} $ has odd cardinality, and twice the order of the image ${\rm det}(G)$ when $\{ {\rm det}(g): g \in G \} $ has even cardinality<|endoftext|> TITLE: Are there structures in a finite signature that are recursively categorically axiomatizable in SOL but not finitely categorically axiomatizable? QUESTION [6 upvotes]: Recall that a structure $\mathcal{M} = \langle M, I^\sigma_M \rangle$ in a signature $\sigma$ is categorically axiomatized by a second-order theory $T$ when, for any $\sigma$-structure $\mathcal{N} = \langle N, I^\sigma_N \rangle$, $\langle N, \mathcal{P}(N), I^\sigma_N \rangle \vDash T$ just in case $\mathcal{N}$ is isomorphic to $\mathcal{M}$. It is fairly easy to find a structure in a finite signature that is categorically second-order axiomatizable but not finitely categorically second-order axiomatizable. Add a single function symbol $f$ to the language of second-order arithmetic, and choose a non-second-order-definable $\zeta: \mathbb{N} \rightarrow \mathbb{N}$. Then consider the theory $T$ that adds to the axioms of second-order arithmetic ($\mathsf{Z}^2$) the sentence $f(\bar{n}) = \overline{\zeta(n)}$ for each natural number $n$, where $\bar{m}$ is the canonical numeral for $m$. (I owe the idea for this example to Andrew Bacon.) This theory $T$, however, is not recursively axiomatizable. Is there a structure in a finite signature that has a recursive categorical second-order axiomatization but no finite categorical second-order axiomatization? I believe that it is possible to find a recursively axiomatizable second-order theory $T$ whose spectrum (i.e., the set $\{\kappa \in \mathsf{Card}: \exists \mathcal{M} (\mathcal{M} \vDash T$ and $\vert \mathscr{M} \vert = \kappa)\}$) is shared by no finitely axiomatizable second-order theory, using partial truth predicates. (Consider the theory with $\mathsf{Z}^2$ relativized to some predicate $N$ and $\{$"The cardinality of the non-$N$s is not $\Sigma^1_n$-characterizable"$: n \in \omega\}$.) But I cannot see how to turn this into a categorical theory. REPLY [2 votes]: There is even an example of a cardinal $\kappa$ and an r.e. categorical second-order theory $T$ such that for no finitely axiomatized second-order theory $U$, the spectrum of $U$ has $\kappa$ as its least element. Let $T_0$ be the theory $\mathsf{ZC}_2+\forall x\exists \alpha(x\in V_\alpha)$, here $\mathsf{ZC}_2$ is the second-order version of Zermelo set theory where in the separation scheme we allow formulas without class quantifiers, but with class variables. It is easy to see that the second-order models of this theory are precisely $(V_\alpha,\in)$, for limit $\alpha>\omega$. Further I denote by $\mathcal{L}$ the set of sentences of the language of this theory. Let $T$ be the extension of $T_0$ by $\forall \alpha (\exists \varphi\in \mathcal{L}) (((V_\alpha,\in)\models_2\varphi)\land \forall \beta<\alpha ((V_\beta,\in)\not\models_2\varphi))$. $\varphi\to\exists \alpha ((V_\alpha,\in)\models_2 \varphi)$, where $\varphi$ ranges over $\mathcal{L}$. Obviously the only second-order model of $T$ is $(V_\alpha,\in)$, where $\alpha$ is the least ordinal such that for any $\varphi\in\mathcal{L}$ if $(V_\alpha,\in)\models \varphi$, then for $(V_\beta,\in)\models \varphi$, for some $\beta<\alpha$. Clearly, $\alpha>\omega^2$ and hence the cardinality of this model is $\beth_\alpha$, which will be our $\kappa$. Suppose for a contradiction that there is a second-order sentence $\varphi$ such that the smallest model of $\varphi$ is in the cardinality $\beth_\alpha$. Let $\varphi'$ be the naturally constructed sentence of second-order set theory expressing that there exists a class-model of $\varphi$. By construction $(V_\alpha,\in)\models_2\varphi'$, but for no $\beta<\alpha$ we have $(V_\beta,\in)\models_2\varphi'$, contradiction.<|endoftext|> TITLE: Lower bounds for class number of function fields with fixed $q$, growing $g$ QUESTION [14 upvotes]: Let $X$ be a smooth project curve of genus $g$ over the finite field with $q$ elements. Let $h$ be $\# \mathrm{Pic}^0(X)(\mathbb{F}_q)$. Weil showed that $h \geq (\sqrt{q}-1)^{2g}$. Lachaud and Martin-Descamps established the bound $h \geq \tfrac{q^g}{g+1} \tfrac{(q-1)^2}{q(q+1)} \approx \tfrac{q^g}{g}$, which is better for $q$ fixed and $g \to \infty$. I was playing around with the estimates I was using in a recent question about finding function fields with class number $1$, and I seem to have come up with an argument that proves a bound of the form $h \geq c \tfrac{q^g}{\log g}$, for a small positive constant $c$. I am trying to figure out if this is interesting. Of course, I'm doing a literature search, but I don't know this area, so maybe someone can tell me. What are the best current bounds in this area, and how does mine compare? Or, I suppose, does someone know a counterexample showing that $\tfrac{q^g}{\log g}$ is too good to be true? REPLY [5 votes]: $\def\FF{\mathbb{F}}$Easy results are usually not original, and this wasn't an exeption. What I was doing was almost exactly the same as On the number of rational points of Jacobians over finite fields, Lebacque and Zykin (2013). The one thing that they are missing is a simple asymptotic analysis, so I'll publish that here rather than trying to make it into a paper. For any $N \geq 1$, Corollary 2.5 of Lebacque and Zykin implies that $$\#J(\FF_q) \geq q^g \exp\left( - \sum_{n=1}^N \frac{1}{n} - \sum_{n=1}^N \frac{q^{-n}}{n} - \frac{2g}{(\sqrt{q}-1)(N+1) q^{N/2}} \right).$$ $$= q^g \exp \left( - \log N + O(1) + O \left( \tfrac{g}{N q^{N/2}} \right) \right).$$ If we choose $N$ such that $q^{N/2} \approx 2g$, then we get a lower bound of $c \tfrac{q^g}{\log g} $.<|endoftext|> TITLE: Is the canonical model structure on strict $\infty$-Cat left proper? QUESTION [5 upvotes]: Is the canonical (or Folk) model structure on the category of (strict) $\infty$-categories as constructed by Lafont, Métayer and Worytkiewicz in A folk model structure on omega-cat left proper ? All its objects are fibrant, so it is right proper, but left properness is not clear at all. I tend to think it is not, but the pushout of a cofibration in $\infty$-Cat are still a relatively well-behaved operation so it might very well be. In any case, I don't know explicit examples of pushouts showing it is not left proper. I would also be interested in answers to this question for the folk model structure for 2 or 3-categories (either strict or semi-strict) as this would very probably shed some light on the question. Motivation: I've stumbled on this for several reasons, but one of them is that the folk model structure has many interesting left Bousfield localizations. For example for each $n$ it can be Bousfield localized so that the fibrant objects of the localization are the $(\infty,n)$-category in the sense that all arrows of dimension $>n$ are weakly invertible (contrary to Ara–Métayer's "generalized Brown-Golanski" model structure in The Brown-Golasinski model structure on strict $\infty$-groupoids revisited, where the arrows of dimension $>n$ are strictly invertible; whether these two things are Quillen equivalent is as far as I know an open problem). But in order for the localization to be an actual Quillen model category (instead of a semi-model category) one need left properness. I'm very curious to know if this localization is an example of a Bousfield localisation of non proper model structure (because those have surprising properties and I don't know really natural examples of this so far), or if it is actually a Quillen model structure (which would also be nice). REPLY [5 votes]: The canonical model structure on 2-Cat is left proper. This is proven in Steve Lack's original paper A Quillen model structure for 2-categories that constructs this model category. The proof involves a long calculation of the pushout of trivial fibrations along the generating cofibrations. (Side note: many of us believe the adjective "folk" is not really appropriate for these model structures, except possibly in the case $n=1$.)<|endoftext|> TITLE: Algebraic K-theory "with proper support" QUESTION [6 upvotes]: I would like to know what is the "correct" algebraic $K$-theory "with proper support". I suppose that the answer should be found in the condensed world, which is mostly inspired the existence of six functors. See Lectures on Condensed Mathematics, Appendix to Lecture VIII. $\DeclareMathOperator\Tor{Tor}\DeclareMathOperator\Perf{Perf}\DeclareMathOperator\Spec{Spec}$More precisely, given a map $f\colon R\to S$ of finitely generated rings, we have the functor $f_!\colon D(S_\blacksquare)\to D(R_\blacksquare)$ of "direct image with proper support" which preserves small colimits. Suppose that $f$ is of finite $\Tor$-amplitude, then $f_!$ induces a functor $\Perf(S_\blacksquare)\to\Perf(R_\blacksquare)$ between compact objects in $D(S_\blacksquare)$ and $D(R_\blacksquare)$ respectively. The ordinary $K$-theoretic machinery leads to a map $K(\Perf(S_\blacksquare))\to K(\Perf(R_\blacksquare))$ which looks like some kind of "integral along fibers" of $\Spec S\to\Spec R$ for compactly supported cohomologies. However, this does not seem to be satisfactory, since both sides are discrete. One way to get a condensed structure might be the following: $\Perf(R_\blacksquare)$ (resp. $\Perf(S_\blacksquare)$) should be a "condensed stable symmetric monoidal $\infty$-category". There might be set-theretic issues, but let's ignore them for a moment. Take the maximal groupoid, we get a map of condensed $E_\infty$-monoids and taking the group completion, we get a map of condensed spectra. Edit: I was mistaken, we should pass to the Waldhausen $S_\bullet$ construction for this, but seemingly this could also be equipped with a condensed structure. It is not immediately clear whether these spectra are solid. Is there alternative characterizations of these condensed spectra, hopefully without reference to "condensed $\infty$-categories"? On the other hand, there is a concept of condensed $K$-theory in Lectures on Analytic Geometry, Prop 10.6. However, this does not seem to coincide with the construction above. Update: As Denis-Charles Cisinski pointed out, the approach sketched above does not work since the corresponding $K$-theory is trivial by Eilenberg's Swindle applied to the infinite product). It is not immediate for me what is a replacement, in view of the compactly supported topological $K$-theory, for example. REPLY [5 votes]: Actually, the possibility of defining such a thing was one of my motivations for studying this condensed mathematics in the first place. That said, the story is far from complete. First, I guess a reasonable definition is the following. For $R\rightarrow S$ a map of finitely generated commutative rings, recall that there is an idempotent commutative algebra object $S_{\infty/R}$ in derived solid $(S,R)$-modules such that modules over $S_{\infty/R}$ are exactly those modules which die on localization to solid $S$-modules. For example, if $S=R[t]$ then $S_{\infty/R}=R((t^{-1}))$, but in general $S_{\infty/R}$ lives in several cohomological degrees. Now the K-theory wth proper support should be the homotopy fiber of $$K(S)\rightarrow K(S_{\infty/R}).$$ However, this leaves open the question of how $K(S_{\infty/R})$ should be defined. one could do it naively by just taking perfect complexes, but this leads to a messy theory. Peter Scholze and I realized that the correct way to define such K-theory is instead to take the continuous K-theory, in the sense of Efimov, of a certain dualizable category of modules which contains, but is bigger than, the ind-category of perfect complexes. This is the category of nuclear modules briefly described in Analytic.pdf. This probably defines what is the "correct" object. That said, I view this actual definition as unsatisfactory, because I would like a natural dualizable category of modules whose continuous K-theory directly gives the K-theory with proper supports. I don't know how to do this; the best I've managed so far is one that gives the suspension. Another thing I'd like to mention is that the reason I was interested in this K-theory with proper supports is that it shows up in the approach I worked out to Artin maps, https://arxiv.org/abs/1703.07842 . If $R$ is a finite type $\mathbb{Z}$-algebra, the source of the Artin map as I define it is the K-theory of the category of $R$-modules in $D^b(LCA)$, the bounded derived category of locally compact abelian groups. If $R$ is a regular $\mathbb{F}_p$-algebra, I conjecture that this K-theory is equivalent to the suspension of the compactly supported K-theory of $R$ over $\mathbb{Z}$ in the above sense, if you interpret $K(S_{\infty/R})$ in the naive way with perfect complexes. By the way, the hypotheses on $R$ are not crucial: one can remove the regularity assumption by working with G-theory, and the $\mathbb{F}_p$-algebra assumption by adding in the appropriate archimedean part at $\infty$ as well. When $R$ is one-dimensional I believe these conjectures have been proved (though with slightly different definitions) by Braunling, https://arxiv.org/abs/1710.10819 .<|endoftext|> TITLE: Line graphs called "graph derivatives": any intuition? QUESTION [5 upvotes]: Short version: in several papers, line graphs (and closely related graphs) are called graph derivatives or derived graphs; is there any intuition for such terminologies, in connection with the classical concept of derivative? Full version of the question. In a 1961 paper entitled Graph Derivatives, Gert Sabidussi defines the graph derivative $\delta G$ of a graph $G$; it is nothing but the nowadays classical line graph of $G$. Similarly, in 1970, Lowell W. Beineke calls this graph the derived graph of $G$ in his paper entitled Characterizations of derived graphs. In 2017, a PhD dissertation entitled Structures of Derived Graphs was awarded to Khawlah Hamad Alhulwah at Western Michigan University. A large part of this work is about line graphs, and it gives the following intuition: "a derived graph of G is a graph obtained from G by a graph operation of some type." It however seems to me that Sabidussi and Beineke (and, probably, others) had in mind a deeper connection between line graphs and the classical concept of (function) derivative. In another line of works, the derived graph of $G$ has the same vertices as $G$ but they are connected by an edge if the distance between them in $G$ is $2$. This terminology was introduced in the 2009 paper Bounds On The Second Stage Spectral Radius Of Graphs by Ayyaswamy, Balachandran and Kannan, and further studied for instance in the 2012 paper Derived graphs of some graphs by Jog, Hande, Gutman, and Bozkurt. However, I can't find any explanation for this terminology in these papers. My questions are: is there any reason to call line graphs graph derivatives? are they related to some extension of classical derivatives in a way that I miss? are there other definitions of graph derivatives that may be seen as such extensions? REPLY [2 votes]: I just found the answer to this question in the paper Synthesis and analysis in total variation regularization by Francesco Ortelli and Sara van de Geer. In the abstract, they write "We give a definition of the discrete graph derivative operator based on the notion of line graph", and they do so in Section 1.2. They use (a directed version of) the incidence matrix of a (directed) graph, and notice it may be used to differentiate a signal on the vertices of the graph along its edges, like gmvh suggests in the answer above. Then, in order to iterate such derivations, they introduce the (directed) line graph and say: The idea behind using the line graph is the following: when computing differences along the direction of the graph, we obtain a value for each edge of the graph. Since we follow the direction of the edges, these values represent the first discrete derivative of the signal with respect to the graph, and constitute the vertices of the line graph. They finally define the k-th order discrete graph derivative operator from the incidence matrix of iterated line graphs (meaning that they take the line graph of the line graph of the line graph etc). It turns out that yes, line graphs and derivatives are strongly related!<|endoftext|> TITLE: Maximum of the weighted binomial sum $2^{-r}\sum_{i=0}^r\binom{m}{i}$ QUESTION [8 upvotes]: Let $m$ be a positive integer and let $f_m(r)=2^{-r}\sum_{i=0}^r\binom{m}{i}$. Clearly $f_m(0)=f_m(m)=1$ and $f_{2r+1}(r)=2^{2r}$. Conjecture: If $m>12$, then the maximum value of $f_m(r)$ for $r\in\{0,1,2,\dots,m\}$ occurs when $r=\lfloor m/3\rfloor +1$. Such a weighted sum arises in Coding Theory and in Information Theory, so experts may know the answer to this conjecture. (Computation shows that it is true for $m\leqslant1000$.) The following seem helpful and relevant: (1) the entropy function $H(p)=-p\log p-(1-p)\log(1-p)$, (2) Stirling's approximations, and (3) the MO questions Lower bound for sum of binomial coefficients?, and Estimating a partial sum of weighted binomial coefficients. Alas, I couldn't see how to prove such a precise conjecture. Proving that the maximum is "near" $m/3$, in some sense, may be more tractable. REPLY [4 votes]: I'll give a crude calculation on the back of this envelope. Assume that $c'm0$ means that $i=o(r^{1/2})$ is enough terms to get most of the sum.) Consequently, $$\frac{f_m(r-1)}{f_m(r)} = (1+o(1)) \frac{2r}{m-r},$$ which is increasing in $r$ and equals 1 when $r=\bigl(\frac13 +o(1)\bigr) m$. To get more precision, find a first approximation to the actual error terms. The calculations become more difficult of course. ADDED. Now I found a bigger envelope and can explain how to get the full result at least for sufficiently large $m$. Having proved that the maximum is close to $m/3$, we will home in on it more accurately. Put $r=m/3+k$ where $k$ is small and $r$ is an integer. I'll leave out lots of details (often because I don't have them). First, $$ \binom{m}{r-i} = 2^{-i}\binom{m}{r}\biggl(1 - \frac{3i(3i-6k-1)}{4m} + \cdots\biggr),$$ where I think that is enough terms but I'm not sure (the next term is $O((i^4+i^2k^2)/m^2)$). Summing over $i\ge 0$, $$ f_m(r) = 2^{-r}\binom{m}{r}\biggl(2 + \frac{9k-12}{m} + O((1+k^2)/m^2)\biggr).$$ Dividing two values, $$ \frac{f_m(m/3+k+1)}{f_m(m/3+k)} = 1 - \frac{3(3k-1)}{2m} + O((1+k^2)/m^2). $$ Now consider the class of $m$ modulo 3. For $m=3M$, integer $M$, $$ \frac{f_m(M+k+1)}{f_m(M+k)}=\begin{cases} 1+\frac{1}{2M}+O(M^{-2}), & \text{ for $k=0$};\\ 1-\frac{1}{M}+O(M^{-2}), & \text{ for $k=1$}, \end{cases} $$ so the maximum occurs for $r=m/3+1$ provided the $O(M^{-2})$ is small enough to not interfere. The same phenomenon occurs if $m$ is 1 or 2 modulo 3 (use $k=-1/3,2/3$ for 1, $k=-2/3,1/3$ for 2). There is always a gap of $\Omega(m^{-1})$ that will dominate the $O(m^{-2})$ extras if $m$ is large enough. To turn this into a theorem, make explicit bounds on the error terms. It could be that easily obtained bounds don't work when $m$ is small, but hopefully that will overlap the computational range.<|endoftext|> TITLE: On which closed Riemannian manifolds are geodesics always recurrent? QUESTION [5 upvotes]: Let $M$ be a closed Riemannian manifold. What are the necessary and sufficient conditions on $M$ to ensure that for every point $p \in M$, and every geodesic $\gamma: [0, \infty) \to M$ with $\gamma(0) = p$, we have that $\liminf_{t \to \infty} d(\gamma(t), p) = 0$? Note: Here d denotes the usual Riemannian distance. REPLY [3 votes]: Just so that it doesn't get lost in the comments: in this paper Nadirashvili shows that a a $C^2$ Riemannian metric in the 2-torus for which the geodesic flow is recurrent (all points are Poisson stable) must be flat. N. S. Nadirashvili, “Conditions of stability in the sense of Poisson of a geodesic flow on a torus”, Mat. Zametki, 44:1 (1988), 147–149<|endoftext|> TITLE: Volume of a divisor on a smooth projective surface QUESTION [5 upvotes]: Let $X$ be a smooth projective surface (over complex numbers). Let $D$ be a divisor on $X$. Then we know that its volume is defined as $$\text{vol}_X(D):= \lim \sup_{m \rightarrow \infty} \frac{h^0(X, \mathcal O_X(mD))}{{m^2}/2}.$$ Suppose that, for a divisor $D$ on $X$, it is known that $\text{vol}_X(D)=D^2$, where $D^2$ stands for its self-intersection number. Question. What does this signify algebro-geometrically? And what are some interesting properties of $D$ that one can deduce once the volume is explicitly known? REPLY [7 votes]: At least for effective divisors, the answer is strongly related to Zariski decomposition. If $D$ is an effective divisor on a smooth surface $X$, Zariski proved in [Z62] that there exists a unique decomposition $D=P + N$, where $P$ is a nef $\mathbb{Q}$-divisor $N$ is an effective $\mathbb{Q}$-divisor $PC=0$ for every curve $C$ appearing in $\operatorname{Supp}(N)$ if $N \neq 0$ and $\operatorname{Supp}(N)=C_1 \cup \ldots \cup C_k$, then the intersection form $I(C_1, \ldots, C_k)$ is negative defined. Furthermore, one also shows that, for all positive integers $m$, the natural map $$H^0(X, \left \lfloor{mP}\right \rfloor) \to H^0(X, \, mD) $$ is an isomorphism. This means that "$P$ carries all sections of $D$" and so $$\operatorname{vol}_X(D)=\operatorname{vol}_X(P)= P^2,$$ where the last equality is a consequence of the asymptotic form of Riemann-Roch theorem, because $P$ is nef. Thus, $\operatorname{vol}_X(D)=D^2$ is equivalent to $D^2=P^2$; since $PN=0$, this happens if and only if $N^2=0$. But the intersection form must be negative defined on $N$ when $N \neq 0$, hence the only possibility is $N=0$, namely, $D=P$. Summing up: given an effective divisor $D$ on a smooth surface $X$, we have $\operatorname{vol}_X(D)=D^2$ if and only if $D$ is nef. More generally, the difference $$D^2-\operatorname{vol}_X(D)$$ equals the self-intersection $N^2$ of the negative part of the Zariski decomposition of $D$, so it can be seen as a measure of how much "$D$ fails to be nef". References. [Z62] O. Zariski The theorem of Riemann-Roch for high multiples of an effective divisor on an algebraic surface, Ann. Math. (2) 76, 560-615 (1962). ZBL0124.37001.<|endoftext|> TITLE: Quantization of normal distribution QUESTION [5 upvotes]: For $n\in\mathbb{N}$, denote by $\mathcal{Q}_n$ the set of all probability measures on $\mathbb{R}$ that are supported on at most $n$ points. Question: Is it known which element in $\mathcal{Q}_n$ is closest to the standard normal distribution with respect to the $p$-Wasserstein distance (for some $p \geq 1$)? I.e., if $\mathcal{N}$ is the standard normal distribution, can we calculate $\arg\min_{\mu \in \mathcal{Q}_n} W_p(\mu, \mathcal{N})$? Remarks: In particular, I am interested in specific optimizers for small $n$ (around $n=10$). If there are provably reliable numerical methods that can calculate the above quantizers, that would also answer my question. I found some papers in this direction, but they all appeared to look for quantizers of the form $\mu = \frac{1}{n}\sum_{i=1}^n \delta_{x_i}$, while I am really interested in the general case $\mu = \sum_{i=1}^n w_i \delta_{x_i}$ for $0 \leq w_i \leq 1$, $\sum_{i=1}^n w_i = 1$. REPLY [6 votes]: $\newcommand{\N}{\mathcal N}\newcommand{\vpi}{\varphi}$For $\mu=\sum_{i=1}^n w_i\delta_{x_i}$ with $0\le w_i\le 1$ and $\sum_{i=1}^n w_i=1$, we have the following (see e.g. page 4): \begin{align*} W_p(\mu,\N)^p&=\int_0^1 du\,|F^{-1}(u)-G^{-1}(u)|^p \\ &=S(x,u):=\sum_{j=1}^n\int_{u_j}^{u_{j+1}} du\,|x_j-g(u)|^p, \end{align*} where $F$ is the cdf of $\mu$, $G$ is the cdf of $\N$, $F^{-1}$ is the generalized inverse of $F$, $g:=G^{-1}$ is the inverse of $G$, $x:=(x_1,\dots,x_n)$, $-\infty TITLE: The Mumford-Tate conjecture QUESTION [6 upvotes]: The Mumford-Tate conjecture asserts that, via the Betti-étale comparison isomorphism, and for any smooth projective variety $ X $, over a number field $ K $, the $ \mathbb{Q}_{ \ell } $-linear combinations of Hodge cycles coincide with the $ \ell $-adic Tate cycles. Question. Would that mean that if the Hodge conjecture and the Tate conjecture hold, then the Mumford-Tate conjecture holds as well ? REPLY [11 votes]: Yes. Under the Hodge conjecture, the Hodge cycles are the algebraic cycles, so the $\mathbb Q_\ell$-linear combinations of Hodge cycles are the $\mathbb Q_\ell$-linear combinations of algebraic cycles. Under the Tate conjecture, the $\ell$-adic Tate classes are the $\mathbb Q_\ell$-linear combinations of algebraic cycles. So under the Hodge and Tate conjectures, these are both equal. This then implies that the identity component of the $\ell$-adic monodromy group is isomorphic over $\mathbb Q_\ell$ to the Mumford-Tate group, by a Tannakian argument. REPLY [5 votes]: One can deduce the Tate conjecture for every abelian variety which satisfies the Mumford-Tate and the Hodge conjecture, and vice versa: (MT) + (H) $\Leftrightarrow$ (T) see section 6 of A survey around the Hodge, Tate and Mumford-Tate conjectures for abelian varieties<|endoftext|> TITLE: Representations of products of symmetric groups QUESTION [10 upvotes]: I'm writing a paper and want to cite some references to efficiently prove that over any field $k$ of characteristic zero, every irreducible representation of a product of symmetric groups, say $$ S_{n_1} \times \cdots \times S_{n_p} $$ is isomorphic to a tensor product $\rho_1 \otimes \cdots \otimes \rho_p$ where $\rho_i$ is an irreducible representation of $S_{n_i}$. I have an open mind about this, but I'm imagining doing it by finding references for these two claims: If $k$ is an algebraically closed field of characteristic zero, every irreducible representation of a product $G_1 \times G_2$ of finite groups is of the form $\rho_1 \otimes \rho_2$ where $\rho_i$ is an irreducible representation of $G_i$. If $k$ has characteristic zero and $\overline{k}$ is its algebraic closure, every finite-dimensional representation of $S_n$ over $\overline{k}$ is isomorphic to one of the form $\overline{k} \otimes_k \rho$ where $\rho$ is a representation of $S_n$ over $k$. Serre's book Linear Representations of Finite Groups states the first fact for $k = \mathbb{C}$ but apparently not for a general algebraically closed field of characteristic zero. (It's Theorem 10.) It could be true already for any field of characteristic zero, which would simplify my life. The second fact should be equivalent to saying that $\mathbb{Q}$ is a splitting field for any symmetric group, which seems to be something everyone knows - yet I haven't found a good reference. REPLY [8 votes]: To summarize the situation given in the other answers (no real new content here) it is classical theory going back to Young that the complex irreducible representations of $S_n$ can be defined over $\mathbb Q$ (i.e., written with $\mathbb Q$-coefficients, or written as $\mathbb C\otimes_{\mathbb Q}V$ with $V$ a $\mathbb QS_n$-irreducible module) and references were given; i.e., the $\mathbb Q$-irreducibles are absolutely irreducible. This can be done via Young symmetrizers and anti-symmetrizers, polytabloids, or a number of other approaches and I have nothing to add to the discussion. What this means concretely is that $\mathbb QS_n\cong \prod_{i=1}^{p_n}M_{d_i}(\mathbb Q)$ where $p_n$ is the number of partitions of $n$ and $d_i$ is the dimension of the $i^{th}$-irreducible representations (and of course all these $d_i$ are well known through tableaux combinatorics and involve counting semi-standard Young tableaux). One way to see this is to use that if $V$ is a finite dimensional $\mathbb QS_n$-module, then $\mathrm{End}_{\mathbb CS_n}(\mathbb C\otimes_{\mathbb Q} V)\cong \mathbb C\otimes_{\mathbb Q}\mathrm{End}_{\mathbb QS_n}(V)$ by standard arguments and so by Schur's lemma, if $\mathbb C\otimes_{\mathbb Q}V$ is irreducible, then $\mathrm{End}_{\mathbb CS_n}(\mathbb C\otimes_{\mathbb Q} V)$ one-dimensional over $\mathbb C$ and hence $\mathrm{End}_{\mathbb QS_n}(V)$ is one-dimensional over $\mathbb Q$ and so apply Wedderburn-Artin to $\mathbb QS_n$ to get the statement. Now, let's just handle $\mathbb Q[S_n\times S_m]\cong \mathbb QS_n\otimes_{\mathbb Q}\mathbb QS_m$. Then by the above, we have that this tensor product is isomorphic to $$\prod_{i=1}^{p_n}\prod_{j=1}^{p_m}M_{d_i}(\mathbb Q)\otimes_{\mathbb Q} M_{c_j}(\mathbb Q)$$ where I introduced $c_j$ for the dimensions of the $S_m$-irreducibles over $\mathbb Q$. Obviously $$M_{d_i}(\mathbb Q)\otimes_{\mathbb Q}M_{c_j}(\mathbb Q)\cong M_{d_i}(M_{c_j}(\mathbb Q))\cong M_{d_ic_j}(\mathbb Q)$$ and hence $M_{d_i}(\mathbb Q)\otimes_{\mathbb Q}M_{c_j}(\mathbb Q)$ is simple with a unique simple module (up to isomorphism) which has dimension $d_ic_j$ (and this dimension characterizes the simple module). Consequently the tensor product of the unique simple modules of the two tensor factors is the unique simple module for this tensor product, e.g., by dimension consideration. Putting it all together, we get that the simple $\mathbb Q[S_n\times S_m]$-modules are the tensor products of the simple $\mathbb QS_n$-modules and $\mathbb QS_m$-modules. Now since $K\otimes_{\mathbb Q} M_r(\mathbb Q)\cong M_r(K)$ and $K\otimes_{\mathbb Q}\mathbb Q^r\cong K^r$, the situation doesn't change when we extend the scalars. We get the same number of irreducibles and they are obtained by extending the scalars from those of $\mathbb Q[S_n\times S_m]$. Of course the argument is the same for any finite number of factors. Tiny update. Although John didn't ask for this, it is also well known that the $p$-element field $\mathbb F_p$ is a splitting field for the symmetric group in characteristic $p$. From this one can again deduce that all irreducible representations of products of symmetric groups are tensor products of irreducible representations of the factors over any field. Likely there is a %100 direct proof using tableaux and the like. But here is one possible proof. First note that if $G$ is any finite group, $K$ is an algebraically closed field of characteristic $p$ and $|G|=p^nm$ with $\gcd(p,m)=1$, then one can show that each character $\chi$ of $G$ over $K$ takes values that are sums of $m^{th}$-roots of unity since $1$ is the only $p^{th}$-root of unity in $K$. Hence the character field of $\chi$ is a finite field. The theory of Schur indices together with Wedderburn's theorem that there are no finite division rings, then tells you that $\chi$ is realizable over the character field $\mathbb F_p(\chi)$. Moreover, a character of $G$ is well known to be determined by its values on the $p$-regular elements of $G$ (the elements of order prime to $p$). Now in $S_n$, every element $g$ of order prime to $p$ is conjugate to $g^p$. Hence $\chi(g)=\chi(g^p) = \Phi(\chi(g))$ where $\Phi$ is the Frobenius automorphism $x\mapsto x^p$. Therefore, $\chi$ takes values in $\mathbb F_p$ and so is realizable over $\mathbb F_p$.<|endoftext|> TITLE: The universal multiset for a finite scheme - reference request QUESTION [5 upvotes]: If $X$ is a finite set of size $n$, then by listing the elements of $X$ we get a canonical element of the symmetric power $X^n/\Sigma_n$, which we can call the universal multiset for $X$. Now let $X$ instead be an affine scheme $\text{spec}(A)$ over a base scheme $S=\text{spec}(k)$, and suppose that $A$ is a free (or just projective) module of rank $n$ over $k$. There is then a canonical section $S\to X^n/\Sigma_n$, which is strongly analogous to the universal multiset as described above. The corresponding ring map $\nu\colon(A^{\otimes n})^{\Sigma_n}\to k$ is almost characterised by the fact that $\nu(a^{\otimes n})=\text{norm}_{A/k}(a)$ for all $a\in A$. When I thought of this the other day, I was amazed by the fact that it had never occurred to me before and that I did not remember seeing it in the literature, although it is adjacent to things that I have thought about many times. I think that the closest thing that I have seen is the notion of a "full set of sections" as described in Section 1.8 of Katz and Mazur's "Arithmetic moduli of elliptic curves", but even that is not all that close. My question is: does this map $\nu$ appear in the literature, and if so, under what name? For completeness, here is the actual definition. Let $P$ be the graded symmetric algebra over $k$ generated by $P_1=A^\vee=\text{Hom}_k(A,k)$, and let $u\in P_1\otimes A$ correspond to the identity under the usual isomorphism $P_1\otimes A=A^\vee\otimes A=\text{Hom}_k(A,A)$. Put $\tilde{u}=\text{norm}_{(P\otimes A)/P}(u)\in P_n$. There is a standard perfect pairing between $P_n=(A^\vee)^{\otimes n}_{\Sigma_n}$ and $(A^{\otimes n})^{\Sigma_n}$, and we define $\nu(a)=\langle\tilde{u},a\rangle$. With a little work one can check that this is a ring homomorphism, and that it satisfies $\nu(a^{\otimes n})=\text{norm}_{A/k}(a)$. REPLY [6 votes]: I have seen the morphishm $\nu\colon (A^{\otimes n})^{\Sigma_n}\to k$ in a couple of places: On page 81 of A. Suslin, V. Voevodsky, Singular homology of abstract algebraic varieties It is defined when $A$ is finite locally free over $k$, but also when $k$ is a normal, $A$ is an integral domain, and $\operatorname{Spec}(A)\to\operatorname{Spec}(k)$ is finite and surjective. The latter construction is used there to show that symmetric powers of schemes represent finite correspondences in the sense of Voevodsky (the "motivic Dold-Thom theorem"), which is a standard result in motivic homotopy theory. It is used to construct the norm functor $N_{A/k}\colon \mathrm{Mod}_A \to \mathrm{Mod}_k$ in Section 3.1 of D. Ferrand, Un foncteur norme I suspect this construction also appears somewhere in David Rydh's thesis, which is related to both of the above papers, although I couldn't immediately locate it.<|endoftext|> TITLE: Exposition of concrete constructions QUESTION [8 upvotes]: I am frequently interested to find less technical proofs of results which already appear in the literature, at least in some special cases of these results. Sometimes a published proof shows that an object with some properties exists, but actually the proof does not (at least not without additional work) construct that object in a concrete way. (Here object is understood to be in a very general sense, so functors for instance also qualify as objects.) Imagine for example a theorem stating that a certain functor is representable, and the proof uses some adjoint functor theorem: this does not really tell us much about the internal structure of a representing object, and it is often (not always) very useful to write down that representing object directly, including a direct proof that it actually represents the functor which also does not use much machinery. I believe that it is worthwile to write down and maybe also publish a more direct, easy and constructive argument in case it is available and presumably not published so far. My question is basically how to fit this into the common "definition-theorem-proof" scheme of mathematical publications. I cannot just have a theorem stating "There is an object with the property ...", since that existence statement is already known and I am actually more interested in the specific construction of that object. Another solution might be to put the construction of that object into a separate section, or a long definition, and then have a theorem stating "The object constructed above has the property ...". Another idea might be to find logical properties of a "concrete construction" which the published existence proof does not have; but I have to admit that this task is nontrivial for someone who is not an expert in mathematical logic, and it might become a bit artificial, and it might turn out that these logical properties actually already follow formally from the published proof. Basically, what I want is a theorem "There is an explicit construction of an object with the property ..." and give the construction in its proof, but the problem is that "explicit construction" is not a well-defined notion, as far as I know, so that it is unclear what qualifies as a proof of it. Do you have other ideas or guidelines for a good exposition for a "concrete construction"? What are good examples of papers which have a good exposition of this type? REPLY [9 votes]: I think the following scheme is quite reasonable, which paraphrases your second option: Def. Object X is... Def. Property Y is... Thm. Object X has Property Y. Rmk. Object X is the first explicit example of Property Y. You might need to put in quite a bit of effort to prove the Theorem, or to construct of X itself. Moreover, the Remark you are making is more about the state of the literature, rather than a formal statement, which is of course allowed in a remark. If you can also motivate why having an explicit example is important, the nontrivial mathematical content and the motivation make a perfectly reasonable recipe for a paper.<|endoftext|> TITLE: A measurable set that acts as a speedometer QUESTION [5 upvotes]: Definitions and some motivation: Say a car is driving on a straight road. All we know is where it starts, and how much time it spends in certain stretches of the road. With just this much information, can we pinpoint the exact trajectory of the car? Translating this loosely to math - does there exist a measurable subset $S$ of $\mathbb R$ such that any absolutely continuous $f: [0, \infty) \to \mathbb R$ with $f(0) = 0$ is determined entirely by how much time it spends in $S$? More precisely, denote by $\mathcal C_0$ the subset of absolutely continuous, real valued functions $f$ on $[0, \infty)$ such that $f(0) = 0$. For any measurable subset $S$ of $\mathbb R$, and any $f \in \mathcal C_0$, define the function $T(f): [0, \infty) \to [0, \infty)$ by $T(f)(x) := \mu(\{t|\ t \in [0, x), f(t) \in S \})$. Question: Does there exist a measurable subset $S$ of $\mathbb R$ such that the operator on $\mathcal C_0$ sending $f$ to $T(f)$ is injective? Note: Here $\mu$ denotes the usual Lebesgue measure. REPLY [3 votes]: No, $T$ cannot be injective. If $S \cap (a, b)$ has zero Lebesgue measure, then $T$ will assign the same value to $f_1$ and $f_2$ if, say, $f_1(t) = f_2(t)$ for $t < 1$ and $f_1(t), f_2(t) \in (a, b)$ for $t \geqslant 1$, and $f_1'(t) \ne 0$ and $f_2'(t) \ne 0$ for $t \geqslant 1$. Suppose that $S \cap (a, b)$ has a positive Lebesgue measure for every interval $(a, b)$, and write $$\phi(x) = \int_0^x \mathbb 1_S(y) dy$$ (that is, $\phi(x) = \mu(S \cap [0, x))$ for $x \geqslant 0$ and $\phi(x) = -\mu(S \cap [x, 0))$ for $x < 0$). Then $\phi$ is strictly increasing, and therefore the inverse function $\psi = \phi^{-1}$ is well-defined. We claim that $\psi(t) \in S$ for almost all $t \in \mathbb R$. This shows that if $f_1(t) = \psi(t)$ and $f_2(t) = \psi(-t)$, then $T(f_1) = T(f_2)$. Thus, it remains to prove our claim. Observe that $$ t = \phi(\psi(t)) = \int_0^{\psi(t)} \mathbb 1_S(y) dy = \int_0^t \mathbb 1_S(\psi(s)) d\psi(s) . $$ Therefore, $\mathbb 1_S(\psi(s)) \ne 0$ for almost all $s$ with respect to $d\psi$. Furthermore, since $0 < \phi(y) - \phi(x) \leqslant y - x$ when $x < y$, we have $\psi(t) - \psi(s) \geqslant t - s$ when $s < t$. In particular, the Lebesgue measure is absolutely continuous with respect to $d\psi$. It follows that $\mathbb 1_S(\psi(s)) \ne 0$ for almost all $s$ also with respect to the Lebesgue measure, as claimed.<|endoftext|> TITLE: Is the elementary transformation of a conic bundle a flip or a flop QUESTION [7 upvotes]: Let $\pi: V\to S$ be a standard conic bundle of a threefold $V$ to a surface $S$, i.e., $\pi$ is relative minimal. Assume that everything is nonsingular and is over $\mathbb{C}$. We may assume that $V$ is embedded in a $\mathbb{P}^2$-bundle $\mathbb{P}(\mathcal{E})$ over $S$, where $\mathcal{E}$ is a rank $3$ vector bundle on $S$, and $V$ is defined as a zero of a section $\sigma\in H^0(\mathbb{P}(E),\mathcal{O}_{\mathbb{P}(E)}(2)\otimes\tau^*\mathcal{L})$, where $\mathcal{L}$ is an invertible sheaf on $S$ and $\tau$ is the standard projection of $\mathbb{P}(\mathcal{E})$ to $S$. Now suppose that $C$ is a curve on $V$ which is isomorphic to its image $\pi(C)$, such that $\pi(C)\cap\Delta=\varnothing$, where $\Delta$ is the degenerate divisor on $S$ for the conic bundle $(V, S,\pi)$, that is the locus of points whose fiber is a degenerate conic, disjoint of two lines or a double line. Then we can apply an elementary transformation to $V$, as described in this paper On conic bundle structures--V. G. Sarkisov, which first blows up $C$ then blows down the strict transform of $B=\pi^{-1}(\pi(C))$. The resulting variety is denoted by $V'$, and $C'$ is the strict transform of $C$. Question: Is this elementary transformation a flop or a flip in the sense of the MMP? I try to answer this question positively, in other words, does the intersection numbers hold $K_V.C<0$ and $K_{V'}.C'>0$? But I meet some trouble when I verify this inequality, by 1.5 of On conic bundle structures--V. G. Sarkisov, we know that $\mathcal{L}$ is isomorphic to $\omega_S\otimes\mathrm{det}(\mathcal{E})$, so $\omega_{\mathbb{P}(\mathcal{E})}\otimes\mathcal{O}_{\mathbb{P}(\mathcal{E})}(V)\cong\mathcal{O}_{\mathbb{P}(\mathcal{E})}(-1)$. Therefore the intersection number of $K_V.C$ should be equal to $(K_{\mathbb{P}(\mathcal{E})}+V).C=\mathrm{deg}_C\mathcal{O}_{\mathbb{P}(\mathcal{E})}(-1)$, as $C$ contained in $V$. We only konw that the divisor associate to $\mathcal{O}_{\mathbb{P}(\mathcal{E})}(-1)$ is relatively ample, but here the curve $C$ is not in a fiber, we can not conclude that $\mathrm{deg}_C\mathcal{O}_{\mathbb{P}(\mathcal{E})}(-1)<0$ directly. On the other hand, it seems that the situation for $(V',S,\pi')$ is the same as $(V,S,\pi)$. Since $(V',S,\pi')$ is also standard. So we have $K_{V'}.C'=\mathrm{deg}_{C'}\mathcal{O}_{\mathbb{P}(\mathcal{E})}(-1)$. Note that $B$ (resp. $B'=\pi'^{-1}(\pi'(C'))$) is a ruled surface over $\pi(C)$ (resp. $\pi'(C)$), if my question has a positive anwser, it seems that $C$ is the negative section of $B\to\pi(C)$ and $C'$ is a positive section of $B'\to \pi'(C')$, i.e., $C^2<0$ and $C'^2>0$. Let $W\to V$ be the blowing up along $C$, and let $\tilde{B}$ and $\tilde{B'}$ be the strict transforms of $B$ and $B'$ respectively. It seems that $\tilde{B}$ and $\tilde{B'}$ are guling along a positive section of $\tilde{B}$ with a negative section of $\tilde{B'}$ in $W$. But I don't know how to prove this. However, it may be false. REPLY [7 votes]: It is neither flip nor flop, because the exceptional locus both on $V$ and on $V'$ is a divisor.<|endoftext|> TITLE: Jones' theorem for non-simply-connected spaces? QUESTION [7 upvotes]: Let $X$ be a smooth manifold. Jones' theorem says that $H^\bullet(\mathcal{L}X)\cong HH_\bullet(\Omega^\bullet_X)$, where $\mathcal{L}X$ is the free loop space of $X$. Is there a modification of this theorem that works for $X$ which is not simply connected? Maybe when $\pi_1(X)$ is abelian? REPLY [2 votes]: One possibility to understand the difference between the cohomologies of $X$ and $LX$ is to add "fusion"-conditions on the loop space side. This works for general manifolds $X$, and the key word here is "transgression". For example, Kottke and Melrose have proved in https://arxiv.org/pdf/1309.7674.pdf an isomorphism $$\check H^k(X,A) \cong \check H^{k-1}_{lf}(LX,A), $$ where the index "lf" denotes a version of Cech cohomology adapted to loop spaces by including fusion.<|endoftext|> TITLE: How to prove there are exactly $8$ integer points on the elliptic curve $y^2 = x^3 + 17$ QUESTION [6 upvotes]: Consider the elliptic curve $y^2 = x^3 + 17$. I know that there are exactly $8$ integer points $(x,y)$ with $y>0$. But how do I prove it? Is there any specific approach to it or any proof for it? REPLY [14 votes]: Using the mathematical programming language Sage, we can run the E.integral_points() command to get a (proof verified) confirmation that there are only $8$ integral points. The points are $(-2,3)$, $(-1,4)$, $(2,5)$, $(4,9)$, $(8,23)$, $(43, 282)$, $(52, 375)$, and $(5234,378661)$. Note that some would call this $16$ points, since all of these points remain solutions to your curve when $(x,y)$ is replaced with $(x,-y)$. If you wish to learn more about the algorithm implemented by Sage, I do believe that they are transparent about their algorithms on their website. NOTE: This case is much easier than the general one, since there is a theorem of Stark (see Theorem 7.2 in Silverman's AEC) which states that there is an effectively computable constant $C_{\epsilon}$ for every $\epsilon$ such that all integral solutions of the curve $y^2=x^3+D$ satisfy $\log(\max\{x,y\})\leq C_{\epsilon}|D|^{1+\epsilon}$, so for $D=17$ there are really not too many values to check. The Hall-Lang conjecture states that this bound can be reduced to one of the form $|x|\leq C_{\epsilon}D^{2+\epsilon}$ but seeing as that is still a conjecture it cannot be used for finding rational points on your curve.<|endoftext|> TITLE: When does Haar measure decompose into products of such? QUESTION [5 upvotes]: We get a "nice" Haar measure on $G=SL(2,R)$ in Iwasawa coordinates $G=NAK$ as follows: $dg=dx {dy\over y^2} dk$. Here $N=\{ n_x\}$, $A=\{a_y\}$ and $K=SO(2)$. Note that $dg=dn\, da\, dk$ is a product of Haar measures on the three subgroups. Question: In what generality does this happen? If I have some nice Lie group $G$, and I have, say, a local bijection $H\times K\to G$ (in a neighborhood of $e$), is $dg=dh\, dk$? (No in general; e.g. $\bar N\times A\times N\to G=SL(2,R)$. Haar measure is a mess involving all three variables.) Thanks! REPLY [2 votes]: FYI, the answer to my question is this Stack Exchange post (h/t Stephen D Miller): https://math.stackexchange.com/a/2325533<|endoftext|> TITLE: Can the opposite of an elementary topos be an elementary topos? QUESTION [15 upvotes]: This question is not really about elementary topoi, it is much more about a category $(\mathcal{E}, \Omega)$ admitting a subobject classifier, or about a category with power objects, you can choose the context that inspires you the most. Of course, elementary topoi are the go-to example. Q: Can we prove, or provide a counter-example, that the opposite of such a category cannot have a subobject classifier, or have power objects, or be an elementary topos? Rem. It is well known that the opposite of a Grothendieck topos cannot be a Grothendieck topos, but the proof relies on the fact that the opposite of a locally presentable category cannot be locally presentable. This completely obscures the importance of $\Omega$, which is what I would like to know about. Rem. Even in the most trivial case, why do we know that $\mathsf{Fin}^\circ$, or $(\mathsf{Fin}^C)^\circ$ does not have a subobject classifier? Clarification. The title mentions elementary topoi because "category with a subobject classifier" would not sound as good. REPLY [24 votes]: The answer to the question in the title is no, assuming you want to exclude the trivial case of the terminal category. Let $E$ be an (elementary) topos whose opposite is also a topos. The initial object of a topos is strictly initial (any map into it is an iso), so the terminal object $1$ of $E$ is strictly terminal. Since there is a map from $1$ to the subobject classifier $\Omega$ of $E$, this implies that $\Omega$ is terminal. Hence every object of $E$ has precisely one subobject. In particular, $1$ has only one subobject. But the initial object $0$ of $E$ is a subobject of $1$. So $0 \cong 1$ in $E$, so $E$ is trivial.<|endoftext|> TITLE: How do I remember which power of the Lefschetz operator $L$ corresponds to the $k$th Primitive cohomology group? QUESTION [5 upvotes]: Let $X$ be a compact Kähler manifold with $L$ denoting the Lefschetz operator $L(\bullet) = \bullet \wedge \omega$. The primitive cohomology groups are defined, for $k \in \mathbb{N}$, by $$P^k(X, \mathbb{C}) = \ker(L^{n-k+1}: H^{k-1}(X, \mathbb{C}) \longrightarrow H^{2n-k+1}).$$ How do I remember that the $k$th primitive cohomology is the kernel of the $(n-k+1)$th power of the Lefschetz operator? REPLY [8 votes]: The formula you give for the primitive cohomology group is not correct. I think you meant to write $$P^k(X, \mathbb{C}) = \ker(L^{n-k+1}: H^{k}(X, \mathbb{C}) \longrightarrow H^{2n-k+2}(X, \mathbb C)).$$ What's going on here is that the hard Lefschetz theorem says that the map $L^{n-k}: H^k \to H^{2n-k}$ is an isomorphism (as is the map you wrote down, which is the same map shifted in $k$ by $1$). To get an interesting kernel, you want to take one more power of $L$ than that, or $L^{n-k+1}$. How do we remember that exactly $n-k$ powers gives an isomorphism? Well Poincare duality says $H^k $ and $H^{2n-k}$ have the same dimension, so it's very plausible that they are isomorphic. It takes $n-k$ powers of $L$ to get from $H^k$ to $H^{n-k}$. So: Think of Poincare duality, then remember the hard Lefschetz isomorphism, then shift by 1. (This is not the fastest route, but remembering how hard Lefschetz works is good for a number of other things.) Why shift only by $1$? We want to take only the most interesting cohomology classes, so we want our condition to be as strict as possible while still containing nonzero vectors. Since we're taking a kernel, we want to take the kernel of a map that just barely fails to be injective, so we shift by $1$ instead of a larger number. Why increase by $1$ and not decrease? The cohomology groups increase towards the center and decrease towards the end, so applying one fewer Lefschetz operator would give a map that is injective but usually not surjective, and applying one more Lefschetz operator would give a map that is surjective but usually not injective. We're looking at the kernel, so we want the surjective but not injective one. The other approach is to look via the representation theory of $SL_2$. The operator $L$ is one part of an action of the Lie algebra $\mathfrak{sl}_2$ on $\bigoplus_{i=0}^{2n} H^i ( X, \mathbb C)$. Specifically, it's a nilpotent element. We can break $\bigoplus_{i=0}^{2n} H^i ( X, \mathbb C)$ into a sum of irreducible representations, and think about how $L$ acts on each one. The $r+1$-dimensional representation of $SL_2$ is the sum of $r+1$ copies of $\mathbb C$, where $L$ sends each copy to the next one in the sequence. Viewed as cohomology classes, the degree of each copy is shifted from the last by $2$. Since representations of $SL_2$ are self-dual, they're symmetric around the central degree $n$. So the degrees are $$n-r, n+2-r, \dots, n+r-2, n+r.$$ The goal of the primitive cohomology is to detect a single representation. Specifically, in degree $k$, we want to detect the representation that just barely intersects degree $k$, which is why we take $r= n-k$ so that $n-r = k$. The other representations that contribute to cohomology in degree $k$ have larger values $r>n-k$. We want to choose a map that zeroes out or chosen representation and then take the kernel. $L^{r}$ is an isomorphism on lowest-degree piece of the $r$-dimensional representation, so we take $L^{r+1} = L^{n-k+1}$, which vanishes.<|endoftext|> TITLE: Does $\pm A \leq B$ imply that $B^{-1} A$ is bounded? QUESTION [8 upvotes]: Lately I have to use a lot of functional calculus. A question that keeps popping up and that I don't manage to resolve is the following: Let $A,B$ be self-adjoint (not necessarily bounded) operators such that $\pm A\leq B$. Is it true that $B^{-1} A$ is a bounded operator? In case this is false, would the result be implied by the stronger condition $A^2\leq B^2$ and $0\leq A\leq B$? REPLY [8 votes]: The first question has been answered by Michael Renardy, and the answer is no. The second answer should have a positive answer. To avoid any domain issues, let me explain rather why $A^2 \leq B^2$ implies that $A B^{-1}$ has norm $\leq 1$ (for bounded operators $B^{-1} A$ is the adjoint of $A B^{-1}$ so it also has norm $\leq 1$, but I never work with unbounded operators so not sure whether this is true always). For $\xi$ in the domain of $B$, if $\eta = B \xi$, $$\|A B^{-1} \eta \|^2 = \| A \xi\|^2 = \langle A^2 \xi,\xi\rangle \leq \langle B^2 \xi,\xi\rangle = \|\eta\|^2.$$ This shows that $A B^{-1}$ extends to a norm $\leq 1$ operator on the closure of the image of $B$, which (I guess since you write $B^{-1}$) is assumed to be the whole space. Another way to write the same proof is, for $A,B \geq 0$, $$ A^2 \leq B^2 \iff B^{-1} A B^{-1} \leq 1 \iff \|A B^{-1}\| \leq 1.$$ The first equivalence is just the fact that the operation $ X \mapsto B X B$ preserves positive operators.<|endoftext|> TITLE: Non-unital Russo-Dye Theorem QUESTION [5 upvotes]: Let $A$ be a C$^*$-algebra and let $\phi$ be a positive linear map from $A$ to $B(H)$ (bounded linear operators on Hilbert's space). If $A$ is unital, then the Russo-Dye Theorem implies that $\|\phi\|=\|\phi(1)\|$, from where it immediately follows that $$ \|\phi\| = \sup\big \{\|\phi(a)\|: a\geq 0,\ \|a\|\leq 1\big \}. \tag 1 $$ Question. Is (1) still valid in case $A$ is non-unital? REPLY [3 votes]: As I mentioned in the comments, I'm hoping to find a one-liner based on Cauchy-Schwartz, as suggested by user Mikael de la Salle. However, while this is not available, let me present a proof I just found based on extending $\phi$ to the unitization of $A$. Let $$ S = \sup\big \{\|\phi(a)\|: a\geq 0,\ \|a\|\leq 1\big \}, $$ and let $\Phi $ be the extension of $\phi$ to the unitization $\tilde A$ defined by setting $\Phi (1)=SI_H$. I claim that $\Phi $ is positive. To see this we must check that $$ \Phi \big ((a-\lambda 1)^*(a-\lambda 1)\big )\geq 0, \tag{$\star$} $$ for every $a$ in $A$, and every $\lambda \in {\mathbb C}$. The case $\lambda =0$ is clearly true, so we may assume that $\lambda \neq 0$. In the latter case, we may change variables by replacing $a$ with $\lambda a$, and then divide everithing by $|\lambda |^2$, leading to the following equivalent form of $(\star)$: $$ \Phi \big ((a-1)^*(a-1)\big )\geq 0, \tag{$\star\star$} $$ Observing that $$ 0\leq (a-1)^*(a-1) = a^*a-a^*-a+1, $$ and fixing an approximate identity $\{u_i\}_i$ for $A$, we have for all $i$ that $$ u_i(a^*+a -a^*a)u_i \leq u_i^2, $$ so $$ \phi\big (u_i(a^*+a -a^*a)u_i\big ) \leq \phi(u_i^2) \leq \|\phi(u_i^2)\|I_H \leq SI_H = \Phi (1). $$ Taking the limit as $i\to \infty $, the above yields $$ \phi (a^*+a -a^*a) \leq \Phi (1), $$ which is equivalent to $(\star\star)$, proving the claim. By [1, Theorem 1.3.3], (which Størmer proves using Russo-Dye and Cauchy-Schwartz), we then deduce that $$ S = \|\Phi (1)\| = \|\Phi \| \geq \|\phi\| \geq S, $$ concluding the proof. [1] Størmer, Erling, Positive linear maps of operator algebras, Springer 2013.<|endoftext|> TITLE: Malgrange preparation theorem with less regularity QUESTION [8 upvotes]: (This question was previously posted on MSE and I decided to post it here too.) I am studying the proof of the Malgrange preparation theorem given in the book "Stable mappings and their singularities" written by Golubitsky and Guillemin (see Chapter IV). The statement is the following. Let $F\in C^\infty(\mathbb{R}\times\mathbb{R}^n; \mathbb{R})$ be defined on a neighborhood of $0\in\mathbb{R}\times\mathbb{R}^n$ and such that $F(t, 0) = g(t)t^k $ where $g(0)\neq0$ and $g$ is $C^\infty$ and defined on a neighborhood of $0\in\mathbb{R}$. Then there exist $C^\infty$ functions $q$, $\lambda_0$, ..., $\lambda_{k-1}$ such that $(qF)(t, x) = t^k + \sum_{i=0}^{k-1}\lambda_i(x)t^i$ on a neighborhood of $0\in\mathbb{R}\times\mathbb{R}^n$ and $q(0)\neq0$. After a careful reading of the proof, it seemed to me that the result is preserved even if the assumption "$F\in C^\infty(\mathbb{R}\times\mathbb{R}^n; \mathbb{R})$" is replaced by "$F(t, \cdot)$ is $C^1$ for all $t$ and $F(\cdot, x)$ is $C^\infty$ for all $x$". (Obviously, the regularity of $q$, $\lambda_0$, ..., $\lambda_{k-1}$ must also be weakened in the same manner.) Indeed, all complex analysis arguments (inspired by Weierstrass preparation theorem) are used on functions depending on $t$, while only the implicit function theorem is used on a function depending on $x$. My question is: Does anyone know a reference where this less regular version of the Malgrange preparation theorem is stated/proved/discussed? Thank you. REPLY [11 votes]: Edit: Here's a better reference, which actually states and proves more clearly the version that you are looking for. This one is due to Peter Michor (who is also on MO and may drop by?) P. MICHOR, The division theorem on Banach spaces, Österrich. Akad. Wiss. Math.- Natur. Kl. Sitzungsber II,189 (1980), pp. 1–18. (link from Michor's website) I think if you apply his theorem with $E$ trivial, $F = C^1(\mathbb{R}^n)$, you should get what you are looking for. See Pierre Milman, "The Malgrange-Mather Division Theorem", Topology 16:395-401 (1977) End of first paragraph reads: In this paper we give a short, rather elementary proof of the Malgrange-Mather Division Theorem, which applies to analytic, $C^\infty$ or $C^m$ functions.<|endoftext|> TITLE: Modern proofs for simplicial localizations QUESTION [10 upvotes]: I know that the references usually regarded as standard for simplicial localizations are the Dwyer and Kan's three articles from the 80's. I would be interested in a more modern approach to the subject, and in particular the following two issues keep surfacing in my work, for which I'd appreciate to be able to cite something that is more compact and student-friendly than the original works. Define the $\infty$-category $L(\mathcal{C}, \mathcal{W})$ associated to a relative category $(\mathcal{C}, \mathcal{W})$ as the homotopy pushout of the span $\coprod_{\mathcal{W}} J \leftarrow \coprod_{\mathcal{W}} \Delta^1 \rightarrow N(\mathcal{C})$ in $\textbf{sSet}$ (or anything equivalent to it). I would like to prove that for a simplicial model category $\underline{\mathbf{M}}$ then there is an equivalence of $\infty$-categories $$ \text{N}_{\Delta}(\underline{\mathbf{M}}^{cf}) \simeq L(\mathbf{M}, \mathcal{W}) $$ where $\mathbf{M}$ is the underlying (unenriched) category of $\underline{\mathbf{M}}$. Ideally, I would like to avoid the three above mentioned articles, preferring instead later developed and thoroughly studied technologies such as the equivalence between Joyal and Bergner model structures on $\mathbf{sSet}$ and $\mathbf{sCat}$, or that between Barwick-Kan and Rezk model structures on $\mathbf{RelCat}$ and $\mathbf{ssSet}$, or the like. I suspect that, for an arbitrary relative category, there should be an equivalence $$\text{N}_{\Delta}(L^H(\mathcal{C}, \mathcal{W})) \simeq L(\mathcal{C}, \mathcal{W})$$ where $L^H$ denotes the hammock localization, and $L$ is the associated $\infty$-category as defined above, but I can't find a proof of this. Again, a modern approach using high technology is preferable to the original paper where the hammock localization was introduced. REPLY [7 votes]: For Question 1. It is documented in Corolary 4.2.4.8 in Lurie's Higher Topos theory that $N_\Delta(\underline{M}^{cf})$ is an $\infty$-category with small limits (and small colimits). Moreover, the inclusion $M^{cf}\subset \underline{M}^{cf}$ induces a functor $$(*)\qquad L(M,W)\cong L(M^{cf},W\cap M^{cf})\to N_\Delta(\underline{M}^{cf})\, .$$ The homotopy $1$-category of the $\infty$-category $N_\Delta(\underline{M}^{cf})$ is the ordinary $1$-localization of $M^{cf}$ by weak equivalences. This means that the comparison functor above induces an equivalence on homotopy categories. On the other hand, since, by virtue of Proposition 7.7.4 of Higher categories and homotopical algebra, $L(M,W)$ also has small (co)limits, it is sufficient to prove that the functor $(*)$ above commutes with finite limits, by Theorem 6.7.10 from loc. cit. The fact that the terminal object is preserved is obvious. Using Proposition 7.5.6 (as well as Theorem 7.5.18) from loc. cit., this amounts to check that pullback squares along fibrations between fibrant objects do provide pulback squares in $N_\Delta(\underline{M}^{cf})$, which follows from Theorem 4.2.4.1 in Higher Topos theory. In fact, with the arguments above, one may characterize $L(M,W)$ as follows: a finitely complete $\infty$-category $M_\infty$ equipped with a functor $\gamma: N(M)\to M_\infty$ is the localization of $M$ by $W$ if and only if the following conditions are verified: $\gamma$ sends any element of $W$ to an invertible map in $M_\infty$; the induced functor $M\to ho(M_\infty)$ is the $1$-localization of $M$ by weak equivalences; the functor preserves terminal objects and sends any pullback square of fibrant objects along fibrations in $M$ to pullback squares in $M_\infty$. (the latter characterization can be extended to all kind of variations on model structures that I know of: semi-model structures, categories of fibrant objects, and so on). For Question 2. I insist that checking that the hammock localization of Dwyer and Kan has the universal property of $L(C,W)$ really is obvious from the original construction via the Quillen equivalence relating relative categories, simplicial categories, and quasi-categories, just by reading the original sources. But there are documented more recent references such as Hinich's Dwyer-Kan localization revisited, for instance.<|endoftext|> TITLE: Explicit construction of division algebras of degree 3 over $\mathbb{Q}$ QUESTION [6 upvotes]: In his book Introduction to arithmetic groups, Dave Witte Morris implicitly gives a construction of central division algebras of degree 3 over $\mathbb{Q}$ in Proposition 6.7.4. More precisely, let $L/\mathbb{Q}$ be a cubic Galois extension and $\sigma$ a generator of its Galois group.If $p \in \mathbb{Z}^+$ and $p \neq t\sigma(t)\sigma^2(t)$ for all $t \in L$, then $$ D=\left\{ \begin{pmatrix} x & y & z\\ p\sigma(z) & \sigma(x) & \sigma(y)\\ p\sigma^2(y) & p\sigma^2(z) & \sigma^2(x) \end{pmatrix} :(x,y,z)\in L^3 \right\} $$ is a division algebra. On page 145, just before Proposition 6.8.8, Morris claims that it is known that every division algebra of degree 3 arises in this manner. This should follow from the fact that every central division algebra of degree 3 is cyclic. I could not find this explicit construction in my references (e.g. Pierce - Associative Algebras, though maybe I missed something) and I would like to know if there is a reference or a quick way to see that this exhausts all central division algebras of degree 3 over $\mathbb{Q}$. REPLY [4 votes]: Looking more carefully in Pierce - Associative algebras, I found the answer I was looking for, which I'm going to describe here for future reference. The algebra $D$ in the question is a realisation of a special type of crossed product algebra, which gives precisely the central division algebras of degree 3 over $\mathbb{Q}$. The first step is to use the powerful result mentioned in the comments, which states that all central simple algebras over number fields are cyclic. For degree 3 algebras, this was proved by Wedderburn. Recall that an algebra $A$ over a field $F$ of degree $n$ is cyclic if there is a cyclic Galois extension $E/F$ of degree $n$ such that $E$ is a maximal subfield inside $A$. Proposition a in Pierce, section 15.1, states that a cyclic algebra $A$ as above, where $\sigma$ generates the Galois group of $E/F$, contains an invertible element $u$ such that $A=\bigoplus_{0 \leq j TITLE: Analytical origins of the Stone duality QUESTION [12 upvotes]: I've asked this question in the HSM community, but by the nature of my question, some user told me to ask this question here. This is the original post https://hsm.stackexchange.com/q/13087/14296 What I want to know is how M.H.Stone, by his research in functional analysis came to the Stone duality. My knowledge of funtional analysis is not very advanced, what I have done is only a semester of bounded operators, that's my formation in it, so please if you can explain your answer a little bit, it would be great. What I have been told (in the cited post) is that: Hermitian projection operators that commute with a given self-adjoint operator (or a family thereof) form a Boolean algebra Also that: It plays a key role in the spectral representation of the operator in terms of projection-valued measures and the related functional calculus What I want to also know is the technical relations between Hilbert space spectral theory and Stone spaces. If you consider my tags to be wrong, please tell me. Also if you consider that this post, should be in Mathematics Stack-Exchange. REPLY [9 votes]: In addition to the historical component, the question also asked about the relation between spectral theory and Stone spaces. $\def\Z{{\bf Z}} \def\R{{\bf R}} \def\C{{\bf C}} \def\Spec{\mathop{\rm Spec}} \def\Proj{\mathop{\rm Proj}} \def\Cont{{\rm C}} \def\MSpec{\mathop{\rm MSpec}} \def\Li{{\rm L}^∞}$ Given an algebra-like object $A$, we assign to it its poset of ideals (typically defined as kernels of homomorphisms $A→B$), which is interpreted as the poset of opens of some space $S$. The technical term for such posets is locale, which is a notion very closely related to topological spaces. In particular, from any locale one can canonically extract a topological space, and this is the topological space $S$ produced in many classical Stone-type dualities. The points of $S$ are ideals corresponding to morphisms $A→k$, where $k$ is often a particularly simple algebra. These often turn out to be maximal ideals in $A$. Conversely, given a space-like object $S$, we assign to it the algebra of morphisms $S→k$, where $k$ is often the “same” algebra $k$ as above, only this time its underlying object is a space, not just a set. Some examples from general topology, measure theory, differential geometry, algebraic geometry, and complex geometry (the list is very much incomplete): algebra homomorphism $k$ ideal space maps Boolean algebra homomorphism $\Z/2$ ideal compact totally disconnected Hausdorff continuous map complete Boolean algebra complete homomorphism $\Z/2$ closed ideal compact extremally disconnected Hausdorff open continuous map localizable Boolean algebra complete homomorphism $\Z/2$ closed ideal hyperstonean space open continuous map localizable Boolean algebra complete homomorphism $\Z/2$ closed ideal compact strictly localizable enhanced measurable space measurable map commutative von Neumann algebra normal *-homomorphism $\C$ closed *-ideal compact strictly localizable enhanced measurable space measurable map commutative unital C*-algebra *-homomorphism $\C$ closed *-ideal compact Hausdorff space continuous map commutative algebra over $k$ homomorphism $k$ ideal coherent space / affine scheme continuous map / morphism of schemes finitely generated germ-determined C$^∞$-ring C$^∞$-homomorphism $\R$ germ-determined ideal smooth locus (e.g., smooth manifold) smooth map finitely presented complex EFC-algebra EFC-homomorphism $\C$ ideal globally finitely presented Stein space holomorphic map The duality relevant to the spectral theory is the duality between commutative von Neumann algebras and compact strictly localizable enhanced measurable spaces. Given a normal operator $T$ on a Hilbert space $H$, $T$ generates a commutative von Neumann algebra $A$ inside $B(H)$, i.e., bounded operators on $H$. (This is precisely the point where normality is crucial; without the relation $T^*T=TT^*$ the algebra generated by $T$ will be noncommutative.) More concretely, $A$ can be described as the closure in the ultraweak topology on $B(H)$ of the set of polynomials with complex coefficients in variables $T$ and $T^*$, i.e., finite sums $∑_{i,j}a_{i,j}T^i(T^*)^j$ for arbitrary $a_{i,j}∈{\bf C}$. By the cited duality, the commutative von Neumann algebra $A$ is dual to a compact strictly localizable enhanced measurable space $\Spec A$. This is indeed the spectrum of $T$ in the usual sense. Under this equivalence, the element $T∈A$ corresponds to the measurable map $\Spec A→\C$ given by inclusion of $\Spec A$ into $\C$. More concretely, we can extract all projections from $A$ (defined as self-adjoint idempotents, $T^2=T$ and $T^*=T$), and these form a complete Boolean algebra $\Proj A$. The easiest way to see this is to observe that Boolean algebras are precisely rings in which $x^2=x$. The ring identities are trivially satisfied by definition of projections. Completeness is implied by the fact that $A$ is closed in the ultraweak topology. The Stone duality converts the Boolean algebra $\Proj A$ into a topological space, its Stone spectrum $\Spec A$. Points in the Stone spectrum are precisely the maximal ideals $P$ in the Boolean algebra $\Proj A$. Open sets correspond to ideals $I$ of $A$: a point $P$ belongs to the open set corresponding to $I$ if $P$ does not contain $I$ as an ideal. In fact, the topological space $\Spec A$ also coincides with the Gelfand spectrum of $A$, interpreted as a commutative unital C-algebra; so points in $\Spec A$ can also be interpreted as maximal closed -ideals $I$ in $A$ as a commutative C-algebra, or, equivalently, as homomorphisms of unital C-algebras $A→\C$. Furthermore, we have a canonical isomorphism of von Neumann algebras $$R\colon A→\Cont(\Spec A,\C),$$ where $\Cont(\Spec A,\C)$ denotes the algebra of continuous complex-valued functions on $\Spec A$. Concretely, given an element $a∈A$ and a point $p∈\Spec A$ given by the homomorphism $A→\C$ of C*-algebras, we set $T(a)(p)=p(a)$. This is the Gelfand transform of $a$. The topological space $\Spec A$ belongs to a very special class of topological spaces, the hyperstonean spaces. From such a space one can extract a σ-algebra $M$ of measurable sets and a σ-ideal $N$ of negligible sets (alias sets of measure 0), both on the set $\Spec A$. The resulting triple $\MSpec A=(\Spec A,M,N)$ is an example of an enhanced measurable space. The elements of $N$ are precisely the nowhere dense subsets of $\Spec A$, which coincides with meager subsets (alias sets of first category). The elements of $M$ are precisely the symmetric differences of clopen (closed and open) subsets of $\Spec A$ and elements of $N$. (This is sometimes referred to as the Loomis–Sikorski construction for $\Spec A$; see John C. Oxtoby's book Measure and Category for more on this topic.) Continuing the above line of reasoning, we have a canonical isomorphism of von Neumann algebras $$S\colon A→\Li(\MSpec A,\C),$$ where $\Li(\MSpec A,\C)$ denotes the algebra of equivalence classes of complex-valued measurable functions $\MSpec A→\C$ modulo the equivalence relation of equality on a conegligible set (alias equality almost everywhere). (Indeed, one can prove that $\Cont(\Spec A,\C)$ is canonically isomorphic to $\Li(\MSpec A,\C)$.) Under this correspondence, the original operator $T∈B(H)$ corresponds to an element $S(T)∈\Li(\MSpec A,\C)$. The map $R(T)\colon \Spec A→\C$ allows us to interpret ponts of $\Spec A$ (and therefore also of $\MSpec A$) as complex numbers. This identifies $\MSpec A$ with the usual spectrum of the operator $T∈B(H)$. Furthermore, the isomorphism $$S\colon A→\Li(\MSpec A,\C),$$ is known as the Borel functional calculus of $T∈B(H)$. More precisely, given a bounded Borel-measurable function $f\colon \C\to\C$, the element $S^{-1}(f\circ R(T))∈A⊂B(H)$ is precisely the operator $f(T)∈B(H)$ given by the traditional Borel functional calculus. One may ask whether we can recover the full spectral theorem for a normal operator in this manner. This is possible once Stone duality is upgraded to Serre–Swan-type duality between modules and vector bundle-like objects (including, e.g., sheaves etc.). Given a vector bundle-like object $V→S$, we assign to it its module of sections, which is a module over the algebra of maps $S→k$. Conversely, given a module $M$ over $A$, the corresponding vector bundle-like object $V→S$ over $S=\Spec A$ has as its fiber over some point $s∈S$ the vector space $M/IM$, where $I$ is the ideal corresponding to $s$. (Many details are necessarily omitted in this brief sketch.) Typically, genuine vector bundles correspond to dualizable modules (dualizable with respect to the tensor product over $A$). Non-dualizable module tend to correspond to sheaves that are not vector bundles, e.g., skyscraper sheaves etc. module vector-bundle-like object module over a Boolean algebra sheaf of $\Z/2$-vector spaces Hilbert W*-module over a commutative von Neumann algebra measurable field of Hilbert spaces representations of a commutative von Neumann algebra on a Hilbert space measurable field of Hilbert spaces Hilbert C*-module over a commutative unital C*-algebra continuous field of Hilbert spaces module over a commutative algebra over $k$ sheaf of modules over an affine scheme dualizable module over a commutative algebra over $k$ algebraic vector bundle dualizable module over a finitely generated germ-determined C$^∞$-ring smooth vector bundle dualizable module over finitely presented complex EFC-algebra holomorphic vector bundle The duality relevant to the spectral theory is the duality between representations of a commutative von Neumann algebras on a Hilbert space and measurable fields of Hilbert spaces. Given a normal operator $T$ on a Hilbert space $H$, $T$ generates a commutative von Neumann algebra $A$ inside $B(H)$, whose spectrum $\Spec A$ is a compact strictly localizable enhanced measurable space. Furthermore, the inclusion of $A$ into $B(H)$ is a representation of $A$ on $H$. As such, it corresponds under the Serre–Swan-type duality to a measurable field of Hilbert spaces over $A$. This is precisely the measurable field produced by the classical spectral theorem. Under the duality, the operator $T$ corresponds to the operator that multiplies a given section of this measurable field of Hilbert spaces by the complex-valued function $\Spec A→\C$ produced above. Thus, we recovered the entire content of the classical spectral theorem. In fact, the above considerations work equally well to establish the spectral theorem for an arbitrary family (not necessarily finite) of commuting normal operators. See also the nLab article duality between algebra and geometry, which may contain additional updates.<|endoftext|> TITLE: A specific coset decomposition of $\mathrm{GL}_n(\mathbb{C})$ QUESTION [11 upvotes]: Disclaimer: I am a theoretical chemist (not a mathematician). I have tried asking this question at Math SE with no luck (https://math.stackexchange.com/questions/4080696/a-specific-coset-decomposition-of-mathrmgl-n-mathbbc). I am reading an old paper [1] where they introduce (without proof) a specific decomposition of a unitary matrix. I would like (if possible) to generalize this decomposition to complex, invertible matrices. The claim I would like to prove goes as follows: Let $\mathbf{M} \in \mathrm{GL}_n(\mathbb{C})$ be an element of the general linear group (i.e. an $n \times n$, complex, invertible matrix). Then $\mathbf{M}$ can be written as \begin{equation} \tag{1} \mathbf{M} = \exp(\mathbf{m}) = \exp(\mathbf{m'}) \exp(\mathbf{m''}) \end{equation} where $\mathbf{m} \in \mathfrak{gl}_n(\mathbb{C})$ is an element of the general linear Lie algebra (i.e. an $n \times n$, complex matrix). The matrices $\mathbf{m}'$ and $\mathbf{m}''$ are $n \times n$, complex, block-matrices of the form \begin{align} \mathbf{m}' &= \begin{bmatrix} \mathbf{0}'_{00} & \mathbf{m}'_{01} & \mathbf{m}'_{02} \\ \mathbf{m}'_{10} & \mathbf{0}'_{11} & \mathbf{m}'_{12} \\ \mathbf{m}'_{20} & \mathbf{m}'_{21} & \mathbf{0}'_{22} \end{bmatrix}\tag{2} \\ \mathbf{m}'' &= \begin{bmatrix} \mathbf{m}''_{00} & \mathbf{0}''_{01} & \mathbf{0}''_{02} \\ \mathbf{0}''_{10} & \mathbf{m}''_{11} & \mathbf{0}''_{12} \\ \mathbf{0}''_{20} & \mathbf{0}''_{21} & \mathbf{m}''_{22} \end{bmatrix} \tag{3}. \end{align} The diagonal blocks are square and of matching dimensions ($\mathbf{0}'_{00}$ has the same dimensions as $\mathbf{m}''_{00}$, say $n_0 \times n_0$, and so on). In essence, $\mathbf{m}'$ has zero blocks on the diagonal while $\mathbf{m}''$ is block-diagonal. I realize that the number of blocks is not essential for the problem; I'm using three by three blocks for illustration. I know that the exponential map of the general linear group is surjective, meaning that for every $\mathbf{M} \in \mathrm{GL}_n(\mathbb{C})$ there exists some $\mathbf{m} \in \mathfrak{gl}_n(\mathbb{C})$ such that $\mathbf{M} = \exp(\mathbf{m})$. This is standard group theory. I can also see that block-diagonal matrices of the form (3) form a Lie algebra, say $\mathfrak{b}_n(\mathbb{C})$, which generates a Lie group of $n \times n$, complex, invertible, block-diagonal matrices, say $\mathrm{B}_n(\mathbb{C})$, which is a subgroup of $\mathrm{GL}_n(\mathbb{C})$. The paper suggests that the factorisation in (1) should be viewed as a coset decomposition. As far as I understand, the group $\mathrm{GL}_n(\mathbb{C})$ is the union of the left cosets \begin{equation} g \, \mathrm{B}_n(\mathbb{C}) = \{ g \, h \;|\; h \in \mathrm{B}_n(\mathbb{C}) \}, \quad g \in \mathrm{GL}_n(\mathbb{C}). \end{equation} I also know that some cosets may be identical so one doesn't need all cosets in order get the whole group (so to speak). What I don't understand is how to prove or disprove the specific form of the matrix $\mathbf{m}'$. Following a comment on my original question on Math SE, we could ask more generally if the general linear Lie group is a product of exponentials of a Lie subalgebra and the complementary vector space of that subalgebra. Any help or references would be much appreciated. [1] J. Linderberg and Y. Öhrn, Int. J. Quantum Chem. 12(1), 161–191 (1977). State vectors and propagators in many-electron theory. A unified approach. EDIT: Any ideas for the less general unitary case are also very welcome. REPLY [7 votes]: Let $$\mathfrak g = \left\{ \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix} \mid a,b \in \mathbb C\right\}, \qquad \qquad \mathfrak p = \left\{ \begin{bmatrix} 0 & b \\c & 0\end{bmatrix} \mid b,c \in \mathbb C\right\}.$$ Then for $x = \begin{bmatrix} 0 & b \\ c & 0 \end{bmatrix} \in \mathfrak p$, $$\exp(x) = 1 + \begin{bmatrix} 0 & b \\ c & 0 \end{bmatrix} + \frac{1}{2} \begin{bmatrix} bc & 0 \\ 0 & bc \end{bmatrix} + ...$$ Note that every odd power of $x$ is zero on the diagonal, while every even power of $x$ has equal diagonal elements. Hence, the diagonal entries of $exp(x)$ are equal. This shows that if $y \in \mathfrak g$, then the two diagonal entries of $\exp(x)\exp(y)$ are either both zero or both nonzero. But there are $A \in GL_2(\mathbb C)$ not of this form, for instance $$ A = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}.$$ Thus $A$ (for example) is not of the form $\exp(x)\exp(y)$ for $(x,y) \in \mathfrak g \oplus \mathfrak p$. This is at least a counterexample for $(m_1,m_2,m_3) = (1,1,0)$.<|endoftext|> TITLE: Isoperimetric type inequality in $\mathbb{R}^2$ QUESTION [7 upvotes]: Fix $L \in (0,\infty)$ and consider $\mathcal{C}_L$ defined as follows: \begin{align*} \mathcal{C}_L := \{ \gamma:[0,1] \rightarrow \mathbb{R}^2 |~ \gamma \text{ is smooth and length($\gamma$)$=L$ }\}. \end{align*} I am interested in the "blow-up" of $\gamma$, denoted $\gamma_{+r}$, defined as follows: For any set $S \subseteq \mathbb{R}^2$ and $r>0$ \begin{align*} S_{+r} := \cup_{z\in S}(z+r\mathbb{D}), \end{align*} where $\mathbb{D}$ is the unit disc in $\mathbb{R}^2$ which is centred at the origin. So $\gamma_{+r}$ is a bounded open set in $\mathbb{R}^2$. My question is for which $\gamma \in \mathcal{C}_L$ is $m(\gamma_{+r})$ maximised? Here $m(\cdot)$ is the Lebesgue measure in $\mathbb{R}^2$. I feel that it should be maximised by the line segment with length $L$. If this is a version of some well known result, please do indicate it. The reason for this title is that sometimes the isoperimetric inequality in $\mathbb{R}^2$ is stated as follows: For any Borel subset $A \subseteq \mathbb{R}^2$ with $m(A) < \infty$ and for every $\epsilon >0$, we have $m(A_{+\epsilon}) \geq m(B_{+\epsilon})$. Here $B$ is a Euclidean ball with $m(A) = m(B)$. Thanks! REPLY [5 votes]: The problem of determining the volume of the tubular neighborhood of a submanifold of Euclidean spaces was studied by Hotelling and Weyl (and then others). See https://www.jstor.org/stable/2371513?seq=1 The amazing result of Weyl is that, as long as the tubular neighborhood has no "overlaps" (in the sense of non-injective normal exponential map) the volume of the tubular neighborhood is independent of (isometric) embedding. (So in the one dimensional case the maximizer is non-unique, since all embeddings are isometric to each other.) His theorem applies when the submanifold is closed (without boundary); but in your setting of a curve, the end-caps can be explicitly added.<|endoftext|> TITLE: Probability of $\ell_1$-norms of vertices of the rotated Hamming cube QUESTION [12 upvotes]: Let $O$ be a $d$-dimensional rotation matrix (i.e., it has real entries and $OO^T = O^TO = I$). Let $\mathbf{x}$ be a uniformly random bitstring of length $d$, i.e., $\mathbf{x} \sim U(\{0,1\}^d)$. In other words, $\mathbf{x}$ is a vertex of the Hamming cube, selected uniformly at random. I would like to show that there exists a $C > 0$ such that $$\mathbb{P}\left[\|O\mathbf{x}\|_1 \leq \frac{d}{4}\right] \leq 2^{-Cd}.$$ I am horribly stuck, any ideas on how to approach this problem would be very much appreciated. Below are some of my own attempts. This question is cross-posted at math stack exchange here. Observation 1: If $O = I$, then the statement holds. If $O = I$, then $\|O\mathbf{x}\|_1 = \|\mathbf{x}\|_1$ is simply the number of ones in the bitstring. Among the $2^d$ choices for $\mathbf{x}$, the number of choices that satisfies $\|\mathbf{x}\|_1 \leq d/4$ is $$1 + \binom{d}{1} + \binom{d}{2} + \cdots + \binom{d}{\lfloor d/4\rfloor} \leq 2^{dH(\lfloor d/4\rfloor/d)} \leq 2^{dH(1/4)},$$ hence the probability is upper bounded by $2^{-d(1-H(1/4))}$. Here, $H(\cdot)$ is the binary entropy function, i.e., $H(p) = -p\log_2(p) - (1-p)\log_2(1-p)$. Observation 2: Numerical experiments support this result. Below is a plot of the probability versus the dimension, where $O$ is selected at random: The blue line is the probability. The orange line is the bound derived in the case where $O = I$. For comparison, here is the same numerical experiment, but with $O = I$: Thus, it appears that the introduction of $O$ decreases the probability. Both plots are obtained by sampling $100000$ $\mathbf{x}$'s at random. The code is here: import numpy as np import matplotlib.pyplot as plt import random from scipy.stats import ortho_group H = lambda p : -p * np.log2(p) - (1-p) * np.log2(1-p) C = 1 - H(1/4) print(C) N = 100000 ds,Ps = [],[] for d in range(2,40): O = ortho_group.rvs(dim = d) # O = np.eye(d) P = 0 for _ in range(N): x = random.choices(range(2), k = d) if np.linalg.norm(O @ x, ord = 1) <= d/4: P += 1/N print(d,P) ds.append(d) Ps.append(P) fig = plt.figure() ax = fig.gca() ax.plot(ds, Ps) ax.plot(ds, [2**(-C*d) for d in ds]) ax.set_yscale('log') ax.set_xlabel('d') ax.set_ylabel('P') plt.show() REPLY [4 votes]: Here is an attempt to the problem for a worst-case $O$, with worse constants. So fix $O$, letting $o_i$ denote its $i$th row, and take $X$ random in $\{0,1\}^d$. We claim that $E |\langle o_i, X\rangle| \ge cst$. To see this, write $$\langle o_i, X\rangle = \langle o_i, \frac{{\bf 1}}{2}\rangle + \langle o_i, (X - \frac{{\bf 1}}{2})\rangle$$ and assume WLOG the first term on the RHS is non-negative. The second term on the RHS is a weighted sum of Rademacher random variables, and so with probability at least $\frac{1}{20}$ it is above its standard deviation, which is $\Omega(1)$ (see for example this paper of Oleszkiewicz) Adding over all $i$'s, the result holds in expectation: $E \|OX\|_1 \ge cst\cdot d$. But since the function $x \mapsto \|Ox\|_1$ is $\sqrt{d}$-Lipschitz wrt $\ell_2$ (and convex), we should be able to use concentration to say that the probability that we get below this mean minus $\frac{cst \cdot d}{2}$ is at most $e^{-cst' \cdot d}$ (see for example Corollary 4.23 of van Handel's notes). This gives the result. REPLY [3 votes]: We prove the weaker bound $$ \mathbf{P} \left[ \|O \mathbb{x}\|_1 \leq \frac{cd}{\sqrt{\log d}} \right] \leq 2^{-Cd} $$ for some constants $C, c$. Define the Gaussian mean width of a compact subset $A \subset \mathbf{R}^d$ as $$ w(A) = \mathbf{E} \sup_{x \in A} \langle G,x \rangle $$ where $G$ is a standard Gaussian vector in $\mathbf{R}^d$. We use the following properties If $\pi$ is an orthogonal projection, then $w(\pi(A))\leq w(A)$. If $A = \{0,1\}^k \subset \{0,1\}^n$, then $w(A)=k/\sqrt{2\pi}$. If $A=B_1^d$ (the unit $\ell_1$ ball), then $w(A) \leq \sqrt{2\log d}$. $w$ is rotation invariant Let $A$ be the set of $x \in \{0,1\}^d$ such that $\|Ox\|_1 \leq cd/\sqrt{\log d}$. We have $w(A) \leq w(cd/\sqrt{\log d} \cdot B_1^d) \leq c\sqrt{2}d$. If $\mathrm{card}(A) \geq 2^{dH(1/4)}$, then the Sauer--Shelah lemma implies that there is a coordinate projection $\pi$ of rank $k=d/4$ such that $\pi(A)$ identifies with $\{0,1\}^k$. Therefore, $w(A) \geq w(\pi(A))=d/4\sqrt{2\pi}$. If we choose $c=1/8\pi$, combining both estimates gives the bound $\mathrm{card}(A) < 2^{dH(1/4)}$, as needed.<|endoftext|> TITLE: Naturally occurring examples of badly behaved categories QUESTION [20 upvotes]: What are some examples of naturally occurring badly behaved (possibly higher) categories? When working with a specific category like ${\bf Set}$ or ${\bf Cat}$, we usually understand/explain them by lauding structural properties they posess -- ${\bf Set}$ is an autological topos, ${\bf Cat}$ is Cartesian closed, etc. What structures can be arranged naturally/canonically into (possibly higher) categories, despite the resulting categories having few/none of the structural properties we would usually like to have in order to carry out category-theoretic-type proofs? A natural example is ${\bf Field}$, the category of fields and field homomorphisms, since it has no terminal object, no initial object, no finite products, is not algebraic and is not presentable. (It is, however, accessible with a multi-initial object given by the set of prime fields). I suspect that it gets worse than this, but I can't think of anything further off the top of my head. Any examples are appreciated. REPLY [4 votes]: The category $\mathbf{Met}$ of metric spaces and metric maps is another example. There are finite limits, but no infinite products. Countable products exist at least when we use continuous maps as the morphisms instead. $\mathbf{Met}$ has no binary coproducts, and this can be seen as the starting point of the Gromov-Hausdorff distance, where we consider all possible metrics on a disjoint union. Coequalizers do not exist either. It is worth pointing out that the injective objects of $\mathbf{Met}$ have gathered some interest.<|endoftext|> TITLE: Unital *-homomorphisms between matrices QUESTION [5 upvotes]: It is mentioned on Wikipedia that every unital *-homomorphism $\Phi:M_i\to M_j$ is necessarily of the form $\Phi(a)=U^*(a\otimes I_r)U$ for some unitary $U$ and some $r$. (Here $M_i$ are the $i\times i$ complex matrices and $I_r$ is the $r\times r$ identity.) No proof or reference is given. How is this proven? (A reference to a textbook is also OK for me.) I will need to formalize the proof in a computer-aided theorem prover, so an elementary proof would be preferred. (By elementary I mean not more than introductory graduate-level textbook level, maybe.) Notes: This answer seems to touch the question but I don't understand it. (It uses "Morita equivalence" which as far as I can tell is quite a power-tool in this context.) Information about whether something similar holds also for the bounded operators on a larger Hilbert space would be appreciated. REPLY [5 votes]: Theorem: For Hilbert spaces $H,K$, every normal unital *-homomorphism $\Phi:\mathcal B(H)\to\mathcal B(K)$ is of the form $\Phi(a)=U(a\otimes 1_{K_0})U^∗$ for some Hilbert space $K_0$ and some unitary $U$. Here is a super down-to-earth proof, from a functional analysis / operator-algebras perspective. I'll start by working in the infinite-dimensional setting. Let $H,K$ be Hilbert spaces, write $\newcommand{\mc}{\mathcal}\mc B(H)$ for the algebra of all bounded operators on $H$, and $\mc K(H)$ for the compact operators. Let $\Phi:\mc B(H)\rightarrow\mc B(K)$ be a unital $*$-homomorphism which is normal (aka weak$^\ast$-continuous). As $\mc K(H)$ is weak$^\ast$-dense in $\mc B(H)$, $\Phi$ is completely determined by its restricton to $\mc K(H)$, That $\Phi$ is unital corresponds to $\Phi:\mc K(H)\rightarrow\mc B(K)$ being non-degenerate meaning that $\{ \Phi(\theta)(\xi) : \theta\in\mc K(H), \xi\in K \}$ has a dense linear span in $K$. If $H$ is finite-dimensional, of course $\mc B(H) = \mc K(H) \cong M_i$ where $i$ is the dimension of $H$. So, consider a non-degenerate $*$-homomorphism $\Phi:\mc K(H)\rightarrow\mc B(K)$. For $\xi,\eta\in H$ write $\theta_{\xi,\eta}$ for the rank-one operator $\alpha\mapsto (\alpha|\eta) \xi$. Then $\theta_{\xi,\eta}^* = \theta_{\eta,\xi}$ and $\theta_{\xi,\eta} \theta_{\xi_1,\eta_1} = (\xi_1|\eta) \theta_{\xi,\eta_1}$. Here I write $(\cdot|\cdot)$ for the inner-product on $H$. Fix a unit vector $\xi_0\in H$, and consider $\theta_{\xi_0, \xi_0}$ which is a projection (self-adjoint idempotent). So $p = \Phi(\theta_{\xi_0, \xi_0})$ is also a projection. Let $K_0\subseteq K$ be the (closed) subspace forming the image of $p$. Define $U:H\odot K_0 \rightarrow K$ by $$ U(\xi\otimes\alpha) = \Phi(\theta_{\xi,\xi_0})(\alpha), $$ and extend by linearity. I write $\odot$ for the algebraic tensor product, and will write $\otimes$ for the (completed) Hilbert space tensor product. Then \begin{align*} ( U(\xi\otimes\alpha) | U(\eta\otimes\beta) ) &= (\Phi(\theta_{\xi,\xi_0})(\alpha) | \Phi(\theta_{\eta,\xi_0})(\beta)) \\ &= (\Phi(\theta_{\xi_0, \eta} \theta_{\xi,\xi_0})(\alpha) | \beta ) \\ &= (\xi|\eta) (\Phi(\theta_{\xi_0,\xi_0})(\alpha) | \beta ) = (\xi|\eta) (p(\alpha) | \beta ) \\ &= (\xi|\eta) (\alpha | \beta ). \end{align*} Thus $U$ is an isometry, and so extends to $H\otimes K_0$. Now compute \begin{align*} U^* \Phi(\theta_{\xi,\eta}) U(\xi_1\otimes\alpha) &= U^* \Phi(\theta_{\xi,\eta}) \Phi(\theta_{\xi_1,\xi_0})(\alpha) \\ &= (\xi_1|\eta) U^*\Phi(\theta_{\xi,\xi_0})(\alpha) \\ &= (\xi_1|\eta) U^*U(\xi\otimes\alpha) \\ &= \theta_{\xi,\eta}(\xi_1) \otimes \alpha. \end{align*} So $U^*\Phi(\theta_{\xi,\eta})U = \theta_{\xi,\eta}\otimes 1$ and so by linearity and continuity, $U^*\Phi(\theta)U = \theta\otimes 1$ for each $\theta\in\mc K(H)$. If we can show that $U$ has dense range, it must be onto (as it's an isometry), and so will be a unitary, and so $UU^*=1$ and so $\Phi(\theta) = U(\theta\otimes 1)U^*$ as required. If $\xi_1$ is another vector, we see that $$ \Phi(\theta_{\xi,\xi_1})(\alpha) = \Phi(\theta_{\xi,\xi_0})\Phi(\theta_{\xi_0,\xi_1})(\alpha) = U(\xi \otimes \beta), $$ say, where $\beta = \Phi(\theta_{\xi_0,\xi_1})(\alpha)$. Letting $\xi, \xi_1,\alpha$ vary, taking linear span, and using non-degeneracy, we see that $U$ does indeed have dense range.<|endoftext|> TITLE: Is the Petersen graph a "Cayley graph" of some more general group-like structure? QUESTION [12 upvotes]: The Petersen graph is the smallest vertex-transitive graph which is not a Cayley graph. Is it the "Cayley graph" of some slightly more general group-like structure? REPLY [9 votes]: We are working on these questions currently with Ignacio Garcia-Marco. As a postive answer I can tell you that the multiplication table of S codes a semigroup. If you take elements 1,6 as connection set, then the right Cayley graph is Cay(S,{1,6}). Its underlying undirected simple graphs is the Petersen graph. From the negative side we can also show, that any directed Cayley graph (of a semigroup) whose underlying undirected simple graph is the Petersen graph has loops.<|endoftext|> TITLE: What are possible applications of deep learning to research mathematics? QUESTION [75 upvotes]: With no doubt everyone here has heard of deep learning, even if they don't know what it is or what it is good for. I myself am a former mathematician turned data scientist who is quite interested in deep learning and its applications to mathematics and symbolic reasoning. There has been a lot of progress recently, and while it is exciting to machine learning experts, the results so far are probably not useful for research mathematicians. My question is simple: Are there areas of research math, that if one had access to a fully trained state-of-the-art machine learning model (like the ones I will describe below), it would make a positive impact in that field? While math is mostly about proofs, there is also a need sometimes for computation, intuition, and exploration. These are the things that deep learning is particularly good at. Let me provide some examples: Good intuition or guessing Charton and Lample showed that Transformers, a now very standard type of neural network, are good as solving symbolic problems of the form $$ \mathit{expr}_1 \mapsto \mathit{expr}_2 $$ where $\mathit{expr}_1$ and $\mathit{expr}_2$ are both symbolic expressions, for example in their paper $\mathit{expr}_1$ was an expression to integrate and $\mathit{expr}_2$ was its integral. The model wasn't perfect. It got the right answer about 93-98% of the time and did best on the types of problems it was trained on. Also, integration is a 200 year old problem, so it is hard to outcompete a state-of-the-art CAS. However, there are some things which make this interesting. Symbolic integration is important, difficult, and (somewhat) easy to check that the solution is correct by calculating the derivative (and checking symbolically that the derivative is equivalent to the starting integrand). Also it is an area where “intuition” and “experience” can definitely help, since a trained human integral solver can quickly guess at the right solution. Last it is (relatively) easy to compute an unlimited supply of training examples through differentiation. (This paper also uses other tricks as well to diversify the training set.) Are there similar problems in cutting edge mathematics, possibly in algebra or combinatorics or logic where one would like to reverse a symbolic operation that is easy to compute in one direction but not the other? Neural guided search Some problems are just some sort of tree or graph search, such as solving a Rubik's cube. A neural network can, given a scrambled cube, suggest the next action toward solving it. A good neural network would be able to provide a heuristic for a tree or graph search and would prevent exponential blow-up compared to a naive brute-force search. Indeed, a paper in Nature demonstrated training a neural network to solve the Rubik's cube from scratch this way with no mathematical knowledge. Their neural network-guided tree search, once trained, can perfectly solve a scrambled cube. This is also similar to the idea behind AlphaGo and its variants, as well as the idea behind neural formal theorem proving—which is really exciting, but also not up to proving anything useful for research math. Puzzle cubes and board games are not cutting edge math, but one could imagine more interesting domains where just like a Rubik's cube one has to manipulate one expression into another form through a series of simple actions, and the ability to do that reliably would be of great interest. (Note, the neural guided tree search I've described is still a search algorithm and much slower than a typical cube solving algorithm, but the emerging field of program synthesis could possibly one day soon learn from scratch a computer program which solves the Rubik's cube, as well as learn computer programs to solve more interesting math problems.) Neural information retrieval Suppose you have a database of mathematical objects (say, integer sequences, finite groups, elliptic curves, or homotopy groups as examples) and you want a user to be able to look up an object in this database. If that object is in the database, you would like the user to find it. If it is not, you would like the user to find similar objects. The catch is that "similar" is hard to define. Deep learning provides a good solution. Each object can be associated with an embedding, which is just a vector in, say, $\mathbb{R}^{128}$. The measure of similarity of two objects is just the inner product of their embeddings. Moreover, there are a number of techniques using self-supervised machine learning to construct these embeddings so that semantically similar objects have similar embeddings. This has already shown a lot of promise in formal theorem proving as premise selection where one wants to pick the most relevant theorems from a library of theorems to use in the proof of another theorem. For example, I think such a neural database search could reasonable work for the OEIS, where one can use a neural network to perform various prediction tasks on an integer sequence. The inner layers of the trained network will compute a vector embedding for each sequence which can be used to search through the database for related sequences. Geometric intuition Neural networks are pretty good at image recognition and other image tasks (like segmentating an image into parts). Are there geometry tasks that it would be useful for a deep learning agent to perform, possibly in dimensions 4 or 5, where human geometric intuition starts to fail us since we can't see in those dimensions. (It would be hard to make, say a convolutional neural network work for a 4 dimensional image directly, but I could imagine representing a 3D surface embedded in 4D as say a point cloud of coordinates. This could possibly work well with Transformer neural networks.) Build your own task Neural networks are very flexible when it comes to the choice of input and output, as well as the architecture, so don’t let the specific examples above constrain your thinking too much. All you need are the following things: A type of mathematical object. One that you care about. It should be representable in some finite way (as a formula, an initial segment of a sequence, an image, a graph, a computer program, a movie, a list of properties). A task to perform on your operation. It can be well specified or fuzzy. It can be solvable, or (like integration) only partially solvable. It can be classifying your objects into finitely many buckets. It could be computing some other related object or property of the object. It could be finding the next element in a sequence. It could be coming up with a prediction or conjecture of some sort. It could be turning that object into a some 2D cartoon image even, just to think outside the box. Lots of training data. Either you need to be able to synthetically generate a lot of training examples as in the integration example above, or like OEIS have a large dataset of 10s of thousands or examples (more examples is always better). Clean data is preferred but neural networks handle messy data very well. Another “data free” solution is reinforcement learning like in the Rubik’s Cube example or AlphaGoZero, where the agent learns to explore the problem on its own (and generates data through that exploration). Patterns in the data, even if you can’t see what they are. Your task should be one where there are patterns which help the machine learning agent solve the problem. (For example, I’m not convinced that factoring large integers would be a good task.) Motivation. Why would this be useful to the field? What purpose would having this trained model have? Would it make it easier to conjecture facts, explore new areas of math, wrap ones head around a bunch of confusing formulas? Or do you have a way to turn a learned heuristic into a proof, such as with a search algorithm (as in the Rubik’s cube example above) or by checking the solution (as in the integration example above)? REPLY [5 votes]: One application that we urgently need is a "smart" catered TeX-math search engine (eg.an interesting one is https://approach0.xyz/search/ ). achieving a collaboration with the people in arxiv to have the engine search through the TeX code of math papers. (So maybe even coming up with some template of math-TeX glossary that we should follow if we want to ease searchability). So this would require an engine that is flexible and even present us with search-results that are close but from different areas, establishing interesting bridges. I imagine the AI would also be useful here in forming a loose family of theorems and definitions based on topic. Thus giving us a sense of the various directions in math. Another interesting application would be to help in PDEs with sharpening the various constants. I imagine a really smart AI would be great for running through a wide range of candidate functions and having the cost function just be based on the sharpening of the constant.<|endoftext|> TITLE: An almost complex structure on the real $n$-sphere $S^n$ QUESTION [5 upvotes]: If $R\mathrel{:=}\mathbb{R}[x_1,\dotsc,x_{n+1}]/(x_1^2+\dotsb+x_{n+1}^2-1)$ and $S^n\mathrel{:=}\operatorname{Spec}(R)$ is the real $n$-spere, a classical result of Borel and Serre says that the only real spheres with an almost complex structure is $S^2$ and $S^6$. An almost complex structure is an endomorphism of the tangent bundle $$J: T_{S^2} \rightarrow T_{S^2}$$ with $J^2=-\operatorname{Id}$. In the case of the real 2-sphere it follows the tangent bundle $T_{S^2}$ is a real algebraic vector bundle of rank 2. Question 1: I'm looking for an example of a real algebraic (even dimensional) manifold $M$ with an almost complex structure $J:T_M \rightarrow T_M$, where $J$ is not algebraic. The problem of constructing a holomorphic structure on $S^6$ — is this still an open problem? Note: If you let $k\mathrel{:=}\mathbb{R}$ and $K\mathrel{:=}\mathbb{C}$, it follows there is an isomorphism $$\operatorname{Spec}(K\otimes R)\cong \operatorname{Spec}(B)\mathrel{:=}S^2_K$$ with $$B\mathrel{:=}K[u,v,w]/(uv-(w^2+1)).$$ If $J_K$ is the pull-back of $J$ to $S^2_K$ it follows that $$\phi\mathrel{:=}\frac{1}{2}(I+iJ) \in \operatorname{End}(T_{S^2_K})$$ is an idempotent: $\phi^2=\phi$ and you get a direct sum $$T_{S^2_K} \cong L_1\oplus L_2$$ with $L_i \in \operatorname{Pic}(S^2_K)$. If $J$ is algebraic it follows $L_i$ are algebraic, and I'm interested in this decomposition. In the case of the real 6-sphere $S^6$ it follows the endomorphism bundle $\operatorname{End}(T_{S^6})$ is a real algebraic vector bundle of rank $36$. We may consider the subvariety $$I(S^6)\mathrel{:=}\{J \in \operatorname{End}(T_{S^6}): J^2=-\operatorname{Id}\}$$ and the group scheme $G\mathrel{:=}\operatorname{GL}(T_{S^6})$. There is a canonical action $$\sigma: G \times I(S^6) \rightarrow I(S^6)$$ and a "parameter space" $I(S^6)/G$ parametrizing algebraic almost complex structures on $S^6$. Is this construction used in the study of the Hopf problem — the problem of constructing a holomorphic structure on $S^6$? If there is a holomorphic structure on $S^6$ — is this neccessarily algebraic? I ask for references. Note: We may also consider the ring $R\mathrel{:=}C^{\infty}(S^6)$ and the $R$-module $T^{\infty}_{S^6}$ of smooth sections of $T_{S^6}$, and the projective $R$-module $\operatorname{End}_R(T^{\infty}_{S^6})$. We may consider a smooth almost complex structure $J\in \operatorname{End}_R(T^{\infty}_{S^6})\cong \Omega^{1,\infty}_R\otimes_R T^{\infty}_{S^6}$. We get a topological subspace $I^{\infty}(S^6)$ of smooth almost complex structures on $S^6$ equipped with an action of the group of smooth automorphisms of $T^{\infty}_{S^6}$. This is a topological subspace of a smooth vector bundle of rank $36$, and the orbit space parametrize all smooth almost complex structures on $S^6$. Hence if there is a holomorphic structure on $S^6$ it "lives" in this orbit space. REPLY [11 votes]: Let $M=SU_3$, the compact semisimple Lie group. By request of the OP, for those unfamiliar with Maurer-Cartan form, let me define it. Write each point of $SU_3$ as a matrix $g$. Left translation by $g^{-1}$ takes $g$ to $I$, so takes $T_g SU_3$ to $T_I SU_3$. Consider any tangent vector $A\in T_g SU_3$ as a $3\times 3$ complex matrix, since $SU_3$ sits in the vector space of $3\times 3$ complex matrices. We have a linear map $A\in T_g SU_3\mapsto g^{-1}A\in T_I SU_3$, just the derivative of left translation (as the derivative of linear map on a vector space is that same linear map, and then restrict from all $3\times 3$ matrices to the linear subspace $T_g SU_3$). We define the left invariant Maurer Cartan form $\omega$ to be the linear map which takes each tangent space $T_g SU_3$ to $T_I SU_3$ by $\omega_g(A)=g^{-1}A$. So $\omega$ is a $1$-form valued in $\mathfrak{su}_3=T_I SU_3$. (This is all very standard, in every Lie group textbook.) Naturally, one writes this form not as $\omega$ but as $g^{-1}dg$. As requested: for more information about the Maurer-Cartan form (for any Lie group, real or complex or $p$-adic): Chevalley, Claude, Theory of Lie Groups. I., Princeton Mathematical Series, vol. 8. Princeton University Press, Princeton, N. J., 1946. ix+217 pp. 152 section V.IV. Sharpe, R. W., Differential geometry. Cartan's generalization of Klein's Erlangen program. With a foreword by S. S. Chern. Graduate Texts in Mathematics, 166. Springer-Verlag, New York, 1997. xx+421 pp. ISBN: 0-387-94732-9, p. 96. Write the left invariant Maurer--Cartan form $g^{-1}dg$ as $\omega=(\omega_{\mu\bar\nu})$ for $\mu,\nu=1,2,3$. The coframing $\omega_{1\bar{1}}+i\omega_{2\bar{2}},\omega_{1\bar{2}},\omega_{1\bar{3}},\omega_{2\bar{3}}$ is complex linear for a unique left invariant complex structure. Take a function $f(g)=e^{-|g_{1\bar{2}}|^2/\varepsilon}$, or your favourite nonzero nonalgebraic function. The coframing $\omega_{1\bar{1}}+if\omega_{2\bar{2}},\omega_{1\bar{2}},\omega_{1\bar{3}},\omega_{2\bar{3}}$ continues to have linearly independent real and imaginary parts, so is still complex linear for a unique almost complex structure. But since $f$ is not holomorphic and is not constant, there is a nonzero Nijenhuis tensor. You could also take $f$ to be instead some smooth function with support in some small compact set, and then you get complex structure on some open set turning smoothly almost complex, not complex, on some other open set, so clearly not algebraic or even analytic. To clarify, as requested: as $SU_3$ is a Lie group, its tangent bundle is trivial as a smooth real vector bundle. Consider the linear map $\Omega \colon TSU_3\to \mathbb{C}^4$ given by $\Omega(v)=(\omega_{1\bar{1}}+i\omega_{2\bar{2}},\omega_{1\bar{2}},\omega_{1\bar{3}},\omega_{2\bar{3}})$ and $\Omega_{\varepsilon} \colon TSU_3\to \mathbb{C}^4$ given by $\Omega(v)=(\omega_{1\bar{1}}+if\omega_{2\bar{2}},\omega_{1\bar{2}},\omega_{1\bar{3}},\omega_{2\bar{3}})$. Let $J_0$ be the standard almost complex structure on $\mathbb{C}^4$. Let $J(v)=\Omega^{-1}(J_0\Omega(v))$. Let $J_{\varepsilon}(v)=\Omega_{\varepsilon}^{-1}(J_0\Omega_{\varepsilon}(v))$. These are my almost complex structures: elements of $\operatorname{End}T_{SU_3}$. You can compute from the Maurer-Cartan equations that $J_0$ is a complex structure, and that $J_{\varepsilon}$ has zero Nijenhuis tensor (so arises from a complex structure) on any open set where $f=0$, but not near any point where $df\ne 0$. The complex structure on $SU_3$ is homogeneous under left action of $SU_3$. It is not Kähler, as $b_2(SU_3)=0$, so all closed 2-forms are exact, and hence no symplectic form on $SU_3$. In particular, $SU_3$ does not have a complex structure under which it could become a Kähler manifold, and a fortiori is not a complex algebraic variety. There are infinitely many nonbiholomorphic complex structures on $SU_3$ invariant under left $SU_3$ action. Note that $SU_3$, in any complex structure, is not a complex Lie group, i.e. its multiplication is not holomorphic. Proof: Indeed $SU_3$ is compact. Take a compact complex Lie group $G$. Take a complex linear finite dimensional representation of $G$. The representation is a map to complex matrices, which form an affine space. Complex affine space has no compact complex subvarieties except points. So $G$ has discrete image. So $G$ has discrete adjoint representation. Hence the identity component of $G$ is an abelian group, compact, so a complex torus. So compact complex Lie groups are precisely finite group extensions of complex tori. In particular, since $SU_3$ is connected and nonabelian and compact, it is not a complex Lie group for any complex structure. If its multiplication were holomorphic for some complex structure, then the holomorphic implicit function theorem would ensure that its inverse operation was also holomorphic, so it would be a complex Lie group. For more information about this and other $SU_3$-invariant complex structures on $SU_3$ (and all other compact and simply connected homogeneous complex manifolds) see Hsien-Chun Wang, Closed manifolds with homogeneous complex structure, American Journal of Mathematics, vol. 76, no. 1 (January 1954), pp. 1-32. Phillip Griffiths, On certain homogeneous complex manifolds, Proceedings of the National Academy of Sciences of the United States of America, vol. 48, no. 5 (May 15, 1962) pp. 780-783.<|endoftext|> TITLE: Is the Besov space $B_{\infty,1}^0(\mathbb{R}^d)$ a multiplication algebra? QUESTION [6 upvotes]: Let $s\in\mathbb{R}$ and $1\leq p,q\leq\infty$. Consider the Besov scale of spaces $B_{p,q}^s(\mathbb{R}^d)$ defined by the norm $$\|f\|_{B_{p,q}^s} := (\sum_{j=0}^\infty \|P_{j} f\|_{L^p}^q)^{1/q},$$ where $\{P_j\}_{j=0}^\infty$ is an inhomogeneous Littlewood-Paley partition of unity with the convention that $P_0$ is the low-frequency projector. I am interested in when $B_{p,q}^s$ is a multiplication algebra: $$\|fg\|_{B_{p,q}^s} \lesssim \|f\|_{B_{p,q}^s} \|g\|_{B_{p,q}^s}.$$ It is well-known (for example, see Corollary 2.86 in Bahouri, Chemin, and Danchin's Fourier Analysis and Nonlinear Partial Differential Equations) that if $s>0$, then $L^\infty \cap B_{p,q}^s$ is an algebra with $$\|fg\|_{B_{p,q}^s} \lesssim_s \|f\|_{L^\infty}\|g\|_{B_{p,q}^s} + \|g\|_{L^\infty}\|f\|_{B_{p,q}^s}.$$ Note that this result does not cover the case $B_{\infty,1}^0$. In the following article, Triebel claims (see Theorem 2) that $B_{\infty,1}^0$ is a multiplication. But if you look at the proof (the last block of equations before Remark 3 on pg. 41), there seems to be error in obtaining the second line given that $$\sum_{k=1}^\infty\sum_{l=k}^\infty 2^{ksq} \|b_l\|_{L^p}^q \neq \sum_{l=1}^\infty 2^{lsq} \|b_l\|_{L^p}^q$$ if $s=0$. So my question is the following: Question. Is the space $B_{\infty,1}^0$ a multiplication algebra? REPLY [5 votes]: You may want to take a look at Herbert Koch and Winfried Sickel, "Pointwise multipliers of Besov spaces of smoothness zero and spaces of continuous functions", Rev. Mat. Iberoamericana 18 (2002), 587–626. They established that the set of all distributions $f$ such that $g \mapsto fg$ is a bounded linear map from $B^0_{\infty,1}$ to itself is a strict subspace of $B^0_{\infty,1}$. Based on their result I think Triebel made an error there. In particular, see their Remark 15, which states a variant of what I wrote in my first comment (that a sufficient condition is for $f\in B^0_{\infty,1}$ and that $\sup_{j \in \mathbb{N}} j \|P_j f|_\infty < \infty$). The explicit function given in part (i) of their Lemma 16 should be a function $f\in B^0_{\infty,1}$ such that $f^2 \not\in B^0_{\infty,1}$.<|endoftext|> TITLE: Publishing solution but temporarily holding back solution method QUESTION [9 upvotes]: https://www.quantamagazine.org/mathematician-disproves-group-algebra-unit-conjecture-20210412/ Above is an article about a researcher disproving an open conjecture in algebra (Kaplansky's unit conjecture, which I was unfamiliar with). It says: Gardam declined to tell the audience just how he had found the long-sought-after counterexample (except to confirm that it involved a computer search). He would share more details in a few months, he told Quanta. But for now, he said, “I’m still optimistic that maybe I have enough tricks left to get some more results.” Is that a usual thing in math? I have seen some cryptography results announced that way, where someone demonstrates an attack on some cryptosystem but temporarily withholds details. The intention there is different though: it's to give people using the broken system some time to fix their stuff before revealing the attack to the wider world. In the math case, I know something like this happened with solutions for cubic and quartic equations in the 16th(?) century but I had the impression that since then, if you've got a general method to solve a given type of problem, that's a bigger deal than cranking a few more specific solutions from it, so you might as well publish early. Don't want to go too much into whether it's good or bad, but just wondering if anyone has seen stuff like this before. REPLY [3 votes]: Apparently, it was told about C. F. Gauss that "He is like the fox, who effaces his tracks in the sand", because of his elliptic style of exposition. In fact, he liked to write down only very polished results, without explaining how he had found them. An informative discussion about the quote above his here: https://hsm.stackexchange.com/questions/3610/what-is-the-original-source-for-abels-quote-about-gausshe-is-like-the-fox-wh<|endoftext|> TITLE: Infinite group for which it is still unknown if it is simple QUESTION [10 upvotes]: Is there an example of an infinite group for which it is still unknown whether or not it is simple? Is there a criterion for determining the simplicity of infinite groups? REPLY [32 votes]: Below I will present an example of a finitely presented group which is conjecturally simple. Thank you for the opportunity to communicate this example to a wider community. Consider the group $$\Gamma=\langle s,t,u \mid s^7=t^7=u^7=1, u=s^3t^3, u^3=st \rangle. $$ Then $\Gamma$ itself is not simple: it has a non-trivial homomorphism to $\mathbb{Z}/7\mathbb{Z}$, defined by $s\mapsto 1$, $t\mapsto -1$, $u\mapsto 0$. However, the kernel $\Gamma_0$ of this homomorphism, is conjecturally simple. Clearly, $\Gamma_0$ is finitely presented, as it is of finite index in $\Gamma$. Much is known about the groups $\Gamma$ and $\Gamma_0$. For example: These groups are infinite groups and they have Kazhdan's property (T). Every nontrivial normal subgroup in either group is of finite index. These groups are non-linear over any field. These groups act cocompactly on an exotic $\tilde{A}_2$-building of thickness 3, so they are somehow related to the Fano plane, the projective plane over the field with two elements, which hints about the role of the power 3 and 7 in the presentation of $\Gamma$. The group $\Gamma$ is briefly discussed in section 10.4 here. It is closely related to the group $$\langle s,t,u \mid s^7=t^7=u^7=1, u=st, u^3=s^3t^3 \rangle $$ which happens to be an index 3 subgroup of an arithmetic lattice in $\mathrm{SL}_3(\mathbb{F}_2(\!(x)\!))$, thus residually finite. The above example is in fact a part of the infinite family of groups acting cocompactly on affine buildings. It is conjectured that if the corresponding building is not a Bruhat-Tits building then the acting group contains a simple subgroup of finite index.<|endoftext|> TITLE: When does a spherical curve equal its tangent indicatrix? QUESTION [5 upvotes]: Given a smooth regular curve $\gamma$ in $\mathbb{R}^{3}$, one defines the tangent indicatrix of $\gamma$ to be the spherical curve $\gamma'/\lVert \gamma'\rVert$. It is then natural to look for spherical curves $\gamma$ whose tangent indicatrix is the same as the original curve, modulo an isometry of $\mathbb{R}^{3}$. Question: Is the circle the only example? I bet that this question has been answered already, but I could not find any reference to it, at least in the English literature. It was also posted years ago on Mathematics Stack Exchange (link here). REPLY [5 votes]: Your question is not very clearly phrased, which may explain why you didn't get any answers on MSE. When you say, "spherical curve $\gamma$ whose tangent indicatrix is the same as the original curve, modulo an isometry of $\mathbb{R}^3$", the simple interpretation is this: Assume that $\gamma$ is unit speed, so that $\|\gamma'(t)\|=1$ for all $t$. Then you seem to want to understand the conditions under which one might have $$ \gamma'(t) = R\,\gamma(t) $$ for all $t$, where $R:\mathbb{R}^3\to\mathbb{R}^3$ is an isometry of $\mathbb{R}^3$. Now this can only happen when $\gamma$ parametrizes an arc of a great circle. For, if $\gamma$ is a unit speed curve on the sphere, then we have the usual result that $$ \gamma''(t) = -\gamma(t) + \kappa(t)\,\,\gamma(t)\times\gamma'(t), $$ where $\kappa(t)$ is the (geodesic) curvature of $\gamma$ as a spherical curve. Thus, $\|\gamma''(t)\|^2 = 1 + \kappa(t)^2$. Meanwhile, if $\gamma'(t) = R\,\gamma(t) = A\,\gamma(t) + b$ where $A$ is a constant orthogonal matrix and $b$ is a constant, then $\gamma''(t) = A\,\gamma'(t)$, which would imply that $\|\gamma''(t)\|^2 = \|\gamma'(t)\|^2 = 1$, since $A$ is orthogonal. Consequently, we would have to have $\kappa(t)\equiv 0$, which implies that $\gamma$ is indeed a great circle on the unit sphere. However, you might have wanted to ask whether it was possible for the image of $\gamma$ to be congruent, as a curve in the $2$-sphere, to the image of $\gamma'$. (Again, I'm assuming that $\gamma$ has been parametrized so as to have unit speed.) This is a more interesting question, and somewhat subtle. Essentially, what this is asking, supposing that $\gamma:I\to S^2$ be a unit speed curve and $I\subset \mathbb{R}$ be an open interval, is whether there exists a function $f:I\to I$ and a rotation $A\in \mathrm{SO}(3)$ such that $$ \gamma'(t) = A\,\gamma\bigl(f(t)\bigr). $$ Differentiating this equation, we get $$ -\gamma(t) + \kappa(t)\,\,\gamma(t)\times\gamma'(t) = \gamma''(t) = A\,\gamma'\bigl(f(t)\bigr)\,f'(t), $$ which implies the equation $1+\kappa(t)^2 = f'(t)^2$. Moreover, since $\gamma'$ and $\gamma{\circ}f$ are congruent curves (although not unit speed), they have to have equal geodesic curvature at corresponding points. Using the usual Frenet apparatus to compute this, one sees that this is equivalent to requiring that $$ \kappa'(t) = f'(t)^3\,\kappa\bigl(f(t)\bigr). $$ Conversely, if $f:I\to I$ and $\kappa:I\to\mathbb{R}$ are, say, differentiable functions satisfying the conditions $$ f'(t)^2 = 1+\kappa(t)^2\quad\text{and}\quad \kappa'(t) = f'(t)^3\,\kappa\bigl(f(t)\bigr),\tag1 $$ then there will exist a unit speed curve $\gamma:I\to S^2\subset\mathbb{R}^3$ with geodesic curvature $\kappa$ such that the curves $\gamma'$ and $\gamma{\circ}f$ are congruent, and such a $\gamma$ will be unique up to isometry. The functional-differential system (1) is somewhat interesting. It is not obvious that there is a solution $(I, f, \kappa)$ that is nontrivial, i.e., that has $\kappa\not=0$. It's clear that, if $I\subset\mathbb{R}$ is a bounded interval, then $\kappa=0$ is the only possibility. It's also easy to see that if $f$ has a fixed point at $a\in I$, i.e., $f(a) = a$ and $f'(a)= b$ where $|b|>1$, then there is an essentially unique formal power series solution $(f,\kappa)$ in a neighborhood of $a$ with $f(a) = a$ and $f'(a) = b$. (If $(f,\kappa)$ is a solution then $(f,-\kappa)$ is also a solution.) (One can easily reduce to the case $a = 0$ by translating $I$. There cannot be more than one fixed point unless $\kappa\equiv0$ between them.) Unfortunately, it is not clear when this formal power series solution has a positive radius of convergence, though. Perhaps there is a solution on $I = (0,\infty)$ or $I=\mathbb{R}$ with, say, $f(t)>t$ for all $t$ (i.e., no fixed points), but nothing obvious suggests itself to me.<|endoftext|> TITLE: How many finitely-generated-by-elements-of-finite-order-groups are there? QUESTION [11 upvotes]: I do not know where this question is on the trivial to intractable spectrum. Consider the set of isomorphism classes of groups finitely generated by elements of finite order. What is the cardinality of this set? REPLY [6 votes]: The cardinality of the set of groups generated by three elements of order two is equal to the cardinality of the set of countable groups, i.e., the cardinality of the continuum. Every countable group embeds in a 2-generator group by the Higman-Neumann-Neumann embedding theorem, and the two generators can be chosen to have infinite order. Since a 2-generator free group embeds into the free product of three cyclic groups of order 2, it is easy to show (starting from the HNN-embedding theorem) that every countable group embeds in a group generated by three elements of order two.<|endoftext|> TITLE: Are the Surreals a cogenerator in the category of ordered fields? QUESTION [5 upvotes]: A cogenerator in a category $\mathcal{C}$ is an object $\Omega$ such that for any pair of parallel arrows $f,g:X\rightrightarrows Y$ in $\mathcal{C}$ we have $$ \forall h:Y\to\Omega\big(h\circ f=h\circ g\big)\implies f=g, $$ so morphisms into $\Omega$ are sufficient to delineate between distinct arrows. My question is Are the Surreals ${\bf N_0}$ a cogenerator in the category ${\bf \leq-Field}$ of ordered fields and ordered field homomorphisms? Obviously we have some size issues here; I will offer one remedy below, but feel free to use whatever tools you like to get the surreals into a category. We will work in $ZFC+\text{there exist two}$ Grothendieck universes $U\subset V$, carying out our discourse in $V$ about the category ${\bf \leq-Field}_U$ of (possibly large in $U$) $U$-ordered fields. The surreals in $U$ are a weakly terminal object in this category; are they a cogenerator? EDIT: To address some potentially confusing language that was initially pointed out by Pace Nielsen (thank you) and resolved in the comments below Kieth Kearnes' answer, by fields that are 'possibly large in $U$' I mean fields whose underlying classes are subclasses of $U$, not fields whose underlying classes are larger than all of $U$ or fields that are 'in $U$' in a membership sense. REPLY [8 votes]: On page 21 of Conway's Field of Surreal Numbers Norman L. Alling Transactions of the American Mathematical Society Jan., 1985, Vol. 287, No. 1, pp. 365-386. one finds the statement that, on page 43 of J. H. Conway, On numbers and games, Academic Press, London, 1976, Conway states that ``As an abstract Field, ${\bf N}_0$ is the unique universally embedding totally ordered Field.'' I don't have a copy of Conway's book, so don't know how he proved it. If the claim is true, then it is sufficient to answer your question, since given distinct $f, g\colon X\to Y$ one can take an embedding $h$ of $Y$ into ${\bf N}_0$ to separate them. **Edit** I am editing my answer to respond to the comments. Originally I thought the main question was in the last sentence of the question: The surreals in U are a weakly terminal object in this category; are they a cogenerator? The answer is affirmative, since in any category where all morphisms are monic, any weakly terminal object is a cogenerator. Now the comments lead me to think that I trivialized the problem. It seems to be this, instead: Given that every $U$-small ordered field embeds in ${\bf N}_0(U)$, does it follow that every $U$-large ordered field in $U$ embeds in ${\bf N}_0(U)$? I say Yes in NBG with Global Choice. Let $\mathbb G$ be a $U$-large ordered field. Use Global Choice to enumerate $\mathbb G$ by the ordinals in $U$. For each cardinal $\lambda$, let $\mathbb G_{\lambda}$ be the $U$-small ordered subfield of $\mathbb G$ generated by the elements enumerated by ordinals $<\lambda$. This yields a well-ordered (class size) filtration of $\mathbb G$. Starting with the embedding of $\mathbb G_0 = \mathbb Q$ into ${\bf N}_0(U)$, extend this embedding recursively to each ${\mathbb G}_{\alpha}$ using the following fact at successor stages: Fact. Any embedding of $\mathbb G_{\alpha}$ into ${\bf N}_0$ can be extended to $\mathbb G_{\alpha+1}$. This follows from the fact that ${\bf N}_0$ is a $\kappa$-universally extending model for all $\kappa$ (see An alternative construction of Conway's ordered field ${\bf N}_0$ by Philip Ehrlich, Algebra Universalis, 25 (1988) 7-16). Here I am relying on the fact that $\mathbb G_{\alpha}$ and $\mathbb G_{\alpha+1}$ are $U$-small in order to reference the $\kappa$-universally extending model property. At limit stages take unions. I believe that this process yields a $U$-large embedding of $\mathbb G$ into ${\mathbf N}_0(U)$. Hence ${\mathbf N}_0(U)$ is weakly terminal even with respect to $U$-large ordered fields in $U$, hence is a cogenerator even for the class of $U$-large ordered fields in $U$.<|endoftext|> TITLE: Polar decomposition in abstract von Neumann algebra QUESTION [5 upvotes]: Probably an easy question, but here goes: In a concrete von Neumann algebra $M \subseteq B(H)$, every element $m \in M$ has a polar decomposition $m= p|m|$ where $p$ is a partial isometry and $|m|= \sqrt{m^*m}$. Imposing extra conditions on $p$ ensures that $p$ is unique. For example, one can ask that $\ker p = \ker m$. Is there a way to describe the unique partial isometry $p$ without referring to the underlying Hilbert space $H$? In other words, in an abstract von Neumann algebra (= $W^*$-algebra) $M$, how would one describe the partial isometry $p$ in the decomposition $m = p|m|?$ Ideally, I hope there is some algebraic characterisation of $p$. REPLY [7 votes]: I would say that the polar decomposition of $m \in M$ is the unique pair $(v,a)$ of elements in $M$ satisfying the following (algebraic) properties. $m = va$. $v$ is a partial isometry and $a$ is positive. $a^2 = m^* m$. Whenever $p \in M$ is a projection satisfying $mp = 0$, we have $vp=0$.<|endoftext|> TITLE: Why does Drinfeld Unitarization work? QUESTION [18 upvotes]: In Drinfeld's paper "Quasi-Hopf Algebras" he illuminates a process by which you can replace the $R \in A \otimes A$ associated to a quasi-Hopf QUE-algebra $(A, \Delta, \varepsilon, \Phi)$ over $k[[h]]$ with an $\overline{R}$ which satisfies the unitary condition. I describe the process here: $\overline{R} = R*(R^{21}*R)^{-\frac{1}{2}}$ Using Sweedler notation if $R = \Sigma (r_1 \otimes r_2)$ then $R^{21} := \Sigma(r_2 \otimes r_1)$. The unitary condition states $RR^{21} = 1$. His argument for why $\overline{R}$ satisfies this condition is as follows: $\overline{R}^{21}\overline{R} = R^{21}*(R*R^{21})^{-\frac{1}{2}}*R*(R^{21}*R)^{-\frac{1}{2}} = R^{21}*(R*R^{21})^{-1}*R = 1$ My question is about this second equality here. How does he arrive at this? Keep in mind that $A$ is not a commutative algebra. If there are any clarification questions please ask! Thank you so much! REPLY [7 votes]: The short answer to your question is that if $x,y$ are elements in an algebra in topologically free $k[[\hbar]]$-modules whose constant term is 1, then they have a unique square root whose constant term is also one, and if $x,y$ commute then say the square root of $x$ also commutes with $y$. Indeed if $a$ is the square root of $x$, then $$(yay^{-1})^2=ya^2y^{-1}=x$$ and because $yay^{-1}$ also has constant term equal to 1, we get $yay^{-1}=a$. This shows at once that $(RR^{2,1})^{\frac12}$ commutes with $R$. One way to think about it is as follow (this is also explained in Joel's paper). Any finitely generated group $G$ has a so-called pro-unipotent aka malcev aka rational completion $G(\mathbb{Q})$. One of its definition is that it is the univrsal uniquely divisible group having a morphism from $G$. In other words, it is the universal group in which images of elements of $G$ have a unique $n$th root for any $n$. So roughly elements of this groups are the $x^{\lambda}$ where $x \in G$ and $\lambda \in \mathbb{Q}$. Now the same argument as above shows if $x,y$ commute, then so do any possibly rational power of their image in $G(\mathbb{Q})$ (uniqueness is again key here). This has a relative version, where in the case at hand you roughly speaking apply this construction to the pure braid group $P_n$ inside of the braid group $B_n$: you get a certain group $B_n(\mathbb{Q})^{rel}$ fitting into an exact sequence $$1 \rightarrow P_n(\mathbb{Q}) \rightarrow B_n(\mathbb{Q})^{rel} \rightarrow S_n\rightarrow 1. $$ Long story short you get this way a morphism from the so-called cactus group $\Gamma_n$ (the group of which coboundary categories give representations) into $B_n(\mathbb{Q})^{rel}$ by taking square roots of the generators of $P_n$ inside there. Now for any quantized quasi-Hopf algebra, of more generally in any braided tensor category over $k[[\hbar]]$ in which the braiding satisfies $$\beta_{U,V}\beta_{V,U} =id_{U\otimes V} +O(\hbar)$$ the representations of $B_n$ you get factor through $B_n(\mathbb{Q})^{rel}$, hence restrict to representations of $\Gamma_n$.<|endoftext|> TITLE: Proofs that the classifying space of Connes' cycle category $\Lambda$ is $\mathbb C \mathbb P^\infty$ QUESTION [9 upvotes]: Connes showed in Cohomologie cyclique et foncteurs $Ext^n$ (1983) that the classifying space of his cycle category $\Lambda$ is $\mathbb C \mathbb P^\infty = B(S^1) = K(\mathbb Z,2)$. Connes' proof is not quite as conceptual as one might like. He shows that $|\Lambda|$ is simply-connected, and then computes its cohomology via an explicit resolution of the constant functor $\Lambda \to Ab$, $[n] \mapsto \mathbb Z$ via a double complex which is cooked-up by gluing together the usual way of resolving cyclic groups and the usual way of resolving simplicial objects, with a few modifications. He's able to get the ring structure on the cohomology using an endomorphism of this double complex. The result follows because $\mathbb C\mathbb P^\infty$ is characterized among simply-connected spaces by its cohomology ring. Surely in the last nearly 40 years new proofs that $|\Lambda| \simeq \mathbb C \mathbb P^\infty$ have been found. Question 1: What are some alternate proofs that $|\Lambda| \simeq \mathbb C \mathbb P^\infty$? Question 2: (somewhat vague) In particular, is there a proof which somehow uses an explicit $S^1$ action on some category or simplicial set related to $\Lambda$? REPLY [4 votes]: Another proof appears as Lemma 8.1.1.15 and Remark 8.1.1.16 of Igusa's Higher Franz-Reidemeister torsion. It uses spaces of graphs to verify that $|\Lambda|$ classifies oriented circle bundles.<|endoftext|> TITLE: Is $\mathbb{Z}[i,\varphi]$ a Euclidean domain? QUESTION [9 upvotes]: I've already asked this question on Math StackExchange but having gotten no responses this may be more obscure than I had initially believed. Here $\varphi=\frac{1+\sqrt{5}}{2}$. It's true that $\mathbb{Z}[\varphi]=\mathcal{O}_{\mathbb{Q}(\sqrt{5})}$ is Euclidean since $\mathbb{Q}(\sqrt{5})$ is norm Euclidean, and I've read that $A=\mathbb{Z}\left[\frac{1+i}{\sqrt{2}}\right]$ is Euclidean as well, though I'm not certain what the Euclidean function is there (the reasonable candidate being $x\mapsto N_{K/\mathbb{Q}}(x)$, $K$ being $A$'s fraction field). So, my questions are if anything is known about: Is $R=\mathbb{Z}[i,\varphi]$ Euclidean? All I know is that $R=\mathcal{O}_L$ ($L$ being $R$'s fraction field), by computing its discriminant and comparing with $D_L$ given by LMFDB, so $R$ is a PID. If so, what is the Euclidean function? The reasonable candidate is $x\mapsto N_{L/\mathbb{Q}}(x)$, but this has proved difficult to work with. If not, is $R$ a (finite-index) subring of a (nice) Euclidean domain? My main goal is to compute $\gcd$s in $R$, so if all of the above don't have known/affirmative answers, a Euclidean algorithm in $R$ (sans a Euclidean function) would be just as great instead. Thanks in advance for any answers. REPLY [5 votes]: It turns out that $K=\mathbb{Q}[i,\varphi]$ is norm-Euclidean. The proof of this fact appears as Appendix A in https://arxiv.org/abs/2205.03007. I'll explain the heft of the argument here. Let $R=\mathbb{Z}[i,\varphi]$ and $N=N_{K/\mathbb{Q}}$. We use this formulation of norm-Euclidean: for all $\alpha\in K$ there exists $\beta\in R$ with $N(\alpha+\beta)<1$. Model $K\cong\mathbb{Q}^4$ via $w+x\varphi+yi+zi\varphi\mapsto(w,x,y,z)$ whence $R$ identifies with $\mathbb{Z}^4$. If $K$ were "extra nice" we could simply verify that for all $\alpha\in\left[-\frac{1}{2},\frac{1}{2}\right)^4\cap K=C$, we have $N(\alpha)<1$ whence for $\alpha\in K$ we could simply let $\beta=-[\alpha]$ where $[\alpha]$ is $\alpha$ with each coordinate rounded to the nearest integer. (This is how the standard proof goes for the Gaussian number field being norm-Euclidean.) Unfortunately, this approach fails, particularly near the corners where exactly three components have the same sign. Notice, however, that $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$ has very small norm, so that if we shift only some points in $C$ then we may be okay, since this would resolve e.g. $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},-\frac{1}{2}\right)$, and all points nearby, since $N$ is a polynomial in the coordinates, so it is definitely continuous. The trick then is to exactly quantify continuity. We will consider $C_n=\left[-\frac{1}{2},\frac{1}{2}\right)^4\cap(\frac{1}{n}\mathbb{Z}^4)$ with $n$ to be decided later. If we can show for each $\alpha\in C_n$ that there exists $\delta\in\mathbb{Z}^4$ such that $N(\alpha+\delta+B_n)<1$ where $B_n$ is the radius-$\frac{1}{n}$ $L^\infty$ ball in $\mathbb{Q}^4$, then we are done. We consider now finding an explicit bound for $N(\alpha+\beta)$ where $\alpha=w+x\varphi+yi+zi\varphi\in K$ is any point and $\beta=d_1+d_2\varphi+d_3i+d_4i\varphi$ has $\|\beta\|_\infty<\varepsilon$, i.e. $\lvert d_1\rvert<\varepsilon$. $N(\alpha+\beta)$ is quartic in the eight variables, so write it out and group terms by "$d$-type," that is, $N(\alpha+\beta)=\sum_{e\in\mathbb{N}^4}d^eP_e(\alpha)$ for polynomials $P_e\in\mathbb{Q}[w,x,y,z]$. (e.g., $d_1d_2^2x+d_2d_3d_4z+2d_1d_2^2y+8d_4wyz$ would have $P_{1200}=x+2y$, $P_{0111}=z$, $P_{0001}=8wyz$, and $P_e=0$ otherwise.) Then, apply the triangle inequality (and positivity of $N$): $N(\alpha+\beta)=\left\lvert\sum_{e\in\mathbb{N}^4}d^eP_e(\alpha)\right\rvert\le\sum_{e\in\mathbb{N}^4}\lvert d\rvert^e\lvert P_e(\alpha)\rvert\le\sum_{e\in\mathbb{N}^4}\varepsilon^{\sum e}\lvert P_e(\alpha)\rvert$. (For the same example, the bound becomes $8\varepsilon\lvert wyz\rvert+\varepsilon^3(\lvert x+2y\rvert+\lvert z\rvert)$.) This now depends entirely on $\alpha$ and $\varepsilon=\frac{1}{2n}$. Let that bound be $f(\alpha,n)$ (also a polynomial in $\alpha$'s coordinates). The strategy is then to check with a computer all $n^4$ elements $\alpha\in C_n$, and for each evaluate $f(\alpha,n)$. If $f(\alpha,n)<1$ then we move on, otherwise check $f(\alpha\pm e_j,n)$ for $e_j$ one of the four Kronecker basis vectors. It turns out that when $n=6$ we only ever need to check $\alpha$ and $\alpha\pm e_j$ (i.e. 0 or 1 steps) for this number field. Curiously, it would seem that this approach could work to show that other biquadratic number fields $\mathbb{Q}\left[i,\sqrt{m}\right]$ of relatively small discriminant are norm-Euclidean, but this approach does not yield any results for $5 TITLE: Can you do math without knowing how to count? QUESTION [10 upvotes]: Are there mathematical theories that do not use intuitive integers? [That is, do not use integers to write statements.] Can you propose a theory that describes natural integers, without using intuitive integers to state axioms, definitions, and propositions? REPLY [3 votes]: There is a famous anecdot concerning Grothendieck and a certain integer. In the sixties, Alexander Grothendieck was developping a complex algebro-arithmetic framework in order to prove a set of conjectures due to Weil concerning varieties over finite fields. One motivation was to get a better grasp of integer solutions of polynomial equations using reduction mod p. The lectures given by Grothendieck were very abstract and one of the auditor asked for an example. "Sure," said Grothendieck, "let's take some prime number p, for example 57". And so now 57 is known as the Grothendieck prime. In modern algebraic geometry, an prime integer is a maximal ideal in the ring $\bf Z$. I don't remember what $\bf Z$ is exactly. Some algebraic geometer may come forward and correct me if I am wrong but I think that this is the terminal object in the category of scheme. Or maybe the antediluvian one. Another fashionable topic at the moment is the field with one element. The number 1 of course should not be understood in its intuitive meaning. All jokes taken aside, some parts of arithmetics nowadays are pretty sophisticated and do not rest on an intuitive notion of integers. 6 for example should be interpreted as a global section of the canonical sheaf of $spec({\bf Z})$, that is as a kind of function that associates to each point p in $spec({\bf Z})$ (a prime number actually) an element of ${\bf Z} / p{\bf Z}$ (the stalk of the sheaf at p) obtained by reducing 6 mod p. This allows to draw a parallel with the polynomial ring ${\bf C}[X]$, which happens to be a ring of functions over the complex line, and use idea coming from analytic geometry to study integers. At that point, you should be confused enough to understand that integers are not what they used to be.<|endoftext|> TITLE: Comparison of two monodromies QUESTION [6 upvotes]: Let us consider a smooth projective curve $\Sigma_b$ of genus $b$, and assume that there is a surjective group homomorphism $$\varphi \colon \pi_1 (\Sigma_b \times \Sigma_b - \Delta) \to G,$$ where $\Delta \subset \Sigma_b \times \Sigma_b $ is the diagonal and $G$ is a finite group. Assume moreover that the homomorphisms $$\psi_i \colon \pi_1(\Sigma_b-\{p \}) \to G, \qquad i=1,\, 2,$$ induced by the inclusion of the two factors, are also surjective. Then, by Grauert-Remmert Extension Theorem, there exist a holomorphic $G$-cover $X \to \Sigma_b \times \Sigma_b$, branched over the diagonal, that, after composing with the first (or the second) projection, yields a Kodaira fibration, namely, a non-isotrivial holomorphic fibration $f \colon X \to \Sigma_b$ with smooth and connected fibres. In fact, everything turns out to be algebraic. Calling $g$ the fibre genus of $f$, the fundamental group $\pi_1(X)$ sits into a split exact sequence $$1 \to \pi_1(\Sigma_g) \to \pi_1(X) \to \pi_1(\Sigma_b) \to 1,$$ that gives, by conjugation, monodromy representations $$\rho \colon \pi_1(\Sigma_b) \to \operatorname{Out} \, \pi_1(\Sigma_g), \quad \bar{\rho} \colon \pi_1(\Sigma_b) \to \operatorname{Aut} \, H^1(\Sigma_g, \, \mathbb{Q}).$$ On the other hand, the homomorphism $\psi_1$ (or $\psi_2$) realizes the fibre $\Sigma_g$ of $f$ as a $G$-cover $\Sigma_g \to \Sigma_b$, branched over exactly one point, and so there is a short exact sequence $$1 \to \pi_1(\Sigma_g) \to \pi_1(\Sigma_b-\{p\})^{\operatorname{orb}} \to G \to 1.$$ Here the group in the middle is the quotient of $\pi_1(\Sigma_b-\{p\})$ by the normal closure of the subgroup $\langle \gamma^s \rangle$, where $\gamma$ is a generator that loops around $p$ and $s$ is the order of $\psi_1(\gamma) \in G$. This sequence gives, by conjugation, two monodromy representations $$\rho_G \colon G \to \operatorname{Out} \, \pi_1(\Sigma_g), \quad \bar{\rho}_G \colon G \to \operatorname{Aut} \, H^1(\Sigma_g, \, \mathbb{Q}).$$ Question. How are $\bar{\rho}$ and $\bar{\rho}_G$ related? In particular, is it true that their invariant subspaces in $H^1(\Sigma_g, \, \mathbb{Q})$ are equal (or, at least, that they have the same dimension)? Edit. Will Sawin's answer shows that the invariant subspace of $\bar{\rho}$ always contains the invariant subspace of $\bar{\rho}_G$. Can one provide some conditions ensuring that also the reverse inclusion holds? Note that, since I am assuming non-trivial ramification of the $G$-cover $X \to \Sigma_b \times \Sigma_b$, then $\psi_1(\gamma) \in G$ is non-trivial. If this can help, in my specific situation $G$ is an extra-special group of order $32$ and $\psi_1(\gamma)$ is the generator of the center $Z(G) \simeq \mathbb{Z}_2$. REPLY [7 votes]: The $\tilde{\rho}$-invariants contain the $\tilde{\rho}_G$-invariants, at least. The map $\Sigma_g \to \Sigma_b$ defines a pullback map $ H^1( \Sigma_b, \mathbb Q) \to H^1(\Sigma_g, \mathbb Q) $ (i.e. cohomology is a contravariant functor). The image of this pullback map is the $G$-invariants in $H^1(\Sigma_g, \mathbb Q)$. The image of this pullback map consists of $G$-invariants as a formal consequence of the fact that $G$ acts by automorphisms of $\Sigma_g$ over $\Sigma_b$. It consists of all the $G$-invariants because, given any $1$-form whose cohomology class is $G$-invariant, we can average to get a $1$-form that is itself $G$-invariant, which necessarily descends from $\Sigma_b$. (Or take the trace / integration map $H^1(\Sigma_g, \mathbb Q) \to H^1(\Sigma_b, \mathbb Q)$, divide by $|G|$, and pull back.) We can check that this pullback map is invariant under the monodromy action of $\Sigma_g$. To do this, one way is to factor $\Sigma_g \to \Sigma_b$ as $\Sigma_g \to X \to \Sigma_b \times \Sigma_b \to \Sigma_b$, with the last map projection onto the left factor, and to note that the image of $H^1(X, \mathbb Q) \to H^1(\Sigma_g, \mathbb Q)$ is the $\pi_1(\Sigma_b)$-invariants of $H^1(X, \mathbb Q)$. So the image of this map is contained in the monodromy invariants.<|endoftext|> TITLE: What's the average order of the reduction of a section of an elliptic curve QUESTION [8 upvotes]: Suppose $E$ is an elliptic curve over $\mathbb Q$ and $x \in E(\mathbb Q)$ is not torsion. We can reduce $x \pmod p$ for a prime $p$ of good reduction and it will have some order $n_p$ in the group $E(\mathbb F_p)$. Has there been any work on the asympotitcs of the average of $n_p$ for $p < X$ as $X \to \infty$? More generally, suppose $x,y \in E(\mathbb Q)$ are two linearly independent sections and let them generate subgroups $G_x(p),G_y(p) \subset E(\mathbb F_p)$ for a prime of good reduction. Have the asymptotics of the average of $G_x(p)\cap G_y(p)$ been studied? This question seems tangentially related. REPLY [4 votes]: Let $E/\mathbb{Q}$ be an elliptic curve. There exist positive integers $d_p$ and $e_p$, with $d_p|e_p$, such that group $E(\mathbb{F}_p)$ is isomorphic to $\mathbb{Z}/d_p\mathbb{Z} \times \mathbb{Z}/e_p\mathbb{Z}$. Kowalski conjectured that there exists a constant $c_E>0$ such that $\sum_{p\leq x}d_p\sim c_E f_E(x)$, where $f_E(x)=x$ if $E$ has CM and $f_E(x) = \mathrm{Li}(x)$ otherwise. Before Kowalski's paper, Duke showed that if $f$ is any function with $\lim_{t\to\infty}f(t)=\infty$, then for almost all primes $p$, $E(\mathbb{F}_p)$ contains a cyclic group of order $p/f(p)$ (where "almost all" is quantified in his paper). He used GRH for certain Dedekind zeta functions to handle the non-CM case, but not the CM case. Freiberg and Pollack unconditionally proved an $\asymp$ version of Kowalski's conjecture in the CM case. We appear to be far off from such a result in the non-CM case. Cojocaru has extensively studied the distribution of $p$ such that $E(\mathbb{F}_p)$ is cyclic, in which case $d_p=1$. Again, progress towards GRH is usually key for the non-CM case, but not the CM case. Duke, William, Almost all reductions modulo (p) of an elliptic curve have a large exponent., C. R., Math., Acad. Sci. Paris 337, No. 11, 689-692 (2003). ZBL1048.11045. Kowalski, E., Analytic problems for elliptic curves, J. Ramanujan Math. Soc. 21, No. 1, 19-114 (2006). ZBL1144.11069. Freiberg, Tristan; Pollack, Paul, The average of the first invariant factor for reductions of CM elliptic curves mod (p), Int. Math. Res. Not. 2015, No. 21, 11333-11350 (2015). ZBL1398.11088. Cojocaru, Alina Carmen, Primes, elliptic curves and cyclic groups, Bucur, Alina (ed.) et al., Analytic methods in arithmetic geometry. Arizona winter school 2016, the University of Arizona, Tucson, AZ, USA, March 12–16, 2016. Providence, RI: American Mathematical Society (AMS); Montreal: Centre de Recherches Mathématiques (CRM). Contemp. Math. 740, 1-69 (2019). ZBL1452.11069.<|endoftext|> TITLE: Polynomial bijections $\mathbb{Z}^2\to\mathbb{Z}^2$ are almost linear QUESTION [13 upvotes]: Let $M$ be the monoid of bijections $\phi:\mathbb{Z}^2\to\mathbb{Z}^2$ such that $\phi(x, y)=(P(x, y), Q(x, y))$ for some polynomials $P, Q\in \mathbb{Z}[x, y]$. Does $\mathrm{GL}_2(\mathbb{Z})\subset M$ together with the maps of the form $(x, y)\to (x+R(y), y)$ where $R\in \mathbb{Z}[y]$ generate $M$? That would imply that $M$ is secretly a group. REPLY [7 votes]: Elements of $M$ have inverses which are integer-valued polynomial maps, so the version with integer-valued polynomial maps forms a group. In fact, we only need to assume surjectivity to guarantee the existence of a polynomial inverse, so the monoid of surjective integer-valued polynomials from the plane to the plane is also a group. I don't know if this implies your desired classification result. $(P,Q)$ defines a map $\mathbb A^2_{\mathbb Q} \to \mathbb A^2_{\mathbb Q}$, which since the source and target have the same dimension must be generically finite. Because the source is irreducible, Hiblert's irreducibility theorem implies that the fibers over most integral points are irreducible, which if the degree is $>1$ contradicts the claim that $(P,Q)$ is surjective. So this map must have degree $1$, i.e. it is a birational transformation, so there are rational functions $A,B$ with $A( P(x,y), Q(x,y))=x, B(P(x,y), Q(x,y))=y$ for $x$, $y$ on a dense open set. Because $P,Q$ take every integer value, on a dense open set they take every integral value on a dense open set, so $P$ and $Q$ must take integral values on a dense open set. If $P$ and $Q$ are rational functions but not polynomials this is impossible, so $P$ and $Q$ must be integer-valued polynomial functions.<|endoftext|> TITLE: Orientation reversal and restriction to submanifold of lower dimension QUESTION [5 upvotes]: Let $M$ be a connected closed oriented manifold with at least one orientation-reversing homeomorphism $M\to M$. Let $S\subset M$ be a connected closed embedded submanifold of lower dimension. Let $f:M\to M$ be an orientation-preserving homeomorphism. Is there an orientation-reversing homeomorphism $g:M\to M$ such that $f|_S=g|_S$? REPLY [7 votes]: If I understand the question correctly, the lens space $L(5,2)$ provides a counterexample. Let $f$ be the identity, and let $S \subset L$ be a circle that generates the first homology. For concreteness sake, take $L$ to be the union of two solid tori of the form $S^1 \times D^2$, and let $S = S^1 \times pt$ in one of the solid tori. So the restriction of $f$ to $S$ is the identity. From the classification of lens spaces, there is an orientation reversing self-homeomorphism $g:L \to L$. Any such $g$ acts by multiplication by $\pm 2$ on $H_1(L)$. In particular, its restriction to $S$ can't be the identity.<|endoftext|> TITLE: Trying to understand "a refinement of the Peter–Weyl theorem" by Lusztig QUESTION [8 upvotes]: "A refinement of the Peter–Weyl theorem" is the title of Chapter 29 in Lusztig's "Introduction to quantum groups" (Birkhäuser 2010, reprint of the 1994 edition). This chapter is inside Part IV ("Canonical basis of $\dot{\mathbf U}$", chapters 23–30). If I understand correctly, the said refinement occurs in Theorem 29.3.3. Here is what I was able to understand from this chapter. Lusztig works with certain modification $\dot{\mathbf U}$ of the (generic?) quantum universal enveloping algebra $\mathbf U$ obtained by removing all the $K$ generators and the unit, and replacing them with the weight space; his motivation is that every $\mathbf U$-module with a weight decomposition can be viewed as an $\dot{\mathbf U}$-module. Then he constructs a canonical basis $\dot{\mathbf B}$ of $\dot{\mathbf U}$ and partitions it into disjoint subsets $\dot{\mathbf B}[\lambda]$ by weights. Then for each $\lambda$ he constructs two-sided ideals $\dot{\mathbf U}[{\geqslant\lambda}]\supseteq\dot{\mathbf U}[{>\lambda}]$ of $\dot{\mathbf U}$, such that the image $\pi(\dot{\mathbf B}[\lambda])$ of $\dot{\mathbf B}[\lambda]$ in $\dot{\mathbf U}[{\geqslant\lambda}]/\dot{\mathbf U}[{>\lambda}]$ is a basis. He proves that the quotient $\dot{\mathbf U}[{\geqslant\lambda}]/\dot{\mathbf U}[{>\lambda}]$ has a unique direct sum decomposition into a bunch of simple left $\dot{\mathbf U}$-modules (call this bunch, say, $\mathscr L_\lambda$) and into another bunch of simple right $\dot{\mathbf U}$-modules (say, $\mathscr R_\lambda$). And then he proves that for each $L\in\mathscr L_\lambda$ and each $R\in\mathscr R_\lambda$ the intersection $L\cap R$ is spanned by a unique element of the basis $\pi(\dot{\mathbf B}[\lambda])$ and this gives a bijection between $\mathscr L_\lambda\times\mathscr R_\lambda$ and the basis $\pi(\dot{\mathbf B}[\lambda])$. There is some additional piece of valuable information which I omit, maybe it is still relevant? I see that this is a beautiful theorem, and I see very vaguely connection with the Peter–Weyl theorem (which says that matrix elements of representations of a compact topological group $G$ are dense in $C(G)$). Can you help me out? In what sense exactly is this a refinement of the Peter–Weyl theorem? For example (as a subquestion) does this require, as a step, to switch from the commutative noncocommutative function algebra $C(G)$ to (some version of) the cocommutative noncommutative group algebra, or the refinement works without that? Does it (another subquestion) also refine the analytic aspects like the key question whether matrix elements separate points, or is it confined to the purely algebraic version? Final subquestion — are you aware of any followup work on that? Relevant previous questions (but I don't think they render this a duplicate; if somebody explains how some of them (or any others) do, I will happily close this one): Peter-Weyl theorem (compact quantum groups) Peter-Weyl vs. Schur-Weyl theorem Canonical basis for the extended quantum enveloping algebras Peter-Weyl theorem as proven in Cartier's Primer Is there analogue of Peter-Weyl theorem for non-compact or quantum group REPLY [5 votes]: For a complex reductive algebraic group $ G $, the Peter-Weyl theorem gives an isomorphism of $ G \times G $ representations $$ \mathbb C[G] \cong \oplus_{\lambda} V(\lambda)^* \otimes V(\lambda) $$ Here $ V(\lambda) $ is the irreducible representation with highest weight $\lambda $, $ \mathbb C[G] $ is the ring of regular functions on $ G $ and we have a map from the right to the left by forming matrix coefficients. This is the direct analog of the Peter-Weyl theorem for compact groups. Next, we have a $ G \times G$-equivariant perfect pairing $ U \mathfrak g \otimes \mathbb C[G] \rightarrow \mathbb C $, by regarding $ U \mathfrak g $ as differential operators at the origin. This allows us to get a Peter-Weyl theorem for $ U \mathfrak g $ and a similar statement holds for the quantum group $ \mathbf U $. I didn't look at the chapter in Lusztig's book, but I believe that his result is a refinement of this statement: his canonical basis of $ \mathbf U $ is compatible with this decomposition (at least after taking associated graded).<|endoftext|> TITLE: Applications of Zorn’s lemma that aren’t chain-complete/directed-complete? QUESTION [27 upvotes]: Zorn’s Lemma applies to posets in which every chain has an upper bound. However, in all applications I know, the poset is also evidently chain-complete — chains have least upper bounds. A few classic such applications: AC, using a poset of “partial choice functions” the Well-Ordering Principle, using a poset of “partial well-orderings” (ordered by end-extension, not simply $\subseteq$) Hahn–Banach, using a poset of “partial functionals defined on subspaces” extension of filters to ultrafilters, using a poset of filters. Are there any natural applications of Zorn’s Lemma where the poset isn’t chain-complete, or where chain-completeness is less obvious than existence of upper bounds? (Indeed, all examples I know are equally obviously directed-complete. Under AC, this is equivalent to chain-complete, but constructively, directed-completeness is stronger. For those wondering why I’m caring about use of AC while considering Zorn’s Lemma, note that constructively, ZL does not imply AC — this is shown in John Bell’s very nice analysis Zorn's lemma and complete Boolean algebras in intuitionistic type theories, JSL 1997, https://doi.org/10.2307/2275642. His surprising (to me) insight is that ZL itself doesn’t imply LEM or AC — most applications, including the classical proof of ZL=>AC, use LEM for the final step “a maximal partial object must be total”.) REPLY [15 votes]: Although this is not an application of Zorn's lemma, the Hausdorff order of almost inclusion on subsets of $\mathbb{N}$ is a natural order where every chain has an upper bound, but many chains do not have least upper bounds. Namely, the almost-inclusion relation $A\subseteq^*B$ for subsets $A,B\subseteq\mathbb{N}$ holds when all but finitely many elements of $A$ are in $B$. One can take equivalence classes with respect to the almost equal relation $A=^*B$, and induce the order on the quotient. Every chain has an upper bound in its union, but nontrivial countable chains never have least upper bounds, as Hausdorff proved. So this is a natural order where every chain has an upper bound, but many chains do not have least upper bounds. Another type of example occurs with the Turing degrees. Every countable chain in the Turing degrees has an upper bound, simply by encoding the whole chain, but no nontrivial countable chain has a least upper bound in light of the exact pair phenomenon. If one were simply to add a node atop the whole structure, then every chain would have an upper bound, but no nontrivial countable chains would have least upper bounds, providing another example. I suppose another trivial example would be the rational unit interval $\mathbb{Q}\cap[0,1]$. Clearly that isn't what you are looking for.<|endoftext|> TITLE: Homotopy type of continuous/smooth/analytic loop spaces? QUESTION [7 upvotes]: Apologies in advance if this is well-known; a google search did not produce anything useful. Let $(M,p)$ be a pointed real analytic manifold. Are the (free or pointed) loop spaces of continuous, smooth and analytic loops in $M$ all homotopy equivalent? REPLY [15 votes]: Suppose $S$ and $M$ are smooth manifolds. For simplicity let us also suppose that $S$ is compact. Then the inclusion $C^\infty(S, M)\hookrightarrow C^0(S, M)$ is a weak equivalence. This is a "standard" fact that follows from the Whitney approximation theorem. A proof can be found the book of Hirsch. There is a discussion here. If $S$ and $M$ are real analytic manifolds my guess is that an analogous result holds for the inclusion of the space real analytic maps, because there exists an analogue of Whitney's approximation theorem for real analytic maps. But I have not seen this implication derived explicitly.<|endoftext|> TITLE: Visualizing holomorphic differentials on a compact Riemann surface? QUESTION [9 upvotes]: It is a classical result that the vector space of holomorphic differentials on a compact Riemann surface of genus $g$ has dimension $g$. I am wondering if there is a way of visualizing this wonderful result. I should perhaps say a bit more about what I mean by "visualizing." I found another MO question on visualizing the Riemann–Roch theorem, and what I am asking for here could be viewed as the simplest case of Riemann–Roch. There are some interesting answers to that question, but I am hoping for something more literally visual, along the lines of Dan Piponi's note On the visualisation of differential forms and/or the paper On the geometry, flows and visualization of singular complex analytic vector fields on Riemann surfaces by Alvarez-Parrilla et al. The ultimate dream would be some kind of animation that allows one to get some intuitive feeling for what holomorphic (or harmonic, if that is easier) differentials are constrained to be like, and why adding an extra hole adds an extra degree of freedom. REPLY [6 votes]: This is not really an answer, it is a comment with images. The paper "Quadrilateral Mesh Generation II : Meromorphic Quartic Differentials and Abel-Jacobi Condition" (arXiv 2019) contains (imo) nice exposition of theoretical background and several illuminating illustrations. REPLY [5 votes]: There is a very nice hydrodynamic interpretation, going back to Riemann and Klein, and popularized by Courant (Hurwitz and Courant, Function theory). For those who do not read German, there is a modern exposition in French: E. Ghys and D. Smai, Six lecons autour des surfaces de Riemann Edit. Klein's book is "On Riemann's theory of algebraic functions", the only one of the three which has been translated into English.<|endoftext|> TITLE: Fourier series analysis QUESTION [5 upvotes]: \begin{equation} F_{r}(\theta)=\sum_{k=1}^{\infty} \frac{\cos (k \theta)}{k^{r}} \qquad 0\leq \theta\leq 2\pi \end{equation} \begin{equation} F_{2}(\theta)=\frac{1}{6} \pi^{2}-\frac{1}{2} \pi \theta+\frac{1}{4} \theta^{2} \qquad 0\leq \theta\leq 2\pi \end{equation} When $r>1$, how to prove that $F_{r}$ is monotonically decreasing on $[0,\pi]$ and is monotonically increasing on $[\pi,2\pi]$? \begin{equation} F_{1}(\theta)=-\log|2\sin(\frac{\theta}{2})| \qquad 0\leq \theta\leq 2\pi \end{equation} When $00$ is opposite to the sign of $\sin t$. So, the desired monotonicity follows. Also, $$f(t,x)=\frac{\sinh x}{2 (\cosh x-\cos t)}-\frac12.$$ So, by monotone convergence, for $r\in(0,1)$ we have $$\int_0^1 dx\,x^{r-1}f(t,x)\to\int_0^1 dx\,x^{r-1}\Big(\frac{\sinh x}{2 (\cosh x-1)}-\frac12\Big)=\infty$$ as $t\to2n\pi$, for any integer $n$, since $\sinh x\sim x$ and $\cosh x-1\sim x^2/2$ as $x\to0$. On the other hand, for real $x>1$, $$|f(t,x)|\le\sum_{k=1}^\infty e^{-k x}\le 2e^{-x}$$ and hence for $r\in(0,1)$ we have $$\int_1^\infty dx\,x^{r-1}|f(t,x)|\le2\int_1^\infty dx\,e^{-x}<1.$$ Thus, $$\Gamma(r)F_r(t)\ge\int_0^1 dx\,x^{r-1}f(t,x)-\int_1^\infty dx\,x^{r-1}|f(t,x)|\to\infty$$ as $t\to2n\pi$, for any integer $n$. So, the claims about the limits follow as well. (Justification of the convergence of the series defining $F_r(t)$, as well as of the interchange of the summation and the integration in (1).) Take any real $r\in(0,2)$. Take any $t\in(0,2\pi)$. Let us show that the limit \begin{equation*} F_r(t)=\lim_{n\to\infty}F_{r,n}(t) \end{equation*} exists, where \begin{equation*} F_{r,n}(t):=\sum_{k=1}^n\frac{\cos kt}{k^r}. \end{equation*} Indeed, for all natural $m$ and $n>m$, \begin{equation*} \begin{aligned} F_{r,n}(t)-F_{r,m}(t)&=\sum_{k=m+1}^n\cos kt\; \frac1{\Gamma(r)}\int_0^\infty dx\,x^{r-1}e^{-k x} \\ &=\frac1{\Gamma(r)}\int_0^\infty dx\,x^{r-1}\sum_{k=m+1}^n\cos kt\, e^{-k x}. \end{aligned} \end{equation*} Next, for $x>0$, \begin{align*} \Big|\sum_{k=m+1}^n\cos kt\, e^{-k x}\Big| & =\Big|\Re\sum_{k=m+1}^n e^{-k(it+x)}\Big| \\ & \le\Big|\sum_{k=m+1}^n e^{-k(it+x)}\Big| \\ &=e^{-(m+1)x}\,\frac{|1-e^{-(n-m)(it+x)}|}{|1-e^{-it-x}|} \\ &\le e^{-(m+1)x}\,\frac2{2\sin(t/2)\,e^{-x/2}} \\ &\le \,\frac{e^{-mx}}{\sin(t/2)}. \end{align*} So, $F_{r,n}(t)$ is Cauchy-convergent in $n$ and hence convergent in $n$. Also, by dominated convergence, we can interchange the summation and the integration in (1).<|endoftext|> TITLE: Are there simplicial spheres with "non-geometric symmetries"? QUESTION [7 upvotes]: Let $\Delta$ be a simplicial sphere, that is, a finite (abstract) simplicial complex whose canonical geometric realization $|\Delta|$ is homeomorphic to a sphere $\mathbf S^d\subset\Bbb R^{d+1}$. Question: Can the homeomorphism $\phi :|\Delta|\to\mathbf S^d$ be chosen in a way, so that all combinatorial symmetries of $\Delta$ are realized geometrically? That is, if $\sigma :\Delta\to\Delta$ is a combinatorial automorphism of $\Delta$ (a bijective simplicial map) I want there to be an isometry $\smash{f_\sigma:\mathbf S^d\to\mathbf S^d}$ so that $$\phi\circ \sigma = f_\sigma\circ \phi.$$ You can think of this as a subdivision of $\mathbf S^d$ that has the same symmetries as the abstract simplicial complex $\Delta$. If we consider the sphere embedded in $\smash{\Bbb R^{d+1}}$, the isometries are exactly the orthogonal transformations restricted to $\smash{\mathbf S^d}$. REPLY [8 votes]: The answer is negative. Already in dimension 4 there are fake real-projective spaces, which are smooth 4-manifolds homotopy-equivalent but not homeomorphic to $RP^4$. These correspond to smooth free involutions $\sigma: S^4\to S^4$ which are not topologically conjugate to orthogonal transformations. Similar examples exist in higher dimensions. See manifold atlas for references. However, in dimensions $n\le 3$ indeed, every finite group of PL homeomorphisms of $S^n$, is PL conjugate to a finite subgroup of the orthogonal group. This is easy in dimension 2 and hard in dimension 3 (a consequence of the "orbifold geometrization theorem"). Edit. A nice, although dated, survey is M.Davis, A survey of results in higher dimensions, In "The Smith Conjecture", (editors: J. W. Morgan and H. Bass), Academic Press, New York, 1984. dealing with examples of "exotic" actions of compact (in particular, finite) groups on spheres. Some of the examples he discusses are PL. Note that smooth actions of finite groups can be always made PL.<|endoftext|> TITLE: Does every countable set of Turing degrees have an upper bound, without AC? QUESTION [11 upvotes]: It is easy to see that every countable collection of sets $A_n\subseteq\mathbb{N}$ has an upper bound in the Turing degrees, since we can just take a copy of their disjoint sum $\oplus_n A_n=\{\langle n,m\rangle\mid m\in A_n\}$, using a computable pairing function $\langle\cdot,\cdot\rangle$. From this, we can easily compute each $A_n$. My question is whether we can do that just in the Turing degrees, where we don't have the $A_n$ as sets, but only their degrees $[A_n]_T$. Question. Without using AC, does every countable set of Turing degrees have an upper bound? Of course, if we had countable choice, we could choose a representative from each degree and then apply the earlier disjoint sum operation. It is consistent with ZF that the reals are a countable union of countable sets, but I couldn't see directly how to turn this into an answer to the question. A similar question seems to arise in the Hausdorff order of almost-inclusion $\subseteq^*$, where $A\subseteq^* B$ if and only if all but finitely many elements of $A$ are in $B$. Let's mod out by the corresponding equivalence relation $A=^*B\iff A\subseteq^*B\subseteq^*A$, and let's remove the top element in the quotient. So we are looking at the almost-equal equivalence classes of co-ininfinite sets, under almost inclusion. Hausdorff proved that every countable increasing chain of almost inclusion has a co-infinite upper bound. But without AC, it is not clear how to use this argument when you are initially given only the almost-equality equivalence classes of the sets, rather than the sets themselves. Question. Without AC, is every countable increasing chain of almost-equality equivalence classes of co-infinite sets bounded by another such class? You seem to need countable choice to pick the representatives. This question arose in a comment on another question. REPLY [12 votes]: No. It is quite possible to get a counterexample, even without having the reals as a countable union of countable sets. Indeed, we only need to add $\omega$ Cohen reals in order to find such a symmetric extension! Let's first make a small observation: Suppose that there is a sequence of equivalence classes mod-finite without a choice function, then there is a countable family of Turing degrees without an upper limit. Proof. Note that if $\{a_n\mid n<\omega\}$ is the family of equivalence classes mod finite, then for every $a\in a_n$, $[a]_T=\bigcup\{[a']_T\mid a'\in a_n\}$. This is because $\sim_{\rm fin}$ is a refinement of $\sim_T$. If we had an upper bound to these Turing degrees, let $t$ be a representative, then we can now use $t$ to compute a representative from each $a_n$ by simply finding the smallest program that produces a representative with $t$ as an oracle. $\square$ Therefore, it is enough to find a model where there is a countable sequence of such equivalence classes. For example, if we add $\omega$ Cohen reals, $r_n$, and forget $\langle r_n\mid n<\omega\rangle$, while preserving $\langle [r_n]_{\rm fin}\rangle$. So let's construct a symmetric extension in which this happens. The idea is as follows: find out how to change each Cohen real on a finite set in an arbitrary fashion, then consider permutations that do exactly that on each of them, separately, and without changing the indices of your newly found Cohen reals; and take a good notion of finite support to generate your filter of subgroups. We force with $\operatorname{Add}(\omega,\omega)$, so a condition is a finite partial function $p\colon\omega\times\omega\to2$. We denote by $\dot r_n$ the name of the $n$th real, $\{\langle p,\check m\rangle\mid p(n,m)=1\}$. Now we consider the group of automorphisms of $\operatorname{Add}(\omega,\omega)$ given by $\prod B_{\rm fin}$ or $\bigoplus B_{\rm fin}$, where $B_{\rm fin}$ is the group of "finite bit flips" on each coordinates. That is, if $\sigma\in B_{\rm fin}$, then there is some finite set $X\subseteq\omega$ such that: $$\sigma p(m)=\begin{cases}1-p(m) & m\in X\\ p(m) & m\notin X\end{cases}$$ so we can write $\sigma=\sigma_X$ in this case. Now, $\pi$ is in our group, let's denote it by $\scr G$, if $\pi\colon\omega\to B_{\rm fin}$, and $$\pi p(n,m) = \pi(n) p(n,m).$$ That is, $\pi$ acts on each coordinate separately with an automorphism from $B_{\rm fin}$. So we can write $\pi_n$ instead of $\pi(n)$, and if we write $\pi_n=\sigma_X$ then it is clear now what we mean by that. For this construction it doesn't matter if we use $\prod$ or $\bigoplus$, the latter is countable and absolute so it has some nicer properties overall. Finally, the filter of subgroups is given by finite supports: for $E\subseteq\omega$ we denote by $\operatorname{fix}(E)=\{\pi\in\mathscr G\mid\forall n\in E,\pi_n=\operatorname{id}\}$; now define $\scr F$ as the filter of subgroups generated by $\{\operatorname{fix}(E)\mid E\in[\omega]^{<\omega}\}$. If $\dot x$ is a hereditarily symmetric name, then we say that $E$ is a support for $\dot x$ when every $\pi\in\operatorname{fix}(E)$ satisfies $\pi\dot x=\dot x$. Claim. For each $n$, $\dot r_n$ is hereditarily symmetric. Proof. $\{n\}$ is a support for $\dot r_n$. $\square$ Claim. For each $n$, $[\dot r_n]_{\rm fin}$ is hereditarily symmetric. Proof. Note that every real in $[\dot r_n]_{\rm fin}$ has a name of the form $\pi\dot r_n$ for some $\pi\in\scr G$, and indeed, if $\pi\in\scr G$, then $1\Vdash\pi\dot r_n\sim_{\rm fin}\dot r_n$. Therefore $[\dot r_n]_{\rm fin}$ has a canonical name: $\{\pi\dot r_n\mid n<\omega\}^\bullet$. Therefore for every $\pi\in\scr G$ and every $n<\omega$, $\pi[\dot r_n]_{\rm fin}=[\dot r_n]_{\rm fin}$. $\square$ Corollary. The sequence $\langle[\dot r_n]_{\rm fin}\mid n<\omega\rangle^\bullet$ is also hereditarily symmetric. $\square$ Finally, we show that there is no choice function. Proposition. $1\Vdash^{\rm HS}\{[\dot r_n]_{\rm fin}\mid n<\omega\}^\bullet$ does not have a choice function. Proof. Suppose that $\dot f$ is a hereditarily symmetric name and $p$ forces that its domain is the above family. Let $E$ be a support for $\dot f$, and let $n\notin E$. Suppose that $q\leq p$ was a condition such that $q\Vdash\dot f([\dot r_n])=\dot x$, by extending $q$ if necessary, we may assume that $\dot x$ is actually $\tau\dot r_n$ for some $\tau\in\scr G$: all we need to do is find out what is that finite difference. Moreover, we can assume that $\tau_k=\operatorname{id}$ for all $k\neq n$, since the only coordinate that can change anything about $\dot r_n$ is $n$ itself. But now take some very large $m$ such that $\langle n,m\rangle\notin\operatorname{dom}q$, and simply consider $\pi$ for which: $\pi_n=\sigma_\{m\}$ and for all other $k$, $\pi_k=\operatorname{id}$. We now have that: $\pi q=q$, since no information inside the condition was changed. $\pi\in\operatorname{fix}(E)$, since $n\notin E$. By combining with the information from the claims we have that $$q\Vdash\dot f([\dot r]_{\rm fin})=\tau\dot r_n\iff\pi q\Vdash\pi\dot f(\pi[\dot r_n]_{\rm fin})=\pi\tau\dot r_n\iff q\Vdash\dot f([\dot r_n]_{\rm fin})=\tau\pi\dot r_n.$$ But we also know that $1\Vdash\tau\pi\dot r_n\neq\dot r_n$, since the two must disagree on whether $m$ is inside or outside! Therefore, if no extension of $p$ can force that $\dot f$ is a choice function, so $p$ must force that $\dot f$ is not a choice function. But that means that no condition can force that any name is choice function, and therefore we proved the statement of the proposition. $\square$ Let me also add that this can be simplified even further when considering iterations of symmetric extensions and improper filters of groups, since we can now just consider the whole thing an iteration of length $\omega$, with finite support of course, where at each step we have some finitary permutation, but as long as finite steps are taken, we are allowed to use the trivial subgroup in our filter. The limit step does all the work for us in this case! In fact, this exact example appears as a toy example for iterating symmetric extensions in my first paper on the topic: Karagila, Asaf, Iterating symmetric extensions, J. Symb. Log. 84, No. 1, 123-159 (2019). ZBL1448.03038.<|endoftext|> TITLE: Strict topology between weak and norm topologies QUESTION [7 upvotes]: I want to believe that this has an easy answer, but I’ve never considered it before and can’t seem to answer it now either. Does every infinite-dimensional Banach space admit a locally convex vector topology that is strictly coarser than the norm topology and strictly finer than the weak topology? If this is non-trivial and constructive, I’d be much obliged if an explicit example (or reference) is provided. REPLY [11 votes]: Another example is the topology of uniform convergence on the norm-compact subsets of the dual. It is coarser than the norm topology (which is the topology of uniform convergence on the dual unit ball) because compact sets are bounded and strictly coarser since the dual unit ball is not compact. It is finer than the weak topology since finite sets are compact and it is strictly finer because every linear independent null sequence in the dual together with its limit is compact and not finite dimensional.<|endoftext|> TITLE: Zero-knowledge proof for $P \ne NP$? QUESTION [5 upvotes]: In computational complexity, $P \ne NP$ is a widely believed conjecture. Suppose that someone discovered a proof for it. He wants to publish a proof that he correctly proved the conjecture. I am aware that NP-complete problems have Zero-knowledge proofs. Assuming $P \ne NP$, can we have Zero-knowledge proof (or morally equivalent one) for $P \ne NP$? REPLY [4 votes]: First of all, it does not really make sense to talk about zero-knowledge proofs for a single statement such as $P\ne NP$. You really should be asking about whether there is a zero-knowledge proof for a computational problem, such as: "Given a statement $S$ in the first-order language of set theory, and a positive integer $n$, does there exist a ZFC-proof of $S$ of length at most $n$?" With this reformulation, you have already partially answered your own question, by noting that there is a standard theorem about the existence of zero-knowledge proofs for any $NP$-complete language. However, there are two caveats to be aware of in that standard theorem. The first caveat is that it assumes the existence of a one-way function, which is a stronger assumption than $P\ne NP$. So if you only have the hypothesis $P\ne NP$ available to you, then that's not quite good enough. The second caveat is that the usual proof relies on a commitment scheme, which has two parts to it, the binding part and the hiding part. Ideally we'd like both parts to be statistically secure and not just computationally secure, but it turns out that it is impossible for both parts to be statistically secure unless the polynomial hierarchy collapses to the second level (Fortnow, "The complexity of perfect zero knowledge"). So if you want the strongest possible form of zero knowledge, then the answer to your question is likely no.<|endoftext|> TITLE: How many consecutive forced moves are possible in chess? QUESTION [5 upvotes]: The question concerns chess. I call a move forced if, in a given position, is the unique move consistent with the rules of the game. I wonder what is the largest integer $n$ such that there exists a legal position in which: both black and white have forced moves for $n$ consecutive times; the position is never repeated (with the same color having to move). Notice that without condition 2., the answer is $\infty$. REPLY [10 votes]: This question has been answered at chess.stackexchange.com. It seems that if you allow promoted pieces, the current record is $n=9$.<|endoftext|> TITLE: Do saturated models require choice? QUESTION [9 upvotes]: Let $T$ be a first-order theory, and suppose we want to build a saturated model $\mathbb U$ of $T$. That is, we want a model $\mathbb U$ of cardinality bigger than $|T|$, saturated in its own cardinality. In particular $\mathbb U$ will be universal and homogeneous for models of smaller cardinality. In order to do this, one often assumes GCH or the universe axiom. I believe that GCH implies choice, so I opt in the following to assume that there is an inaccessible cardinal $\kappa$ with $T \in V_\kappa$. This theory (call it ZFCI) proves the existence of a saturated model $\mathbb U$ of cardinality $\kappa$. Question 1: Does ZFI (analogous to above but without choice) prove the existence of a saturated model $\mathbb U$ of $T$? Question 2: If not, can the exsitence of $\mathbb U$ be proven with less than the full axiom of choice? Question 3: In the absence of choice, is there an alternative notion of saturatedness which is preferable to work with? In the above, by "saturated in its own cardinality", I think I mean "saturated with respect to subsets which do not admit a surjection to $\mathbb U$", but perhaps I should mean something slightly different. If it's easier to discuss models saturated in some fixed cardinality, or universal and homogeneous with respect to some fixed cardinality, I'd find results about such constructions interesting too. REPLY [4 votes]: On request, I summarize here some of the results mentioned in the comments (now unfortunately deleted), even though they do not really answer the question. For definiteness, I assume that $M$ is saturated means that for every $A\subseteq M$ that does not surject onto $M$, every partial $1$-type over $A$ which is finitely satisfiable is satisfied in $M$. Proposition. If every theory with a model has a saturated model, then: The Boolean Prime Ideal Theorem (aka Ultrafilter Lemma) holds. For every cardinal $l$, there exists a cardinal $k>l$ such that for all $m\le k$, either $m$ surjects onto $k$, or $k$ surjects onto $k^m$. (Here, none of the cardinals is assumed well ordered.) Proof: For 1, if $X$ is an arbitrary set, endow $X$ with the first-order structure with unary predicates $P_Y$ for each $Y\subseteq X$, let $T$ be the elementary diagram of $X$, and let $M$ be a saturated model of $T$. If $F\subseteq\mathcal P(X)$ is a nontrivial filter, then $p_F(x)=\{P_Y(x):Y\in F\}$ is a finitely satisfiable partial type over $\varnothing$, hence there is $a\in M$ such that $M\models p_F(a)$. Then $\{Y:M\models P_Y(a)\}$ is an ultrafilter extending $F$. For 2, let $T$ be, say, the Vaught set theory (VS), with axioms postulating for each standard natural number $n$ the existence of $\{x_0,\dots,x_{n-1}\}$ for all $x_0,\dots,x_{n-1}$. If $M\models T$ is saturated, let $k=|M|$. Fix $c\in M$. For any $f,u\in M$, we define $$ap(f,u)=\begin{cases}v&\text{if $v$ is unique such that $\langle u,v\rangle\in f$,}\\c&\text{if no such $v$ exists.}\end{cases}$$ If $m\le k$, fix $A\subseteq M$ such that $|A|=m$, and define a function $F\colon M\to M^A$ by $(F(u))(a)=ap(u,a)$. If $A\times2$ does not surject onto $M$, then $F$ is onto: for any $f\colon A\to M$, $p_f(u)=\{ap(u,a)=b:f(a)=b\}$ is a finitely satisfiable partial type over $A\cup f[A]$, and its realization is an $u\in M$ such that $F(u)=f$. [This only gives that $2m$ surjects onto $k$, or $k$ surjects onto $k^m$. However, restricting the argument to $f\colon A\to A$, we also get: $m$ surjects onto $k$, or $k$ surjects onto $m^m$, hence onto $2^{2m}$ (if $m\ge4$). In the latter case, $2m$ cannot surject onto $k$ due to Cantor’s theorem, hence $k$ surjects onto $k^m$ as wanted.] In order to ensure $k>l$, it suffices to expand $T$ with $l$ many pairwise distinct constants. QED One can strengthen condition 2 as follows: for any sequence $\langle A_u:u\in M\rangle$ of subsets $A_u\subseteq M$ that do not surject onto $M$, $M$ surjects onto $\bigcup_{u\in M}M^{A_u}$. To see this, modify $F$ such that if $w=\langle u,v\rangle$, then $F(w)$ is the function $A_u\to M$ defined by $(F(w))(a)=ap(v,a)$ for $a\in A_u$. Instead of VS, we can take any consistent theory with a definable function $ap(-,-)$ that satisfies the axioms $$\forall x_1,\dots,x_n,y_1,\dots,y_n\:\exists z\:\Bigl(\bigwedge_{i\ne j}x_i\ne x_j\to\bigwedge_iap(z,x_i)=y_i\Bigr)$$ for each $n$. Note that in ZFC, condition 2 says that there are arbitrarily large cardinals $\kappa$ such that $\kappa^{<\kappa}=\kappa$, which is equivalent to the existence of saturated models for every theory. Without AC, the conditions are likely only necessary, not sufficient. The question specifically invokes inaccessible cardinals. By the (now unfortunately deleted) comment by Asaf, I’m taking inaccessibility to mean: uncountable regular well-ordered cardinal $\kappa$ such that no $x\in V_\kappa$ surjects onto $\kappa$. Whether the inaccessibility of $\kappa$ implies the existence of some saturated model for any consistent theory $T\in V_\kappa$ is something I strictly speaking cannot answer. However, if I take the question to mean if it implies the existence of saturated models of size $\kappa$ (which is how inaccessibles work in the classical case), this cannot work without assuming, essentially, that choice holds up to $V_\kappa$: Corollary. If $\kappa$ is an inaccessible cardinal, the following are equivalent: Every theory with a countable model has a saturated model of cardinality $\kappa$. (We really just need this to hold for VS.) Every consistent theory $T\in V_\kappa$ has a saturated model of cardinality $\kappa$. $V_\kappa$ is well orderable (hence necessarily of cardinality $\kappa$). Proof: If $V_\kappa$ can be well ordered, this gives enough of choice so that the usual construction of saturated models of size $\kappa$ goes through. On the other hand, by the Proposition, 1 implies that $\kappa$ surjects onto $\kappa^\mu$ for all $\mu<\kappa$; this makes $\kappa^\mu$ well orderable, hence in fact (using the inaccessibility of $\kappa$) $\kappa^\mu=\kappa$, and $2^\mu<\kappa$. A standard argument by induction on $\alpha<\kappa$ then shows that $V_\alpha$ is well orderable of cardinality $<\kappa$ (see e.g. the proof of Theorem 9.1 (b) in Jech, The Axiom of Choice.) In fact, this inductive argument gives the well-orderability of $V_\kappa$ itself if we can prove that $\bigcup_{\alpha<\kappa}\mathcal P(\alpha)$ can be well ordered, such as by showing that $\kappa$ surjects onto $\bigcup_{\alpha<\kappa}\kappa^\alpha$. This follows by the strengthening mentioned below the Proposition. QED<|endoftext|> TITLE: Root system of fixed point Lie sub-algebra QUESTION [7 upvotes]: It is known that a non-simply laced simple root system can be constructed from the simply-laced root system by folding the Dynkin diagram and hence the corresponding non-simply-laced Lie algebra can be constructed by taking the fixed points of a non-trivial diagram automorphism (outer automorphism). Now let $\theta$ be an inner automorphism of order $2$ of a simple Lie algebra $\mathfrak{g}$ over $\mathbb C$ and let $\mathfrak{g}= \mathfrak{g}_0 \oplus \mathfrak{g}_1$ be the eigendecomposition. The fixed point subalgebra $\mathfrak{g}_0$ is reductive. Now is there a way to get the root system of $\mathfrak{g}_0$ from the root system of $\mathfrak{g}$? REPLY [7 votes]: Let ${\frak g}$ be a simple Lie algebra over $\Bbb C$, and let $\theta$ be an inner involution of ${\frak g}$, that is, an inner automorphism of ${\frak g}$ of order dividing 2. Such automorphisms are classified by Kac labelings of the extended Dynkin diagram ${\widetilde D}={\widetilde D}({\frak g})$. We fix a Cartan subalgebra ${\frak t}\subset{\frak g}$ and a Borel subalgebra ${\frak b}\supset {\frak t}$ and consider the Dynkin diagram $D({\frak g})=D({\frak g},{\frak t},{\frak b})$, whose vertices are the simple roots $\alpha_1,\dots,\alpha_\ell$. We consider also the extended Dynkin diagram ${\widetilde D}$ whose vertices are $\alpha_1,\dots,\alpha_\ell$ and $\alpha_0$, where $\alpha_0$ is the lowest root (the opposite to the highest root). There is a unique linear relation $$m_0\alpha_0+m_1\alpha_1+\dotsb+m_\ell\alpha_\ell=0$$ normalized such that $m_0=1$. It is easy to see that the numbers $m_i$ are positive integers; we write them near the vertices of the extended Dynkin diagram. See Table 6 in Lie groups and algebraic groups by Onishchik and Vinberg, or Table 1 in Section 9 of the paper Galois cohomology of real semisimple groups via Kac labelings by Borovoi and Timashev. Example: Here I give the coefficients $m_i$ for the extended Dynkin diagram of type ${\sf E}_7$. The extreme right-hand vertex corresponds to the lowest root $\alpha_0$. A Kac labeling of ${\widetilde D}$ is a family of nonnegative integers ${\bf q}=(q_0,q_1,\dots,q_\ell)$ satisfying $$ m_0q_0+m_1q_1+\cdots+m_\ell q_\ell=2.$$ Clearly we have $q_i\le 2$ for all $i$, and there can be either one vertex (type I) $\alpha_i$ with nonzero $q_i$, or two such vertices (type II). See Table 7, Types I and II, in the book by Onishchik and Vinberg, where the list of all possible Kac labelings is given. For a Kac labeling $\bf q$, the vertices with $q_i\neq 0$ are painted in black in this table (this determines uniquely the values of $q_i$ for all vertices $\alpha_i$). According to Victor Kac (1969), the inner involutions $\theta$ of ${\frak g}$ (up to conjugation by automorphisms of ${\frak g}$) correspond bijectively to the Kac labelings of ${\widetilde D}({\frak g})$ (up to automorphisms of ${\widetilde D}({\frak g})$). The fixed subalgebra ${\frak g}^\theta$ for $\theta=\theta({\bf q})$ is reductive. It is semisimple for Kac labelings ${\bf q}$ of type I, and has one-dimensional center for ${\bf q}$ of type II. The Dynkin diagram of the derived subalgebra of ${\frak g}$ is obtained from the extended Dynkin diagram ${\widetilde D}$ by removing the black vertices. See Table 7 in the book by Onishchik and Vinberg. Example 1. Here $\theta$ corresponds to the real form $\sf EVI$ of ${\sf E}_7$. We have ${\frak g}^\theta={\sf A}_1\oplus{\sf D}_6$. Example 2. Here $\theta$ corresponds to the real form $\sf EVII$ of ${\sf E}_7$. We have ${\frak g}^\theta={\sf E}_6\oplus {\Bbb C}$. Kac classified all automorphisms of ${\frak g}$ (inner and outer) of any finite order $r$. See Chapter 3, Section 3 in: V. V. Gorbatsevich, A. L. Onishchik, and E. B. Vinberg, Structure of Lie groups and Lie algebras, Lie Groups and Lie Algebras III, Encyclopaedia of Mathematical Sciences, Vol. 41, Springer–Verlag, Berlin, 1994. For an outer involution $\theta$ of ${\frak g}$, the Lie algebra ${\frak g}^\theta$ is semisimple. The Dynkin diagram of ${\frak g}^\theta$ is obtained by removing the black vertex corresponding to a Kac labeling from the affine Dynkin diagram of $({\frak g},\theta)$; see Table 7, Type III in the book by Onishchik and Vinberg. In particular, it is not true that for all outer involutions $\theta$, the Dynkin diagram of ${\frak g}^\theta$ is obtained by folding the Dynkin diagram of ${\frak g}$. Acknowledgements. To draw the extended Dynkin diagrams for this answer, the dynkin-diagrams package of @BenMcKay was used. EDIT. I describe an inner automorphism $\nu_{\bf q}$ of $\frak g$ corresponding to a Kac labeling ${\bf q}=(q_0,q_1,\dots,q_\ell)$ of $\widetilde D$. Let $G$ be the connected semisimple $\Bbb C$-group of adjoint type with Lie algebra $\frak g$ (we may take for $G$ the identity component of ${\rm Aut}\, \frak g$). Let $T\subset G$ be the maximal torus of $G$ with Lie algebra $\frak t$. Let $t\in T$ be the element such that $\alpha_i(t)=(-1)^{q_i}$ for all simple roots $\alpha_i\colon T\to{\Bbb C}^\times$. Then we associate to $\bf q$ the conjugacy class of the inner automorphism $\nu_{\bf q}:={\rm Ad}(t)$ of $\frak g$. I describe the action of $\nu_{\bf q}$ on $\frak g$. We have the root decomposition $${\frak g}={\frak t}\oplus\bigoplus_{\beta\in R} {\frak g}_\beta\, ,$$ where $R=R(G,T)$ denotes the root system. Then $\nu_{\bf q}$ acts on $\frak t$ trivially, and it acts on the root subspace ${\frak g}_\beta$ as $(-1)^{s_\beta}$, where $$ s_\beta =c_1 q_1+\dots+c_\ell q_\ell\quad\text{for} \quad \beta=c_1\alpha_1+\dots+c_\ell\alpha_\ell$$ with $c_i\in {\Bbb Z}$. Concerning proofs: see, for instance, Borovoi and Timashev, 2015. This is much easier to read than our 2021 Transformations Group paper referred to above.<|endoftext|> TITLE: Generalization of Bernstein’s inequality QUESTION [5 upvotes]: I'm using Muscalu and Schlag's textbook to study harmonic analysis and I encountered the following claim: Given some function $f \in \mathcal{S}(\mathbb{R}^{d})$, where $\mathcal{S}(\mathbb{R}^{d})$ denotes the Schwartz space of functions. Let $\hat{f}$ denote the Fourier transform of $f$. Assume that there exists some measurable set $E$, such that $\text{supp}(\hat{f}) \subset E \subset \mathbb{R}^d$. Then for any $1 \leq p \leq q \leq \infty$, we have the following inequality: ($|E|$ below denotes the Lebesgue measure of $E$) $$||f||_{L^q} \leq |E|^{\frac{1}{p}-\frac{1}{q}}||f||_{L^p}$$ I have managed to show the special case when $q=+\infty$ and $p=2$ by using Young's inequality and Plancherel identity. However, the hint says that we still need to use duality and interpolation to deduce the general conclusion. Any ideas on this? Moreover, how might this estimate be related to the probability version of Bernstein inequality? Thanks in advance! REPLY [2 votes]: As I have explained in the comment the inequality holds under the assumption $p\leq q\leq \infty$ and $02$. Indeed, let $q=\infty$. Let me construct a counterexample on the unit circle. Euclidean case can be adapted in a similar way. Consider $f=\frac{1}{\sqrt{N}}\sum_{1\leq k \leq N} e^{2^{k}xi}$. Then $\|f\|_{\infty}=N^{1/2}$. Also $\|f\|_p=C(p)$ as for large $N$ our function $f$ behaves as complex gaussian random variable. So $\|f\|_{\infty} \leq |E|^{1/p} \|f\|_{p}$ implies $N^{1/2}\lesssim N^{1/p}$ which is impossible for $p>2$. That being said perhaps there is a typo in this claim, and the assumption $0 TITLE: Is the automorphism group of free group of rank two relatively hyperbolic? QUESTION [8 upvotes]: By Behrstock, Drutu and Mosher [BDM], we know that the (outer) automorphism groups $\mathrm{Aut}(F_n)$ and $\mathrm{Out}(F_n)$ of free group of rank $n$ are not relatively hyperbolic if $n \geq 3$ (Theorem 9.2 of [BDM]). If $n = 2$, then $\mathrm{Out}(F_2)$ is isomorphic to $\mathrm{GL}(2,Z)$ so that it is virtually free. I hope to know what happens to $\mathrm{Aut}(F_2)$. Since the left transvection and right transvection commute, $\mathrm{Aut}(F_2)$ contains a free abelian subgroup of rank two which implies that it is not hyperbolic. Then, is it relatively hyperbolic? Or is it not relatively hyperbolic? REPLY [7 votes]: Here is more direct and elementary argument. Lemma: $\mathrm{Aut}(W_3)$ and $\mathrm{Aut}(\mathbb{F}_2)$ are isomorphic, where $W_3$ denotes the free product $\mathbb{Z}_2 \ast \mathbb{Z}_2 \ast \mathbb{Z}_2$. Sketch of proof. Let $F \leq W_3$ denote the kernel of the morphism $W_3 \twoheadrightarrow \mathbb{Z}_2$ sending all the generators $a,b,c$ to $1$. In other words, $F$ coincides with the elements that can be written as reduced words of even lengths. Because $F$ is a characteristic subgroup freely generated by $\{ab,bc\}$, we find a morphism $\mathrm{Aut}(W_3) \to \mathrm{Aut}(\mathbb{F}_2)$. It can be checked that it is an isomorphism. $\square$ I found this statement in Varghese's article The automorphism group of the universal Coxeter group. In the sequel, I work with $\mathrm{Aut}(W_3)$ instead of $\mathrm{Aut}(\mathbb{F}_2)$, but it is just because I am more confortable with it. As the isomorphism above is completely explicit, you can translate everything in $\mathrm{Aut}(\mathbb{F}_2)$ if you want. Proposition: $\mathrm{Aut}(W_3)$ is not hyperbolic relative to proper subgroups. Proof. Fix a basis $\{a,b,c\}$ of $W_3$. For every $g \in W_3$, let $\iota_g$ denote the inner automorphism given by the conjugation by $g$. Also, set $$\kappa_1 : \left\{ \begin{array}{ccc} a & \mapsto & a \\ b & \mapsto & b \\ c & \mapsto & c^{ab} \end{array} \right. \text{ and } \kappa_2 : \left\{ \begin{array}{ccc} a & \mapsto & a^{cb} \\ b & \mapsto & b \\ c & \mapsto & c \end{array} \right..$$ They are partial conjugations with infinite-order images in $\mathrm{Out}(W_3)$. (I use the notation $g^h=hgh^{-1}$.) Finally, set $$\varphi := \iota_{ab} \circ \kappa_1, \ \psi := \iota_b \circ \varphi \circ \iota_b, \text{ and } \xi := \iota_{cb} \circ \kappa_2.$$ The key observation is that $\varphi$ (resp. $\psi$, $\xi$) fixes $ab$ and $bcb$ (resp. $ab$ and $c$, $bc$ and $bab$). As a consequence, the subgroups $$\begin{array}{l} A:= \langle \iota_{ab}, \iota_{bcb}, \varphi \rangle \\ B:= \langle \iota_{ab}, \iota_c, \psi \rangle \\ C:=\langle \iota_{bc}, \iota_{bab}, \xi \rangle \end{array}$$ split as direct products between a(n infinite) virtually free group (namely, the part in the inner subgroup) and an infinite cyclic group. This implies that, if we assume that $\mathrm{Aut}(W_3)$ is hyperbolic relative to some collection of subgroups $\mathcal{P}$, then there exist $I,J,K \in \mathcal{P}$ such that $A \subset I$, $B \subset J$, and $C \subset K$. Now, we can conclude because $\mathcal{P}$ must be an almost malnormal collection of subgroups. Indeed: Because $I \cap J \supset \langle \iota_{ab} \rangle$, necessarily $I=J$. This implies that $J \supset \langle \iota_{ab}, \iota_{bcb}, \iota_c \rangle \supset \langle \iota_{(bc)^2} \rangle$, hence $J \cap K \supset \langle \iota_{(bc)^2} \rangle$, and finally $J=K$. Thus, $I=J=K$ contains $\langle \iota_{ab} , \iota_{bcb}, \iota_{c}, \iota_{bc}, \iota_{bab} \rangle= \mathrm{Inn}(W_3)$. Finally, from the fact that $\mathrm{Inn}(W_3)$ is a normal subgroup in $\mathrm{Aut}(W_3)$, we conclude that $I= \mathrm{Aut}(W_3)$. Thus, we have proved that $\mathcal{P}$ must contain a subgroup that is not proper. $\square$<|endoftext|> TITLE: n sets, each is large, the intersection of every three is small, what is the size of the union? QUESTION [15 upvotes]: Let $A_1, A_2, \ldots, A_n$ be $n$ sets such that: (1) for each $i\in [n]$, $\frac{n}{3}\leq |A_i|\leq n$; (2) for any $1\leq i\frac{6+3a}{2a+2} = \frac{3}{2}\left(1+\frac{1}{a+1}\right)$, there is a positive probability that none of the intersections is larger than $a$, so a collection of subsets of $[m]$ must exist with no large intersections. This gives an $o(n^2)$ bound for $a \geq 3$.<|endoftext|> TITLE: Non-Abelian Hodge theory QUESTION [12 upvotes]: Let $X$ be a compact Riemann surface. I would like to find a somehow complete reference for the proof of the so called non-Abelian Hodge correspondence relating Dolbeaut, Betti and Higgs bundle moduli spaces. I've tried to read the original articles by Hitchin (1987) or Simpson (1990) but it seems to me that I've not found somehow a complete reference showing a precise proof of all the statements involved (for the case of a curve). Does anyone know a book or notes related to this? REPLY [5 votes]: I recently attented a nice online talk by Pengfei Huang and he indicated two sources: the first chapter of his own phd Non-abelian Hodge theory and some specializations - TEL - Thèses en ligne Introduction to Nonabelian Hodge Theory: flat connections, Higgs bundles and complex variations of Hodge structure by Alberto Garcia-Raboso, Steven Rayan (see Introduction to Nonabelian Hodge Theory | SpringerLink for the published version) I am not sure to which extent they can be called complete references, but as general references on the subject, I think they certainly deserve to be mentionned.<|endoftext|> TITLE: Tangent Space of the Hodge bundle on the moduli space of curves QUESTION [5 upvotes]: Let $k$ be an algebraically closed field and $\mathcal M_g$ denote the moduli space (stack) of smooth curves of genus $g$ over $k$. Using the universal curve $\pi \colon \mathcal C_g \to \mathcal M_g$, there is a natural vector bundle $\pi_* \Omega_{\mathcal C_g / \mathcal M_g}$ of rank $g$ called the Hodge bundle. The fiber over a point $X \in \mathcal M_g$ is $H^0(X,\Omega_X)$. Let $\Omega \mathcal M_g$ denote the total space of this bundle. I would like to understand the tangent space of $\Omega \mathcal M_g$ at a point $(X, \omega)$, where $X$ is a smooth curve and $\omega$ is a global section of $\Omega_X$ (for simplicity let us assume that $g>1$). By definition a tangent vector at $(X, \omega)$ is a morphism from the dual numbers (i.e. $\varepsilon^2=0$) $\operatorname{Spec}k[\varepsilon] \to \Omega \mathcal M_g$ with image $(X,\omega)$. Since $\Omega \mathcal M_g$ is a vector bundle of rank $g$ this map is the same thing as two maps $a \colon \operatorname{Spec}k[\varepsilon] \to \mathcal M_g$ and $b \colon \operatorname{Spec}k[\varepsilon] \to \mathbb A^g_k$ with image $X$ and $\omega$ respectively. By the universal property of $\mathcal M_g$ the map $a$ corresponds to a first order deformation of $X$ and the map $b$ to a tangent vector of $\mathbb A^g_k$ at $\omega$, i.e. just a different point of $\mathbb A^g_k$ or another differential on $X$. I would like to arrive at the following description (that I will sketch below) of the tangent space that is used for example in Hubbard, John; Masur, Howard, Quadratic differentials and foliations, Acta Math. 142, 221-274 (1979). ZBL0415.30038. before proposition 4.5 (for quadratic differentials) or in Möller, Martin, Linear manifolds in the moduli space of one-forms, Duke Math. J. 144, No. 3, 447-487 (2008). ZBL1148.32007. in the proof of theorem 2.1. According to the papers above, the tangent space at $(X, \omega)$ can be identified with the set of Cartesian diagrams $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} (X,\omega) & \ra{f} & (\mathcal X, \tilde{\omega})\\ \da{} & & \da{} \\ \operatorname{Spec}k & \ras{} & \operatorname{Spec}k[\varepsilon] \\ \end{array} $$ where $\tilde\omega$ is a global section of $\Omega_{\mathcal X}$ such that the pullback of $(\mathcal X, \tilde \omega)$ is $(X, \omega)$. I can see that this gives a vector space of dimension $4g-3$ and is therefore isomorphic to the tangent space at $(X,\omega)$ by an isomorphism that does not change the deformation of the curve $X$, i.e. the map $a$. Moreover given one deformation $(\mathcal X, \tilde \omega)$ of $(X, \omega)$ all other deformations of $(X, \omega)$ only differ by an element of $H^0(X, \Omega_X)$. But I can not make this last map canonical. I can not figure out what the deformations of $(X, \omega)$ are, that do not change the differential $\omega$. REPLY [4 votes]: The universal property of the total space $$\mathbf{V}(E^\vee) = \operatorname{Spec}_M \operatorname{Sym} E^\vee $$ of a vector bundle (locally free sheaf) $E$ on some scheme $M$ is: giving a map $T\to \mathbf{V}(E^\vee)$ corresponds to giving a map $f\colon T\to M$ and a section $\omega$ of $f^* E$. We apply this to $T=\operatorname{Spec} k[\varepsilon]$, $M=\mathcal{M}_g$ and $E$ the Hodge bundle, so that $\mathbf{V}(E^\vee) = \Omega\mathcal{M}_g$. Then a tangent vector at $(X, \omega)$ is the same as a first order deformation $\tilde X = f^* \mathcal{C}_g$ over $T$ (corresponding to an $f\colon T\to \mathcal{M}_g$, so that $\tilde X$ is the pullback of the universal family), together with a section $\tilde\omega$ of $$ f^* E = f^* \pi_* \Omega_{\mathcal{C}_g/\mathcal{M}_g} = p_* \tilde f^* \Omega_{\mathcal{C}_g/\mathcal{M}_g} = p_* \Omega_{\tilde X/T}.$$ Here $\tilde f\colon \tilde X = \mathcal{C}_g\times_{\mathcal{M}_g} T\to \mathcal{C}_g$ and $p\colon \tilde X\to T$ are the projections. (For the first equality, one needs to check that the formation of $\pi_* \Omega_{C/S}$ for a smooth projective family of curves $C\to S$ commutes with base change along $S'\to S$.)<|endoftext|> TITLE: The sum (with multiplicity) of the cubes of irreducible character degrees of a finite group QUESTION [16 upvotes]: Throughout $G$ is a finite, non-abelian group. $\DeclareMathOperator\Irr{Irr}\DeclareMathOperator\AD{AD}\DeclareMathOperator\cp{cp}\newcommand\card[1]{\lvert#1\rvert}$ Let $\Irr(G)$ be the set of irreducible characters (working over the complex ground field). Let $$ \AD(G) = \card G^{-1} \sum_{\chi\in\Irr(G)} (d_\chi)^3 \;. $$ (The invariant $\AD(G)$ arose under a different name in work of Johnson on amenablity constants of Fourier algebras, and is a special case of something more general that can be defined for countable virtually abelian groups in terms of Plancherel measure for such groups. But this question is only concerned with character theory of finite groups.) Question. If $\AD(G)\leq 2$, does this imply that $d_\chi\leq 2$ for all $\chi\in\Irr(G)$? Here is my naive reasoning to support my guess that the answer is positive. Let $\Irr_n(G)=\{\chi\in\Irr(G) \mid d_\chi =n\}$. Then $$ 2\card{\Irr_1(G)} + 4\card{\Irr_2(G)} +\card G\AD(G) = \sum_{\chi\in\Irr_1(G)} 3 + \sum_{\chi\in\Irr_2(G)} 12 + \sum_{n\geq 3} \sum_{\chi\in\Irr_n(G)} (d_\chi)^3 $$ and therefore, since $\card G=\sum_{n\geq 1} \sum_{\chi\in \Irr_n(G)} (d_\chi)^2$, $$ 2\card{\Irr_1(G)} + 4\card{\Irr_2(G)} +\card G\AD(G) \geq 3\card G. $$ If we impose the condition that $\AD(G)\leq 2$, this yields $$ 2\card{\Irr_1(G)} + 4\card{\Irr_2(G)} \geq \card G $$ which seems difficult to achieve if there are irreps of degree $\geq 3$. EDIT: the original version of this post asked if there existed any $G$ with irreps of degree $\geq 3$ that satisfied this inequality, with the naïve hope that no such $G$ existed. Victor Ostrik has pointed out in a comment that an extraspecial $2$-group of order $32$ has $16$ linear characters and a single nonlinear character which has degree $4$, so that the inequality above is indeed satisfied. On the other hand the AD of this group is $(16+64) / 32 = 5/2$. Some other background facts which may or may not be useful. If $H\leq G$ then $\AD(H)\leq\AD(G)$. I am not sure if this has a simple proof just using the definition above, but it follows from the alternative description in tems of "amenability constants of Fourier algebras". There is some relationship between $\AD(G)$ and the commuting probability $$ \cp(G) = \frac{\# \hbox{conjugacy classes of $G$}}{\card G} = \frac{\card{\Irr(G)}}{\card G} $$ which shows that for the former to be “small” the latter should be “large”. To be precise, note that Hölder with conjugate exponents 3 and 3/2 gives $$ \sum_{\chi\in\Irr(G)} (d_\chi)^2 \leq \left( \sum_{\chi\in\Irr(G)} 1^3 \right)^{1/3} \left( \sum_{\chi\in\Irr(G)} (d_\chi)^3\right)^{2/3} $$ so that $$ 1 \leq \cp(G) \AD(G)^2. $$ Moreover, equality is strict since we assume $G$ is non-abelian. In particular, if $\AD(G)\leq 2$ this implies $\cp(G) > 1/4$. Can this be leveraged, perhaps by using the good behaviour of $\cp$ with respect to quotients? I had a quick look in the 2006 paper On the commutating probability in finite groups of Guralnick–Robinson and I suspect that results there could be useful, but as I am not a specialist in finite group theory I couldn't see how to make progress. REPLY [11 votes]: Here are a few remarks which may be helpful. I may be able to say more later. The condition that $\operatorname{cp}(G) > \frac{1}{4}$ already severely restricts the possibilities for $G$. If $G$ is a finite group which does not have a normal Sylow $p$-subgroup for a prime $p$, then it is proved in (GR 2006) mentioned in the question, that we have $\operatorname{cp}(G) \leq \frac{1}{p}$. Hence $G$ has a normal Sylow $p$-subgroup for each prime $p >3$. Also, it follows from the results of that paper that $G$ has an Abelian Sylow $p$-subgroup for each prime $p > 3$. It follows that $G$ has an Abelian normal subgroup $A$ with $[G:A]=2^{a}3^{b}$ for non-negative integers $a$, $b$. Notice that such a group $G$ is solvable. However, if $G$ is the direct product of an extra-special group of order $8$ and an extra-special group of order $27$, then we have $\operatorname{cp}(G) = \frac{55}{216} > \frac{1}{4}$. Nevertheless, it is true that $\operatorname{AD}(X \times Y) = \operatorname{AD}(X)\operatorname{AD}(Y)$, and for $G$ above we obtain $\operatorname{AD}(G) = \frac{3}{2} \times \frac{7}{3} = \frac{7}{2} > 2$. Later edit: It is also proved in (GR 2006) that when $G$ is solvable (which is the case now) we have $\operatorname{cp}(G)^{2} \leq \frac{\operatorname{cp}(F)}{[G:F]}$, where $F = F(G)$, the (unique) largest nilpotent normal subgroup of $G$. Since $\operatorname{cp}(G) > \frac{1}{4},$ this forces $[G:F] \leq 12$ if $F$ is Abelian and $[G:F] \leq 9$ if $F$ is non-Abelian. This is because $G/F$ is a $\{2,3\}$-group , while we have $[G:F] < 16$ if $F$ is Abelian, and $[G:F] < 10$ if $F$ is non-Abelian ($\operatorname{cp}(F) \leq \frac{5}{8}$ in the latter case, by a Theorem of W.Gustafson). So far, this does not seem to resolve your problem, but it does at least suggest a small bound on the largest irreducible character degree if $\operatorname{AD}(G) \leq 2$. For it is a theorem of N. Ito that the degree of an irreducible character of a finite group $G$ divides the index of any Abelian normal subgroup of $G$ (in fact, it is mentioned in Isaacs' book on Character Theory that each irreducible character of $G$ has degree at most $n$ if $G$ has an Abelian subgroup of index $n$, even if the Abelian subgroup is not normal). If $F$ is Abelian, we may conclude that every irreducible character of $G$ has degree at most $12$. In fact, if $O_{2}(G) = O_{2}(F)$ is Abelian, but $F$ is non-Abelian, we obtain $[G:F] \leq 6$ because $\operatorname{cp}(F) \leq \frac{11}{27}.$ In that case, we must have $G/F$ Abelian, for otherwise $G/F \cong S_{3}$ and $\operatorname{cp}(G) \leq \operatorname{cp}(G/F)\operatorname{cp}(F) \leq \frac{11}{54} < \frac{1}{4},$ a contradiction. By the way, detailed analysis may seem messy, but Abelian normal subgroups can't be ignored, because we always have $\operatorname{AD}(G \times H) = \operatorname{AD}(G)$ whenever $G$ is a finite group and $H$ is a finite Abelian group. Later edit: I have been looking up some results in the other direction, where a conjecture of D. Gluck seems very relevant. For example, Alex Moreto and Tom Wolf have proved that if $G$ is a finite solvable group, then $G$ has an irreducible character whose degree is at least $[G:F(G)]^{\frac{1}{3}}$. Hence the only finite solvable groups $G$ which have no irreducible character of degree greater than $2$ have $[G:F(G)] \leq 8.$ Also, Cossey, Halasi, Maroti and Nguyen have proved that if $G$ is a finite solvable group with $|F(G)|$ of order coprime to $6$, then $G$ has an irreducible character whose degree is at least $[G:F(G)]^{\frac{1}{2}}$. If the conjecture of Gluck is true, then the restriction that $|F(G)|$ has order coprime to $6$ could be removed. In any case, at present it is known that the only finite solvable groups $G$ which have $|F(G)|$ of order coprime to $6$ and no irreducible character of degree greater than $2$ have $[G:F(G)] \leq 4.$ To conclude: if your question has a positive answer, then it must be the case that any finite group $G$ with $\operatorname{AD}(G) \leq 2$ has $[G:F(G)] \leq 8$ (recall that such a group $G$ has been shown to be solvable). Note that this is a little stronger than I managed to prove above.<|endoftext|> TITLE: Groups acting on products of hyperbolic spaces QUESTION [7 upvotes]: I am interested in groups acting properly and cocompactly by isometries on finite products of Gromov-hyperbolic metric spaces. I am mostly interested in the case where the group itself is not virtually a product of Gromov-hyperbolic groups. Classical examples include irreducible uniform lattices in products or rank $1$ simple Lie groups, and irreducible uniform lattices acting on products of trees. These examples are all $\operatorname{CAT}(0)$. Question. Are there examples of such groups which are not $\operatorname{CAT}(0)$, or not known to be $\operatorname{CAT}(0)$? REPLY [3 votes]: This question for a product of one Gromov hyperbolic space is a famous open problem (see for example Bestvina's problem list) and as far as I am aware the general case is also open. REPLY [3 votes]: Though it does not answer this question, the paper https://arxiv.org/abs/1509.03748 contains a weaker notion of nonpositive curvature than CAT(0) which all groups acting geometrically on product of Gromov hyperbolic groups satisfy and such that there are examples of non-CAT(0)-groups which also satisfy this weaker notion.<|endoftext|> TITLE: Non orientable, closed manifold covered by two simply-connected charts QUESTION [8 upvotes]: This question arose during my Differential Geometry course. Possibly there is an obvious answer, but I do not see it, and I could not find it in the literature. The same question was asked yesterday on MSE, but it did not get much attention. Question. Does it exist a closed, non-orientable smooth manifold that can be written as the union of exactly two simply-connected charts? If so, what is a reference? Of course, the intersection of the two charts must be disconnected. As noted in the linked MSE question, the open Möbius band and the closed Möbius band give examples in the open case and in the non-empty boundary case, respectively. Moreover, the Klein bottle is covered by two charts, both homeomorphic to cylinders, but they are not simply-connected. REPLY [22 votes]: Take a non-orientable $S^n$ bundle over $S^1$ with $n \geq 2$ (*), (sometimes called generalised Klein bottles) then covering $S^1$ by two intervals and taking preimages should work. (*) Let $\tau : S^n \rightarrow S^{n}$ be a non-orientable diffeomorphism of $S^n$ given by $(x_{1},\ldots,x_{n+1}) \mapsto (x_1,\ldots,-x_{n+1}) $. Then such a bundle is given by taking the quotient of $[0,1] \times S^{n}$ by the equivelance closure of the relation $$(0,p) \sim (1,\tau(p)). $$ Such bundles are discussed in the following paper https://academic.oup.com/plms/article-abstract/s3-4/1/196/1497906?redirectedFrom=PDF The non-orientable $S^2$-bundle over $S^1$ appear's in Hatchers 3-manifold notes, since it plays a role in the prime decomposition theorem for non orientable $3$-manifolds. See page 8 of https://pi.math.cornell.edu/~hatcher/3M/3Mdoublepage.pdf<|endoftext|> TITLE: Advice: What topics to study now in analytic number Theory( And if there are video lectures( Open Online course) / Course notes available on website) QUESTION [6 upvotes]: I am a person living in a 3rd world country and completed my masters in mathematics in July 2020. Then I began to study some additional topics in Pure Mathematics as I was applying for Ph.D. abroad( In Number Theory) as I felt that I should study more topics before applying in December2020- January 2021. Background in analytic number theory: I have studied Number Theory from Elementary number theory by Burton, Introduction to Analytic Number Theory (Tom Apostol) , Modular Functions and Dirichlet Series in Number Theory ( Tom Apostol). I am thinking of studying more Analytic number theory in the free time I am having. But instead of studying from textbook I thought of studying from Lecture notes ( If available on course page of course) as when studying from textbook one covers many more topics than required in a course. But if the course is not available in lecture notes form then textbook recommendations are also most welcome. So, Can you please tell which courses in analytic number theory should I study now and give some suggestions of course pages/ textbooks? I am thinking of covering material equivalent to 2 courses. In rest of time I am planning to study some topics from other branches of pure mathematics( one of which is Elliptic curves, is it part of analytic number theory?) depending on time available. Should I cover more analytic number theory and less of other topics in pure mathematics? I am badly in need of advice/ guidance on this and will be really thankful! If you want any more detail about anything just comment. REPLY [18 votes]: (I'm posting this as an answer since I prefer to mantain my anonymity, thus I cannot comment.) I'm also from a third-world country. At the age of 24, when I was finishing my master's thesis, I was diagnosed with bipolar disorder, and I was drinking way too much. It took me years to recover from alcoholism and to handle my depression. There were three things that kept me alive: the few people that kept around me and cared about me, teaching (part-time), and studying number theory (and that is why I'm posting this comment). Basically, it was the only thing that I was passionate about at that time. It helped me a lot to focus on positive things and to avoid depressing thoughts. Now I'm in my 30s, and finishing my phd on number theory, and I couldn't be happier. So, dear anon, please stay positive about the future, stay away from illegal drugs and alcohol, be responsible with your treatment (I hope you find a good doctor), and keep studying what you love, in this case number theory. More important, respect your recovery processes. It is way more important to be healthy mentally (and physically, as far as possible), than anything else.<|endoftext|> TITLE: $\infty$-natural transformations and adjunctions QUESTION [6 upvotes]: I'm having troubles proving these two related statements, which are immediate for 1-categories and should of course be true for $\infty$-categories: Given a natural transformation $\alpha: f \Rightarrow g: \mathcal{C} \to \mathcal{D}$, for every two objects $X,Y \in \mathcal{C}$ there is a homotopy commutative square $\require{AMScd}$ \begin{CD} \mathcal{C}(X,Y) @>f>> \mathcal{D}(fX,fY)\\ @VgVV & @VV \alpha_{Y*} V\\ \mathcal{D}(gX,gY) @>> \alpha_X^* > \mathcal{D}(fX,gY) \end{CD} Given two functors $L: \mathcal{C} \leftrightarrows \mathcal{D}: R$ and two natural transformation $\eta: 1_{\mathcal{C}} \Rightarrow RL$ and $\varepsilon: LR \Rightarrow 1_{\mathcal{D}}$, these data constitute an adjunction if and only if the homotopy triangular identities hold, that is, $(R \ast \varepsilon) \circ (\eta \ast R) \simeq 1_R$ and $(\varepsilon \ast L) \circ (L \ast \eta) \simeq 1_L$. Any insight would be welcome. REPLY [9 votes]: 1- Your natural transformation can be seen as a functor $C \to D^{\Delta^1}$, which therefore induces a commutative square of $\infty$-categories $\require{AMScd} \begin{CD} C_{x/} @>>> D^{\Delta^1}_{\alpha_x/} \\ @VVV @VVV \\ C @>>> D^{\Delta^1}\end{CD}$ and thus a morphism of fibers over $y\in C$. The fiber of the leftmost vertical map is $map_C(x,y)$ , and the fiber of the rightmost vertical map over the image of $y$, i.e. $\alpha_y$ sits in a (cartesian) square $\require{AMScd} \begin{CD} map(\alpha_x,\alpha_y)@>>> map(fx,fy) \\ @VVV @VVV \\ map(gx,gy) @>>> map(gx,fy) \end{CD}$ For your purposes, you don't even need to know the square is cartesian, and then it just comes from the fact that $\Delta^1\times\Delta^1$ is a commutative square. Here's maybe more detail : consider an arrow $g:x_1\to y_1$ in an $\infty$-category $E$, then $Fun(\Delta^2,E)\times_{Fun(\Delta^1,E)} \{g\}\simeq E_{/g}$, where we look at evaluation at $\Delta^{\{1,2\}}$. Also recall that the forgetful map $E_{/g}\to E_{/x_1}$ is an equivalence (because the space of fillings for a given inner horn is contractible). Now consider the following sequence of inclusions $\Delta^1\to \Delta^2 \to\Delta^1\times \Delta^1$ : the first map is inclusion at $\Delta^{\{0,1\}}$,and the second one is the diagonal arrow. This provides you with functors $Fun(\Delta^1\times\Delta^1,E)\to Fun(\Delta^2,E)\to Fun(\Delta^1,E)$, where the second one induces an equivalence upon taking the fiber over $g$ and $x_1$ respectively. So you have functors $Fun(\Delta^1\times\Delta^1,E)\times_{Fun(\Delta^1, E)} \{g\}\to Fun(\Delta^2,E)\times_{Fun(\Delta^1,E)}\{g\}\to Fun(\Delta^1,E)\times_E \{x_1\}$, where the second one is an equivalence. Now do the same for the other nondegenerate $2$-simplex in $\Delta^1\times\Delta^1$, where this time you'll look at another arrow $f: x_0\to y_0$ in $E$. This gives you a big diagram as follows,witnessing the fact that $\Delta^1\times\Delta^1$ is a commutative square: (because AMScd does not support diagonal arrows, this is hard to draw on MO, so I just drew it and took a picture) It all commutes, and the "wrong way" morphisms become invertible upon taking the appropriate fibers. On fibers (I'll let you figure out which fibers I mean), this then gives you the following, still commutative diagram : which induces the desired commutative diagram, relating $map(f,g), map(x_0,x_1), map(y_0,y_1)$ and $map(x_0,y_1)$. For your question 2-, the answer is not as easy because it depends on what your original definition of adjunction is. If you're just following Higher topos theory, then your statement is essentially 5.2.2.12 in that book, and the proof requires a number of preliminaries. If that's not what your definition is, you're going to need to be more precise about what you mean.<|endoftext|> TITLE: Homotopy groups of $K(n)$-localization of the Brown-Peterson spectrum QUESTION [6 upvotes]: We fix $p$ prime and $n$ a natural number. We let $K(n)$ be the $2(p^{n}-1)$-periodic Morava $K$-theory, i.e. $K(n)_*=\mathbb{F}_p[v_n^{\pm 1}]$ with $|v_n|=2(p^n-1)$. I distinctly recall that we should have $\pi_*(L_{K(n)}BP)\cong (v_n^{-1}BP_*)^{\wedge}_{I_n}$, yet I am unable to find an explicit reference in the literature to this fact. Do you have any idea where I can find the proof of such computation? Also, I was wondering if we apply additional localizations with respect to these Morava $K$-theories this behavior continues. E.g. for $m TITLE: Find all Non-isomorphic good drawings of $K_{3,3}$？ QUESTION [8 upvotes]: Sometimes I look at all non-isomorphic good drawings of graphs on a plane or sphere. Good drawing means that no edge crosses itself, no two edges cross more than once, and no two edges incident with the same vertex cross each other. Non-isomorphic means that there is no isomorphism of the drawings as embedded graphs. My questions are: How many good drawings are there of $K_{3,3}$? Is there any website or program that lists all good drawings of graphs with at most $n$ vertices? The three requirements on a good graph are illustrated below: For small graphs, there are the following results. For complete graphs with at most 5 vertices, the list of good drawings is basically clear. They are also summarized in Marcus Sahefer's book on Crossing Numbers of Graphs (CRC Press, 2018): There are two non-isomorphic good drawings of $K_4$ on the sphere. There are five non-isomorphic drawings of $K_5$ on the sphere. Meanwhile $K_{2,3}$ has $4$ good drawings, as described in Michal Stas, "Join Products [K. sub. 2, 3]+[C. sub. n]" (Mathematics v8.6, 2020). They are illustrated in the graphic below: We can get lists of non-isomorphic graphs from websites (e.g. http://users.cecs.anu.edu.au/~bdm/data/graphs.html) or from software such as maple or geng. How can we get a similar list of non-isomorphic good drawings? REPLY [5 votes]: The list of nonisomorphic good drawings of $K_{m,n}$ with $2\le m,n \le 3$ appears in the following paper: Heiko Harborth, Parity of numbers of crossings for complete n-partite graphs, Mathematica Slovaca, Vol. 26 (1976), No. 2, 77-95. Persistent URL: http://dml.cz/dmlcz/136111 The number of nonisomorphic good drawings of $K_{3,3}$ obtained is $102$. For $K_{2,3}$, there are six nonisomorphic drawings. The two that are missing in Figure 1 referenced in the question may not be realizable with straight-line edges. The list of nonisomorphic good drawings of graphs with up to five vertices appears in the paper H-D. O. F. Gronau and H. Harborth, Numbers of nonisomorphic drawings for small graphs, Proceedings of the Twentieth Southeastern Conference on Combinatorics, Graph Theory, and Computing (Boca Raton, FL, 1989). Congr. Numer. 71 (1990), 105–114.<|endoftext|> TITLE: Does any subset of a finitely generated group with positive upper density contain three points in arithmetic progression? QUESTION [7 upvotes]: Let $G$ be a countable finitely generated group, with word metric $d_S$ induced by some generating set $S$. Let $B_r$ be the ball of radius $r$ around the identity under $d_S$. We say that $A \subset G$ has positive upper density if $\limsup_{n\to\infty} \frac{|A\cap B_n|}{|B_n|}>0$ for some, hence any word metric $d_S$. Question: For any $A\subset G$ with positive upper density, does it hold that there exists some $a \in A$, and nonzero $g \in G$ such that $a, a + g, a + 2g$ are in $A$? Edit: The original question asked about abelian groups, which is answered in the positive by Roth’s theorem as stated by Sean Eberhard in the comments. REPLY [3 votes]: I hope this might be of interest, I'll consider a simpler problem and present a couple $(G,A)$ where one can't find a progression of length four: $a, ag, ag^2, ag^3$. Example. Let $G$ be the free group $F_2$ on $2$ generators with generating set $x_1, x_1^{-1}, x_2, x_2^{-1}$. Take $A$ as the union of all words of length $\varphi(n)=4^n$ for $n\ge 2$. It is easy to see that $A$ has positive upper density. However one can not find elements $a, ag, ag^2,ag^3\in A$ with $g\ne 1$. The reason here is the following. Observation. Fix $a$ and $g$ and consider the sequence of words $\ldots, ag^{-1},a,ag,\ldots, ag^n,\ldots$. Let $l(n)$ be equal to the length of $ag^n$. Then $l(n)$ is given by formula $d|n-x|+b$, where $d>0$, $x$ a fixed (half integer) number $b\ge 0$. I don't want to write a complete proof of this observation. But the idea is that the multiplication by $g$ is acting by isometry of the Cayley graph of $F_2$ and it sends a unique geodesic (in the graph) to itself, and points $ag^{-n}$ are on the same distance to this geodesic (maybe there is a more obvious reasoning). From this observation it is very easy to deduce the claim. Consider for example the case when $l(a)=l(ag) TITLE: Deformation of a diagram preserve the homotopy limit QUESTION [6 upvotes]: I have been a bit sloppy in the title, but let me be specific. I stepped again into the subtle difference between homotopy limit and limit in the homotopy category, in the following version. Suppose you have two diagrams of spaces $X_{\bullet}, Y_{\bullet}$ and a homotopy equivalence $f_i : X_i \to Y_i $ such that for any morphism $i\to j$, the morphisms $Y(i\to j)\circ f_i$ and $f_j\circ X(i\to j): X_i\to Y_j$ are homotopic. Is it true that $\text{holim} X_{\bullet} \simeq \text{holim} Y_{\bullet} $ ? My favorite diagram is $\Delta$, i.e. cosimplicial objects, but I suspect the result should not depend on this. Let me remark that if the $f_i$ would commute strictly the result would be true. Bonus question. If the answer is "No, they can be different", are there a higher compatibility constraints one can impose so that it becomes true? For example, by specifying what the homotopy $f_{ij}$ between $Y(i \to j) f_i $ and $f_j X(i \to j)$ should respect, and then bla bla.. Remark. I suspect that a "not so higher" version should be true for the following reason. If I have a cosimplicial object, I take the chain complex pointwise, and I apply the Dold-Kan correspondence, I get a bicomplex $C_*(X_\bullet)$: the horizontal differential is the differential of chains, while the vertical differential is the alternated sum of the cosimplicial (co)faces. I have tried to apply the definition of multicomplex homotopy equivalence between $C_*(X_{\bullet})$ and $C_*(Y_{\bullet})$, but there a few more maps to find and I don't want to mess with so many equations. In particular I should find an homotopy inverse to $f$. I hope that the "not so high" is low enough that one homotopy suffices, without higher constraints. REPLY [5 votes]: Let me try to complement Dylan Wilson's great answer with an example which is easier to analyze at the point-set level. Take $I = B\mathbb N^2$, so that a functor $I\to\mathcal S$ is a space $X$ together with two commuting endomorphisms $\alpha_{0},\alpha_1:X\to X$. An explicit model for the homotopy limit is the space of tuples $(x,\gamma_i,\Gamma)$, where $x\in X$ is a point, $\gamma_i$ is a path from $x$ to $\alpha_i(x)$, and $\Gamma:[0,1]^2\to X$ is a square with sides $\gamma_0,\alpha_0\circ\gamma_1,-\alpha_1\circ\gamma_0,-\gamma_1$. For the counterexample, set $X = S^1\times\mathbb R, \alpha_0(z,t) = (z,t+1),\alpha_1(z,t) = (e^{2\pi i t}z,t)$; one can check that for any choice of $(x,\gamma_i)$ the winding number of the concatenation of the four paths $\gamma_i$ and $\alpha_i\circ\gamma_{1-i}$ is $1$, so that the square $\Gamma$ can't exist, i.e. the homotopy limit is empty. Then set $Y = S^1$ with both endomorphisms the identity; since the limit of this diagram is nonempty, so is the homotopy limit (in fact, one easily computes the homotopy limit as $S^1\times \mathbb Z^2$); in particular the homotopy limits of $X$ and $Y$ are not homotopyequivalent. The homotopy equivalence $f$ is the projection to the first factor, and the required homotopy between $(z,t)\mapsto z$ and $(z,t)\mapsto e^{2\pi i t}z$ is conctructed by rescaling the second variable. The reason that this counterexample works is that $X$ is a point-set model for the $\infty$-functor $B\mathbb N^2\to \mathcal S$ which sends the point to $S^1$ and the morphisms to the identity, but fills the resulting commutative square with a nontrivial self-homotopy of the identity of $S^1$.<|endoftext|> TITLE: Empty preimage under homomorphism of finitely presented groups independent of ZFC QUESTION [5 upvotes]: Is there a homomorphism of finitely presented groups $f:G\to H$ and an element $h\in H$ such that the statement "$f^{-1}(h)$ is empty" is independent of ZFC? REPLY [5 votes]: The answer is yes, as a consequence of my answer to your other question. Namely, in that answer, we have a finite group presentation $H$ and a word $h$ such that the question $h=1$ in $H$ is independent of ZFC. So if we take $G$ to be trivial and $f:G\to H$ the unique homomorphism, we have the statement "$f^{-1}(h)$ is nonempty" being independent of ZFC. The general lesson of that answer supplies also an answer to Benjamin Steinberg's comment here concerning a confusion between computable undecidability and ZFC or logical undecidability. The general lesson, which I argue on the other post, is that every computably undecidable enumerable decision problem is saturated with logical undecidability. So the two notions of undecidability are actually intimately connected.<|endoftext|> TITLE: How far to find a well-order? QUESTION [10 upvotes]: Consider a transitive set $M$. Let's call the well-ordering number for $M$ the smallest ordinal $\alpha$ so that $L_\alpha(M)$ contains as an element a well-order of $M$, and denote it as $\upsilon = \upsilon(M)$. My question is which ordinals are the $\upsilon(M)$ for some $M$. This is phrased in a general format, and maybe it's helpful to narrow in on the specific case I have in mind. My primary interest is when $M$ is a countable model of ZFC. The best possible answer would be to completely characterize such $\upsilon(M)$, but that's likely difficult. The main thing I'd like to know (but haven't been able to figure out on my own) is what small ordinals are possible. Of course $\upsilon(M) = 1$ is possible—see below—but what about other finite ordinals? Or, say, $\omega+1$? Here's some easy observations: It's possible that $\upsilon(M)$ is undefined, i.e. $L(M)$ thinks that $M$ cannot be well-ordered. For example, if $\lambda$ is an $I_0$ cardinal then $M = V_\lambda$ is like this. $\upsilon(M)$ cannot be a limit ordinal, just from the definition of the $L(M)$ hierarchy. $\upsilon(M) = 1$ is possible, which happens just in case $M$ has a definable well-order. It's possible $\upsilon(M)$ is defined and $>1$. Suppose $M$ is countable in $L$. Then a condensation argument yields that $L_\alpha(M)$ sees that $M$ is countable for some large enough countable $\alpha$, giving an upper bound on $\upsilon(M)$. If $M$ doesn't have definable well-order, then $\upsilon(M) > 1$. But I don't see how to turn this into a more precise calculation. REPLY [10 votes]: If there is a transitive set model $M$ of ZFC, then all finite successor ordinals $n+1$ and all $\omega+n+1$ are $\upsilon(M)$ for some such $M$, and also many other ordinals: Claim: Let $\Omega,\alpha<\omega_1^L$ be such that $L_\Omega\models\mathrm{ZFC}$ and $0<\alpha$ and there is a surjection $f:\omega\to\Omega+\alpha$ in $L_{\Omega+\alpha+1}$. Then there is a transitive model $M\models\mathrm{ZFC}$ of height $\Omega$ such that $\upsilon(M)=\alpha+1$; moreover, in $L_{\alpha+1}(M)$ there is a surjection $g:\omega\to M$. Remark: Note there calso be $\beta<\alpha$ such that $L_{\Omega+\beta+1}$ has a surjection $f':\omega\to\Omega+\beta$, and hence a surjection $f'':\omega\to\Omega=\mathrm{OR}\cap M$, but there won't be any wellorder of $M$ in $\mathcal{J}_{\Omega+\beta+1}$. The claim gives in particular that if there is a transitive $M\models\mathrm{ZFC}$ then all finite ordinals, $\omega+1$, and many "small" successor ordinals are $\upsilon(M)$ for some $M$. (E.g. if $\Omega$ is least such that $L_\Omega\models\mathrm{ZFC}$ then for each of these "small" ordinals $\alpha$ we can find an $M$ of height $\Omega$ with $\upsilon(M)=\alpha+1$, and also for cofinally many $\alpha<\omega_1^L$ there is $M$ of height $\Omega$ with $\upsilon(M)=\alpha+1$. On the other hand, if there is say a transitive model $M$ of ZFC of cardinal height $\kappa$, then $L_\kappa$ also models ZFC, and for whatever ordinal $\alpha>0$, we get that $L_{\kappa+\alpha}\models$``$\kappa$ is a cardinal'', and by taking the $\Sigma_{n+1}$-hull in $L_{\kappa+\alpha}$ of the single parameter $\kappa$, we get $L_{\bar{\kappa}+\bar{\alpha}}$ modelling the same about $\bar{\kappa}$, and $L_{\bar{\kappa}}\models\mathrm{ZFC}$, and there is a surjection $f:\omega\to\bar{\kappa}+\bar{\alpha}$ in $L_{\bar{\kappa}+\bar{\alpha}+1}$.) Proof: I'm going to work with the $\mathcal{J}$-hierarchy, not the $L$-hierarchy, since particularly for dealing with successor levels, the $\mathcal{J}$-hierarchy is smoother. This doesn't affect the outcome, because we will have $\mathcal{P}(M)\cap\mathcal{J}_\alpha(M)=\mathcal{P}(M)\cap L_\alpha(M)$ for each ordinal $\alpha$, and hence they agree about when a wellorder of $M$ is constructed. (It is easy enough to see that in fact $L_\alpha(M)\subseteq \mathcal{J}_\alpha(M)$ for each $\alpha$; for the converse, one must code $\mathcal{J}_\alpha(M)$ over $L_\alpha(M)$, uniformly enough in $\alpha$.) If one prefers to work only with the $L$-hierarchy, one should be able to translate everything that follows into it (swapping $\mathcal{J}_\beta$ for $L_\beta$ and $\mathcal{J}_\beta(M)$ for $L_\beta(M)$), but then the proof would directly involve more coding work. Recall that if $\beta$ is sufficiently closed then $\mathcal{J}_\beta=L_\beta$; this holds in particular for $\Omega$, since $L_\Omega\models\mathrm{ZFC}$. The plan is to obtain $M$ as a class generic extension of $\mathcal{J}_\Omega$, via generic $G$, setting $M=\bigcup_{\alpha<\Omega}\mathcal{J}_\Omega[G\upharpoonright\alpha]$ (in particular, $G$ itself is not included as a predicate of $M$), and with enough symmetry that by taking $G$ generic over $\mathcal{J}_{\Omega+\alpha}$, $\mathcal{J}_\alpha(M)$ cannot have a wellorder of $M$; but using that $\mathcal{J}_{\Omega+\alpha+1}$ has a surjection $f:\omega\to \mathcal{J}_{\Omega+\alpha}$, we will take $G\in \mathcal{J}_{\Omega+\alpha+1}$, and we therefore get a surjection $g:\omega\to M$ in $\mathcal{J}_{\alpha+1}(M)$. It will be similar to the standard construction of models of ZF + $\neg\mathrm{AC}$ of the form $\mathrm{HOD}(\mathbb{R}^*)^{V[G]}$, for some reasonably symmetric set of reals $\mathbb{R}^*$, and some generic $G$ for some homogeneous forcing. But it is finer in detail, since we are aiming for choice to fail right up to some point, and then to hold right after that. So, work for the moment in $\mathcal{J}_\Omega$; we specify the forcing. For $\kappa$ regular let $\mathbb{P}_\kappa$ be the forcing to add a subset of $\kappa$ via conditions $f:\beta\to 2$ for $\beta<\kappa$, and let $\mathbb{P}$ be the class product of all such $\mathbb{P}_\kappa$, with set-sized supports in the indices $\kappa$. (That is, conditions are functions $f:R\to V$ where $R$ is a set of regular cardinals and $f(\kappa)\in\mathbb{P}_\kappa$ for each $\kappa\in R$, and the ordering is by extension.) Note that for each regular $\gamma$, we can factor $\mathbb{P}$ as $(\mathbb{P}\upharpoonright\gamma)\times(\mathbb{P}\upharpoonright[\gamma,\infty))$, and $\mathbb{P}\upharpoonright[\gamma,\infty)$ is ${<\gamma}$-closed. Using the enumeration $f:\omega\to \mathcal{J}_{\Omega+\alpha}$ in $\mathcal{J}_{\Omega+\alpha+1}$, let $G\in \mathcal{J}_{\Omega+\alpha+1}$ be such that $G$ is $(\mathcal{J}_{\Omega+\alpha},\mathbb{P})$-generic (here I mean generic for all dense sets which are elements of $\mathcal{J}_{\Omega+\alpha}$; recall $\alpha>0$, so this includes all dense sets which are definable from parameters over $\mathcal{J}_\Omega$.) Let $M$ be defined as indicated above. Then $M\models\mathrm{ZFC}$. Because we have a surjection $h:\omega\to L_{\Omega}$ with $h\in\mathcal{J}_{\Omega+\alpha+1}$, and $G\in\mathcal{J}_{\Omega+\alpha+1}$, we get a surjection $g:\omega\to M$ with $g\in \mathcal{J}_{\Omega+\alpha+1}\subseteq \mathcal{J}_{\alpha+1}(M)$. So we just need to see there is no wellorder of $M$ in $\mathcal{J}_\alpha(M)$. This takes some work regarding the forcing details, but other than that, it is like in the constructions of models of $\neg\mathrm{AC}$ like mentioned above. One first needs to work through the forcing details, to see level-by-level that $\mathcal{J}_{\Omega+\beta}$ has names for all elements of $\mathcal{J}_\beta(M)$, and the forcing relation is level-by-level and quantifier-by-quantifier appropriately definable. (Such level-by-level forcing calculations are commonly used in inner model theory, so there are some papers where they can be found). Here is a sketch of what we need in this case. At the level of $\mathcal{J}_\Omega$ and $M$ itself, note that all elements of $M$ have names in $\mathcal{J}_\Omega$, and the $\Sigma_n$-forcing relation is $\Sigma_{n+k}$, for some fixed $k$, uniformly in $n$. Then proceed inductively on $\beta>0$. Define canonical names for elements of $\mathcal{J}_\beta(M)$, corresponding to how things are defined in the $\mathcal{J}$-hierarchy; these are natural kinds of symmetric names. E.g. the subsets of $M$ which are in $\mathcal{J}_1(M)$ are just those which are definable from parameters over $M$, so we can take names for such sets as being tuples $\sigma=(\varphi,\tau)$ where $\tau\in\mathcal{J}_\Omega$ is a $\mathbb{P}$-name and $\varphi$ is a formula, and then this name gets interpreted as $\sigma_G=\{y\in M\bigm|M\models\varphi(y,\tau_G)\}$. These canonical names should, moreover, each have a certain support in $\mathcal{J}_\Omega$, corresponding to some bound on the parameters in $M$ needed to define the object in the generic extension $\mathcal{J}_\beta(M)$. That is, if a set $A\subseteq M$ is defined over $M$ from $x\in M$, then we can say that $\gamma$ is the support of $A$, where $\gamma$ is least with $x\in V_\gamma^M$. Higher up, if $A\in\mathcal{J}_{\beta+1}(M)$ is defined over $\mathcal{J}_\beta(M)$ from parameters $x_1,\ldots,x_n\in\mathcal{J}_\beta(M)$, then these $x_i$'s already have some supports $\gamma_i<\Omega$, then their supremum gives a support for $A$. The canonical names are just codes for these kinds of definitions of objects from finitely many ordinals and elements of $M$, and hence can also be assigned some support $<\Omega$. For $\beta>0$, the $\Sigma_n$-forcing relation over $\mathcal{J}_{\Omega+\beta}$ (for $\Sigma_n$ facts about $\mathcal{J}_\beta(M)$) will be $\Sigma_{n+k}$-definable over $\mathcal{J}_{\Omega+\beta}$ in the parameter $\Omega$, with some fixed $k$, uniformly in $\beta$. (Actually you can state it much more precisely than this, but one has to think a bit about what exactly the "$\Sigma_n$ forcing relation" should mean.)) Now suppose for a contradiction that we have a wellorder $<^*$ of $M$ in $\mathcal{J}_\alpha(M)$. Then we get a canonical name $\tau\in\mathcal{J}_{\Omega+\alpha}$ such that $\tau_G={<^*}$ (and here note the canonical name $\tau$ gets interpreted naturally in the $\mathcal{J}_\beta(M)$ hierarchy). Since $M\models\mathrm{ZFC}$, from $<^*$, it is easy to recover a surjection $g:\Omega\to M$ (consider the $<^*$-least wellorder of $V_\alpha^M$, for each $\alpha<\Omega$, and use these wellorders (which are each in $M$) to define $g$). So we get such a $g\in\mathcal{J}_\alpha(M)$. Let $\gamma<\Omega$ be a support for $g$. Then $g$ is defined over some $\mathcal{J}_\beta(M)$, where $\beta<\alpha$, from ordinal parameters and parameters in $V_\gamma^M\subseteq \mathcal{J}_\Omega[G\upharpoonright\xi]$ for some $\xi<\Omega$. Let $X=G_\kappa$ be the generic subset of some $\mathcal{J}_\Omega$-regular $\kappa\geq\xi$. Then $X\in M$ but $X\notin\mathcal{J}_{\Omega+\alpha}[G\upharpoonright\xi]=\mathcal{J}_\alpha(\mathcal{J}_\Omega[G\upharpoonright\xi])$, since $G$ is $\mathcal{J}_{\Omega+\alpha}$-generic. Let $\zeta<\Omega$ be such that $X=g(\zeta)$. Now noting that $\mathbb{P}\upharpoonright[\xi,\Omega)$ is sufficiently homogeneous, we see that $X$ is actually definable over $\mathcal{J}_{\Omega+\beta}[G\upharpoonright\xi]$, hence $X\in\mathcal{J}_{\Omega+\alpha}[G\upharpoonright\xi]$, a contradiction.<|endoftext|> TITLE: Element being trivial in a finitely presented group independent of ZFC QUESTION [15 upvotes]: Is there an explicit finitely presented group $G$ and an element $g\in G$ such that the statement "$g$ is equal to the identity" is independent of ZFC? REPLY [21 votes]: The answer is yes. This is just an instance of the general phenomenon that every non-computable decision problem is saturated with logical independence. (See this related MO answer.) Theorem. If $A$ is a computably enumerable undecidable decision problem, such as the word problem for groups, then for any consistent c.e. theory $T$ extending PA, such as ZFC, there must be true instances of $n\notin A$ that are not provable in $T$. Proof. Suppose that $T$ is a consistent c.e. theory extending PA. It follows that all the true instances of $n\in A$ are provable in PA and hence in $T$. If conversely $T$ proved all true instances of $n\notin A$, then we could decide $A$ as follows: on input $n$ we search, during the day, for positive instances of $n\in A$, since $A$ is c.e., and at night, we search for a proof from $T$ that $n\notin A$. Since $T$ is consistent and proves all true existentials, it can be trusted for any proof of $n\notin A$. So if all such true instances of $n\notin A$ were provable in $T$, then $A$ would be decidable, contrary to assumption. $\Box$ In particular, if ZFC is consistent, then there must be specific concrete instances of the word problem where a word $g$ is nontrivial for a specific presentation, but this is not provable in ZFC. If you want to get actual independence of ZFC, then you can do so by assuming that ZFC is $\Sigma_1$-sound, which means that it proves only true existential arithmetic statements. If there is an inaccessible cardinal or even only a worldly cardinal, then ZFC is $\Sigma_1$-sound. But much less is required. Corollary. If $T$ is a consistent c.e. $\Sigma_1$-sound theory, then for any c.e. computable undecidable decision problem $A$, there must be concrete instances $n$ for which $n\in A$ is independent of $T$. Proof. The point is that if we have a true instance of $n\notin A$ that is not provable in $T$, then also by soundness $T$ will not prove $n\in A$, and so $n\in A$ will be independent of $T$. $\Box$ In particular, assuming ZFC is $\Sigma_1$-sound, which is a mild extra assumption about ZFC, then for any nondecidable decision problem, such as the word problem for groups, there will be concrete instances that are independent of ZFC. But finally, since I expect that you might want not just a proof that there is an explicit instance, but the explicit instance itself, let me describe how one might construct a specific instance. We know that the word problem in groups is undecidable because we can code the halting problem into it. Consider the Turing machine that searches for a proof of a contradiction in ZFC, halting only when found. From this instance of the halting problem, we can construct an group presentation $G$ and a word $g$ which is trivial in $R$ just in case the program halts. In other words, $g=1$ is true in $G$ if and only if $\neg\text{Con}(\text{ZFC})$, and this would be provable in a weak theory. If consistent, ZFC will not prove $g\neq 1$, since that would mean proving its own consistency; but if also $\text{Con}(\text{ZFC})$, then ZFC will not prove $g=1$ in $G$. And so it will be independent. This example also reduces the soundness requirement to mere consistency. REPLY [2 votes]: For any finite presentation there is an algorithm that will tell you if a word is equal to the identity: 1. Fix some positive integer $n$. 2. Write down all words in the generators that can be written as a product of at most $n$ conjugates of the given relators by elements that can be written as a product of at most $n$ of the given generators and their inverses. 3. If none of these words is equal (as a reduced word) to your given group element, increase the value of $n$ and repeat. If your word is not equal to the identity, the algorithm will not terminate. If you had a proof that ZFC could not show that your word was equal to the identity, then doesn't the above algorithm imply that it is not equal to the identity?<|endoftext|> TITLE: Is the diagonal of finitely presented groups computable? QUESTION [12 upvotes]: Let $f:G\to H$ be a surjective homomorphism of finitely presented groups. If the kernel of $f$ is finitely generated then is $G\times_H G$ is a finitely presented group? Can one compute an explicit finite presentation? REPLY [17 votes]: The answer is "no". The hypotheses for the fibre product to be finitely presentable are given by the 1-2-3 theorem of Bridson--Baumslag--Miller--Short: 1-2-3 Theorem (BBMS): The fibre product $G\times_H G$ is guaranteed to be finitely presentable as long as: the kernel of $G\to H$ is finitely generated; $G$ itself is finitely presented; $H$ satisfies a higher finiteness condition, namely being of "type $F_3$". However, there is no general algorithm that can compute a finite presentation for $G\times_H G$. Indeed, stronger, there is no algorithm that can compute its abelianisation, and the result follows since the abelianisation can be computed from a finite presentation. This is proved in my paper with Bridson: Bridson & Wilton, "On the difficulty of presenting finitely presentable groups", Groups Geom. Dyn. 5(2), pp. 301–325, 2011 available on the arXiv here. See Theorem A, and note in the proof that the subgroup $\Lambda_n$ is indeed a fibre product. To give a brief sketch of the argument, one takes a sequence of perfect groups $Q_n$ such that the second Betti numbers $b_2(Q_n)$ are known to be impossible to compute. The Rips short exact sequence gives $$1\to K_n\to\Gamma_n\to Q_n\to 1$$ with $K_n$ finitely generated, and then let $\Lambda_n\leq \Gamma_n\times\Gamma_n$ be the fibre product. Finally, an argument with the Stallings exact sequence relates the Betti numbers: $$b_1(\Lambda_n) = 2b_1(\Gamma_n)+b_2(Q_n).$$ Since $b_1(\Gamma_n)$ is computabe and $b_2(Q_n)$ isn't, it follows that $b_1(\Lambda_n)$ is not computable. In particular, no presentation for $\Lambda_n$ can be computed, even though we are given explicit generating sets and know that they are finitely presented.<|endoftext|> TITLE: Completion of an Alexandrov space QUESTION [7 upvotes]: Let $X$ be an incomplete Alexandrov space with sec $\ge -1$ in the sense that for any point in $X$ there exists a small neighborhood in which the four-points criterion is satisfied. Suppose $X$ is convex in the sense that any pair of points $p,q \in X$ there exists a minimizing geodesic connecting $p$ and $q$. Can we prove that the completion of $X$ is also an Alexandrov space with sec $\ge -1$. REPLY [6 votes]: Yes, the conclusion is exactly the main result of Petrunin's paper "A globalization for non-complete but geodesic spaces", Mathematische Annalen volume 366, pages387–393(2016).<|endoftext|> TITLE: Empty preimage under homomorphism of finitely presented groups with decidable word problems QUESTION [5 upvotes]: Let $f:G\to H$ be a homomorphism of finitely presented groups with decidable word problems. Assume you are given explicit finite presentations for both $G$ and $H$ and you are given the words to which $f$ sends the generators of $G$. Question. Is it possible that no algorithm exists deciding, given $w\in H$, whether $f^{-1}(w)$ is empty or not? P.S. Having explicit finite presentations is important https://arxiv.org/abs/1003.5117 REPLY [5 votes]: It is known that there exists a finitely presented group $H$ with solvable word problem that has a finitely generated subgroup $K$ whose subgroup membership problem is unsolvable. For example, Mikhailova has shown that the product $F_2\times F_2$ of non-abelian free groups has such a subgroup (see here). Now let $k_1,\ldots,k_n$ be the generators for $K$, let $G$ be a free group with $n$ generators $x_1,\ldots,x_n$, and let $f\colon G\to H$ be the homomorphism that maps each $x_i$ to $k_i$. Then a given element $h\in H$ has the property that $f^{-1}(h)$ is nonempty if and only if $h\in K$, and therefore it is undecidable in general whether $f^{-1}(h)$ is nonempty.<|endoftext|> TITLE: Non orientable, closed manifold covered by two contractible charts QUESTION [8 upvotes]: This is a follow up of my previous MO question "Non orientable, closed manifold covered by two simply-connected charts." Nick L's nice answer shows that such manifolds actually exist, examples being provided by some non-orientable $S^n$-bundles over $S^1$, with $n \geq 2$. In these examples, the two charts have the homotopy type of $S^n$. So, let me ask the following Question. Does it exist a closed, non-orientable smooth manifold that can be written as the union of exactly two contractible charts? Note. A comment by Denis Nardin to the aforementioned question shows that, by the strong form of Seifert-van Kampen theorem, a manifold covered by two contractible charts $U$, $V$ has the same homotopy type of the suspension of $U \cap V$. REPLY [15 votes]: No. Recall that the Lusternik-Schnirelmann category of a space $X$, denoted $\operatorname{cat}(X)$, is the minumum $k$ such that $X$ may be covered by open sets $U_0,U_1,\ldots, U_k$ such that each inclusion is null-homotopic. The standard lower bound for LS-category is the cup-length of reduced cohomology: if $R$ is a commutative ring, and $x_1,\ldots, x_k\in \tilde{H}^*(X;R)$ are cohomology classes whose cup product $x_1\cdot \cdots \cdot x_k\in \tilde{H}^*(X;R)$ is non-zero, then $\operatorname{cat}(X)\geq k$. The proof of this is a nice exercise in the naturality of relative cup products. Now suppose $N$ is a closed non-orientable $n$-manifold which is covered by two contractible charts $U_0$ and $U_1$. It follows that $\operatorname{cat}(N)=1$ (it can't be $0$, because closed manifolds are never contractible), and all cup products in $\tilde{H}^*(N;\mathbb{Z}/2)$ are trivial. Since $N$ is non-orientable, its first Stiefel-Whitney class $w_1(N)\in H^1(N;\mathbb{Z}/2)$ is non-zero. Poincaré duality gives a non-singular pairing $H^1(N;\mathbb{Z}/2)\times H^{n-1}(N;\mathbb{Z}/2)\to H^n(N;\mathbb{Z}/2)$. In particular $w_1(N)\cdot y\neq 0$ for some $y\in H^{n-1}(N;\mathbb{Z}/2)$. Contradiction.<|endoftext|> TITLE: Is this function concave? QUESTION [12 upvotes]: Let $$h(u):=u^3 \left|\int_u^\infty \frac{e^{-i t}}{t^3} \, dt\right|$$ for $u>0$. Is the function $h$ concave on $(0,\infty)$? (For context, see Proposition 4.4.4 and formula (4.4.21) in this paper or, equivalently, Proposition 4.4 and formula (4.21) in the corresponding arXiv preprint. In particular, according to that proposition, $h(v^{1/6})^2$ is concave in $v>0$.) Here is the graph $\{(u,h(u)):0U$, the integrand in the numerator is multiplied at every point by at most $\frac{U+a}{U_1+a}$ and the integrand in the denominator by at least that amount ($T\mapsto \frac{U+T}{U_1+T}$ is increasing in $T>0$). We now have the Moral If $\frac{T\Psi'(T)}{\Psi(T)}$ is a decreasing function of $T$, then $$ \frac{\int_0^\infty\frac{T^2}{(U+T)^2}\Psi(T)\,dT}{\int_0^\infty\frac{T}{U+T}\Psi(T)\,dT} $$ is a decreasing function of $U$. Step 5: The decreasing property of $\frac{T\Psi'(T)}{\Psi(T)}$ We have $$ \frac{T\Psi'(T)}{\Psi(T)}=\frac12\frac{\sqrt T\Phi'(\sqrt T)}{\Phi(\sqrt T)} $$ so it suffices to check that $\frac{t\Phi'(t)}{\Phi(t)}$ decreases in $t>0$. Now, $$ \frac{t\Phi'(t)}{\Phi(t)}=-t-\frac{\int_0^\infty\frac{\tau^2t}{(t+\tau)^2}e^{-\tau}\,d\tau}{\int_0^\infty\frac{\tau^2}{t+\tau}e^{-\tau}\,d\tau}=-t-1+\frac{\int_0^\infty\frac{\tau^2}{(t+\tau)^2}\tau e^{-\tau}\,d\tau}{\int_0^\infty\frac{\tau}{t+\tau}\tau e^{-\tau}\,d\tau} $$ and it remains to show that the last term is a decreasing function of $t$. By the Moral (with $t$ instead of $U$ and $\tau$ instead of $T$), we just need to check that $$ \frac{\tau\frac{d}{d\tau}(\tau e^{-\tau})}{\tau e^{-\tau}}=1-\tau $$ is a decreasing function of $\tau$, which is sort of obvious. The End<|endoftext|> TITLE: Equality to a power of a given word undecidable in finitely presented group with decidable word problem QUESTION [5 upvotes]: Let $G$ be a group with an explicit finite presentation. Assume $G$ has a decidable word problem. Can there exist an explicit word $w\in G$ such that there is no algorithm deciding if a given word $w'\in G$ is equal to $w^k$ for some $k\in \mathbb{Z}$? REPLY [3 votes]: This question is very similar to Is it decidable to check if an element has finite order or not? and has the same answer. Namely, an example was constructed by McCool. He also constructs in that paper a finitely presented group with solvable word problem where you cannot decide if an element has finite order, known as the order problem. I'm pretty sure that the power problem, which is what you asked for here has also been answered before on Mathoverflow. Note the group is explicitly constructed via a recursive presentation and then uses Higman embedding to get the finite presentation so it may not be as explicit as you would like because you would have to chase the Higman embedding but the word $w$ is explicitly given and fixed modulo the Higman embedding.<|endoftext|> TITLE: "Drinking number" of a graph QUESTION [12 upvotes]: Motivation. A while ago I attended a party and I only knew some, but not all, of the attendees. There were 2 kinds of drinks: beer and soda. I noticed that amongst my acquaintances, more than half drank beer. So in order to go against what I perceived as the mainstream ("drink a beer"), I picked a soda. Then I wondered whether a beer / soda mapping to all the guests is possible so that everyone has a drink such that at least half of his acquaintances have the other drink. Let's formalize this. Formal version. Let $G=(V,E)$ be a simple, undirected graph (not necessarily finite). For $v\in V$, let $N(v) = \{w\in V: \{v,w\}\in E\}.$ If $\kappa > 0$ is a cardinal, we call a map $d: V \to \kappa$ a "drinking map" if for all $v\in V$ with $N(v) \neq \emptyset$ we have $$|N(v)\cap d^{-1}(\{d(v)\})| \leq |N(v) \setminus d^{-1}(\{d(v)\})|.$$ (We imagine $d(v)$ to be the "drink" that $v$ is having, and $v$ does not want that more than half of her friends are having $d(v)$ as well.) Let the "drinking number" of $G$ be the smallest cardinal $\kappa$ such that there is a "drinking map" $d: V\to \kappa$, and denote it by $\text{dr}(G)$. Question. Can $\text{dr}(G)$ become arbitrarily large for finite graphs? How about infinite graphs? REPLY [11 votes]: Known as unfriendly partition conjecture. Open for countable graphs: http://www.openproblemgarden.org/op/unfriendly_partitions.<|endoftext|> TITLE: Monochromatic infinity operads as algebras over the "operad operad" QUESTION [6 upvotes]: In the "ordinary" operad category, it is known that there is a colored operad $Op$ with set of colors $\mathbb{N}$ corresponding to "degrees" of vertices and with operations indexed by trees, such that algebras over $Op$ in $\mathrm{Set}$ (or more generally any symmetric monoidal category) correspond to monochromatic operads. Is it known that $\infty$-algebras over $Op$ are equivalent to monochromatic $\infty$-operads? REPLY [3 votes]: Yes, combine Corollary 9.4.1 and Theorem 7.11 of arXiv:1410.5675, for example. This topic is also examined more explicitly in the work of Chu and Haugseng, arXiv:1707.08049. Corollary 5.1.13 shows that enriched ∞-operads are equivalent to ∞-algebras over the operad of colored operads. Theorem 5.2.10 proves (using arXiv:1410.5675) that the underlying ∞-category of the relative category of ordinary enriched colored operads is equivalent to the ∞-category of enriched colored ∞-operads.<|endoftext|> TITLE: Enriched coends which preserve equivalences QUESTION [5 upvotes]: Although this question might be formulated in higher generality, let me try to be concrete: Let $(\mathbf{Top},\times,*)$ be the monoidal category of compactly generated weak Hausdorff spaces; and let $\mathbf{C}$ be a small and $\mathbf{Top}$-enriched category (in the easiest case I have in mind: a topological group). Now let $A\colon \mathbf{C}^{\mathrm{op}}\to \mathbf{Top}$, $A'\colon\mathbf{C}^{\mathrm{op}}\to\mathbf{Top}$, and $B\colon\mathbf{C}\to \mathbf{Top}$ be three enriched functors, and let $\vartheta\colon A\Rightarrow A'$ be a natural transformation. I am looking at the induced map between the enriched coends $$\int^{c\in\mathbf{C}}\vartheta_c\times\mathrm{id}_{Bc}\colon \int^{c\in\mathbf{C}}Ac\times Bc\to \int^{c\in\mathbf{C}}A'c\times Bc.$$ Assume that $\vartheta$ is an equivalence, in the sense that each $\vartheta_c\colon Ac\to A'c$ is a weak equivalence. Under which extra conditions can I conclude that also the induced map $\int^c\vartheta_c\times\mathrm{id}_{Bc}$ is a weak equivalence? Some conditions I could imagine and which I would be happy with could be: $A$ and $A'$ are free in the sense that for each $c\in\mathbf{C}$, the group $\mathrm{Aut}_{\mathbf{C}}(c)$ acts freely on $Ac$ and $A'c$, for each morphism $f\colon c\to c'$ in $\mathbf{C}$, the map $Bf\colon Bc\to Bc'$ is an (equivariant?) cofibration. Let me finish the question with some examples I had in mind: If $\mathbf{C}=G$ is just a topological group, and we write $A$, $A'$ and $B$ for the corresponding $G$-spaces, then we just look for the induced map $A\times_G B\to A'\times_G B$, and here if is e.g. suffient that $G$ acts freely on both $A$ and $A'$. If $\mathbf{C}$ is the semisimplex category $\mathbf{\Delta}^+$ of finite ordered sets $[n]=\{0<\dotsb TITLE: Is there a $4$-manifold which Immerses in $\mathbb{R}^6$ but doesn't Embed in $\mathbb{R}^7$? QUESTION [24 upvotes]: I'm interested in both version of the question in the title, i.e. in the topological category and in the smooth category. By a topological immersion I mean a local embedding. I was asking in relation to this question: https://math.stackexchange.com/questions/1801318/dimensions-of-immersions-vs-embeddings In the very nice answer given in that thread, they work out almost all of the low-dimensional cases for smooth, compact manifolds and smooth immersions/embeddings. The only 'smooth, compact' case not covered by their answer is the one in the title of this question, so it would be interesting to know if there is a compact example, separately. As a side-question, for topological $4$-manifolds the case of immersion in $\mathbb{R}^5$ is what remains to be fleshed out in the answer to that previous thread. Some relevant questions for that case: Is every compact $4$-manifold which immerses in $\mathbb{R}^5$ smoothable? Is there a $4$-manifold that immerses in $\mathbb{R}^5$ but doesn't embed in $\mathbb{R}^6$ (resp. $\mathbb{R}^7$)? By results of Quinn, every open $4$-manifold is smoothable so it's sufficient to prove the smooth case for non-compact manifolds. I cross-listed one of these on MSE expecting a quick counterexample, but still no takers. Someone gave a partial answer to the effect of "in high dimensions, codimension-$1$ locally flat immersion implies smoothability": https://math.stackexchange.com/questions/4115222/is-every-compact-n-manifold-that-immerses-in-mathbbrn1-smoothable REPLY [21 votes]: Edit: The answer below is incorrect. In fact, $\bar{w}_3(\mathbb{R}P^2\times\mathbb{R}P^2)=0$ (thanks to Rafal Walczak for pointing this out) so by the cited result $\mathbb{R}P^2\times\mathbb{R}P^2$ does embed in $\mathbb{R}^7$! So please strip me of my points and consider the question unanswered... Edit 2: In fact what the below shows is that the answer to the question in the title is no in the smooth compact case. If $M^4$ immerses in $\mathbb{R}^6$ then $\bar{w}_3(M)=0$, which by the quoted result of Fang implies that $M$ embeds in $\mathbb{R}^7$. The manifold $\mathbb{R}P^2\times \mathbb{R}P^2$ smoothly immerses in $\mathbb{R}^6$, as a product of Boy's surfaces. However, the main result of Fang, Fuquan, Embedding four manifolds in (\mathbb{R}^ 7), Topology 33, No. 3, 447-454 (1994). ZBL0824.57014 asserts that a closed smooth $4$-manifold $M$ embeds in $\mathbb{R}^7$ if and only if the normal Stiefel-Whitney class $\bar{w}_3(M)$ vanishes. A quick calculation shows that this is not the case for $M=\mathbb{R}P^2\times \mathbb{R}P^2$, which therefore doesn't embed in $\mathbb{R}^7$. The cited article also gives necessary and sufficient conditions for topological embeddability of $4$-manifolds in $\mathbb{R}^7$. For example, a closed smooth non-orientable $4$-manifold $M$ admits a locally flat embedding in $\mathbb{R}^7$ if and only if $\bar{w}_3(M)=0$ and $KS(M)=0$, where $KS$ is the Kirby-Siebenmann invariant.<|endoftext|> TITLE: No canonical isomorphism QUESTION [35 upvotes]: I thought that it would be interesting to collect into a big list various instances of isomorphic structures with no preferred isomorphism between them. I expect the examples to be interesting since it seems that in such situations choice of a particular isomorphism is frequently an important kind of structure. For some reason all examples that I could come up with are related to some sort of self-duality, although there must be others not related to any duality, and I am especially curious about the latter. So let me do this: I will ask some questions about these self-dualities. If most of the answers are about these, I will not add the big list tag. If there are many examples of some other kind, then I will. The simplest and most ubiquitous one, you have already guessed it: isomorphism between a finite-dimensional vector space and its dual. The choice of isomorphism amounts to a non-degenerate bilinear form. While we are at that, let me ask this: at a first glance, the fact that such forms can be (at least in characteristic 0) decomposed into the sum of a symmetric and a skew-symmetric form is just the consequence of the fact that eigenvalues of an involution are $\pm1$. But initially, given just a nondegenerate form, there is no involution present, unless we have another such form. So how to explain that such a decomposition still exists? Or does it in fact not, and one has to speak about pairs of such forms?? My subsequent examples will be just generalizations of the first. A finite abelian group and its Pontryagin dual. If it is an elementary $p$-group, this is a particular case of the above (vector spaces over prime fields). What about the general case? Is a choice of isomorphism, i.e. a nondegenerate pairing $A\otimes A\to\mathbb Q/\mathbb Z$ a structure that is actually used somewhere? I've heard about Weil pairings but know too little about them to figure out whether they are an instance of such a thing. Conjugacy classes of a finite group and its irreducible representations. Again, does choice of a bijection between these two sets come up somewhere in mathematics? What comes next are examples when an isomorphism need not exist. An isomorphism between an abelian variety and its dual. Is this used somewhere? A diffeomorphism/PL-isomorphism/homeomorphism between a manifold and its dual. Here I don't even know what I am asking. Does this make sense at all? Here I know what I am asking: a homotopy equivalence between a finite CW-complex and its Spanier-Whitehead dual. Do such equivalences have a name? Related questions: Are there situations when regarding isomorphic objects as identical leads to mistakes? Equality vs. isomorphism vs. specific isomorphism Later Excuses to those who contributed extremely interesting answers and comments, but as it has been pointed out there actually exists a very similar question (with equally interesting answers). Besides, although closed, all of the answers will be accessible to everybody, right? REPLY [8 votes]: Algebraic closures of any given field are isomorphic, but there is no preferred isomorphism (unless the given field is already algebraically closed).<|endoftext|> TITLE: Quasiconformal maps in arbitrary dimensions QUESTION [7 upvotes]: I am aware that a quasiconformal map satifies the formula $$ \frac{\partial f}{\partial \overline{z}} = \mu(z) \frac{\partial f}{\partial z} $$ where $\sup\{\mu(z):z \in \text{Domain}\{f\}\}<1$ imposes a bound on the eccentricity of the ellipses in the image of $f$. For a multicomplex function $F(z_1, z_2, \dots, z_n)$, one could impose the quasiconformality condition for each variable $z_k$. I have read computer science articles about three-dimensional quasiconformal mappings, but no explicit formula relating the conjugate partial derivative to the partial derivative is given. Question: Does a generalization of the formula $\frac{\partial f}{\partial \overline{z}} = \mu(z) \frac{\partial f}{\partial z}$ apply to quasiconformal mappings in $\mathbb{R}^{2n+1}$? REPLY [4 votes]: The correct definition of higher-dimensional quasiconformal maps does not use complex variables. The correct condition is $$ |Df(x)|\le K |J_f(x)| $$ where $J_f$ is the Jacobian determinant. An orientation-preserving homeomorphism $f$ between two domains $U, V$ in $R^n$ is called quasiconformal if it belongs to the Sobolev class $W^{n,1}_{loc}(U)$ (hence, has 1st order partial derivatives a.e. in $U$) and there exists a constant $K$ such that the above inequality is satisfied a.e. in $U$. There are many other alternative definitions. For instance, instead of working with Sobolev spaces, you can assume that $f$ is absolutely continuous on almost every coordinate line segment in $U$ (hence, has partial derivatives a.e. in $U$) and satisfies the same inequality as above. if you do not know about Sobolev spaces or absolute continuity, just think of $f$ as a diffeomorphism (this is not enough, but suffices for the intuition). Other definitions are in terms of conformal moduli, conformal capacities, quasisymmetry,... Some references: Iwaniec, Tadeusz; Martin, Gaven, Geometric function theory and nonlinear analysis, Oxford Mathematical Monographs. Oxford: Oxford University Press (ISBN 0-19-850929-4/hbk). xv, 552 p. (2001). ZBL1045.30011. Reshetnyak, Yu. G., Space mappings with bounded distortion, Translations of Mathematical Monographs, 73. Providence, RI: American Mathematical Society (AMS). xv, 362 p. (1989). ZBL0667.30018. Väisälä, Jussi, Lectures on (n)-dimensional quasiconformal mappings, Lecture Notes in Mathematics 229. Berlin-Heidelberg-New York: Springer-Verlag. XIV, 144 p. (1971). ZBL0221.30031. Addendum. The analogue of the Beltrami differential of a map $f$ in higher dimensions is $$ M_f(x):= J^{-2n}_f(x) (Df(x))^T Df(x), $$ a field of symmetric positive-definite matrices on $U$. The Beltrami equation $$ M_f(x)=A(x), $$ with $A(x)$ a field of positive-definite symmetric matrices on $U$, is overdetermined if $n\ge 3$ (as Alex noted in his answer). Nevertheless, this equation is sometimes useful when working with quasiconformal maps, although not as useful as the classical Beltrami equation: Most tools in higher dimensions are not analytic but geometric. An interesting thing is that if $A(x)$ is smooth, then there are known necessary and sufficient conditions for solvability of the higher-dimensional Beltrami equation (the condition is a 3rd order nonlinear PDE on $A$ if $n=3$ and a 2nd order nonlinear PDE on $A$ if $n\ge 4$). I do not know if anybody worked out a distributional analogue of this classical result.<|endoftext|> TITLE: Can you use Oseledet's theorem to numerically approximate the Lyapunov spectra? QUESTION [5 upvotes]: Let's say you have a set of first order differential equations with known Jacobian $J$. Let $x_0, x_1, ..., x_n$ be sampled points on the trajectory near the attractor. Let $T_n = J(x_{n-1})J(x_{n-2})...J(x_0)$. Oseledet says that $$\frac{1}{n} \log(\mbox{the singular values of } T_n)$$ converge to the Lyapunov spectra as $n \to \infty$. However, can this be used in practice to numerically approximate the Lyapunov spectra (say just to the 2nd decimal place)? Is it known how big $n$ must be to do so? REPLY [4 votes]: Once you have a method to estimate the top Lyapunov exponent, you can use the action on wedge products to estimate the other exponents (This should be clear if you examine the proof of Oseledet's theorem.) Estimating the top exponent is difficult, in general. Doing it via the definition is very slow as you observed. More efficient approaches are usually based on Furstenberg's foundational paper [1] that preceded Oseledets' work. The key is to estimate the stationary measure on projective space constructed in [1]. See also [2]. Sometimes this measure is discrete which allows for efficient computation [3]. A sophisticated complex analysis approach is in [4]. [1] Furstenberg, Harry. "Noncommuting random products." Transactions of the American Mathematical Society 108, no. 3 (1963): 377-428. [2] Gol'dsheid, I. Ya, and Grigorii Aleksandrovich Margulis. "Lyapunov indices of a product of random matrices." Russian mathematical surveys 44, no. 5 (1989): 11-71. [3] Kenyon, Richard, and Yuval Peres. "Intersecting random translates of invariant Cantor sets." Inventiones mathematicae 104, no. 1 (1991): 601-629. [4] Pollicott, Mark. "Maximal Lyapunov exponents for random matrix products." Inventiones mathematicae 181, no. 1 (2010): 209-226.<|endoftext|> TITLE: Optimality of the Riemann Hypothesis QUESTION [10 upvotes]: The Riemann hypothesis is equivalent to the assertion that the prime counting function $\pi(x) := \sum_{p \le x} 1$ deviates from the logarithmic integral $Li(x) = \int_2^x \frac{dx}{\log x}$ in the order $O(\sqrt{x} \log x)$. Since $\log x = O(x^\alpha)$ for any $\alpha>0$, the Riemann hypothesis implies that: $$\forall \alpha > \frac12, |\pi(x) - Li(x)| = O(x^\alpha)$$ From what I understand, this is the best possible power bound available, meaning that for any $\alpha \le \frac12$, we don't have $|\pi(x) - Li(x)| = O(x^\alpha)$. (I'm not sure whether the bounds like $O(\sqrt{x} \log \log x)$ could be possible.) Where can I read a proof that this indeed is the best possible power bound? The answer on this question claims that any textbook on analytic number theory will do, but I'd like to know an explicit textbook reference that I can look into. More concretely, I followed the steps outlined in the question referenced above, and could not progress at one point. For the Chebyshev function $\psi(x) = \sum_{p^n \le x}\log p$, I derived that \begin{align*} \left| \psi(x) - x \right| = \left| \log 2\pi + \sum_{\rho} \frac{x^\rho}{\rho} \right| = \left| \log 2\pi + \frac12 \log (1- \frac1{x^2}) + \sum_{\rho \text{ ntv.}} \frac{x^\rho}{\rho} \right| \sim \left| \sum_{\rho \text{ ntv.}} \frac{x^\rho}{\rho} \right| \end{align*} where $f(x) \sim g(x)$ iff $\lim_{x \rightarrow \infty} f(x)/g(x) = 1$, the summation over $\rho$ is the summation over the zeroes of the Riemann zeta, and the qualifier "ntv." refers to counting only the nontrivial zeroes. Since $\psi(x) \sim \pi(x) \log x$, at this point I'd like to show two things: The Riemann hypothesis is equivalent to the assertion that $|\psi(x) - x| < \sqrt{x} \log ^2 x$ for large enough $x$ (Schoenfield 1976, according to Wikipedia) Regardless the Riemann hypothesis, one cannot have $|\psi(x) - x| = O(x^\alpha)$ for any $\alpha \le \frac12$. It looks as if both of the equivalences should follow straightforwardly from the expression $$|\psi(x) - x| \sim \left| \sum_{\rho \text{ ntv.}} \frac{x^\rho}{\rho} \right|$$ However I don't know how to work out the phase factors appearing in the following expression: $$\frac{x^\rho}{\rho} = \left( \frac{e^{it \log x} (\sigma - it)}{\sigma^2+t^2} \right) \cdot x^\sigma \text{, where $\rho = \sigma + it$}$$ to obtain a lower bound or an upper bound to the error term. Further analysis indeed depends on the distribution of the imaginary part of the nontrivial zeroes, which will control how much the weights $\frac1\rho$ contribute in the end. REPLY [14 votes]: See Chapter 15 ("Oscillation of error terms") in Montgomery-Vaughan: Multiplicative number theory I. See especially Theorems 15.2-15.3 and 15.11. Added by Steven Clark and GH from MO. For convenience, we copied below the relevant theorems and some additional text from the book. As usual, $$M(x):=\sum\limits_{n\le x}\mu(n)$$ is the Mertens function. Theorem 15.2. Let $\Theta$ denote the supremum of the real parts of the zeros of the zeta function. Then for every $\varepsilon>0$, $$\psi(x)-x=\Omega_\pm(x^{\Theta-\varepsilon})\tag{15.1}$$ and $$\pi(x)-\mathrm{li}(x)=\Omega_\pm(x^{\Theta-\varepsilon})\tag{15.2}$$ as $x\to\infty$. Theorem 15.3. Suppose that $\Theta$ is the supremum of the real parts of $\zeta(s)$, and there is a zero $\rho$ with $\Re\rho=\Theta$, say $\rho=\Theta+i\gamma$. Then $$\underset{x\to\infty}{\text{lim sup}}\ \frac{\psi(x)-x}{x^{\Theta}}\geq \frac{1}{| \rho |}\tag{15.4}$$ and $$\underset{x\to\infty}{\text{lim inf}}\ \frac{\psi(x)-x}{x^{\Theta}}\leq -\frac{1}{| \rho |}.\tag{15.5}$$ Theorem 15.11. As $x\to\infty$, $$\psi(x) -x = \Omega_{\pm} \bigl(x^{1/2} \log \log \log x\bigr),$$ and $$\pi(x) - \mathrm{li}(x) = \Omega_{\pm} \bigl(x^{1/2}(\log x)^{-1}\log \log \log x\bigr).$$ Then in the manner of the proof of Theorem 15.3, we find that if $\Theta+i\gamma$ is a zero of $\zeta(s)$, then $$\underset{x\to\infty}{\text{lim sup}}\ \frac{M(x)}{x^{\Theta}}\geq \frac{1}{| \rho\,\zeta'(\rho) |},\tag{15.11}$$ and $$\underset{x\to\infty}{\text{lim inf}}\ \frac{M(x)}{x^{\Theta}}\leq -\frac{1}{| \rho\,\zeta'(\rho) |}.\tag{15.12}$$<|endoftext|> TITLE: Induction and restriction of unitary representations QUESTION [7 upvotes]: $\DeclareMathOperator\Rep{Rep}\DeclareMathOperator\Ind{Ind}\DeclareMathOperator\Res{Res}$Given a locally compact group $G$ and a closed subgroup $H\subset G$, let $\Rep(G)$ and $\Rep(H)$ denote their categories of (strongly continuous) unitary representations on Hilbert spaces. Let $NG$ be the group von Neumann algebra of $G$ (the weak closure of the left action of $G$ on $L^2G$), and let $NH$ be the group von Neumann algebra of $H$. We have full subcategories $\Rep(NG)\subset \Rep(G)$, and $\Rep(NH)\subset \Rep(H)$. (The category $\Rep(NG)$ can also be described as the full subcategory of $\Rep(G)$ consisting of reps which are direct sums of direct summands of the regular representation.) It is well-known that the restriction and induction functors $\Res: \Rep(G) \to \Rep(H)$and $\Ind: \Rep(H) \to \Rep(G)$            defined by $\Ind(V) = L^2(G/H , G\times_HV)$ are NOT each other's adjoints in any sense (see for example this previous MO question). But maybe I'll be more lucky if I restrict these functors to the subcategories $\Rep(NG)$ and $\Rep(NH)$? $\Res: \Rep(NG) \to \Rep(NH)$and $\Ind: \Rep(NH) \to \Rep(NG)$ Is there any relationship between the two latter functors which can be formulated in category-theoretic terms? More precisely, is there an $NH$-$NG$-bimodule $X$ such that $\Res: \Rep(NG) \to \Rep(NH)$ is given by tensoring by $X$, and $\Ind: \Rep(NH) \to \Rep(NG)$ is given by tensoring by the complex conjugate of $X$? One result in that direction is this old theorem of Mackey (from this paper): REPLY [3 votes]: Yes, you can take $\overline X = {}_{N(G)}L^2(G)_{N(H)}$ using the canonical inclusion of $N(H)$ into $N(G)$ to restrict the "trivial" correspondence $L^2(G)$ to a $N(G)-N(H)$ correspondence. (I am assuming here "bimodule" is a "correspondence" in the sense of Connes, i.e. a Hilbert space with commuting, normal actions of $N(G)$ and $N(H)^o$. The (tensor) product is the Connes fusion, aka the Sauvageot relative tensor product (Sur le produit tensoriel rélatif d’espaces de Hilbert). See also Takesaki, Volume 2, Chapter IX, Section 3.) Restriction is easy to check. As $N(G)$ acts in standard form on $L^2(G)$, we know that ${}_{N(G)}L^2(G)_{N(G)}$ is self-dual, so $X \cong {}_{N(H)}L^2(G)_{N(G)}$. From Takesaki, Prop 3.19, we know that if ${}_{N(G)}H$ is any left $N(G)$-module then ${}_{N(G)}L^2(G)_{N(G)} \otimes {}_{N(G)}H \cong {}_{N(G)}H$. So ${}_{N(H)}X_{N(G)} \otimes {}_{N(G)}H \cong {}_{N(H)}H$, which is restriction. Checking that we get induction seems harder. I think if $G$ is discrete, or more generally $H$ open, this can be done with a long but elementary computation. I want to give an indication of the general case, which involves analytic difficulties. First a sanity check: if we induce the left-regular representation $\lambda_H$ we get $\lambda_G$. So we wish to induce $L^2(H)$ to $L^2(G)$ as $N(H)$ and $N(G)$ modules. But $\overline X_{N(H)} \otimes {}_{N(H)}L^2(H)$ can indeed be shown to be isomorphic to ${}_{N(G)} L^2(H)$ (see below). As any left module of $N(H)$ is obtained by direct sum and subrepresentation of $L^2(H)$, and as induction respects direct sums, this would seem to indicate that we get what we want. However, I see no "soft" way of checking that all the isomorphisms involved really do commute. That is, if $\pi_1 \oplus \pi_2 \cong \lambda_H$ then both "induction" methods give modules of $N(G)$ which sum to $\lambda_G$, but I do not see why the individual components are automatically isomorphic. Perhaps there is some way to use imprimivity, but I also do not see this. Instead, I want to build "models" of induction, and tensoring by $\overline X$, which are manifestly the same. I will use the model of induction which works by using a (suitably measurable) cross section $G/H \rightarrow G$. This is "Realization III" in Kaniuth and Taylor's book (page 79-80). We can deal with any locally compact $G$ by using results of Kehlet - Cross sections for quotient maps of locally compact groups. Following Kehlet and page 3494 of Daws et al. - Closed quantum subgroups of locally compact quantum groups (I thank in particular my coauthor Pawel here) we pick a quasi-invariant measure $\lambda$ on $G/H$ and then we find a unitary $$ T: L^2(G/H,\lambda) \otimes L^2(H) \rightarrow L^2(G), $$ with $T(1\otimes \rho^H(h))T^* = \rho^G(h)$ for $h\in H$, where $\rho^H$ and $\rho^G$ are the right-regular reps of $H$ and $G$, respectively. If you perform the calculation, you'll also find that $$ T U^{\lambda_H}(g) T^* = \lambda^G(g) \qquad (g\in G), $$ where $\lambda^G$ is the left-regular representation of $G$, and $U^{\lambda_H}$ is the induced representation of the left-regular representation of $H$, up to $G$. Thus we have our "model" of induction. Clearly, if $\pi$ is a sub-representation of $\lambda^H$ then we get the "same" model by restricting $T$ to $L^2(G/H,\lambda) \otimes K$ where $K\subseteq L^2(H)$ is the $N(H)$-invariant subspace associated to $\pi$. (Really "clearly" here means follow the isomorphisms through.) What about tensoring with $\overline X$? I don't want to compute with the Sauvageot tensor product, as that would involve consideration of vectors in $L^2(G)$ bounded with respect to the Plancheral measure on $N(H)$ etc. Instead, I'll use a realisation of correspondences as self-dual Hilbert $C^*$-bimodules (see work of Reiffel and Paschke). For von Neumann algebras $M$, $N$ this identifies ${}_M H_N$ with $$ \hom_N(L^2(N), {}_MH_N) $$ the bounded linear maps $t:L^2(N)\rightarrow H$ with $t(\xi n) = t(\xi)n$ for $n\in N$. Composition of operators defines a left $M$ and right $N$ module structure, and we get an $N$-valued inner-product by setting $(t|s) = t^*s$. Note that $t^*s$ is an operator on $L^2(N)$ which commutes with the right $N$-action on $L^2(N)$, that is, can be identified with an element of $N$. Conversely, given a bimodule ${}_MY_N$ we take the balanced tensor product $Y_N \otimes_N L^2(N)$ which gives a Hilbert space, and this space inherits the left $M$ action from ${}_M X$ and the right $N$ action from $L^2(N)$, so becoming a correspondence. Our $\overline X = {}_{N(G)} L^2(G)_{N(H)}$ is identified with $$ \hom_{N(H)}(L^2(H), L^2(G)) $$ and if $K\subseteq L^2(H)$ is a sub-rep of $\lambda_H$, then $\overline X \otimes K$ is identified with the left $N(G)$-module $$ \hom_{N(H)}(L^2(H), L^2(G)) \otimes_{N(H)} K. $$ This, by definition, is the separation-completion of the algebraic tensor product for the scalar-valued inner-product $$ (t_1\otimes\xi_1 | t_2\otimes\xi_2) = (\xi_1 | t_1^*t_2 \xi_2)_{K}, $$ where $t_1^*t_2 \in N(H)$ and so acts on $K$. However, this is simply $$ (t_1(\xi_1)| t_2(\xi_2))_{L^2(G)}, $$ which makes sense as $K\subseteq L^2(H)$. Thus $\overline X \otimes K$ is identified with a subspace of $L^2(G)$; the left $N(G)$-action works as expected. The space we get is the closed linear span of $\{t(K) : t\in \hom_{N(H)}(L^2(H), L^2(G)) \}$. We want to understand $\hom_{N(H)}(L^2(H), L^2(G))$. Apply the unitary $T$ to identify $L^2(G)$ with $L^2(G/H)\otimes L^2(H)$. Due to the above commutation of the right-regular representations, the right $N(H)$ action is intertwined by $T$. So we get $$ \hom_{N(H)}(L^2(H), L^2(G/H)\otimes L^2(H)) = \{ t:L^2(H) \to L^2(G/H)\otimes L^2(H) : tn' = (1\otimes n')t \ (n'\in N(H)')\}. $$ As $K$ is the range of a projection $e\in N(H)'$ we see that $t(K) = te(L^2(H)) = (1\otimes e)t(L^2(H))$ and as $t$ varies we obtain all of the image of $1\otimes e$, that is, $L^2(G/H) \otimes K$. Conclude: both operations of "induction" respect sub-representations of $\lambda_H$ in the same way. By taking direct sums, this shows that both versions of induction agree on all $N(H)$-modules, as required.<|endoftext|> TITLE: Volume of singular Kahler metric QUESTION [5 upvotes]: Let $X$ be a compact complex manifold of complex dimension $n$ and let $\omega$ be a smooth Kahler form on it. Let $Y \subset X$ be a complex (possibly singular) hypersurface and let $u: X \setminus Y \to \mathbb{R}$ be a smooth function. Question: If $ \sup_{X \setminus Y} |u| < \infty$ and $\omega + i \partial\bar{\partial} u>0$ on $X\setminus Y$, is it true that $$ \int_{X \setminus Y} (\omega + i \partial\bar{\partial} u)^n = \int_X \omega^n \hspace{8mm} ? $$ REPLY [5 votes]: Yes, this is true. By the boundedness assumption, $u$ extends to a (bounded) $\omega$-psh function on $X$. Bedford and Taylor defined in '82 the Monge-Ampère operator of a bounded psh function, which has later been extended to the global quasi-psh case. Almost by definition of that operator, you have $\int_X(\omega+dd^c \varphi)^n=\int_X \omega^n$ for any bounded $\omega$-psh function $\varphi$ (cf last paragraph), and BT showed that the measure $(\omega+dd^c \varphi)^n$ puts not mass on pluripolar sets (in particular, it puts no mass on analytic subsets like $Y$). You then get $$\int_{X\setminus Y}(\omega+dd^c \varphi)^n=\int_X(\omega+dd^c \varphi)^n=\int_X \omega ^n.$$ To understand why the mass is cohomological, look at $n=2$ where $(dd^c \varphi)^2:=dd^c(\varphi dd^c \varphi)$ by the very definition of BT. In particular, $\int_X(\omega+dd^c\varphi)^2=\int_X\omega^2+2 \int_X\omega\wedge dd^c \varphi+\int_X(dd^c\varphi)^2$ and the last two terms vanish by Stokes theorem.<|endoftext|> TITLE: Top cohomology of profinite Poincaré duality group QUESTION [6 upvotes]: The paper "Cohomology of p-adic analytic groups" by Symonds and Weigel is considered one of the main references for continuous cohomology of profinite groups. There is a passage I do not understand: the context is the following. We consider $G$ to be a pro-$p$ group Poincaré dual: this is defined in §4.4 to be a group of finite cohomological dimension $n$ such that $\mathbb{Z}_p$ (considered as a trivial module) admits a projective resolution of finite length. Moreover we have $H^k(G; \mathbb{Z}_p [\![G]\!] )=0$ for $k \neq n$ and $H^k(G; \mathbb{Z}_p [\![G]\!] )=\mathbb{Z}_p$. The first two conditions are only for finiteness and well behavior of the associated cohomology groups, the key property is the third which ensures $H^*(G;-)$ and $H_{n-*}(G;{-})$ are related by an appropriate duality. In fact we can show the existence of a right $\mathbb{Z}_p [\![G]\!] $-module $D_p(G)$ whose underlying abelian group is $\mathbb{Z}_p$ and such that there are natural bjiections \begin{equation} H^*(G;M)\cong H_{n-*}(G; D_p(G)\otimes M) \end{equation} for $M$ in an appropriate category of left $\mathbb{Z}_p [\![G]\!] $-modules. My question regards Corollary 4.5.2 of this paper: it states that for such $G$ and any module $M$ which is isomorphic to $\mathbb{Z}_p$ as abelian group then the following two conditions are equivalent $H^n(G;M)\neq 0$ $D_p(G)\widehat{\otimes}_{\mathbb{Z}_p}M \cong \mathbb{Z}_p$ as $\mathbb{Z}_p [\![G]\!]$-modules. I do not see why 1 should imply 2: the duality implies $H^n(G;M)\cong H_{0}(G; D_p(G)\widehat{\otimes}_{\mathbb{Z}_p}M)\neq 0$ and we know $H_0(G;N)\cong N_G=N/_{ \overline{\langle g.n-n \mathrel: g \in G, \ n \in N \rangle}}$. It seems to state that $(D_p(G)\widehat{\otimes}_{\mathbb{Z}_p}M)_G \neq 0$ implies $(D_p(G)\widehat{\otimes}_{\mathbb{Z}_p}M)_G=\mathbb{Z}_p$ and by duality the top cohomolgy $H^n(G;M)$ can only be $0$ or $\mathbb{Z}_p$. But this should be false: a priori we could have that $\overline{\langle g.x-x \mathrel: g \in G, \ x \in D_p(G)\widehat{\otimes}_{\mathbb{Z}_p}M}\rangle$ is a proper subgroup of $D_p(G)\widehat{\otimes}_{\mathbb{Z}_p}M$ (which is just $\mathbb{Z}_p$ as abelian group). Also it is easy to provide a counterexample to the statement that $H^*(G;M)$ is either $0$ or $\mathbb{Z}_p$. Take $G=1+p\mathbb{Z}_p$ and $M=\mathbb{Z}_p$ with the action by multiplication, we can compute $H^1(G;M)=\mathbb{Z}/p$ and $H^k(G;M)=0$ for $k \neq 1$. REPLY [4 votes]: I privately asked Symonds about the matter and he seems to agree with me that the corollary in this form is wrong. My counterexample should be valid. The point is that this corollary 4.5.2 is used in corollary 5.2.5 of the cited paper to prove that the right action on the dualizing module of a Poincarè group is given by the determinant of the adjoint action on $L(G)$, the logarithm algebra associated to $G$. In the proof we actually show that that for a precise module $M$ we have $H^n(G;M)\otimes \mathbb{Q}_p \neq 0$ and we use the implication $1)\Rightarrow 2)$ to conclude. Since we show the rationalization is not trivial, modifying $1)$ to ask $H^n(G;M)$ to be torsion-free or $\mathbb{Z}_p$ does not break the proof of corollary 5.2.5. I am not aware of any instance in the literature where this corollary 4.5.2 is used in this wrong form, if you know such a case maybe reporting it here could be useful to the discussion.<|endoftext|> TITLE: Whitehead product and a homotopy group of a wedge sum QUESTION [8 upvotes]: Note : this is a crosspost from the Mathematics StackExchange, as suggested by this meta post. Let $X$ be an $n$-connected ($n\geqslant1$) CW-complex and $Y$ be a $k$-connected ($k\geqslant1$) CW-complex. My goal is to prove the following isomorphism : $$\pi_{n+k+1}(X\vee Y)\cong\pi_{n+k+1}(X)\oplus\pi_{n+k+1}(Y)\oplus[\pi_{n+1}(X),\pi_{k+1}(Y)],$$ with $[\;\cdot\;,\;\cdot\;]$ denoting the Whitehead product (here, it is understood that we take the whitehead product of the subgroups $\pi_{n+1}(X)<\pi_{n+1}(X\times Y)$ and $\pi_{k+1}(Y)<\pi_{k+1}(X\times Y)$). So far, I have done the following. (Do let me know if I have done any mistake !) We can always assume, up to a homotopy equivalence, by the hypothesis on $X$ and $Y$, that their respective $n$ and $k$ skeletons are of the following form : $$\text{Sk}_nX=\{\ast\}\qquad\text{and}\qquad\text{Sk}_kY=\{\ast\}.$$ In particular, $X$ and $Y$ only have cells in dimensions $\geqslant n+1$ and $\geqslant k+1$ respectively. Therefore, the product $X\times Y$ has only cells starting in dimension $n+1$ or $k+1$, accordingly to which one is the smallest, and that cells in dimensions $\leqslant n+k+1$ come from cells of either $X$ or $Y$, but not both. Therefore, we get : $$\text{Sk}_{n+k+1}(X\times Y)\subset X\vee Y,$$ and thus the pair $(X\times Y,X\vee Y)$ is $(n+k+1)$-connected. I then tried using a part of the exact sequence of the pair : $$\dots\longrightarrow\pi_{n+k+2}(X\times Y,X\vee Y)\overset{\partial_\ast}{\longrightarrow}\pi_{n+k+1}(X\vee Y)\overset{\imath_\ast}{\longrightarrow}\pi_{n+k+1}(X\times Y)\overset{\text{rel}_\ast}{\longrightarrow}\pi_{n+k+1}(X\times Y,X\vee Y)\longrightarrow\dots$$ We can use the $(n+k+1)$-connectedness of the pair to re-write the sequence as : $$\dots\longrightarrow\pi_{n+k+2}(X\times Y,X\vee Y)\overset{\partial_\ast}{\longrightarrow}\pi_{n+k+1}(X\vee Y)\overset{k}{\longrightarrow}\pi_{n+k+1}(X)\oplus\pi_{n+k+1}(Y)\overset{\text{rel}_\ast}{\longrightarrow}0,$$ with $k$ being given by the composite of $\imath_\ast$ and of the isomorphism $\pi_\bullet(X\times Y)\cong\pi_\bullet(X)\oplus\pi_\bullet(Y)$. Now, the sequence splits at $\pi_{n+k+1}(X\vee Y)$, since we have $p\circ\imath=\text{id}$ and $q\circ\imath=\text{id}$ in : $$X\vee Y\overset{\imath}{\longrightarrow}X\times Y\overset{p}{\longrightarrow}X\subset X\vee Y\qquad\text{and}\qquad X\vee Y\overset{\imath}{\longrightarrow}X\times Y\overset{q}{\longrightarrow}Y\subset X\vee Y,$$ by functoriality and by using that $\pi_\bullet$ sends products to products. We shall denote as $p_\ast\oplus q_\ast:\pi_{n+k+1}(X)\oplus\pi_{n+k+1}(Y)\to\pi_{n+k+1}(X\vee Y)$ the splitting retraction. Therefore, by an algebraic lemma (not exactly the Splitting lemma, but something rather similar), we obtain : $$\pi_{n+k+1}(X\vee Y)\cong\text{Im}(p_\ast\oplus q_\ast)\oplus\ker(k).$$ Now, I recognized that $\text{Im}(p_\ast\oplus q_\ast)\cong\pi_{n+k+1}(X)\oplus\pi_{n+k+1}(Y)$ by construction, so I am left with computing $\ker(k)$. And here, I am completely stuck... How to recognize the Whitehead product as the kernel I am missing ? REPLY [9 votes]: Here are some details which are related to Tyler's comment. I recommend looking at the paper "Induced Fibrations and Cofibrations" by Tudor Ganea (1967). For connected based spaces $X$ and $Y$, there is a fibration up to homotopy $$ \Sigma (\Omega X) \wedge (\Omega Y) \to X\vee Y \to X\times Y $$ where the first map in the display is a kind of generalized Whitehead product (see below). After looping once, the sequence splits, so $$ \Omega (X\vee Y) \simeq \Omega X \times \Omega Y \times \Omega \Sigma ((\Omega X) \wedge (\Omega Y))\, . $$ Your isomorphism will follow by applying $\pi_{n+k}$ to this splitting--we only need to identify the term on the right. To this end, note that if $X$ is $n$-connected and $Y$ is $k$-connected ($n,k\ge 1$), then $\Omega \Sigma ((\Omega X)\wedge (\Omega Y))$ is $(n+k-1)$-connected (here I am using the Hurewicz theorem). Moreover, the map $$ (\Omega X)\wedge (\Omega Y)\to \Omega \Sigma (\Omega X)\wedge (\Omega Y) $$ is $(2n+2k-1)$-connected. In particular, it will induce an isomorphism on $\pi_{n+k}$. As $(\Omega X)\wedge (\Omega Y)$ is $(n+k-1)$-connected, the Hurewicz theorem says that $$ \pi_{n+k} ((\Omega X)\wedge (\Omega Y)) \cong H_{n-k} ((\Omega X)\wedge (\Omega Y)) $$ and the Künneth formula provides an isomorphism $$ H_{n+k} ((\Omega X)\wedge (\Omega Y)) \cong H_n((\Omega X) \otimes H_k(\Omega Y)\, . $$ Another application of the Hurewicz theorem shows that $$ H_n(\Omega X) \otimes H_k(\Omega Y) \cong \pi_{n+1}(X) \otimes \pi_{k+1}(Y)\, . $$ Putting this all together, we obtain an isomorphism $$ \pi_{n+k+1} (X\vee Y) \cong \pi_{n+k+1} (X) \oplus\pi_{n+k+1} (Y) \oplus \, \, \pi_{n+1}(X) \otimes \pi_{k+1}(Y)\, . $$\ It remains describe the generalized Whitehead product $\Sigma (\Omega X) \wedge (\Omega Y) \to X\vee Y $. Taking the adjoint, we seek a map $$ (\Omega X) \wedge (\Omega Y) \to \Omega(X\vee Y)\, . $$ Now, there are evident inclusions $ \Omega X \to \Omega(X\vee Y)$ and $\Omega Y \to \Omega(X\vee Y)$. Very roughly, the idea is to map a pair of loops $(\gamma,\omega) \in (\Omega X) \wedge (\Omega Y) $ to the commutator $$ [\gamma,\omega] \in \Omega(X\vee Y) $$ where care is required to make sense of the commutator. I will refrain from writing down the formula here. I believe that the details may be found in Ganea's paper.<|endoftext|> TITLE: Intuition/elegant reason for why Langevin diffusion converges to $\exp(-U)$? QUESTION [6 upvotes]: Given a potential function $U: \mathbb{R}^n \to \mathbb{R}$, Langevin diffusion is gradient descent plus a Brownian motion term: $X' = -\nabla U(X) + \sqrt{2} \text{ }dW$. It happens that the stationary distribution of Langevin diffusion is very nice: it is proportional to $\exp(-U)$. I can verify this by plugging $\exp(-U)$ into a PDE and checking that some second-order terms miraculously cancel. But seeing as $\exp(-U)$ is such a simple solution, I expect an elegant reason for it... or at the very least, some intuition for why the stationary distribution at $x$ should depend only on $U(x)$. Is there any intuition to be found here? REPLY [6 votes]: The reason a Langevin diffusion leaves $\nu(x)=e^{-U(x)}$ invariant is because it is symmetric or reversible with respect to $\nu$. In comparison to general diffusion processes, the ergodic properties of symmetric diffusions are not only easier to analyze, but it is often possible to obtain an explicit formula for the stationary density of a symmetric diffusion. This symmetry property is a bit more transparent when the Langevin equation is written as $$ dX_t = \nabla \log \nu(X_t) dt + \sqrt{2} dW_t \;. \tag{$\star$} $$ The symmetry property can then be easily verified by checking that the infinitesimal generator of the Langevin diffusion is symmetric in an $L^2$-inner product weighted by $\nu$. This property of the Langevin diffusion is analogous to the detailed balance condition of a Markov chain. It is also the basis of standard MCMC methods based on Langevin diffusions. In fact, many random-walk based MCMC methods (like random walk Metropolis) with target probability density proportional to $\nu(x)$, can weakly approximate the corresponding Langevin diffusion ($\star$). This is remarkable because, e.g., the random walk Metropolis algorithm does not involve the gradient of $U$. To read more about this connection, see the introduction and Theorem 5.2 of this paper. REPLY [4 votes]: Here is one way to think about this. Let $\rho(X,t)$ denote the transient joint density. Consider the free energy functional $$F(\rho) := \mathbb{E}\left[U + \log \rho\right] = D_{\rm{KL}}\left(\rho||\exp(-U)\right) \geq 0,$$ where $D_{\rm{KL}}$ denotes the Kullback-Leibler divergence. This shows that $\exp(-U)$ is the minimizer of $F$. Let $\zeta := 1+\log\rho + U$. The evolution of $\rho(X,t)$ is governed by the Fokker-Planck or Kolmogorov's forward PDE $$\frac{\partial\rho}{\partial t} = \nabla\cdot\left(\rho\nabla\zeta\right).$$ Notice that $$\frac{{\rm{d}}F}{{\rm{d}}t} = -\mathbb{E}\left[\|\nabla\zeta\|^{2}\right],$$ where the expectation operation $\mathbb{E}$ is with respect to the joint density $\rho(X,t)$. Therefore, $F$ is a Lyapunov functional for the above PDE, i.e., $F$ decreases along the solution trajectory $\rho(X,t)$, and is equal to zero at the stationary density $\propto \exp(-U)$. Another way to say the same is that the above PDE is gradient flow of $F$ with respect to the 2-Wasserstein metric; see e.g., Ch. 8.3 in Villani's book "Topics in Optimal Transportation".<|endoftext|> TITLE: Schwartz regularity for the density of a stochastic process QUESTION [5 upvotes]: Let $B$ be a standard Brownian motion in $\mathbb R$. Define the variables $$\begin{align*} X &= B_1, & Y &= \int_0^1B_s\mathrm ds, & Z&= \int_0^1B_s^2\mathrm ds. \end{align*}$$ It is known, see below, that $(X,Y,Z)$ admits a smooth density. Is is true that the density of $(X,Y,Z)$ is a Schwartz function? I think there should be a simple enough argument, and I'm making it more complicated than it actually is. Let me share two observations about this triple. First of all, all of its components belong to low order Wiener chaos: specifically, $X$ and $Y$ belong to the first Wiener chaos, and $Z$ to the second. This makes it easy enough to compute the Malliavin matrix ($\langle DX,DX\rangle$ and its peers): $$ \Gamma = \begin{pmatrix} 1 & 1/2 & 2\alpha \\ 1/2 & 1/3 & 2\alpha-\beta \\ 2\alpha & 2\alpha-\beta & 4\gamma \end{pmatrix} $$ with $$ \begin{align*} \alpha&=\int_0^1tB_t\mathrm dt, & \beta&=\int_0^1t^2B_t\mathrm dt, & \gamma&=\int_0^1\Big(\int_t^1B_s\mathrm ds\Big)^2\mathrm dt. \end{align*} $$ Using this expression, we get $\det\Gamma=(\gamma-\alpha^2)/3 - (\alpha-\beta)^2$, and it would suffice to prove that $\mathbb P[\det\Gamma\leq\varepsilon]$ decreases faster than any power of $\varepsilon$, or that $1/\det\Gamma$ has moments of all order (the former implies the latter). This would follow from Proposition 2.1.5 in Nualart's The Malliavin Calculus and Related Topics (p. 103), as suggested by Fabrice Baudoin in the comments. As for the second observation, the variable in question is the evaluation at time 1 of a hypoelliptic process started at $(0,0,0)$: if one replaces every 1 by a $t$ in the above definitions, we get a process $(X_t,Y_t,Z_t)$ solution to the hypoelliptic equation $$\begin{align*} \mathrm dX_t &= \mathrm dB_t, & \mathrm dY_t &= X_t\mathrm dt,& \mathrm dZ_t&= X_t^2\mathrm dt \end{align*}$$ such that the value at time 1 is the variable described above. This shows that the density of $(X_1,Y_1,Z_1)$ is smooth. One can find more about this approach (with slightly different conventions) in section IV-1.5 (page 118) of my PhD thesis, but here are a few key insights. We can find a (not so nice) candidate for the characteristic function of $(X_t,Y_t,Z_t)$ (hence for $(X,Y,Z)$ as well) in the form $$\text{constant}\times\exp\big(\text{quadratic form in }(\xi_x,\xi_y)\big)$$ for $t$ and $\xi_z$ fixed. Specifically, $$ \mathbb E\big[\exp\big(-i(\xi_xX_t+\xi_yY_t+\xi_zZ_t)\big)\big] = \frac1{\sqrt{\mathrm{cah}}}\cdot \exp\left( - \frac t2\mathrm{tah}\cdot\xi_x^2 + \frac i{2\xi_z}\Big(1-\frac1{\mathrm{cah}}\Big)\cdot\xi_x\xi_y + \frac{it}{4\xi_z}(1-\mathrm{tah})\cdot\xi_y^2\right), $$ for $\mathrm{cah} = \cosh(\zeta)$, $\mathrm{sah} = \sinh(\zeta)/\zeta$, $\zeta = \sqrt{2i\xi_zt^2}$ (these are quantities of the form $f(\sqrt{(\cdots)})$ for $f$ holomorphic and even, so they are actually holomorphic), $\mathrm{tah} = \mathrm{cah}/\mathrm{sah}$. One can show that the zeros of $\alpha\mapsto\cosh(\sqrt\alpha)$ are real negative, so everything can be defined holomorphically in the above formula. This is a nice candidate because it is one solution of the Feynman-Kac formula seen in Fourier space, in a rather strong sense. If it were the actual characteristic function, then the question would be equivalent to show that the candidate is itself a Schwartz function (in the dual space variables). However, this possibly equivalent result does not seem obvious to me either, since one would have to show that the real part of the quadratic form is bounded below, or something of the sort. REPLY [2 votes]: Using the representation in terms of i. i. d. Gaussians $\xi_1,\xi_2,\dots,$ $$ B_t=\sqrt{2}\sum_{n=1}^\infty (-1)^{n+1}\xi_n\frac{\sin \pi \left(n-\frac12\right)t}{\pi \left(n-\frac12\right)}, $$ we get $$ X=\sqrt{2}\sum_{n=1}^\infty\frac{\xi_n}{\pi \left(n-\frac12\right)},\quad Y=\sqrt{2}\sum_{n=1}^\infty \frac{(-1)^{n+1}\xi_{n}}{\pi^2 \left(n-\frac12\right)^2},\quad Z=\sum_{n=1}^\infty\frac{\xi^2_n}{\pi^2 \left(n-\frac12\right)^2}. $$ The Fourier transform of a square of a Gaussian can be computed explicitly: $$ \frac{1}{\sqrt{2\pi}}\int_\mathbb{R} e^{-\frac{x^2}{2}+itx^2}\,dx= \frac{1}{\sqrt{2\pi}} \frac{1}{\sqrt{1-2it}}\int_{z\in\sqrt{1-it}\mathbb{R}}e^{-\frac{z^2}{2}}dz=\frac{1}{\sqrt{1-2it}}, $$ by considering the contour integration over the "pizza slice" domain. Hence, we have $$ \varphi_Z(t)=\prod^\infty_{n=1}\left(1-\frac{2it}{\pi^2 \left(n-\frac12\right)^2}\right)^{-\frac12}=\left(\cos(\sqrt{2it})\right)^{-\frac12}, $$ which is clearly of Schwarz class on $\mathbb{R}$, and hence so is the density of $Z$. The joint characteristic function of $(X,Y,Z)$ can be computed similarly. Define independent random variables $$ (X_n,Y_n,Z_n)=\left(\frac{\sqrt{2}\xi_n}{\pi \left(n-\frac12\right)},\frac{\sqrt{2}\xi_{n}}{\pi^2 \left(n-\frac12\right)^2},\frac{2\xi^2_n}{\pi^2 \left(n-\frac12\right)^2}\right) $$ so that they are independent and $$ (X,Y,Z)=\sum_{n=1}^\infty(X_n,Y_n,Z_n). $$ We can compute the characteristic function of $(X_n,Y_n,Z_n)$ using the identity $$ \frac{1}{\sqrt{2\pi}}\int_\mathbb{R}e^{-\frac{x^2}{2}+ax+ibx^2}\,dx=\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}e^{-\frac{\left(\sqrt{1-2ib}x-\frac{a}{\sqrt{1-2ib}}\right)^2}{2}+\frac{a^2}{2(1-ib)}}\,dx=e^{\frac{a^2}{2(1-2ib)}}\frac{1}{\sqrt{1-2ib}}. $$ This yields, denoting $\varphi_n=\varphi_{(X_n,Y_n,Z_n)},$ $$ \varphi_n(t_1,t_2,t_3)= \exp\left(-\frac{\left(t_1+\frac{(-1)^{n+1}t_2}{\pi\left(n-\frac12\right)}\right)^2}{\pi^2 \left(n-\frac12\right)^2\left(1-\frac{2it_3}{\pi^2 \left(n-\frac12\right)^2}\right)}\right)\left(1-\frac{2it_3}{\pi^2 \left(n-\frac12\right)^2}\right)^{-\frac12} $$ After multiplying this, the idea is to show that the product of the exponentials gives a function that decays exponentially with all derivatives outside any cone containing the $t_3$ axis (and each of its derivatives is bounded by a polynomial in the cone), while the rest gives $\varphi_Z(t_3)$ which decays fast enough in the cone, and thus conclude that we get a Schwarz class function. To elaborate, denote $$ \Phi(t_1,t_2,t_3)=\sum_n\frac{\left(t_1+\frac{(-1)^{n+1}t_2}{\pi\left(n-\frac12\right)}\right)^2}{\pi^2 \left(n-\frac12\right)^2\left(1-\frac{4it_3}{\pi^2 \left(n-\frac12\right)^2}\right)}; $$ it suffices to show that, on the one hand, $\Phi$ and all its derivatives grow at most polynomially as $(t_1,t_2,t_3)\to\infty$ (which is straightforward, just differentiate and estimate the absolute value of the (power of) the last term in the denominator as $\geq 1$), and that outside the above-mentioned cone, $$ \Re\Phi(t_1,t_2,t_3)>c(|t_1|+|t_2|)$$ for some constant $c$, which seems to follow by looking at the first two terms in the sum and noticing that the real part of the others is non-negative. Indeed, with these observation at hand, we have $$ \varphi_{(X,Y,Z)}=\exp(-\Phi(t_1,t_2,t_3))\cos(\sqrt{2it_3})^{-\frac12} $$ The derivatives of this guy will have the form $$ Poly(t_1,t_2,t_3)e^{-\Phi(t_1,t_2,t_3)}, $$ where $Poly$ is some polynomial in derivatives of $\Phi(t_1,t_2,t_3)$ and in $\cos(\sqrt{2it_3})^{-\frac12}$ and its derivatives, in which every monomial is linear in $\cos(\sqrt{2it_3})^{-\frac12}$ or one of its derivatives. Because of this last property, the whole expression will decay faster than polynomial in the cone, and the exponential will kill it off outside the cone.<|endoftext|> TITLE: An invariance property of rational singularities QUESTION [8 upvotes]: Let $X$ be a normal variety over a field of characteristic zero with rational singularities. If $\pi:Y \to X$ is a birational proper morphism with $Y$ also normal, then does $Y$ also have rational singularities? It is easy to see that this is true if $\dim(X) = 2$, but the higher dimensional case seems more difficult and perhaps it is even false. If true, I would also be interested in analogous results in positive or mixed characteristic e.g., for pseudo-rational singularities. REPLY [6 votes]: No, $Y$ need not have rational singularities. See Section III of the paper for an example in dimension three: Cutkosky, A characterization of rational surface singularities, Inventiones Mathematicae, 1990. In that example, $Y$ ($Z$ in the notation of the paper) is normal but not Cohen-Macaulay, so it is not a rational singularity. For an explicit example (in dimension three) with the same property, see Theorem 3.11 in Huckaba and Huneke, Normal ideals in regular rings, Journal fur die Reine und Angewande Mathematik, 1999.<|endoftext|> TITLE: Non-isomorphic algebras becoming isomorphic after adding identities: mistake in an exercise QUESTION [13 upvotes]: In the book: Drozd, Y.A., Kirichenko, V.V.: Finite dimensional algebras. Springer, Berlin (1994), there is an exercise which suggests a positive answer to the next question, and Ryszard R. Andruszkiewicz gives a counterexample: Let $A,B$ be two arbitrary finite-dimensional algebras (spaces with associative laws) over a field $K$, and let $\overset{\sim}{A}:=\{(a,\alpha)\mid a\in A, \alpha\in K\}$, it is a well-known method of adjoining an identity to an arbitrary algebra $A$. $$(a_1, α_1) · (a_2, α_2) = (a_1 · a_2 + α_2a_1 + α_1a_2, α_1α_2)$$ $$(a_1, α_1) + (a_2, α_2) = (a_1 + a_2, α_1 + α_2)$$ $$\forall\beta\in K:\beta· (a, α) = (βa, βα)$$ $\overset{\sim}{A}$ is an algebra with identity $(0, 1)$, in the same way we will define $\overset{\sim}{B}$. The exercise: $A\cong B$ if and only if $\overset{\sim}{A}\cong\overset{\sim}{B}$. Ryszard R. Andruszkiewicz's counterexample: Surprise at adjoining an identity to an algebra. Andruszkiewicz's counterexample is, and if I understand correctly, $A$ is not isomorphic to $B$ because an isomorphism of algebras preserves both the left and right identities, but $A$ has neither a right identity nor an identity while $B$ has a right identity. Which shows that there might be an error in the exercise. Is it true that there is an error in the exercise? Edit: For the discussion in the comments REPLY [12 votes]: The counterexample is correct. I had put the same counterexample on mathoverflow in a comment last year, unaware of the above paper, on a question asking about rings not isomorphic to their opposite. See my comment on Pete Clark's answer to Simplest examples of rings that are not isomorphic to their opposites. Conceptually it is this. If you take a 2 element left zero semigroup $S=\{x,y\}$ with multiplication $ab=a$ for all $a,b\in S$ then obviously $x,y$ are distinct right identities and if $T$ is the opposite semigroup consisting of two right zeroes then $T$ has two distinct left identies. Thus the semigroup algebras $KS$ and $KT$ are not isomorphic over any field. But when you adjoin identities to $KS$ and $KT$ (or equivalently adjoin an identity to the semigroups and take monoid algebras) you get algebras both isomorphic to $2\times 2$ upper triangular matrices. The trick is in $KS^1$ where $S^1$ is the monoid obtained by adjoining an identity to $S$ the elements $1,1-x,1-y$ is a basis but $1-x,1-y$ form a right zero semigroup isomorphic to $T$. So $KS^1\cong KT^1$. Both the algebras live in $2x2$ upper triangular matrices, which is the result of adding and identity. $KS$ consists of those upper triangular matrices whose second row is zero and $KT$ consists of those whose first column is zero. I don't have the book. By any chance were they assuming $A$ and $B$ have identities in the exercise?<|endoftext|> TITLE: Membership to double cosets in free groups QUESTION [7 upvotes]: Is there an elementary and efficient algorithm for testing the membership to a double coset of f.g. subgroups in a free group? Has this membership problem been implemented in GAP/Magma? More precisely, I have a finitely generated free group $F$, two finitely generated subgroups $A,B$, given by their free generating sets, and an element $f\in F$. I would like to determine whether $f \in AB$. I was hoping for something in the spirit of Stallings core graphs, but the standard reference (the paper of Kapovich and Myasnikov [Kapovich, Ilya; Myasnikov, Alexei, Stallings foldings and subgroups of free groups, J. Algebra 248, No. 2, 608-668 (2002). ZBL1001.20015.]) does not seem to treat membership to double cosets. Theoretically the required algorithm certainly exists. In fact, a much stronger property was proved by Grunschlag ( see Theorem 4.10 in [Grunschlag, Zeph, Computing angles in hyperbolic groups, Gilman, Robert H. (ed.), Groups, languages and geometry. 1998 AMS-IMS-SIAM joint summer research conference on geometric group theory and computer science, South Hadley, MA, USA, July 5-9, 1998. Providence, RI: American Mathematical Society. Contemp. Math. 250, 59-88 (1999). ZBL0953.20034.] REPLY [6 votes]: Here is a second answer that is just rephrasing @DerekHolt’s answer based on the comments. So upvote his answer first! Let $X$ be a finite alphabet. An inverse automaton is a finite directed graph labeled over the alphabet $X$ which is folded in the sense of Stallings, meaning you cannot find two edges entering or leaving a vertex with the same label. Equivalently these are immersions over a bouquet of circles. An inverse automaton has a unique initial state and a set of final states. It recognizes a subset of the free group by looking at all reduced words labelling a path from the initial state to a final state, where inverse letters read edges backward. The language accepted by an inverse automaton is a finite union of cosets of finitely generated subgroups. The languages accepted by inverse automata are closed under intersection. You simply build the product automaton whose vertices are pairs of vertices and there is an edge labeled by $x$ from $(p,q)$ to $(p’,q’)$ if and only if there are edges from $p$ to $p’$ and $q$ to $q’$ labeled by $x$. This construction is the fiber product of the two inverse automata viewed as immersions over a bouquet of circles. The initial state is the pair of initial states and the final states are those pairs $(p,q)$ where both $p$ and $q$ are final in their respective automata. This is an inverse automata that will accept precisely those reduced words which both automata accept. In particular, the intersection is empty iff the initial vertex of the product construction cannot reach any terminal vertex. Now to recognize $Hw$ write a linear automaton reading the word $w$ (with inverse letters read backward) and attach at the initial point a generating set for $H$ as loops and apply Stallings foldings. The initial vertex is the intial vertex of $w$ and the final vertex is the end point of $w$. These may collapse after folding. This is essentially the Stallings core of $H$ with a thorn labeled $w$ sticking out of the base point and Stallings uses this automaton in his proof of Marshall Hall's Theorem. We can do a similar thing for $gK$, but we put the generators of $K$ as loops at the end of the path labelled $g$, and then to test if $w\in HgK$ we just form the product automaton for $Hw$ and $gK$ and check if the final state is reachable from the initial state. Note $w=hgk$ iff $h^{-1}w=gk$. So $Hw$ and $gK$ intersect iff $w\in HgK$.<|endoftext|> TITLE: Yoneda Lemma for monoidal functors QUESTION [11 upvotes]: Let $(\mathcal V,\otimes,I)$ be a closed symmetric monoidal category, and let $\mathcal C$ be a $\mathcal V$-enriched category. The (weak) enriched Yoneda Lemma gives us a nice description of the set $Hom(F,G)$ of natural transformations between two $\mathcal V$-enriched functors $F,G\colon\mathcal C\to\mathcal V$ when $F$ is representable: it is in bijection with the set of maps $I \to G(Y)$ in $\mathcal V$ where $Y$ is an object representing $F$. Now suppose that $\mathcal C$ itself is a monoidal category, and that our two functors $F$ and $G$ are monoidal functors. Is there a similarly nice description of the set $Hom^\otimes(F,G)$ of monoidal natural transformations between the functors, again in the case that $F$ is representable? My suspicion is that the following might be true (possibly with extra conditions on $\mathcal C$). The fact that $F$ is a (lax) monoidal functor induces the structure of a comonoid on the representing object $Y$, and so there is an induced comonoid structure on $G(Y)$. My guess would be that monoidal natural transformations $F\to G$ are in bijection with morphisms of comonoids $I\to G(Y)$, but I can't prove this in general. (I can prove this in the case that $\mathcal V$ is the category of sets with cartesian product, but only for trivial reasons: every map in Set is a morphism of comonoids, and every natural transformation between monoidal Set-valued functors is a monoidal transformation.) I would be especially interested in any references where this might be addressed. REPLY [2 votes]: For your suspicion to work you need $G$ to be pseudomonoidal I would think, otherwise I don't see how to obtain a comonoid structure on $G(Y)$ from that on $Y$? With both $F$ and $G$ lax monoidal considering Day-convolution $\hat\oslash$ on $\hat A = \mathcal V^{A^\text{op}}$ gives you a "monoidal yoneda lemma", not as nice as you suspected, as follows. The universal property of Day-convolution is that it creates the yoneda embedding $\text y\colon A \to \hat A$ in the double category of monoidal profunctors $A \nrightarrow B$, that is lax monoidal functors $J\colon A^\text{op} \otimes B \to \mathcal V$, and lax monoidal functors (and you really want a double category here I believe, a 2-category won't work or at least not as nicely). Other properties such as it defining $\hat A$ as the monoidal free cocompletion follow from that. The property that we will use is that it induces an equivalence of between the category of monoidal profunctors $A \nrightarrow B$ and that of lax monoidal functors $B \to \hat A$, mapping $J\colon A \nrightarrow B$ to any chosen $J^\lambda\colon B \to \hat A$ such that $J \cong \hat A(\text y, J^\lambda)$ as monoidal profunctors (e.g. take $J^\lambda(y) = J(-, y)$. Writing $I$ for the unit $\mathcal V$-category with its monoidal structure, lax monoidal functors $A \to \mathcal V$ are precisely monoidal profunctors $A \nrightarrow I$. Your functor $F\colon A^\text{op} \to \mathcal V$ being representable means that there is a lax monoidal functor $y\colon I \to A$, i.e. a monoid $y \in A$, such that $F$ corresponds to the monoidal profunctor $A(-, y)$. Under the equivalence the latter corresponds to $A(-, y)^\lambda = \text yy\colon I \to \hat A$. Monoidal transformations $F \Rightarrow G$ are then transformations $A(-, y) \Rightarrow G$ of monoidal profunctors $A \nrightarrow I$ which, by the equivalence, correspond precisely to monoidal transformations $\text yy \Rightarrow G^\lambda$ of lax monoidal functors $I \to \hat A$. Since $G \cong \hat A(\text y, G^\lambda)$ the latter can be thought of as given by a $\mathcal V$-morphism $\phi\colon I \to G(y)$, but its compatibility with the monoidal structures has to be written as a commuting diagram of two parallel morphisms $I \otimes I \to \hat A(\text yy \hat\oslash \text yy, G^\lambda)$. Only when $F$, and hence $y$, is pseudomonoidal the previous isomorphism restricts to $G(y) \cong \hat A(\text yy, G^\lambda)$ of monoidal profunctors $I \nrightarrow I$, and the condition on $\phi$ reduces to it being a morphism of monoids. I did not think much about $G$ being pseudomonoidal. You might want to look at the double category of monoidal profunctors and colax monoidal functors instead? As for references, the equivalence between monoidal profunctors and lax monoidal functors can be extracted from Section 2 of Pisani's Sequential Multicategories (although at the moment I don't remember exactly how...) The bigger picture of Day-convolution creating monoidal yoneda embeddings is one of the main motivations of my draft paper A double-dimensional approach to formal category theory Edit: Thinking a bit more about this, using the fact that $\text y$ is pseudomonoidal allows us to streamline the condition on $\phi$ above further. All put together we get: Monoidal Yoneda lemma. Let $G\colon A^{\text op} \to \mathcal V$ be lax monoidal and $y \in A$ a monoid, thus making $A(-,y)\colon A^{\text op} \to \mathcal V$ lax monoidal. Monoidal transformations $A(-, y) \Rightarrow G$ correspond precisely to morphisms $I \to Gy$ in $\mathcal V$ such that the composites $$ I \otimes I \to Gy \otimes Gy \xrightarrow{G_\otimes} G(y \otimes y) \qquad \text{and} \qquad I \otimes I \cong I \to Gy \xrightarrow{Gy_\otimes} G(y \otimes y)$$ coincide, as well as (as Alexander points out below) $$ I \to Gy \xrightarrow{G(y_I)} G(I_A) \qquad \text{and} \qquad I \xrightarrow{G_I} G(I_A) $$ where $G_I$ and $y_I$ are the unitors of $G$ and $y$. If $G$ is pseudomonoidal then this recovers the lemma that you suspected and Maxime proved.<|endoftext|> TITLE: Uses for (Framed) E2 algebras twisted by braided monoidal structure QUESTION [5 upvotes]: $\newcommand{\C}{\mathcal{C}}$ $\newcommand{\g}{\mathfrak{g}}$ If $\C$ is a monoidal category (not necessarily a symmetric monoidal category), it's possible to define the notion of an algebra object $A$ in $\C$, with multiplication operations $$A^{\otimes n} (:= A\otimes_\C A\otimes_\C \cdots\otimes_\C A)\to A.$$ Similarly, if $\C$ is a braided monoidal category (resp., a ribbon category), one can define a notion of $E_2$ DG algebra $A$ (resp., framed $E_2$ DG algebra $A$) "twisted" by $\C$, consisting of operations $A^{\otimes n}\to A$ compatible with braiding. (Note: I actually don't know a reference for this, but it follows from standard "homotopy field theory" arguments involving the Ran space.) In particular, if $\C$ is a braided monoidal (or ribbon) category coming from an associator on a Lie algebra $\g$ (with choice of Casimir), there is a whole category of "associator-twisted" $\g$-equivariant $E_2$ (resp., framed $E_2$) algebras. My question is whether algebras of this type have been encountered before. They feel very CFT-ish, and so I'm particularly curious about physics and knot theory applications. In particular, the framed variant should gove some kind of derived 2D TQFT-style invariants. Any references would be useful. Thanks! REPLY [4 votes]: I don't know specific references (the papers - in reverse chronological order - of Liang Kong, Hao Zheng, Ingo Runkel, Christoph Schweigert and Jurgen Fuchs is where I'd start), but the notion is certainly very natural in TFT, in at least two (closely related) ways: if you think of your braided tensor category $C$ as defining a 4d TFT (i.e. as an object in the Morita 4-category thereof), then $E_2$ algebras $A$ in $C$ are a special case of boundary conditions in the TFT (left modules over $A$ form a monoidal category over $C$, or Morita morphism to/from the unit). As such they are actually quite rare in say reps of quantum groups generically, and have to do with coisotropic subalgebras of $\mathfrak g$. if you think of $C$ as the value on the circle of a 3d TFT (eg as defined by a rational vertex algebra), you get $E_2$ algebras $A$ in $C$ as the value on the circle of boundary conditions for the TFT. Or you can look at interfaces of 3d TFTs, giving $E_2$-functors of $E_2$-categories of which I think your setup is a special case by passing to $E_2$-A-modules (maybe I'm getting turned around though). Or less generally - a concrete example of your setup is $C=B-E_2-mod$ and $B\to A$ an $E_2$-morphism - as such this appears all over CFT, when you have a homomorphism of algebras of observables (eg a map of rational vertex algebras). From the TFT POV the difference between $E_2$ and framed $E_2$ is the difference between framed TFT and oriented TFT - the latter naturally produces framed $E_2$-algebras ( as usual the terminology is awful, for TFT definitely seems better to talk about "oriented 2-disc algebras" (framed $E_2$) and "framed 2-disc algebras" ($E_2$).<|endoftext|> TITLE: Acute triangles in "obtuse" polygons? QUESTION [15 upvotes]: Let $P$ be a convex polygon. Suppose every interior angle of $P$ is obtuse. Is it always the case that there exist three vertices $p, q, r$ of $P$ such that $\triangle pqr$ is acute? I conjecture that the answer is yes. I have tried different types of triangulations of $P$ (e.g. fan triangulation, triangulation by repeatedly connecting every other vertex, etc.) However, I haven't come up with a proof yet. Edit: And how about if $P$ has at most one acute angle (others stay obtuse)? I think the result should be the same. REPLY [24 votes]: Take a very obtuse isosceles triangle and chop its acute angles.<|endoftext|> TITLE: Proof of main theorems in étale cohomology theory QUESTION [15 upvotes]: (In this question, $p$ can be $0$.) I'm curious if theorems on étale cohomology can be proved by easier way. For example, proper base change theorem. This theorem can be stated as the following way. Theorem. Let $k$ be a field of charateristic $p$ and $p\ne \ell$. Let's consider that Cartesian diagram $$ \begin{aligned} X'&\overset{g'}{\to} X \\ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!f'\downarrow& \,\,\,\,\,\,\,\,\downarrow f\\ S'&\overset{g}{\to} S \end{aligned} $$ of $k$-schemes. Then the natural map $$ g^*(R^if_*\mathscr{F})\to R^if'_*((g')^*\mathscr{F})$$ is an isomorphism if $f$ is proper and $\mathscr{F}$ is a $\mathbb{Z}/\ell$-constructible sheaf. It is an analogue of proper base change theorem for topological spaces and in topological space case, the proof is relatively easy. See https://stacks.math.columbia.edu/tag/09V4. But for étale cohomology, the proof is so hard! The stages in Freitag-Kiehl are the following. 1. Reduce the case of $S=\mathrm{Spec} A$ for a strictly henselian $A$ and $g=s:X=\mathrm{Spec} k\to \mathrm{Spec} A$ for $k=A/\mathfrak{m}$. 2. Prove that if $\dim X_s\le 1$, the natural map $H^i(X,\mathbb{Z}/\ell)\to H^i(X_s,\mathbb{Z}/\ell)$ is a bijection where $i=0$, a surjection where $i\ge 1$ with standard GAGA argument. 3. Prove the above natural maps are actually isomorphism for all $i$ and for all constructible $\mathrm{Z}/\ell$-sheaves using some dimension argument. 4. To prove general case, use $\mathbb{P}^1\times \cdots \times \mathbb{P}^1\to \mathbb{P}^n$ and reduce the general case to the case of $\mathbb{P}^n\to S$. 5. Then we have an inclusion $X\hookrightarrow \mathbb{P}^n_S\to S$ and we win. This is so complicated! I may understand stage 1,2,3. But I have confused with stage 4,5. To prove the general dimension case, is there the reason for considering $\mathbb{P}^1\times \cdots \times \mathbb{P}^1\to \mathbb{P}^n$? This seems an ad hoc and so different to the proof of proper base change theorem for topological spaces. In case of $p=\ell$, I have found an elegant proof in https://arxiv.org/pdf/1711.04148.pdf (Corollary 10.6.2) which uses only the fact that higher direct images of well-defined finiteness are also finite and do not use somewhat dimension reduction. so I think $p=\ell$ case proper base change is much easier than $p\ne \ell$ proper base change. Another example is Artin vanishing theorem. Theorem. Let $k$ be an algebraically closed field of characteristic $p$ and $X$ be a finite type affine scheme over $k$. Then we have $H^i(X,\mathscr{F})=0$ for all $i>\dim X$ and $\mathbb{Z}/\ell$-constructible sheaf $\mathscr{F}$. The proof is in https://stacks.math.columbia.edu/tag/0F0P and the stratege of stacks project is to use Noether normalization theorem and Leray spectral sequence to $\mathbb{A}^{i+1}\to \mathbb{A}^i$. Nevertheless that strategy has a clear reason for Artin vanishing, I can't make a conception of this proof and it seems a typical reduction technique. In $p=\ell$ case, Artin vanishing theorem can be stated as Theorem. For any finite type affine scheme $X$ over $\mathbb{C}_p$, we have $H^i(X,\mathscr{F})=0$ for $i>\dim X$ and $\mathbb{Z}/p$-constructible $\mathscr{F}$. Proof) (Sketch) Let $X$ be smooth. Then by adjoining $p^n$-th root of unities, we have a perfectoid covering $X_{\infty}\to X$. In $X_{\infty}$, the cohomology dimension of $X_{\infty}$ is $0$ by $$ H^i(X_{\infty},\mathbb{F}_p)\otimes \mathcal{O}^+_{X_{\infty}}/(p)=H^i(X_{\infty},\mathcal{O}^+_{X_{\infty}}/(p))$$ and almost Tate acyclicity on affinoid perfectoid space. Then we can apply Čech-to-derived functor spectral sequence and cohomological dimension property of the group $\mathbb{Z}^{\dim X}_p$ to prove Artin vanishing in case of $X$ smooth and $\mathscr{F}=\mathbb{F}_p$. In general locally constant $\mathscr{F}$, just find a covering that trivializes $\mathscr{F}$ and in general constructible sheaf use dimension induction on $\dim X$ and the definition of constructible sheaf. In general finite type scheme, we can choose a resolution of $X$ and use Leray spectral sequence. The rest is the comparison theorem of étale cohomology with proétale cohomology which is done by Bhatt-Scholze. It is just a special case of original Artin vanishing. But the proof seems clear than the proof of the original version. So I would prefer that proof to the proof or original version. As that examples, I would like to think the proof of "Theorem X" in $p\ne \ell$ case is harder than the proof of "Theorem X" in $p=\ell$ case. Why $p\ne \ell$ is much harder than $p=\ell$? I don't know but maybe it seems to be linked to Scholze's global Hodge theory conjecture, that we have somewhat comparison theorem in $p=\ell$ but there is no well-sensed comparison theorem in $p\ne \ell$! But if there is real statement and proof of global Hodge theory conjecture, then I think maybe we can prove "Theorem X" in $p\ne \ell$ case more easily. The question is Question. Is $p\ne \ell$ case really hard than $p=\ell$ case? Of course, sometimes $p=\ell$ case is more hard than $p\ne \ell$. (Think Poincaré duality in case of $p=\ell$ which is not well-defined yet.) But with sufficient background, $p=\ell$ case seems easy than $p\ne \ell$ case. Is this right? REPLY [9 votes]: You seem to mean two slightly different things by the $p=\ell$ case. Your first theorem is about a field of characteristic $p$, and the second is about $\mathbb C_p$, which is a field of characteristic zero. Let's talk first about the significance of this. For a field of actual characteristic $p$, the proof of "Artin vanishing" for mod $p$ coefficients is even easier than you gave. One doesn't need any perfectoid spaces, as you can just use the Artin-Schreier sequence $0 \to \mathbb F_p \to \mathcal O_X \to \mathcal O_X \to 0$ to handle the case of affine schemes. By a Lefschetz principle argument, your proof over $\mathbb C_p$ can be adapted to work over an arbitrary field of characteristic zero. Of course, by the same argument, and the comparison theorem between etale and analytic cohomology, the Artin affine theorem can also be proved by analytic arguments over fields of characteristic zero (say, using a Morse function). Characteristic zero really is pretty much always easier than characteristic $p$. In actual characteristic $p$, the $\ell=p$ case is not necessarily always easier. I think a good description is that it is more degenerate. The cohomology groups are usually much smaller, in fewer degrees. It's not surprising that vanishing results are easier to prove in this setting! But other results we want, like Poincare duality or Euler characteristic formulas, are simply not true for mod $p$ coefficients. There is certainly something unsatisfying about these somewhat ad hoc arguments, but they're not THAT bad. I think getting over some of your discomfort with them might be easier than finding a better argument. It's perhaps worthwhile to point out that Grothendieck famously wanted to solve problems by setting up theories that make the problems easy rather than the concrete calculations favored by mathematicians like Serre and Deligne, which were later necessary to solve problems like the Weil conjectures. If you find the arguments in the part of etale cohomology worked out by Grothendieck to have too many concrete calculations and not enough big theories then I am tempted to say you have gone too far, but I perhaps shouldn't because many fields of mathematics have trended in that direction since Grothendieck's day, and that perspective does work for at least some mathematicians. Anyways, the use of $\mathbb P^1$ in this argument is certainly not a pure application of general theory, but it is also not ad hoc. It is part of a systematic strategy in the foundations of etale cohomology, which is to first prove the desired statement for curves using the powerful theorems available for curves (precise descriptions of $H^1$ and $H^2$, together with dimension estimates), and then find an inductive approach to the general case based on varieties mapping to a curve with fibers mapping to a curve, etc., suitable to the particular problem. For different problems, it makes sense to try slightly different curves, and put them together in slightly different ways. For the affine vanishing theorem, we of course want to look at affine curves, so we might look first at the simplest such, $\mathbb A^1$, and then observe we can prove the theorem for arbitrary products of $\mathbb A^1$, or $\mathbb A^n$. Can we reduce from the general case to $\mathbb A^n$? Yes, by Noether normalization. For the comparison of etale and analytic cohomology, it's convenient to work with affine curves, and fibrations of affine curves by affine curves, so that all our cohomology will match the cohomology of the fundamental group, as then we can take advantage of the comparison of the etale and analytic fundamental groups. In this case, it suffices to check that smooth varieties are covered by open subsets isomorphic to an iterated fibration of curves, and this we can do. For the proper base change theorem, we of course want to start with proper curves, and so we try the simplest such, $\mathbb P^1$. There's no clear reason to take nontrivial fibrations of $\mathbb P^1$ over $\mathbb P^1$, so we just look at powers, $(\mathbb P^1)^n$. Can we reduce from a general projective variety to $(\mathbb P^1)^n$? Not directly, as we only know that projective varieties map to $\mathbb P^n$, but using the map $(\mathbb P^1)^n \to \mathbb P^n$ we can close the gap. So in some sense it's all the same idea, just adapated in different ways to different contexts. The Weil conjectures are one place this strategy failed, as Grothendieck tried to prove them by showing every smooth projective variety is covered by a product of curves, but this is false. (Deligne was later, in Weil II, able to solve the problem by an inductive argument, using the theory of weights of constructible sheaves.)<|endoftext|> TITLE: Modular forms with finitely many or very few non-zero Fourier coefficients QUESTION [10 upvotes]: I have an elementary question on modular forms, but which I don't know how to solve. a) Is there a congruence subgroup $\Gamma \leq \mathrm{SL}_2(\Bbb Z)$, an integer $k \in \Bbb Z$ and a non-constant modular form $f \in M_k(\Gamma)$ such that $f$ has only finitely many non-zero Fourier coefficients $a_n(f)$ ? b) What about $\{n \geq 0 \mid a_n(f) \neq 0\}$ having zero density? One can easily have a non-zero modular form $f$ such that $a_n(f) = 0$ for every odd integer $n$. For part a), I think the answer should be no : $f$ is just a trigonometric polynomial and I guess one can come up with some elementary argument, but I don't know exactly how. Part b) is maybe a more subtle question, I would be glad to have any information about it! I already asked it here, but got no comment nor any reply. Possibly related: Modular forms with prime Fourier coefficients zero. REPLY [12 votes]: There is a completely elementary way to see that the answer to a) is negative - if $f$ had only finitely many Fourier coefficients, it would extend to an entire function on $\mathbb C$. But for any $\pmatrix{a&b\\c&d}\in\Gamma$ we get $$f\left(\frac{b}{d}\right)=f\left(\frac{a\cdot 0+b}{c\cdot 0+d}\right)=(c\cdot 0+d)^kf(0)=d^kf(0)$$ which means that either $f$ has a dense set on which it is constant (if $k=0$ or $f(0)=0$), and hence $f$ is constant, or $f$ will take arbitrarily high values within interval $(0,1)$, which is impossible. The same argument, together with simple upper bouds on $a_n(f)$, also gives some, though admittedly bad, lower bounds on how many nonzero coefficients there ought to be - the nonzero coefficients cannot be exponentially sparse. GH from MO's answer gives a much tighter bounds in these regards.<|endoftext|> TITLE: Energy levels of double well potential QUESTION [7 upvotes]: Consider the (quantum) Hamiltonian on the real line $$H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x).$$ Let us assume that the potential $V$ is an even smooth functions with exactly two non-degenerate minima, and $\lim_{|x|\to +\infty}V(x)=+\infty$. Such a $V$ is called double well potential. Let $E_1,E_2$ be the first two minimal eigenvalues of $H$. It is known in physics literature (see problem 3 after $\S$ 50 in Landau-Lifshitz, vol. 3) that under some extra assumption which are not quite specified there one has $$E_2-E_1 \approx \frac{\omega\hbar}{\pi} \exp(-\frac{1}{\hbar} C), $$ where $\omega, C$ are positive constant which can be written down explicitly, and the result is understood asymptotically as $\hbar\to 0$. I am looking for a mathematically more rigorous discussion of this result where, in particular, the assumptions are formulated more explicitly. REPLY [9 votes]: This is either Helffer-Sjostrand https://www.tandfonline.com/doi/abs/10.1080/03605308408820335 or Barry Simon https://www.jstor.org/stable/pdf/2007072.pdf?refreqid=excelsior%3A258084917fff9e0c10088abbb2679c55 PS: I cannot resist pointing out that the last paper is in Annals of Mathematics proving, as you notice yourself, an exercise in Landau/Lifshitz.<|endoftext|> TITLE: Are hypergeometric series not taught often at universities nowadays, and if so, why? QUESTION [43 upvotes]: Recently, I've become more and more interested in hypergeometric series. One of the things that struck me was how it provides a unified framework for many simpler functions. For instance, we have $$ \log(1+x) = x\ {_2F_1}\left(1,1;2;-x\right) ;$$ $$ \sin^{-1}(x) = x\ {_2F_1}\left(1/2,1/2;3/2;x^{2}\right) ;$$ $$e^{x} = \lim_{b \to \infty} \ {_2F_1}\left(1,b;1;x/b\right), $$ and many more similar identities. When I saw this for the first time, I was intrigued. Yet at the same time I was also surprised because I hadn't seen it before, and I studied mathematics at a university. I did a quick check, and it seems only a handful of universities in the Netherlands teach hypergeometric series, usually during the late stages of the bachelor's degree or during the master's degree. I am not sure about other countries, but I suspect they're not very often part of the curriculum over there either. Considering the subject's potential to unify many functions and ideas in analysis, I think it could be useful to learn more about this topic. So my question is twofold: Is it true that currently, hypergeometric series are generally not taught often at universities across the world? If so, why is this the case? REPLY [4 votes]: I find some banal answers to this question convincing: Mathematicians prefer to focus on functions of one or two variables. The power series for $_2F_1$ requires an unusual number-theoretic function (the Pochhammer symbol). The position of $_2$ before $F_1$ makes the hypergeometric function almost uniquely awkward to typeset or to read about. As a comparison: Are the Pearson distributions not taught often at universities nowadays, and if so, why? They provide a unified framework for many simpler probability distributions, e.g. \begin{align} B(\alpha,\beta) &= \operatorname{Pearson}_\text I(2-\alpha-\beta, \alpha-1, 1, -1, 0)\\ \Gamma(\alpha,\beta) &= \operatorname{Pearson}_\text{III}(1, \beta-\alpha\beta, 0, \beta, 0)\\ N(\mu,\sigma) &= \operatorname{Pearson}_\text{III}(1, -\mu, 0, 0, \sigma^2). \end{align} The subscripts here indicate the distribution type, which can also be calculated from the parameters. (There are plenty of parallels with the hypergeometric function: The arcsin and exponential distributions are subcases of $B$ and $\Gamma$. The five parameters for the Pearson distribution are projectively invariant, so they have the same degrees of freedom as $_2F_1$. And Pearson was apprarently motivated to construct these distributions by analyzing the hypergeometric distribution, whose cdf uses $_3F_2$, and the corresponding differential equation.) I think the answer is clear: The Pearson distributions are not taught often. While they unify other distributions, it is rare that this unification is useful for proving anything. Occasionally the closest fit to some data is a Pearson IV distribution. Even then, we have so little intuition about the distribution and so little sense of a mechanism that would generate it, that it's usually better not to use such a fit. In short: the Pearson distributions are so ugly that most people prefer to avoid them when they can. The hypergeometric functions may be similar.<|endoftext|> TITLE: When are immanants irreducible? QUESTION [10 upvotes]: For a partition $\lambda$ let $\chi_\lambda$ be the corresponding irreducible representation of the symmetric group $S_n$. Let $\mathrm{Imm}_\lambda(x) = \sum\limits_{\pi \in S_n} \chi_\lambda(\pi) x_{1 \pi(1)} \dotsm x_{n \pi(n)}$ be the immanant corresponding to $\lambda$. Question 1: Is $\mathrm{Imm}_\lambda$ irreducible over $\mathbb{Z}$? If not, in how many irreducible factors does it split depending on $\lambda$? It seems for all partitions of $n$ for $n$ at most 5 those polynomials are irreducible. Question 2: Let $R=\mathbb{Q}(x_{i,j})/I$ where $I$ is the ideal generated by the $\mathrm{Imm}_\lambda$. What is the Krull dimension of $R$? The sequence starts with $2,6,12$ for $n=2,3,4$. REPLY [6 votes]: They are irreducible. Actually the polynomials of the form $$ P:=\sum_{\pi} F(\pi)x_{1\pi(1)}\ldots x_{n\pi(n)} $$ are rarely reducible. Assume that $P=QR$. Consider both sides as the polynomials in the $i$-th column. Since $P$ has degree 1, it follows that one of $Q,R$ has degree 1 and another does not depend on the $i$-th column. That is, all columns are partitioned to $Q$-columns (which are represented in $Q$) and $R$-columns. Analogously the rows. Consider the element $x_{ij}$ where $i$ is the $Q$-row and $j$ an $R$-column (such $i$, $j$ exist since $Q$, $R$ are not-constant). We see that neither $Q$ nor $R$ depends on $x_{ij}$. Therefore we should have $F(\pi)=0$ whenever $\pi(i)=j$. But the irreducible characters can not satisfy this very restrictive property (since they depend only on the conjugacy class of $\pi$, we would get that $\chi({\rm id})=0$ when $i=j$ and $\chi(\pi)=0$ for all other permutations when $i\ne j$. The second option corresponds to a reducible (regular) character, the first is absurd.)<|endoftext|> TITLE: Initial conditions in the Klein-Gordon equation QUESTION [5 upvotes]: I am interested in what are the largest families of initial conditions of the problem (in $\mathbb{R}^{4}$) \begin{equation}\label{kg} \left\lbrace \begin{array}{ll} (\square+m^2)F(x)=0\\ F(0,\vec{x})=g(\vec{x}) \\ \frac{\partial F}{\partial x^{0}}(0,\vec{x})=f(\vec{x}) \end{array} \right. \end{equation} I could find something like "In order to this differential equation be well posed, $g$ must live in $H^{2}(\mathbb{R}^{3})$ and $f\in H^{1}(\mathbb{R}^{3})$" on the Web, but not for my equation. So my question are: What are sufficient conditions over $f$ and $g$ in order to the problem be well posed? And is there any relation between these conditions and some Sobolev space? I know probably the answer is in some classical book but I can find it. Many thanks in advance. REPLY [9 votes]: One must remark that derivatives in Sobolev spaces are usually taken in the sense of distributions: given $k\in\mathbb{N}_0=\{0,1,2,\ldots\}$, $H^k(\mathbb{R}^n)$ is the space of tempered distributions $u$ on $\mathbb{R}^n$ such that all partial derivatives of order $\leq k$ belong to $L^2(\mathbb{R}^n)$, where these derivatives are in the distribution (i.e. weak) sense: for all $f\in\mathscr{S}(\mathbb{R}^n)$ we have that $$u(f)=\int_{\mathbb{R}^n}\overline{u(x)}f(x)dx$$ and $$u_\alpha(f)=(-1)^{|\alpha|}u(\partial^\alpha f)=(-1)^{|\alpha|}\int_{\mathbb{R}^n}\overline{u(x)}\partial^\alpha f(x)dx=\int_{\mathbb{R}^n}\overline{u_\alpha(x)}f(x)dx$$ define $u(x),u_\alpha(x)\in L^2(\mathbb{R}^n)$ for each $$\begin{split}\alpha&=(\alpha_0,\ldots,\alpha_{n-1})\in\mathbb{N}_0^n\ ,\,\partial^\alpha=\partial_0^{\alpha_0}\cdots\partial_{n-1}^{\alpha_{n-1}}\ ,\,\\1&\leq|\alpha|=\|\alpha\|_{l^1}=\alpha_0+\cdots+\alpha_{n-1}\leq k\ .\end{split}$$ Just think of the above formulae as in integration by parts. This means that, as a rule, solutions $u$ of the Klein-Gordon equation in $\mathbb{R}^{n+1}$ that belong to $H^k(\mathbb{R}^{n+1})$ are weak solutions: for all $f\in\mathscr{S}(\mathbb{R}^{n+1})$ we have that $$u((\Box+m^2)f)=0\ .$$ In order to convert weak derivatives to classical derivatives so as to conclude that $u$ is a classical solution of the Klein-Gordon equation, one needs $u$ to be at least $\mathscr{C}^2$, which is the case if $u\in H^k(\mathbb{R}^{n+1})$ for $k>\frac{n+1}{2}+2$ thanks to the Sobolev inequality. That being said, the well-posedness of the Cauchy problem for the Klein-Gordon equation stems from the so-called energy estimates: for all $k\in\mathbb{N}$, $T>0$ we have $C_k>0$ such that for all $u\in\mathscr{S}(\mathbb{R}^{n+1})$ $$\begin{split}\|u(t,\cdot)\|_{H^k(\mathbb{R}^n)}&+\|\partial_0 u(t,\cdot)\|_{H^{k-1}(\mathbb{R}^n)}\leq C_k\left(\|u(0,\cdot)\|_{H^k(\mathbb{R}^n)}+\|\partial_0 u(0,\cdot)\|_{H^{k-1}(\mathbb{R}^n)}\phantom{\int^T_0}\right.\\ &\left.+\int^T_0\|(\Box+m^2)u(s,\cdot)\|_{H^{k-1}(\mathbb{R}^n)}ds\right)\ ,\quad |t|\in[0,T]\ .\end{split}$$ $H^k$ norms of higher-order time derivatives of $u$ for $k\geq 0$ also satisfy similar estimates once we appeal to the identity $\partial_0^2=(\Box+m^2)+\sum^n_{j=1}\partial_j^2-m^2$ and use the bound $$\begin{split}\|\partial_0^2 u(0,\cdot)\|_{H^k(\mathbb{R}^n)}&\leq\|(\Box+m^2)u(0,\cdot)\|_{H^k(\mathbb{R}^n)}+C\|u(0,\cdot)\|_{H^{k+2}(\mathbb{R}^n)}\\ &\leq C\|u(0,\cdot)\|_{H^{k+2}(\mathbb{R}^n)}+D\int^T_0\left(\|(\Box+m^2)u(s,\cdot)\|_{H^k(\mathbb{R}^n)}\right.\\ &\phantom{\leq}\left.+\|(\Box+m^2)\partial_0 u(s,\cdot)\|_{H^k(\mathbb{R}^n)}\right)ds\end{split}$$ for suitable constants $C,D>0$ (the last estimate comes from the one-dimensional Sobolev inequality on $[0,T]$, see e.g. Section 8.2 of the book Functional Analysis, Sobolev Spaces and Partial Differential Equations by H. Brézis (Springer-Verlag, 2011)). Unfortunately, the control of higher-order time derivatives by means of energy estimates is omitted or at best only glossed over in the vast majority of textbooks ("details are left as an exercise to the reader"). Putting all these together allows us to get the following, slightly weaker form for the energy estimates: for all $T>0$, $k\in\mathbb{N}$ we have constants $C'_k>0$ such that for all $u\in\mathscr{S}(\mathbb{R}^{n+1})$ $$\begin{split}\|u\|_{H^k([0,T]\times\mathbb{R}^n)}&\leq C'_k\left(\|u(0,\cdot)\|_{H^k(\mathbb{R}^n)}+\|\partial_0 u(0,\cdot)\|_{H^{k-1}(\mathbb{R}^n)}\right.\\&\phantom{\leq}\left.+\|(\Box+m^2)u\|_{H^{k-1}([0,T]\times\mathbb{R}^n)}\right)\ .\end{split}$$ These estimates by themselves already guarantee (by density of $\mathscr{S}$ in $H^k$ for all $k\in\mathbb{Z}$) uniqueness of $H^k$ weak solutions of the Klein-Gordon equation in $[0,T]\times\mathbb{R}^n$. They can also be extended to $k\in-\mathbb{N}_0$ by appealing to the Green operator ( = fundamental solution) $(1-\Delta)^{-1}$, where $\Delta$ is the ($n$- or $(n+1)$-dimensional) Laplacian, observing that such an operator commutes with all partial derivatives and therefore with the Klein-Gordon operator as well. The existence of weak solutions in $H^k([0,T]\times\mathbb{R}^n)$ for initial data in $H^k,H^{k-1}$ can be achieved by appealing to the duality between positive-order and negative-order Sobolev spaces (up to an appropriate prescription of initial conditions at $t=0$ and $t=T$). Since the Klein-Gordon operator is formally self-adjoint, the $H^k$ energy estimate allows us to define a weak solution as a continuous linear functional in the image of (the formal adjoint of) $\Box+m^2$ in $H^{-k+1}$ for all $k\in\mathbb{Z}$, suitably extended to the whole codomain $H^{-k-1}$ by means of the Hahn-Banach theorem (uniqueness of the extension is guaranteed by the energy estimates in $H^{k+1}$). The importance of extending the energy estimates to $k\in-\mathbb{N}_0$ is that it allows us to obtain weak solutions in $H^k$ for positive $k$ by this duality technique. As mentioned above, if $k>\frac{n+1}{2}+2$ we conclude that the weak solution $u$ is $\mathscr{C}^2$ and therefore is a classical solution of the Klein-Gordon operator, thanks to the Sobolev inequality. (Edit: on the other hand, also due to the Sobolev inequality, one obtains a unique distributional (weak) solution $u\in\mathscr{D}'(\mathbb{R}^{n+1}$) for given distributional sources and initial data, since the energy estimates for (the formal adjoint of) $\Box+m^2$ allow one to control a complete family of seminorms for test functions supported in any given compact subset. Uniqueness of distributional (weak) solutions then follows from the energy estimates by density of $\mathscr{D}$ in $\mathscr{D}'$. Moreover, one can also localize the energy estimates in space - more precisely, within slices of past- or future-directed lightcones -, which then also yield the finite speed of propagation property of solutions that characterizes hyperbolic PDE's such as the Klein-Gordon equation) This duality method for proving well-posedness of the Cauchy problem for the Klein-Gordon equation (as well as other linear PDE's) goes back to Garding and Lax. One can find a terse exposition of this strategy for the (possibly variable-coefficient) wave equation in Chapter I of the book Lectures on Non-Linear Wave Equations by C. D. Sogge (2nd. edition, International Press, 2008).<|endoftext|> TITLE: Infinite oscillation of minimum word length in 2-generated group QUESTION [7 upvotes]: Let $G$ be a group with generators $a, b\in G$. Define $\mathrm{len}:G\to\mathbb{Z}_{\geq 0}$ by sending $g$ to the minimum length of a word in $a, b, a^{-1}, b^{-1}$ equal to $g$. Assume that for all $g\neq e\in G$ there is infinitely many $n\in \mathbb{Z}_{\geq 0}$ such that $\mathrm{len}(g^{n+1})<\mathrm{len}(g^n)$. Must every element of $G$ be of finite order then? REPLY [6 votes]: This answer is partly inspired by HJRW's comment. Definition 1. Let $G$ be a finitely generated group and $g\in G$, and word length $|\cdot|$ with respect to some finite generating subset. Say that $g$ is very distorted if $\liminf |g^n|/\log(n)=0$, or equivalently if $u_g(n)=\sup\{m:|g^m|\le n\}$ satisfies $\limsup\log(u_g(n))/n=\infty$. It is not hard to check that this implies that $|g^n|>|g^{n+1}|$ for infinitely many $n$ (otherwise the growth would be more than exponential). Proposition. Let $G$ be a finitely generated group with finitely many conjugacy classes. Then every element $g$ of $G$ is very distorted. Proof: if $g$ has finite order this is trivial. Otherwise, $g$ has infinite order. Then there exists $n>1$ and $h\in G$ such that $hgh^{-1}=g^n$. Hence $h^kgh^{-k}=g^{n^k}$ for all $k$, and hence $h$ has infinite order. In turn there exists $f\in G$ and $m>1$ such that $fhf^{-1}=h^m$, so $|f^khf^{-k}|=|h^{m^k}|=O(k)$. In turn, $|h^{m^k}gh^{-m^k}|=|g^{n^{m^k}}|=O(k)$. Hence $\liminf |g^k|/\log\log k<\infty$, so $g$ is very distorted. The existence of torsion-free finitely generated groups with finitely many conjugacy classes was proved by Ivanov in the late 80s, and later Osin found examples with two conjugacy classes (I think both can be chosen generated by a pair). So this answers the question.<|endoftext|> TITLE: Is there a large colimit-sketch for topological spaces? QUESTION [12 upvotes]: Question. Is there a large colimit-sketch $\mathcal{S}$ such that $\mathrm{Mod}(\mathcal{S}) \simeq \mathbf{Top}$? In other words, is there a category $\mathcal{E}$ with a class of cocones $\mathcal{S}$ in $\mathcal{E}$ such that topological spaces can be seen as those functors $\mathcal{E} \to \mathbf{Set}$ which map the cocones in $\mathcal{S}$ to colimit cocones? Since the forgetful functor $\mathbf{Top} \to \mathbf{Set}$ creates colimits, I don't see any a priori reason why this cannot be true. Let me explain a bit where this question comes from, since I also have some reference requests about related questions. Burroni has given in Esquisses des catégories à limites et des quasi-topologies (zBMATH review, see also Burroni's 1970 CR note) a large mixed sketch based on filter convergence (see here) whose models are topological spaces. Maybe this thesis contains other sketches as well, perhaps even a colimit-sketch, thus answering my question, but I could not find it online anywhere, and Ehresmann told me that, in fact, there is no digital copy. I was quite surprised to find out that there is actually a limit-sketch for topological spaces. This is explicitly mentioned by Guitart in Toute théorie est algébrique et topologique (Proposition 25, pdf), without proof, but the construction comes out of Edgar's description of topological spaces in The class of topological spaces is equationally definable (doi:10.1007/BF02945113), which in turn is based on Kelley's characterization of topological spaces in terms of net convergence (General topology, Chapter 2, Theorem 9): A topological space can be described as a set $X$ together with a monomorphism $C(P,X) \to X^P \times X$ for every directed set $P$ satisfying four axioms (think of $C(P,X)$ as the set of $P$-indexed convergent nets including their limits). These axioms can be written down in categorical language. For example, the axiom "subnets converge to the same point" is the following: for every cofinal map $Q\to P$ there is a morphism $C(P,X) \to C(Q,X)$ such that $$\begin{array}{cc} C(P,X) & \rightarrow & X^P \times X \\ \downarrow && \downarrow \\ C(Q,X) & \rightarrow & X^Q \times X \end{array}$$ commutes. I have written down all the details of that sketch, but I wonder (Side question) if this limit-sketch is written down somewhere else already? As a byproduct we get the notion of a topological space object internal to any complete category (not just a topos as studied by Macfarlane and Stout for instance). Let me also mention that I found out that $\mathbf{Top}^{\mathrm{op}}$ is the category of models of a large limit-sketch, which means, however, that $\mathbf{Top}$ is the category of $\mathbf{Set}^{\mathrm{op}}$-valued models of a large colimit-sketch. Please let me know if you know any other references about this topic apart from those already mentioned. *Edit. Details of the sketches mentioned above can now be found in Large limit sketches and topological space objects. REPLY [9 votes]: Isbell shows in Function spaces and adjoints (1975) that every cocontinuous functor $\mathbf{Top} \to \mathbf{Set}$ is a coproduct of copies of the forgetful functor. (I have found an alternative proof for this within a more general theory, details will be added here when it's online.) But a necessary condition for a category to be modelled by a (possibly large) colimit sketch is that the cocontinuous functors to $\mathbf{Set}$ are jointly conservative. By picking any bijective continuous map which is not an isomorphism, we get a contradiction. *Edit. An alternative proof can be found in my new paper Large limit sketches and topological space objects (2021), Theorem 8.7.<|endoftext|> TITLE: Exponential convergence of Ricci flow QUESTION [6 upvotes]: I've been trying to understand the asymptotic behavior of Ricci flow, and there are two facts which I am unable to square away. I'm interested in higher dimensional manifolds, but my question is easier to state for Riemann surfaces. I suspect that the solution for surfaces will also solve the case in higher dimensions, so I'll focus on the two-dimensional case. Short version Given a Riemann surface which is topologically a sphere, the normalized Ricci flow is known to converge exponentially to a round metric. However, when we analyze the evolution of the scalar curvature for a small deformation of the round metric, it appears that there is a non-trivial center which seems to prevent the metric from converging exponentially. As such, I'm trying to determine the correct asymptotics for the scalar curvature, and why my calculation is going wrong. Long version Fact 1: In [1], Richard Hamilton proved that given a Riemann surface of positive curvature, if one evolves the metric by normalized Ricci flow \begin{equation} \frac{\partial}{\partial t} g_{i j}=(r-R) g_{i j}, \end{equation} then the metric will converge exponentially quickly to a metric of constant positive curvature (i.e., a round sphere). Here, $R$ is the scalar curvature and $r$ is the average scalar curvature, which acts to normalize the flow. "Fact 2": The round unit sphere is a fixed point for normalized Ricci flow, so we study asymptotics of the flow for small (volume-preserving) deformations of the round metric. Under Ricci flow, the scalar curvature $R$ evolves via the reaction-diffusion equation \begin{equation} \frac{\partial R}{\partial t}=\Delta R+R^{2}-r R, \end{equation} where $\Delta$ is the Laplace-Beltrami operator with respect to $g(t)$. Since we are considering the unit sphere, the Gauss-Bonnet theorem implies that $r = 2$, so we consider the function $\phi = R-2$ and have that \begin{equation} \frac{\partial \phi}{\partial t}=\Delta \phi+R \phi. \end{equation} If we linearize this equation at the round metric, we find that \begin{equation} \frac{\partial \phi}{\partial t}=\Delta_{\mathbb{S}^2} \phi+ 2 \phi, \end{equation} To understand the asymptotic behavior of $\phi$, we can decompose it as $\phi = \sum_{\ell, m} a^{\ell}_m Y^m_{\ell}$, where $Y^m_{\ell}$ are spherical harmonics. Doing so, the flow under the preceding equation is simply \begin{equation} \phi(z,t) = \sum_{\ell, m} a^{\ell}_m e^{-(\ell(\ell+1)+2)t} Y^m_{\ell} \end{equation} Here, this is using the fact that the spectrum of $\Delta_{\mathbb{S}^2}$ is $\ell (\ell+1)$. However, this seems to introduce a problem; the reaction term $2 \phi$ exactly cancels the diffusion effect on the $Y^m_{1}$-components of $\phi$. In other words, it seems like if $\phi$ is a principle eigenfunction of the Laplace-Beltrami operator (on the round sphere), then it is unchanged by Ricci flow, which seems to preclude exponential convergence of the flow. Reintroducing the non-linearity One might suspect that the issue is that I've linearized the flow at the round metric. However, the flow converges exponentially quickly to a round metric and a metric's principle eigenvalue is Lipschitz in the metric (which can be seen using the Rayleigh quotient property). As such, for a small deformation of the round metric where $\phi$ is close to a principal eigenfunction, we have that $ \left| \frac{\partial \phi}{\partial t}-\Delta_{\mathbb{S}^2} \phi- 2 \phi \right | < C e^{-\delta t} \phi $ for some $C$ and $\delta>0$. In other words, it seems that we can construct $\phi$ so that $$ \left| \frac{\partial \phi}{\partial t} \right | < C e^{-\delta t} \phi,$$ which again seems to preclude exponential convergence of the metric. What's wrong with Fact 2? Clearly, something is wrong with the analysis in the second argument. There is something that I have missed or computed incorrectly. However, I can't find the flaw in the argument, and I'd really like to know where I am going astray here. [1] Hamilton, Richard S., The Ricci flow on surfaces, Mathematics and general relativity, Proc. AMS-IMS-SIAM Jt. Summer Res. Conf., Santa Cruz/Calif. 1986, Contemp. Math. 71, 237-262 (1988). ZBL0663.53031. REPLY [2 votes]: Having thought about this a little more, I think a more detailed explanation of the precise issue you are describing is that you are being too cavalier in "Fact 2" about what "linearizing" means. You are treating the scalar curvature as something you can freely vary, but that is not really true since $\phi$ is not an independent variable but is something computed in terms of the underlying metric. In other words, what you can freely vary is the metric and if you do so (within a fixed conformal class) you resolve the issue you described. Indeed, if $g_s=e^{2u_s} g_{\mathbb{S}^2}$ is a variation of the metric (so $u_0=0$), then $$ \phi_s=2e^{-2u_s}-2e^{-2u_s} \Delta_{\mathbb{S}^2} u_s-2\frac{4\pi}{\int_{\mathbb{S}^2}e^{-2u_s} dvol_{\mathbb{S}_2}}. $$ Linearizing at $s=0$, we see that the linearized $\phi=\frac{d}{ds}|_{s=0} \phi_s$ from your Fact 2 must satisfy $$ \phi=-2 \Delta_{\mathbb{S}^2} v-4v $$ where $v=\frac{d}{ds}|_{s=0}u_s$ is the variation of the conformal factor (we used that the average scalar curvature is fixed so $v$ has average zero in order to eliminate the last term). This means $\phi$ is not arbitrary, but is in the image of the operator $\Delta_{\mathbb{S}^2} +2$. By the Fredholm alternative, this means that $\phi$ does not have any of the modes you are worried about in your original post. You could also linearize the the Kazhdan-Warner identity Otis mentioned in his answer to get the same effect.<|endoftext|> TITLE: Degree one self-map of $\Bbb R^2\big\backslash \big\{(n,0):n\in \Bbb Z\big\}$ not homotopic to any self-homotopy equivalence QUESTION [8 upvotes]: Consider the surface $\Sigma=\Bbb R^2\big\backslash \big\{(n,0):n\in \Bbb Z\big\}$. Does there exist a proper map $f\colon \Sigma\to \Sigma$ of degree $1$ and not homotopic to any self-homotopy equivalence of $\Sigma$ ? $\bullet$ Here $H_\mathbf{c}^2(\Sigma;\Bbb Z)=\Bbb Z$, so for any proper map $f\colon \Sigma\to \Sigma$ we can talk about the an interger, denoted by $\deg(f)$ so that the induced map $f^*\colon H_\mathbf{c}^2(\Sigma;\Bbb Z)\to H_\mathbf{c}^2(\Sigma;\Bbb Z)$ is multiplication by $\deg(f)$. This integer is invariant under proper homotopy. $\bullet$ Note that $\pi_1(\Sigma)$ is a free group on countably infinitely many generators, and proper $\deg$ $1$ map is $\pi_1$-surjective, but may not be $\pi_1$-injective as a free group on infinitely many generators is not Hopfian. $\bullet$ Here is a lemma that I am trying to use to construct a degree one map. Lemma: Let $f\colon M\to N$ be a proper map between two non-compact orientable connected manifolds without boundary of same dimension. Let $x_0\in M$ be such that $f^{-1}\big(f(x_0)\big)=\{x_0\}$. Suppose $f$ is a local homeomorphism near $x_0$. Then $$\deg( f)=\begin{cases}+1&\text{ if }f\text{ is orientation-preserving at }x_0,\\-1&\text{ if }f \text{ is orientation-reserving at }x_0. \end{cases}$$ This is already crossposted here and has no answer yet. REPLY [7 votes]: What about something like $f(x,y)=(g(x),y)$ where $$ g(x)=\begin{cases} x\qquad &x\in(-\infty,1/2]\\ 1-x & x\in [1/2,3/2]\\ x-2 & x\in [3/2,\infty) \end{cases} $$ $f$ does not induce the identity mapping on $H_1$, but I do believe it is a proper map of degree $1$. REPLY [7 votes]: We can produce such a map by folding. We define $f(x,y)$ to be $(x,y)$ if $x < 0$, to be $(-x,y)$ if $0 \leq x \leq 1$, and to be $(x-2,y)$ if $x > 1$. This map is proper and degree one, but is not injective at the level of fundamental groups. Edit: Beaten by 45 seconds! I’ll leave this here as (perhaps) it is a (tiny) bit easier to see that my map is not injective on $\pi_1$.<|endoftext|> TITLE: A criterion for loxodromicity in Gromov-hyperbolic spaces QUESTION [6 upvotes]: Recall that an isometry of a Gromov-hyperbolic space $X$ is called loxodromic if it has exactly two fixed points on the Gromov boundary $\partial X$, one being "attracting" and the other "repelling". This terminology is not completely standard but I am following Das–Simmons–Urbański in their book (available on arxiv), see Definition 6.1.2 there. I am wondering whether there is a reference for the following criterion for an isometry $f$ of $X$ to be loxodromic : $f$ is loxodromic iff there is an open subset $U \subset \partial X$ such that $\overline {fU} \subsetneq U$. Moreover if this is the case then $f$ has an attracting fixed point in $fU$. In the "classical" case ($X$ is proper and geodesic) this follows from the classification of isometries together with standard facts (parabolic and elliptic isometries preserve metrics on the boundary or boundary minus one point so this contracting behaviour is impossible). I think this should also work for the spaces which Das--Simmons--Urbański call "strongly hyperbolic" (Definition 3.3.6) where there are also nice metrics on the boundary (Observation 3.6.7 and Proposition 3.6.19). I am unsure about the general non-proper case. The motivation to establish this criterion is to be able to deduce from the usual ping-pong argument the following more precise result : if $f, g$ are two loxodromic isometries of $X$ with disjoint fixed points sets in $\partial X$ and sufficiently large minimal translation then the subgroup $\langle f, g \rangle$ is freely generated by $f, g$, all its non-trivial elements are loxodromic and two elements either have disjoint fixed point sets or they have a common power. If there is a reference for this in full generality I would also be interested. REPLY [2 votes]: An answer was given in Ashot Minasyan's comment, i'm writing it here so the question does not remain unanswered. In Lemma 2.5 of "Group actions on metric spaces: fixed points and free subgroups" by Mathias Hamann (on arxiv : https://arxiv.org/abs/1301.6513) he gives a criterion for an element of a group acting on a "contractive completion" to be hyperbolic (in a sense he defines for these actions). In the next section he proves that the Gromov bordification of a geodesic hyperbolic space is a contractive completion for any group of isometries, and his terminology in the previous section agrees with the usual one (so an element to which the ctiterion applies will be loxodromic in the terminology of my question). The conditions in the criterion are a bit stronger that what i wrote in the question but they are sufficiently lax for the application to ping-pong and in fact Hamann does that in the paper (Theorem 2.7).<|endoftext|> TITLE: Are two quasi-isometric, isomorphic on large enough balls, transitive graphs isomorphic? QUESTION [7 upvotes]: Take two transitive graphs $X,Y$ (potentially directed and edge-labelled, e.g. Cayley graphs). Assume $X,Y$ are quasi-isometric with constant $K$, i.e. there exists a function $f:VX \to VY$ ($VX,\,VY$ are the respective vertex sets) such that: Any vertex of $Y$ is at most at distance $K$ from the image of a vertex of $f(X)$, For any $x,x' \in VX$, we have $$K^{-1}d(x,x')-K \leq d(fx,fx') \leq Kd(x,x') + K.$$ Furthermore, assume that the balls of $X$ and $Y$ of radius $ \leq 2K$ are isomorphic. Question. Does it follow that $X$ and $Y$ themselves are isomorphic? At first, my intuition would say "yes" because the graphs are "the same" both in their coarse structure, and locally, but after further thoughts, maybe the local and global similarities don't match. It seems that this question is natural enough to have been studied before, but Google didn't help. Note. The constant $2K$ has been chosen semi-arbitrarily: if it is taken to be too small, the question can be answered negatively, e.g. by considering two cycle graphs of different length. REPLY [9 votes]: Questions of this flavour have indeed been studied before. There may be earlier references, but I know at least a body of work that was initiated by Benjamini, Ellis and Georgakopoulos (see here for precise references). They were interested in a question that is not exactly yours, but rather whether, given a transitive graph $X$, there is a constant $R$ such that any other graph $Y$ that has the same balls of radius $R$ as $X$ is covered by $X$. Such a graph is called Local-Global rigid. Many graphs are LG-rigid, but not all. In a paper with Romain Tessera, we tried to address some of these questions. And one of our constructions is as follows: for every integer $R$, we construct a continuum of pairwise non-isomorphic vertex-transitive graphs $(X_i)$ that have the same balls of radius $R$ and are all quasi-isometric to a product of two $4$-regular trees with constant $4$. You can replace "product of two $4$-regular trees" by other models, for example $\mathrm{SL}_4(\mathbf Z)$. See Theorem H in the above reference for a precise statement. This answers your question negatively. The situation is probably quite different if you restrict to Cayley graphs (even regarded as unlabeled unoriented graphs). For example, if $X$ is the Cayley graph of a finitely presented group $G$, then there is a constant $R(K,X)$ such that any other Cayley graph that is $K$-QI to $X$ and has the same balls of radius $R(K,X)$ than $X$ is isomorphic to $X$. This is quite obvious: there are only finitely many Cayley graphs that are $K$-QI to $X$. Added later a justification of the last assertion The assertion is that, if $X$ is the Cayley graph of a finitely presented group, there are only finitely many Cayley graphs with given degree (this assumption what forgotten) which are $K$-QI to $X$. This follows from the following lemma, which bounds the length of the relators in a presentation of a group whose Cayley graph is $K$-QI to $X$. Lemma: Let $(G,S)$ and $(H,T)$ be two groups with finite generating sets, such that the associated Cayley graphs $X$ and $Y$ are quasi-isometric with constant $K$. If $G$ has a presentation $G=\langle S \mid R\rangle$ with relators of length $\leq \ell$, then $H$ has a presentation $H=\langle T \mid R'\rangle$ with relators of length $\leq f(\ell,K)$ for some function $f$. Proof: the fact that $G$ admits such a presentation with relators of length $\ell$ is equivalent to $X$ being simply connected at scale $\ell$ (meaning that filling all loops of length $\leq \ell$ in $X$ turns $X$ into a simply connected space). But being simply connected at large scale is QI-invariant, see for example Theorem 2.2 here. So $Y$ is simply connected at scale $f(\ell,K)$, and the lemma follows.<|endoftext|> TITLE: Independent vectors in the permuting coordinates action of $S_n$ on $\mathbb{R}^n$ QUESTION [7 upvotes]: Let $V$ be the hyperplane in $\mathbb{R}^n$ with equation $\sum_i x_i=0$. The symmetric group $S_n$ acts on $V$ by $s\cdot (v_1,\ldots,v_n)=(v_{s^{-1}(1)},\ldots,v_{s^{-1}(n)})$. Consider those $v\in V$ with the following property: For any $n-1$ elements $s_1=\text{id},s_2,\ldots,s_{n-1}\in S_n$, the vectors $v,s_2\cdot v,\ldots,s_{n-1}\cdot v$ are independent. Question: what $v$ have this property? Some simple observations: If $H_{ij}$, for $1\leq i TITLE: Which knots appear as the singular locus of a polyhedral metric on the 3-sphere? QUESTION [18 upvotes]: What can be said about a knot $K\subseteq S^3$ for which there exists a (Euclidean) polyhedral metric (aka Euclidean cone-manifold structure) on $S^3$ whose singular locus is precisely $K$? I'm particularly interested in the case where the conical angle $\alpha$ around $K$ is less than $2\pi$. The classic example of this is the figure-8 knot. If you quotient $\mathbb E^3$ by the crystallographic group $\mathrm P2_13$, you get a Euclidean orbifold structure on $S^3$ whose singular locus is the figure-8 knot with angle $2\pi/3$. Dunbar (pp.82–86) gives a classification of all Euclidean orbifold structures on $S^3$, and this is the only one whose singular locus is a knot. Therefore, we know that if $K$ is not the figure-8 knot, then $\alpha$ cannot equal $2\pi/n$. That's a starting point, at least! Dunbar, William D., Geometric orbifolds, Rev. Mat. Univ. Complutense Madr. 1, No. 1-3, 67-99 (1988). ZBL0655.57008. Edit. Mednykh and Rasskazov construct a very simple family of fundamental polyhedra $\mathcal P(\theta)$ for the polyhedral manifolds $\mathcal C(\theta)$ ($S^3$ with conical angle $\theta$ along the figure-8 knot), for $\theta\in[0,4\pi/3)$. They show that, as $\theta$ increases, $\mathcal P(\theta)$ goes from hyperbolic to spherical, passing through Euclidean at $\theta=2\pi/3$. The reason I mention this is that they note (p.446): ...that [their] approach is general and can be applied to other two-bridge links and knots. And again, they (p.448): ...describe an algorithm for the construction of the fundamental set $\mathcal P$ for the figure-eight orbifold which effectively works in every space of constant sectional curvature. With slight modifications this algorithm can also be used to construct the fundamental set of any 2-bridge link orbifold. Mednykh, Alexander; Rasskazov, Alexey, Volumes and degeneration of cone-structures on the figure-eight knot, Tokyo J. Math. 29, No. 2, 445-464 (2006). ZBL1124.57008. Final Edit. According to a recent paper of Mednykh, the two-bridge knots $4_1,5_2,6_1,6_2,6_3,7_2,7_3,7_4,7_5,7_6$ and $7_7$ all admit polyhedral metrics with a given conical angle $\alpha_0<\pi$. With the exception of $7_5,7_6$ and $7_7$, the value of $\alpha_0$ is given implicitly, but exactly. Do all two-bridge knots have this property? REPLY [9 votes]: The ones that have a Euclidean cone metric with cone angle $\leq \pi$ should only be the spherical Montesinos knots (including 2-bridge knots, excluding the (2,n) torus knots). This should follow from the proof of the orbifold theorem, but one would need to search through the proof to sort out the logic. As indicated in the comments, Porti hinted at this in his paper on 2-bridge links. If there’s a spherical metric, then the $\pi$ orbifold with knot as the singular locus is spherical, and the representation into $SO(4)$ can be deformed to decrease the cone angle until it collapses to a Euclidean cone metric. Actually, the proof proceeds in the other direction: a hyperbolic metric is deformed to increase the cone angle until it collapses to a (rescaled) Euclidean cone metric. Conversely, if there’s a metric with cone angle $\leq\pi$, then the double branched cover admits an metric of non-negative Alexandrov curvature, and therefore must be modeled on $S^3$, $S^2\times R$, or be Euclidean. One can rule out the last two cases in the case of knots (Dunbar classified the knots whose $\pi$ orbifold admits exceptional geometries). For the general case with cone angle $> \pi$, I think this is wide open. There are some restrictions though. The case of cone angle $\geq 2\pi$ is impossible, since this corresponds to a CAT(0) metric on the 3-sphere which doesn't exist. Similarly, if the cone angle is $> \pi$, the double branched cover will admit a CAT(0) metric. This is only possible if it is hyperbolic. If there’s an essential torus that meets the branch locus, then it can be isotoped to be a flat plane invariant under the involution, and there’s an essential Conway sphere for the knot, and the cone angle would be $= \pi$, a contradiction. An essential sphere cannot exist, since there is a CAT(0) metric so $\pi_2$ vanishes. If the torus does not meet the singular locus, then the knot was a satellite. But then the torus may be isotoped to be totally geodesic (again by the flat plane theorem). The pattern knot would then admit a flat metric with totally geodesic torus boundary. But the only such manifold is the twisted I-bundle over a Klein bottle, which is not a knot complement. So one should restrict to knots whose double cover is either spherical (and the cone angle is $\leq \pi$) or hyperbolic (and the cone angle is betweeen $\pi$ and $2\pi$). The knot will be prime and not satellite and have no essential Conway sphere. Most arborescent knots, such as the (-2,3,7) pretzel considered in Sam Nead’s answer, will be ruled out. A possible obstruction would come from the representation of $\pi_1(S^3-K)$ to $Isom^+(\mathbb{E}^3)\cong \mathbb{R}^3 \rtimes SO(3)$. If the action of this group has no global fixed point (so is not conjugate into $SO(3)$), then the projection of this representation into $SO(3)$ gives a representation with non-trivial cohomology with coefficients in $\mathbb{R}^3$ twisted by the representation. Any knot $K$ will have many $SO(3)$ representations of $\pi_1(S^3-K)$, but one could try to look for ones for which this cohomology always vanishes and for which the holonomy of the meridian is elliptic (this means the translational part of the cocycle must be perpendicular to the axis of rotation of the $SO(3)$ holonomy of the meridian). For a Euclidean cone structure, I would suspect that this representation does not have a global fixed point, since otherwise the developing map would have precompact image, which shouldn't happen for a Euclidean cone structure (a geodesic in a generic direction should miss the cone locus by general position and hence have non-compact developing map).<|endoftext|> TITLE: Signs in Chevalley systems for reductive groups QUESTION [5 upvotes]: Let $G$ be a pinned split reductive group. There exists a Chevalley system: For each root $b$ in its root system there are parametrisations $x_b: \mathbb{G}_a \rightarrow U_b$ of the corresponding root subgroup, and for each element $w$ of the Weyl group (of the root system) there are lifts $\dot{w}$, such that $\dot{w} x_b(c) \dot{w}^{-1}$ = $x_{w(b)} (\pm c)$. Are there any explicit statements about these signs? Do they have any plain coherence properties? (There are related questions on MO regarding signs in the commutator formula for root subgroups and signs in the Chevalley basis for Lie algebras; I wondered whether perhaps something else is known here.) REPLY [2 votes]: One usually defines $w_\alpha(\varepsilon) = x_\alpha(\varepsilon) \cdot x_{-\alpha}(-\varepsilon^{-1}) \cdot x_\alpha(\varepsilon)$, which is the image of $$\begin{pmatrix} 0 & \varepsilon \\ -\varepsilon^{-1} & 0 \end{pmatrix}$$ under the mapping of $\operatorname{SL}_2\to G$ defined by $x_\alpha$ and $x_{-\alpha}$. Then there are relations (sometimes called Steinberg relations) on $x_\alpha$, $w_\alpha$ and $h_\alpha(\varepsilon)=w_\alpha(\varepsilon)w_\alpha(1)^{-1}$, among them $$w_\alpha(\varepsilon) \cdot x_\beta(\xi) \cdot w_\alpha(\varepsilon)^{-1} = x_{w_\alpha\beta}\left(\eta_{\alpha\beta} \cdot \varepsilon^{-\langle\beta,\alpha\rangle} \cdot \xi\right).$$ Here the numbers $\eta_{\alpha\beta}=\pm1$ satisfy the following properties: $$\eta_{\alpha\beta}=\eta_{\alpha,-\beta}, \qquad \eta_{\alpha\alpha} = -1, \qquad \eta_{\alpha\beta}\eta_{\alpha,w_\alpha\beta}=(-1)^{A_{\alpha\beta}},$$ where $A_{\alpha\beta} = 2(\alpha,\beta)/(\alpha,\alpha)$ are the Cartan integers. Moreover, $\eta_{\alpha\beta}=1$ if $\alpha\pm\beta\neq0$ and $\alpha\pm\beta\notin\Phi$ (in this case $x_\beta$ commutes with $x_\alpha$ and $x_{-\alpha}$); $\eta_{\alpha\beta} = -\eta_{\beta\alpha}$ if $\langle \alpha,\beta\rangle=\langle\beta,\alpha\rangle=-1$ (the case when $\alpha,\beta$ are the fundamental roots for an $\mathsf{A}_2$ subsystem); $\eta_{\alpha\beta}=-1$ if $\langle \alpha,\beta\rangle=0$ and $\alpha\pm\beta\in\Phi$ (the case when $\alpha,\beta$ are two orthogonal short roots of a $\mathsf{C}_2$ subsystem). When you look at a lift of an arbitrary $w\in W(\Phi)$, the result depends on a particular choice of lifts for $w_\alpha$, which, in turn, depends on the choice of $x_\alpha$. The numbers $\eta_{\alpha\beta}$ can be expressed in terms of the structure constants $N_{\alpha\beta ij}$ (the expression is usually very simple, but in a few cases involves also the lengths of a certain root chain). It is more complicated in $\mathsf{G}_2$, and in my experience in this case it is much easier to just fix some signs in an explicitly chosen matrix representation and work with that. The standard reference for all this is "Simple groups of Lie type" by R. W. Carter (Proposition 6.4.3). Since this book is not easy to get legally, you can also take a look at Chevalley groups over commutative rings: I. Elementary calculations by N. Vavilov and E. Plotkin (Section 13).<|endoftext|> TITLE: Vacuum region with positive measure for the Schrödinger equation QUESTION [6 upvotes]: Let us consider the free Schrödinger equation $(i\partial_t+\Delta_x)\psi=0$ in $\mathbb{R}_t\times\mathbb{R}_x^d$. I'm trying to understand the structure of the vacuum region $$\Omega(\psi):=\{(t,x)\in \mathbb{R}_t\times\mathbb{R}_x^d \;\;s.t.\;\,\psi(t,x)=0\}$$ for solutions with finite energy. In particular, my question is the following: does there exist a non-zero solution $\psi\in \mathcal{C}(\mathbb{R},H^1(\mathbb{R}^d))$ of the free Schröodinger equation such that $\Omega(\psi)\subseteq\mathbb{R}^{1+d}$ has positive measure? If yes, is it even possible that $\Omega(\psi)$ contains an open ball? Thank you for any suggestion. REPLY [7 votes]: The purpose of this answer is to extend Christian Remling's answer to dimension $d = 3$. There are two steps. (N.B. below the cut I show how to replace part 1 by a different argument that works in all dimensions, and so this should answer the question posed.) We assume that we have a solution $\phi$ to the Schrodinger equation such that it vanishes on a (WLOG) $[-a,a]\times B(0,R)\subset \mathbb{R}\times\mathbb{R}^3$ 1. Controlling the radial parts Let $\psi:\mathbb{R}\times\mathbb{R}^+ \to \mathbb{C}$ be the function $$ \psi(t,r) = r \cdot \frac{1}{4\pi} \int_{\mathbb{S}^2} \phi(t, r\omega) ~d \omega $$ the spherical mean of $\phi$ multiplied by the radius. We have that $$ \psi(t,r) \equiv 0, \text{ when } r \leq R , |t| \leq a $$ and that $\psi$ is a solution to the one dimensional Schrodinger equation $$ i \partial_t \psi = \partial^2_{rr} \psi $$ Thus Christian Remling's answer sufficies to imply that $\psi \equiv 0$ everywhere. (Note that if $\phi(t,\bullet)\in L^2(\mathbb{R}^3)$, then $\psi(t,\bullet)\in L^2(\mathbb{R}_+)$). Remark: when dimension $d \neq 3$, or when $d = 3$ but considering other spherical harmonics, we get that the equation being satisfied is $i \partial_t \psi = \partial^2_{rr} \psi + \alpha r^{-2} \psi$; so if Christian Remling's answer can be extended to the one dimensional Schrodinger equation with inverse square potentials, then this would also give a general answer. 2. Controlling the rest Absent the needed result from the previous remark, we can argue thus in $d = 3$. Applying the same argument, but now with center at $x_0 \in B(0,R/2)$, we see that the spherical means of $\phi(t,x)$ vanishes for all radii if we center it at $x_0$. I claim that this is enough to guarantee that $\phi(t,x) \equiv 0$. The following proof is probably not the most straightforward, but that's the one I know. Let $u$ be the solution to the linear wave equation $(-\partial^2_s + \triangle)u = 0$ (I use $s$ for the time parameter to disambiguate from the time parameter in the Schrodinger equation) in $d = 3$ with initial data $u(0,x) = 0$ and $\partial_su(0,x) = \phi(t,x)$ for any fixed $t$. Using the fundamental solution and the spherical mean property, we have that $u(s,x) = 0$ for all $x \in B(0,R/2)$. By the strong version of the Holmgren uniqueness theorem (see chapter IV, section 3 of F. John Partial Differential Equations) we have that $u \equiv 0$. This implies that $\phi(t,x) = 0$ for all $x$. Since $t$ is arbitrary: we are done. Radial part: redux Let's now prove the spherical mean property directly using the fundamental solution of the Schrodinger equation; this argument holds for all dimensions. Let $\phi_0(x) = \phi(0,x)$. Suppose for $(t,x) \in [-a,a]\times B(0,R)$ we have that $\phi(t,x) = 0$, this implies that, using the fundamental solution to the Schrodinger equation, evaluated at $x = 0$, that $$ \int e^{i|x|^2/4t} \phi_0(x) ~dx = 0 $$ for all $t \in [-a,a]\setminus \{0\}$. Let $\tilde{\phi}_0$ be the spherical mean of $\phi_0$. We note that by assumption $\tilde{\phi}_0(r) = 0$ for all $r \leq R$. Our integral identity above implies $$ \int_0^\infty e^{i r^2 / 4t} \tilde{\phi}_0(r) r^{d-1} ~ dr = 0 $$ Change variables you get $$ \int_0^\infty e^{i \rho / 4t} \underbrace{\tilde{\phi}_0(\sqrt{\rho}) \rho^{d/2-1}}_{g(\rho)} ~ d\rho = 0 $$ Since $\tilde{\phi}_0$ is supported away from $r = 0$, we can extend $g(\rho)$ by zero to the negative half line. The vanishing of the above quantity for all $|t| \leq a$ shows that $g$ has compact Fourier support and hence is analytic; but its vanishing for $\rho < R^2$ implies that $g \equiv 0$, and hence the spherical mean vanishes. Combining this with part 2 from above, we extend the uniqueness property to all dimensions.<|endoftext|> TITLE: Is the $\infty$-category $N_{dg}(\mathrm{Ch}(\mathcal{A}))$ presentable? QUESTION [9 upvotes]: (See Jacob Lurie's "Higher Algebra", section 1.3.5 for context.) Let $\mathcal{A}$ be a Grothendieck abelian category. Then the stable $\infty$-category $\mathcal{D}(\mathcal{A})$ is a localisation of the dg-nerve $N_{dg}(\mathrm{Ch}(\mathcal{A}))$ by Higher Algebra Proposition 1.3.5.13, and it is presentable according to Higher Algebra Proposition 1.3.5.21, since it is underlying a combinatorial model category. My question is: what can we say about $N_{dg}(\mathrm{Ch}(\mathcal{A}))$ itself? Is it presentable? As far as I can understand, the introduction of https://arxiv.org/pdf/1710.11388.pdf seems to suggest that there is an underlying model structure which is not combinatorial but accessible, although I am not sure if this is enough to conclude... REPLY [10 votes]: This fails already with the category of abelian groups. If the dg-nerve of the dg-category of chain complexes of abelian groups were presentable, then the associated triangulated category would be well generated in the sense of Neeman in his book Triangulated categories (Annals of Math. Studies, Vol. 148, 2001); this is easy to deduce from Krause's characterization together with section 1.4.4 of Lurie's Higher Algebra. But in appendix E of his book, Neeman proves that the triangulated category of chain complexes up to chain homotopy equivalences does not have a generating set (Lemma E.3.2 page 438), which provides a definitive obstruction against presentability. He also proves that the opposite category is not well generated, by the way (but this is easier).<|endoftext|> TITLE: Filtered 2-colimits commute with finite 2-limits QUESTION [8 upvotes]: Is there an explicit proof anywhere in the literature that filtered 2-colimits commute with finite 2-limits (all meant in the weak bicategorical sense) in the 2-category of groupoids? I have only been able to find a handful of papers about filtered 2-colimits, notably Dubuc and Street's A construction of 2-filtered bicolimits of categories, and have not found this result in any of them. Of course one could try to deduce this from $\infty$-categorical results, but the latter proofs are also often sketchy, and in the case of groupoids it ought to be possible to give a concrete constructive proof. REPLY [8 votes]: Two relevant papers are: Dupont's Interchange of filtered 2-colimits and finite 2-limits. Canevali's 2-filtered bicolimits and finite weighted bilimits commute in Cat. The former proves that finite conical pseudolimits and filtered pseudolimits commute in $\mathbf{Cat}$; whereas the latter proves the analagous result for finite weighted bilimits and filtered bicolimits. It may be that these are enough to recover commutivity in $\mathbf{Grpd}$ for the cases in which you are interested.<|endoftext|> TITLE: Determinant of walk matrix for a skew-symmetric matrix of even order QUESTION [6 upvotes]: Let $S=(s_{ij})$ be a skew-symmetric integral matrix of order $n$. We only consider the case that $n$ is even. Let $e$ be the all-one vector in $\mathbb{R}^n$. Define the walk matrix $$W(S)=[e,Se,\cdots,S^{n-1}e].$$ (the name "walk matrix" comes from graph theory, where $S$ is the adjacency matrix of an undirected graph. Of course, $S$ is not skew-symmetric in the setting of graphs.) It is well-known that $\det(S)$ is always a square number. I find that the integer $\sqrt{\det(S)}$ is always a divisor of $\det W(S)$. But I cannot find any references on this relation. For example, consider $$S=\left( \begin{array}{cccc} 0 & 4 & 0 & -3 \\ -4 & 0 & -2 & -1 \\ 0 & 2 & 0 & 3 \\ 3 & 1 & -3 & 0 \\ \end{array} \right).$$ Then, $$W(S)=\left( \begin{array}{cccc} 1 & 1 & -31 & -3 \\ 1 & -7 & -15 & 165 \\ 1 & 5 & -11 & -87 \\ 1 & 1 & -19 & -75 \\ \end{array} \right).$$ Using Mathematica, we find that $\det(S)=18^2$, $\det(W)=16128=18\times 896$ and $\sqrt{\det(S)}\mid \det(W)$. It seems that the above relation $\sqrt{\det(S)}\mid \det(W)$ always hold for any skew-symmetric integral matrix of even orders. In particular, when $\det(S)=0$ then $\det(W)=0$. This particular case is not hard to show. REPLY [7 votes]: Surely, there is nothing special in the all-ones vector: the claim holds for any integer-valued $e$. Notice that $$ \det W^TW =\det\bigl[e^T (-1)^iS^{i+j}e\bigr], $$ Since $S$ is skew-symmetric, we have $e^TS^ie=0$ for all odd $i$. Permuting now the rows and columns of $W^TW$ we get $$ \det W^TW=\det\begin{bmatrix} A_0&0\\ 0& -A_1\end{bmatrix}, $$ where $$ A_0= \begin{bmatrix} e^Te& e^TS^2e& e^TS^4e& \cdots& e^TS^{n-2}e\\ e^TS^2e& e^TS^4e& e^TS^6e& \cdots& e^TS^ne\\ \vdots& \vdots& \vdots& \ddots& \vdots\\ e^TS^{n-2}e& e^TS^ne& e^TS^{n+2}e& \cdots& e^TS^{2n-4}e \end{bmatrix}, \quad A_1= \begin{bmatrix} e^TS^2e& e^TS^4e& e^TS^6e& \cdots& e^TS^ne\\ e^TS^4e& e^TS^6e& e^TS^8e& \cdots& e^TS^{n+2}e\\ \vdots& \vdots& \vdots& \ddots& \vdots\\ e^TS^{n}e& e^TS^{n+2}e& e^TS^{n+4}e& \cdots& e^TS^{2n-2}e \end{bmatrix}. $$ We show that $$ \det A_1=\pm\det S\det A_0, \qquad(*) $$ so that $\det W=\pm\det A_0\sqrt{\det S}$, which yields what we want. Denote $$ u_i= \begin{bmatrix} e^TS^ie& e^TS^{i+2}e& \cdots& e^TS^{i+n-2}e \end{bmatrix}. $$ The eigenvalues of $S$ are purely imaginary, so by Cayley—Hamilton we get $$ S^{n}=\sum_{j=0}^{n/2-1} \alpha_jS^{2j}, $$ where $\alpha_0=-\det S$. Therefore, $$ u_n= -\det S\, u_0+\sum_{j=1}^{n/2-1} \alpha_ju_{2j}. $$ Plug this into $\det A_1$, expand by linearity, and erase vanishing summands to get $(*)$.<|endoftext|> TITLE: Definition of an arithmetic subgroup of an algebraic group QUESTION [9 upvotes]: I'm struggling with the definition of an arithmetic subgroup of an algebraic group defined over $\mathbb{Q}$. In Wikipedia you can read: If $\mathrm G$ is an algebraic subgroup of $\mathrm{GL}_n(\mathbb{Q})$ for some $n$ then we can define an arithmetic subgroup of $\mathrm G(\mathbb{Q})$ as the group of integer points $\Gamma = \mathrm{GL}_n(\mathbb{Z}) \cap \mathrm G(\mathbb{Q}).$ [...] The subgroup defined above can change when we take different embeddings $\mathrm G \to \mathrm{GL}_n(\mathbb{Q}).$ Thus a better notion is to take for definition of an arithmetic subgroup of $\mathrm G(\mathbb{Q})$ any group $\Lambda$ which is commensurable [...] to a group $\Gamma$ defined as above (with respect to any embedding into $ \mathrm{GL}_n$). With this definition, to the algebraic group $\mathrm G$ is associated a collection of "discrete" subgroups all commensurable to each other. This definition imply that for two embeddings $\mathrm G \to \mathrm{GL}_n(\mathbb{Q})$ the groups $\mathrm{GL}_n(\mathbb{Z}) \cap \mathrm G(\mathbb{Q})$ are commensurable. Why is it true? REPLY [14 votes]: Write the coordinates $a_{ij}$ of one embedding into $GL_n$ as polynomial functions, defined over $\mathbb Q$, in the coordinates $b_{ij}$ of a different embedding into $GL_n$. We can do this because, by definition, embeddings give an isomorphism of algebraic varieties. (I guess we should allow the inverse of the determinant as one of our coordinates. I will still write them as $a_{ij}, b_{ij}$ for notational simplicity.) Let $N$ be the greatest common denominator of all rational numbers appearing as coefficients of these polynomials. Now if $b_{ij} = \delta_{ij}$ for all $i,j$ then the group element goes to the identity under the second embedding, so it is the identity, so $a_{ij} =\delta_{ij}$ for all $i,j$. It follows, by elementary algebra, that if $b_{ij}$ is congruent to $\delta_{ij}$ modulo $N$ for all $i,j$, then $a_{ij}$ is an integer for all $i,j$. Since matrices congruent to the identity mod $N$ are a finite index subgroup, this shows the matrices with $a_{ij}$ integral contain a finite index subgroup of the matrices with $b_{ij}$-integral. Handling the other direction symmetrically, they are commensurable.<|endoftext|> TITLE: Relationship between provable in $RCA_0$ and effectively true QUESTION [8 upvotes]: Question: What is the relationship between provability in $RCA_0$ and effectively true? In other words: Given a problem, if a statement asserting the existence of a solution of the problem is provable in $RCA_0$, does it follow that given a computable instance of the problem we can compute a solution to the problem? Motivation: I'm getting into Reverse Mathematics and Hirchfeldt gives an exercise in his book Slicing the Truth to prove that $RCA_0$ proves the Ramsey theory for singletons $RT^1_k$ for each $k \geq 2$. This together with $REC$ being the intended model for $RCA_0$ implies to me that the homogenous set should be computable. However, $RT^1_k$ seems to be effectively true only in the trivial case of $k=1$. My thinking so far got me the following observations: If the colour in the infinite pigeonhole principle is effectively true, then so is $RT^1_k$, but this doesn't seem to help, since the two are equivalent. Trying to define the property that the homogenous set is infinite leaves me with the solution being $\emptyset''$-computable. To my limited knowledge the most probable solution to this, unless I'm missing some trick that would make $RT^1_k$ effectively true, is that while homogenous sets are computable, we cannot effectively decide which homogenous set is infinite, hence cannot effectively compute the solution to the instance. My naive leap from this is that provable in $RCA_0$ implies effectively true only with finite and trivial instances. But this seems quite a long stretch and the use of trivial is rather vague(to incorporate the $RT^1_k$ case, something along the lines of Rice's theorem) hence the question. (Apologies if this is not a research-level question, my competence to evaluate these is lacking) REPLY [4 votes]: Three (related) approaches that have not been mentioned are as follows: the meta-theorems from proof-mining (see U. Kohlenbach's "Applied Proof Theory") provide the kind of results you are looking for in a (much more) general setting. These meta-theorems are restricted to specific classes of formulas, but there exist counter-examples to most purported generalisations. In other words, one cannot do better in general, esp. when classical logic is used. Makoto Fujiwara has obtained a number of results connecting "provability in constructive math" and "provability in classical mathematics with witnesses". This is a list of some of his papers. The results are perhaps more concrete than proof mining, so not as general, but a better starting point for the novice. In the case of Ramsey's theorem, there is even a constructive version based on the notion of 'almost full' (Veldman and Bezem). This formulation is quite elegant and even has applications in CS, as shown here. Formulated with almost-fullness, there is a computable interpretation 'by default'.<|endoftext|> TITLE: Index of the mapping class group $\Gamma_{g,n}$ inside $\text{Out}(\Pi_{g,n})$ QUESTION [7 upvotes]: Let $\Sigma_{g,n}$ denote an $n$-punctured surface of genus $g$, with $2g+n-2 > 0$. Let $\Pi_{g,n}$ be its fundamental group (for some choice of base point), and let $\Gamma_{g,n}$ denote its pure, orientation preserving mapping class group. Then, by the Dehn-Nielsen-Baer theorem, the outer action of $\Gamma_{g,n}$ on $\Pi_{g,n}$ is faithful and induces an isomorphism between $\Gamma_{g,n}$ and the subgroup $\text{Out}^*(\Pi_{g,n})$ of $\text{Out}(\Pi_{g,n})$ which fixes the conjugacy class of each simple closed curve surrounding a puncture. Question. When does $\text{Out}^*(\Pi_{g,n})$ have finite index inside $\text{Out}(\Pi_{g,n})$? If $(g,\,n)=(1,\,1)$, then the index is $2$. Since $\Pi_{1,1} = \Pi_{0,3}$ and $\Gamma_{0,3}$ is trivial, one gets infinite index for $(g, \, n)=(0,\,3)$. What happens in the other cases? I'm particularly interested in the case where $g\ge 2, \, n = 1$. REPLY [7 votes]: It only has finite index in very low-complexity degenerate cases. Here's a proof that it always has infinite index for $\Sigma_{g,1}$ with $g \geq 2$. This proof generalizes in an obvious way to deal with all the other cases too, but the notation gets worse. Set $\pi = \pi_1(\Sigma_{g,1})$. Let $\{a_1,b_1,\ldots,a_g,b_g\}$ be the usual generating set for $\pi$ and let $$\omega = [a_1,b_1] \cdots [a_g,b_g] \in \pi$$ be the surface relation. It is enough to construct a sequence of elements $f_1,f_2,\ldots$ in $\text{Aut}(\pi)$ such that the conjugacy classes of the $f_k(\omega)$ are all distinct. This will immediately imply that the outer automorphisms associated to the $f_k$ are all in different mapping class group cosets. In fact, you can take the $f_k$ to be powers of almost any random automorphism you write down. For instance, let $f_k$ be the automorphism defined by the formula $$f_k(a_1) = a_1 b_1^k \quad \text{and $f_k$ fixes all other generators},$$ so $f_k$ is the $k$th power of $f_1$. Letting exponentiation denote conjugation, we then have \begin{align*} f_k([a_1,b_1]) &= [a_1 b_1^k, b_1]\\ &= b_1^{-k} a_1^{-1} b_1^{-1} a_1 b_1^k b_1\\ &= b_1^{-k} a_1^{-1} b_1^{-1} a_1 b_1 b_1^k\\ &= [a_1,b_1]^{b_1^k}. \end{align*} and thus $$f_k(\omega) = [a_1,b_1]^{b_1^k} [a_2,b_2] \cdots [a_g,b_g].$$ As long as $g \geq 2$, these are all distinct conjugacy classes. REPLY [6 votes]: This is surely not the most direct answer. But $\mathrm{Out}(\Pi_{g,1}) \cong \mathrm{Out}(F_{2g})$ surjects onto $\mathrm{GL}(2g,\mathbf Z)$, and the image of $\Gamma_{g,1}$ lands in $\mathrm{Sp}(2g,\mathbf Z)$. So the index is infinite for $g \geq 2$. In fact a more careful version of this argument shows that $(g,n)=(1,1)$ is the unique case where you get a finite index subgroup. Let me flesh out the argument. For all $n>0$ we have similarly that $\mathrm{Out}(\Pi_{g,n}) \cong \mathrm{Out}(F_{2g+n-1})$ surjects to $\mathrm{GL}(2g+n-1,\mathbf Z)$. The corresponding representation $H$ of $\Gamma_{g,n}$ is just the action of the mapping class group on the first homology of your favorite genus $g$ surface $\Sigma$ with $n$ punctures. But the action must be compatible with a lot of extra structure coming from geometry: there is the short exact sequence $$ 0 \to \mathbf Z^{n-1} \to H_1(\Sigma,\mathbf Z) \to H_1(\overline \Sigma,\mathbf Z) \to 0 $$ and $\Gamma_{g,n}$ preserves it, where $\overline \Sigma$ is the compact surface obtained by filling in the punctures. Moreover, $H_1(\overline \Sigma,\mathbf Z)$ is of rank $2g$ and carries a symplectic form preserved by $\Gamma_{g,n}$; the action of $\Gamma_{g,n}$ on $\mathbf Z^{n-1}$ is trivial. These conditions define an infinite index subgroup of $\mathrm{GL}(2g+n-1,\mathbf Z)$ unless $(g,n)=1$.<|endoftext|> TITLE: Riemann rearrangement theorem for $L^1$ functions QUESTION [11 upvotes]: Let $c_n$ be a sequence of real numbers with $\sum c_n$ converging conditionally but not absolutely. Suppose $\delta_n > 0$ is another sequence with $\delta_n \to 0$, and $\sum c_n \delta_n$ converging also conditionally but not absolutely. Does there exist, for every $L^1$ function $f: [0, 1] \to \mathbb R$, a bijection $\gamma: \mathbb N \to \mathbb N$, and a sequence of measurable sets $A_n$ with $\mu(A_n) = \delta_n$ such that $\sum c_{\gamma(n)}1_{A_{\gamma(n)}} \to f$, in $L^1$ and pointwise a.e? Note: Here $\mu$ denotes the Lebesgue measure. REPLY [8 votes]: The problem is trickier than I initially thought, but with the corrected condition it can be done. I need to assume that $\sum c_n \delta_n$ is conditionally but not absolutely convergent, $0 < \delta_n < 1$ with $\delta_n \to 0$ and $c_n \to 0$ (which is currently only implied by the conditional convergence of $\sum c_n$). Wlog. we can assume $f: [0,1] \to \mathbb{R}$ to be non-negative (otherwise split the series in two and use them to approximate positive and negative part separately) and non-increasing (by rearrangement, just to simplify notation). Now proceed as following: Denote the current partial sum by $\tilde{f}_k$. Take the next unused $n\in \mathbb{N}$ such that $c_n > 0$ and consider the sets $$A_\lambda := \{x \in [\lambda,1]: f(x)- \tilde{f}_k(x) \geq c_n \}$$ for $\lambda \in [0,1]$. If there is a $\lambda$ such that $|A_\lambda| = \delta_n$, then choose $\gamma(k) := n$, $A_n := A_\lambda$, add $c_n 1_{A_n}$ to the partial sum and iterate. (The "positive process") If not, take the next unused $m \in \mathbb{N}$ such that $c_m < 0$, choose $\gamma(k) := m$ and find a set $A_m \subset [0,2\max(\delta_n,\delta_m)]$ with $|A_m| =\delta_m$, for which the approximation is best, i.e. $x \notin A_m$ implies $f(x)-\tilde{f}_k(x) \geq f(y)- \tilde{f}_k(y)$ for all $y \in A_m$, add $c_m 1_{A_m}$ to the partial sum and iterate. (The "negative process") The positive process cannot continue indefinitely, since it increases the integral by $c_n \delta_n$ and $\tilde{f}_k \leq f$. The negative process cannot continue indefinitely, since at some point it will achieve $f(x) - \tilde{f}_k(x) \geq c_n$ on $[0,\delta_n]$, at which point the positive process will take over again. Thus we use all indices and have constructed a bijection. Now fix $\epsilon > 0$. At some point, all future $\delta$ are smaller than $\epsilon/2$, so after that on the interval $[\epsilon,1]$ the approximation $\tilde{f}_k$ is monotone increasing in $k$. Furthermore, whenever the negative process takes over, we know that $f(x) - \tilde{f}_k < c_n$ except for a set of measure $< \delta_n$. Thus as $f$ is decreasing, $$\int_\epsilon^1 f - \tilde{f}_k dx < c_n (1-\epsilon) + \delta_n \sup_{[\epsilon,1]} (f-\tilde{f}_k).$$ Here, the first term converges to $0$ as $c_n \to 0$, while for the second term $\sup_{[\epsilon,1]} f \leq f(\epsilon)$ as $f$ is non-increasing and $\sup_{[\epsilon,1]} (- \tilde{f}_k)$ is bounded as there were only a finite number of negative $c_m$ until we reached monotonicity on $[\epsilon,1]$. Then since $\delta_n \to 0$, that term converges as well and we have that $\tilde{f}_k \to f$ in norm on $[\epsilon,1]$, which together with monotonicity implies pointwise convergence a.e. on $[\epsilon,1]$ and as $\epsilon$ was arbitrary, we then have pointwise convergence a.e. on $[0,1]$. Finally consider the negative process. We know that $|\{x\in [0,2 \max(\delta_n,\delta_m)]: \tilde{f}_k(x) < -c_n \}| <\delta_n \leq \max(\delta_n,\delta_m)$, otherwise the positive process would already have taken over. But this means that for the set $A_m$ we choose, we have $\tilde{f}_k(x) \geq -c_n$ for all $x\in A_m$ and hence at all points modified, we will have $\tilde{f}_{k+1} (x) \geq -c_n +c_m$. But since this is true for all negative steps, we have $\tilde{f}_k \geq - 2\max_{n\in \mathbb{N}} |c_n|$ for all $k\in\mathbb{N}$. But than we have an integrable lower bound which allows us to conclude that the pointwise limit $f$ is also the $L^1$ limit of $\tilde{f}_k$ by dominated convergence.<|endoftext|> TITLE: A convolution-type identity for the "major index" QUESTION [6 upvotes]: For a permutation $\pi\in\frak{S}_n$, define the number of descents of $\pi$ as $$\text{des}(\pi)=\vert\{i: \pi(i)>\pi(i+1)\}\vert.$$ The following is a well-known (and interesting) identity: $$\binom{k\ell+n-\text{des$(\pi)$}-1}n=\sum_{\sigma\tau=\pi} \binom{k+n-\text{des$(\sigma)$}-1}n\binom{\ell+n-\text{des$(\tau)$}-1}n.$$ This motivated me to ask: QUESTION. Is there a similar identity for the major index? What about other "statistic" on $\frak{S}_n$? REPLY [6 votes]: Richard's identity $(*)$ can be found as Theorem 11 (though not stated exactly this way) in my paper Multipartite  P-partitions and inner products of skew Schur functions, Combinatorics and algebra (Boulder, Colo., 1983), 289–317,Contemp. Math., 34, Amer. Math. Soc., Providence, RI, 1984. (It's actually $\tau\sigma=\pi$ rather than $\sigma\tau=\pi$.) You can find this paper at http://people.brandeis.edu/~gessel/homepage/papers/multipartite.pdf. I don't think that setting $x_i=q^{i-1}$ and $y_j=t^{j-1}$ works except when $\pi$ is an identity permutation or the reverse of an identity permutation when $F_{D(\pi)}$ is symmetric. However we can get a $q$-analogue if we keep the descent number in addition to the major index, thought it's not as simple as one might like: we take $x_i = q^{i-1}$ for $i=1,2,\dots, M$ with $x_i=0$ for $i>M$ and $y_j= q^{m(j-1)}$ for $j=1,2,\dots, N$ (where $N$ may be $\infty$). See Section 4 of T. Kyle Petersen, Cyclic descents and P-partitions, Journal of Algebraic Combinatorics 22 (2005) 343-375, https://arxiv.org/abs/math/0405479. Some related formulas can be found in D. Krob, B. Leclerc, and J.-Y.Thibon, Noncommutative symmetric functions. II. Transformations of alphabets, Internat. J. Algebra Comput. 7 (1997), no. 2, 181–264.<|endoftext|> TITLE: Does a suitable famlly of eigenvectors of non self-adjoint operators, sufficiently close to an adjoint one, form a basis? QUESTION [5 upvotes]: It is very well known that if $A\in L^\infty(B_1;\mathbb R ^{d\times d})$ is a positive definite symmetric matrix, the eigenvalue of the self adjoint operator $H^2(B_1)\cap H^1_0(B_1)\to L^2(B_1)$ $$T:u\to\text{div}(A Du)$$ are all real, and the corresponding eigenvectors can be chosen to form an orthonormal eigenbasis of $L^2(\Omega)$ and an $A$-orthogonal basis of $H^1_0(\Omega)$. My question is whether it is known if this extends to the case when the operator is perturbed by a lower order term, namely $$T^\prime:u\to\text{div}(A Du)+C\cdot Du$$ with $C\in L^\infty(\Omega;\mathbb R^d)$? It could have complex eigenvalues and eigenvectors, but nevertheless would the span of the real and imaginary part of such vectors still be a generating family? Giorgio Metafune has pointed out that the question is answered positively when $A$ and $C$ are constant, which was the formulation of this question before this edit. In that case, the eigenvalues and eigenvectors of $T^\prime$ are connected to that of $T$ thanks to a transformation. Indeed, note that $$ \text{div}(A Du)+C\cdot Du = \lambda u \Leftrightarrow \text{div}(A Dv)= \lambda v $$ where $v=u\exp(-\frac12 A^{-1}C \cdot x)$. It therefore immediately follows that the answer is positive. The very same argument carries over (as Giorgio Metafune pointed out) when $A$ and $C$ are non constant, but $$ A^{-1}C =D\phi $$ for some function $\phi$. The question remains in general. Decomposing the problem by this method, I suppose the next simple case is $$ T^\prime : u\to \Delta u+ \phi \cdot Du $$ where $\text{div} \phi =0$. But if a "lower order terms therefore perturbation therefore yes" argument exist, it will avoid this case by case dissection. REPLY [3 votes]: In one space dimension, the answer is yes, and the eigenvalues are real and simple (Sturm-Liouville theory). In higher space dimension, the answer is negative, because the operator needs not be diagonalisable. If you let the data $(A,C)$ depend upon a parameter, you can pass from a situation where all the eigenvalues are real to one where there is a pair of complex conjugate eigenvalues. At the transition point, the operator is not semi-simple, admitting a double eigenvalue whose eigenspace is only one-dimensional. An explicit construction is given in this answer. Just rewrite the operator $A:D^2$ as ${\rm div}(AD)+C\cdot D$ where $-C$ is the row-wise divergence of $A$.<|endoftext|> TITLE: Fusing conjugacy classes QUESTION [7 upvotes]: Consider a finite group $G$ and two conjugacy classes $H$ and $I$ of isomorphic subgroups of $G$. Question. Is there some finite overgroup of $G$ which fuses $H$ and $I$ into a single conjugacy class? REPLY [7 votes]: The answer is yes, as noted in the comments above by Will Sawin. Let $H$ and $I$ be isomorphic subgroups of the finite group $G$. Consider the regular permutation representation $G \leq \operatorname{Sym}(\Omega)$, where $\Omega = G$. Then for $H$ and $I$ the action on $\Omega$ splits into $[G:H] = [G:I]$ orbits. Furthermore, for both groups the action on every orbit is regular. So since $H$ and $I$ are isomorphic, their actions on $\Omega$ are equivalent, which means that they are conjugate in $\operatorname{Sym}(\Omega)$. In general, for any group $G$, with HNN extension you can find an infinite overgroup $G^*$ of $G$ such that $H$ and $I$ are conjugate in $G^*$.<|endoftext|> TITLE: How many category structures are possible on two sets? QUESTION [14 upvotes]: For two sets $O$ and $A$, we will call a category structure a collection of functions ${\sf dom}:A\to O,\ {\sf cod}:A\to O,\ {\sf 1}:O\to A,\ \circ:A\times_OA\to A$ satisfying the usual axioms for a category. Can we parametrize the number of category structures (up to iso or equivalence) on two sets $O$ and $A$ in terms of their cardinalities? If a closed-form solution is too much to ask, can we get asymptotics for finite sets? Denote by ${\sf Cat}^\cong(n,m)$ the number of isomorphism classes of category structures on two sets $O$ and $A$ as above with $|O|=n$ and $|A|=m$. We trivially have that $n\leq m$. For $n=m=0$ we have exactly one category, and for $n=m=1$ we have a unique up to iso category. In general for $n=m$ we have ${\sf Cat}^\cong(n,m)=1$, but all these observations are trivial. For $n=1$ and $1\leq m$ we are counting the number of monoids on a set with $m$ elements, which I tried to search but was unable to find -- I did find this related question, and the comments by Qiaochu Yuan give a lower bound of $B^\leq_m\leq{\sf Cat}(1,m)$ where $B^\leq_m$ is the $m^{th}$ ordered Bell number. As the comment by Douglas Zare suggests, this indicates that asymptotics are the best we should hope for since the ordered Bell numbers grow faster than exponential in $m$. The case for semigroups is cutting edge for a set with $12$ elements by a paper linked in the answer to the linked question, so any asymptotics will have to use the existence of units to hopefully shave things down. The second linked paper gives a closed-form solution for the number of nilpotent semigroups of degree $3$, so adding mild restrictions seems to potentially allow for more tractable counting. Denote by ${\sf Cat}^\simeq(n,m)$ the number of equivalence classes of category structures on two sets $O$ and $A$ as above. We trivially have ${\sf Cat}^\simeq(n,m)\leq{\sf Cat}^\cong(n,m)$ with equality holding for $n=m$, but beyond this I don't see much concrete to say. It may be useful to use the cardinalities of the hom-sets between objects in the category structures instead of the cardinality of the overall set of arrows, but beyond more obvious observations nothing jumps out at me using this approach either. For $1 TITLE: Stable homotopy groups of complex projective plane QUESTION [9 upvotes]: We know that there is a cofiber sequence $S^3\xrightarrow{\eta}S^2\to\mathbb{C}\mathbb{P}^2$. It's easy to know that $\pi_3^s(\mathbb{C}\mathbb{P}^2)=0$ so there is a surjection $$\partial:\pi_7^s(S^2\wedge\mathbb{C}\mathbb{P}^2)\to\pi_7^s(\mathbb{C}\mathbb{P}^2\wedge\mathbb{C}\mathbb{P}^2)$$ by the long exact sequence of $\eta$. The former group $\pi_7^s(S^2\wedge\mathbb{C}\mathbb{P}^2)$ is $\mathbb{Z}/12$, generated by the second hopf element. I wonder whether $\partial$ is an isomorphism? REPLY [7 votes]: $\newcommand{\Z}{\mathbb Z}\newcommand{\cA}{\mathcal A}\newcommand{\Sq}{\mathrm{Sq}}\newcommand{\CP}{\mathbb{CP}}\newcommand{\Ext}{\mathrm{Ext}}$It's possible to run the Adams spectral sequence directly to show that the $2$-torsion subgroup of $\pi_7^s(\CP^2\wedge\CP^2)$ is isomorphic to $\Z/2$, so that $\partial$ can't be an isomorphism. The calculation is a little longer than Lennart Meier's answer above, but it is more elementary. In what follows, all cohomology is with $\Z/2$ coefficients. As an $\cA$-module, $\tilde H^\ast(\CP^2\wedge\CP^2)$ looks like $\tilde H^\ast(\CP^2)\oplus \Sigma^2\tilde H^\ast(\CP^2)$, with the lowest- and highest-degree elements joined by a $\Sq^4$. There is a short exact sequence of $\cA$-modules \begin{equation} 0\longrightarrow \Sigma^4\tilde H^\ast(\CP^2)\longrightarrow \tilde H^\ast(\CP^2\wedge\CP^2)\longrightarrow \Sigma^2 \tilde H^\ast(\CP^2)\longrightarrow 0. \end{equation} Here's a picture of that short exact sequence. A short exact sequence of $\cA$-modules induces a long exact sequence of Ext groups. Often one computes this long exact sequence by drawing an Adams chart for the Ext of both the sub and the quotient; the boundary map has bidegree $(-1, 1)$. Beaudry and Campbell, section 4.6, give some examples of this technique. To use this, we need to know $\Ext_\cA(\tilde H^\ast(\CP^2), \Z/2)$; you can look it up in Hood Chatham and Dexter Chua's Adams spectral sequence calculator (this is for $\Sigma^{-2}\tilde H^\ast(\CP^2)$, so shift everything to the right by $2$), but it is also possible to compute it by hand in the range needed using a similar long exact sequence associated to the short exact sequence $0\to\Sigma^4\Z/2\to\tilde H^\ast(\CP^2)\to\Sigma^2\Z/2\to 0$. Thus the long exact sequence in Ext looks like this: We're done once we figure out whether the boundary map in red is nonzero: either way, there are no differentials to or from the $7$-line, and there are no possible hidden extensions on the $7$-line. We will show the red boundary map is nonzero by showing the map $$i^\ast\colon \Ext_\cA^{1,9}(\tilde H^\ast(\CP^2\wedge\CP^2), \Z/2) \longrightarrow \Ext_\cA^{1,9}(\Sigma^4\tilde H^\ast(\CP^2), \Z/2),$$ which is induced by $i\colon \Sigma^4\tilde H^\ast(\CP^2)\to\tilde H^\ast(\CP^2\wedge\CP^2)$, vanishes; exactness then forces the red boundary map to be injective, so the $7$-line of the $E_2$-page for $\tilde H^\ast(\CP^2\wedge\CP^2)$ has only a single $\Z/2$. Now let's show $i^\ast$ vanishes. $\Ext_\cA^{1,9}(\Sigma^4\tilde H^\ast(\CP^2), \Z/2)\cong\Z/2$ and is generated by the upside-down question mark extension: Therefore if $i^\ast\ne 0$, there is some extension $0\to\Sigma^9\Z/2\to M\to \tilde H^*(\CP^2\wedge\CP^2)\to 0$ whose pullback along $i$ is the upside-down question mark extension. The only way for this to work would be the following diagram: However, this choice of $M$ is not a valid $\cA$-module: in $M$, $\Sq^5x = \Sq^1\Sq^4x \ne 0$, but $\Sq^5 = \Sq^4\Sq^1 + \Sq^2\Sq^1\Sq^2$, and $\Sq^4\Sq^1x = \Sq^2\Sq^1\Sq^2x = 0$. Therefore $i^\ast = 0$.<|endoftext|> TITLE: Which categories are injective with respect to fully faithful functors? QUESTION [8 upvotes]: Recall that a poset $K$ is a complete lattice if and only if $K$ is injective with respect to poset embeddings in that sense that for any poset $B$, any embedded subposet $A \subseteq B$, and any order-preserving map $A \to K$, there exists an extension $B \to K$. I'm wondering what sorts of generalizations or analogs this fact has when passing from posets to categories. Ultimately I'd be interested to know if (co)completeness conditions can be characterized by injectivity, but I'm not sure what the correct question is in this direction, so let's start with something concrete: Question: Which locally small categories $\mathcal K$ are injective with respect to fully faithful functors $\mathcal A \to \mathcal B$ between small categories? There is some ambiguity about what "injective" means here. In general, "injective with respect to fully faithful functors" means that given a fully faithful functor $\mathcal A \to \mathcal B$ and a functor $\mathcal A \to \mathcal K$, there exists a functor $\mathcal B \to \mathcal K$ making the requisite diagram commute. But I can think of at least 4 possible meanings of "commute" -- we could ask for the diagram to commute strictly (giving (1) "strict-injectivity"), up to isomorphism (giving (2) "pseudo-injectivity"), or up to a natural transformation (giving (3) "lax-injectivity" or (4) "oplax-injectivity" depending on the direction of the transformation). So really there are at least 4 questions here. I think the most interesting versions are the "strict" and "pseudo" versions, and I suspect the answers in these two cases should be rather close. Notes: At any rate, it seems that by duality, the answer can't be "the complete categories" except perhaps in case (3) or (4). If it makes a difference to change from considering fully faithful functors $\mathcal A \to \mathcal B$ to just considering full replete subcategories $\mathcal A \subseteq \mathcal B$ or something like that, I'd be happy with an answer to any such small variation. I would also be interested in knowing the answer when considering fully faithful functors between locally small categories, rather than just between small categories. REPLY [3 votes]: There's a nice answer if we take a slightly stronger notion of "injectivity", namely where we take Kan extensions instead of extensions. I would expect this to coincide in some nice cases with the non-Kan notion of injectivity. This is a question for which the theory of (co)KZ doctrines is well-suited, in which (co)completeness properties are characterised by the existence of certain extensions. I'll choose to work with KZ doctrines, and thus cocompleteness properties. A good reference is Walker's Distributive laws via admissibility, and the full definitions of the various concepts I make use of below can be found there. Let $(\mathbb P, \eta)$ be a KZ doctrine on a 2-category $\mathcal K$. An object $X \in \mathcal K$ is $\mathbb P$-cocomplete if for all $g \colon B \to X$, there exists a left extension $\mathrm{lan}_{\eta_B} g$, exhibited by an invertible 2-cell, such that the left extension respects those defined by the KZ doctrine in an appropriate sense (q.uiver link). From Marmolejo–Wood's Kan extensions and lax idempotent pseudomonads, we know that the $\mathbb P$-cocomplete objects are equivalently the pseudoalgebras for the pseudomonad $\mathbf P \colon \mathcal K \to \mathcal K$ induced by $(\mathbb P, \eta)$, so that when we take $\mathbf P$ to be a pseudomonad for a cocompletion under a class of weights $\Phi$, then $\mathbb P$-cocompleteness corresponds to admitting all $\Phi$-colimits. In particular, for $\mathbb P$ the small presheaf construction on locally small categories, a $\mathbb P$-cocomplete object is just a small-cocomplete locally small category. To relate this to your question, we also need to introduce the notion of admissibility for a KZ doctrine. A 1-cell $f \colon A \to B$ is $\mathbb P$-admissible if, for any $h \colon A \to X$ for $X$ a $\mathbb P$-cocomplete object, there exists a left extension $\mathrm{lan}_f h$, such that the left extension is preserved by $\mathbb P$-cocontinuous 1-cells into any $\mathbb P$-cocomplete object (q.uiver link). Crucially, as Walker observes in Remark 25 ibid., a $\mathbb P$-admissible 1-cell is $\mathbb P$-fully faithful (meaning $\mathbf Pf$ is representably fully faithful) if and only if every left extension as in the diagram above is exhibited by an invertible 2-cell. Taking $\mathbb P$ to be the small presheaf construction, the $\mathbb P$-admissible 1-cells are the functors $f \colon A \to B$ for which $B(f{-}, b)$ is a small presheaf for all $b \in B$. We shall call these small functors. (Ivan Di Liberti mentions several equivalent conditions for a functor to be small in this answer.) The $\mathbb P$-fully faithful 1-cells are precisely the fully faithful functors. So, if a locally small category $X$ is small-cocomplete, every functor $A \to X$ admits a left extension along small functors $A \to B$ (hence $X$ is "Kan lax-injective" with respect to small functors). Furthermore, these left extensions are exhibited by invertible 2-cells precisely for those small functors $A \to B$ that are fully faithful (hence $X$ is "Kan pseudo-injective" with respect to small fully faithful functors). Conversely, if a locally small category $X$ admits left extensions of functors $A \to X$ along small functors, it in particular admits a left extension along the Yoneda embedding of $A$ (since the unit $\eta$ of a KZ doctrine is $\mathbb P$-admissible), and hence is small-cocomplete. Therefore, the locally small categories that are Kan lax-injective with respect to small functors are precisely the small-cocomplete categories. One question remains, which regards the case when $X$ is only known to admits left extensions along $\mathbb P$-fully faithful $\mathbb P$-admissible 1-cells. Here, we may restrict our consideration to the fully faithful KZ doctrines, which are those for which the components of the unit $\eta_B \colon B \to \mathbb P B$ are representably fully faithful. In this case, $X$ once again admits extensions along $\eta_B$, and hence is $\mathbb P$-cocomplete. Since the Yoneda embedding is fully faithful, the small presheaf construction is a fully faithful KZ doctrine. Therefore, the locally small categories that are Kan pseudo-injective with respect to small fully faithful functors are precisely the small-cocomplete categories. (Let me know if something doesn't quite look right – I could well have made a mistake somewhere along the line!)<|endoftext|> TITLE: Non-isomorphic smooth affine varieties dominating each other QUESTION [8 upvotes]: Can non-isomorphic smooth affine varieties dominate each other? In the projective case one can take isogenous abelian varieties. REPLY [10 votes]: Let $X = \mathbb A^3$ and let $Y = SL_2$. The map $X \to Y$ sending $(a,b,c)$ to $$\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ b & 1 \end{pmatrix} \begin{pmatrix} 1 & c \\ 0 & 1 \end{pmatrix}= \begin{pmatrix} 1+ba & a+c + bac \\ b & 1 +bc\end{pmatrix} $$ is dominant because its image contains all matrices with lower-left corner nonzero. The map $Y \to X$ sending $\begin{pmatrix} x & y \\ z & w \end{pmatrix}$ to $(x,y,z)$ is dominant because its image contains all triples with $x \neq 0$. But $X$ and $Y$ are not isomorphic, for example because their complex points are nonisomorphic manifolds. REPLY [8 votes]: To complement the nice answer of Will Sawin, I give you an example in dimension $2$. Let $X=\mathbb{A}^2$ and let $Y=\mathbb{P}^1\times \mathbb{P}^1\setminus \Delta$, where $\Delta$ is the diagonal. If you take $\ell\subset \mathbb{P}^1\times \mathbb{P}^1$ to be a fibre of one projection, then $Y\setminus \ell$ is isomorphic to $\mathbb{A}^2$, so you have an open embedding $X\hookrightarrow Y$, which is then a dominant morphism. Conversely, you choose a point $p\in \Delta$ and consider the pencil of curves $C$ of $\mathbb{P}^1\times \mathbb{P}^1$ of bidegree $(1,1)$ such that $C\cap \Delta=\{p\}$, i.e, being tangent to $\Delta$. This pencil gives a rational map $\mathbb{P}^1\times \mathbb{P}^1\dashrightarrow \mathbb{P}^1$ and restricts to a morphism $Y\to \mathbb{A}^1$. Choosing two different points of $\Delta$, you obtain two morphisms $Y\to \mathbb{A}^1$ and thus a morphism $Y\to \mathbb{A}^2$. It is dominant because the general fibres of the two intersect into finitely many points. You can also see this by embedding $Y$ into $\mathbb{A}^3$, using the Segre embedding of $\mathbb{P}^1\times \mathbb{P}^1\hookrightarrow \mathbb{P}^3$, and project onto two factors. The two surfaces $X$ and $Y$ are not isomorphic, because the Picard group of $X$ is trivial but the one of $Y$ is not.<|endoftext|> TITLE: Irreducible representations of the symmetric group on homology of simplicial complex QUESTION [6 upvotes]: I am following Wall's paper A note on symmetry of singularities and I have some questions regarding representation theory and the homology of some objects: Consider an action of $\Sigma_k$ on a finite simplicial complex $X$ of dimension $n$, such that the action is simplicial and a simplex is fixed iff it is point-wise fixed. Consider $\sigma\in\Sigma_k$, then $$\chi_{Top}(X^\sigma)=\chi_{\Sigma_k}(X)(\sigma):= \sum_i(-1)^i \text{trace } \sigma^*:H_i(X)\rightarrow H_i(X), $$ where $X^\sigma$ is the fixed subcomplex fixed by $\sigma$ and $\chi_{Top}$ the Euler-Poincaré characteristic. My questions are: 1.- Is the sum of characters $\chi_{\Sigma_k}(X)$ called the equivariant Euler characteristic? Because I believe the equivariant Euler characteristic is the Euler characteristic of $H_*(X)^G$ in other texts. 2.- Can one say anything about the character on a single homology group instead of the alternated sum of all the groups? For example, compare the trace of $\sigma$ on one $H_i$ and $H_i(X^\sigma)$. I have a lot of troubles trying to isolate one homology group from this equality (with the extra structure that I have in my problem). 3.- Can one say something about the isotypes corresponding to a single irreducible representation of $\Sigma_k$? For example, compare the number of copies of the $sign$ representations and $H_*(X^\sigma)$ for some $\sigma$. If it is useful, in my problem I know that every $X^\sigma$ has lower dimension than $X$, if $\sigma\neq 1_{\Sigma_k}$. Edit: I forgot to say that I am only interested on the free part of the homology groups, so everything makes sense if we take complex coefficients. REPLY [2 votes]: Given a friendly self map $\sigma: X \rightarrow X$ of a finite complex, the alternating sum of the traces you write down defines $L(\sigma)$, the Lefschetz number of $\sigma$. It is standard that this number is 0 if $\sigma$ has no fixed points, and the proof of this involves all the homology groups together. It is also clear that if $\sigma$ is homotopic to the identity, then $L(\sigma)$ is just the Euler characteristic. Finally, if $A \subset X$ is a subspace taken to itself by $\sigma$, then $$ L(\sigma:X \rightarrow X) = L(\sigma: A \rightarrow A) + L(\sigma: (X,A) \rightarrow (X,A)).$$ Letting $A=X^{\sigma}$, one learns that $L(\sigma: X \rightarrow X) = \chi(X^{\sigma})$, which is the displayed equality in your post. The fact that your $\sigma$ is part of a group action doesn't really seem that relevant (except that the fixed point space will still be a finite complex).<|endoftext|> TITLE: Possible number of components of anticanonical sections of projective manifolds QUESTION [7 upvotes]: Let $M$ be a connected projective complex manifold with a smooth anticanonical divisor $D$ ($D \sim -K_M$). Let $k$ be the number of components of $D$. Some cheap thoughts give: If $M$ is a Fano manifold of dimension higher than one, $k=1$. If $M=\mathbb P^1$ or $M= \mathbb P^1 \times X$, where $X$ is a projective manifold with trivial canonical class, then $k=2$. Is it possible to have $k$ bigger than two? Is $k$ bounded? REPLY [4 votes]: (Let me turn my comment into an answer to give a correct attribution.) Yes, in the setup you describe, the maximum number of connected components of the divisor $D$ is 2. This is related to the so-called connectedness principle in birational geometry. The statement that you want follows from Proposition 5.1 of Log canonical singularities are Du Bois by Kollár and Kovács. They prove in particular that if $(Y,\Delta)$ is an lc pair such that $K_Y + \Delta$ is $\mathbf Q$-linearly equivalent to 0, then the non-klt locus $\operatorname{nklt}(Y,\Delta)$ has 1 or 2 components. In your case, taking $\Delta=D$, then $\operatorname{nklt}(Y,\Delta)$ is just (the support of) $D$, so this set has at most 2 connected components.<|endoftext|> TITLE: Necessary and sufficient curvature condition for a regular planar curve to be simple and closed QUESTION [6 upvotes]: Given a smooth, $2\pi$-periodic function $\kappa(s)$, the associated planar curve $\gamma(s)$ for which $\kappa(s)$ is the (signed) curvature, is uniquely determined up to Euclidean invariance: a canonical parametrization is for example given by $$ \gamma(s) = \left( \int_0^{s} \cos \phi(\sigma)\,\text{d}\sigma,\,\int_0^{s} \sin \phi(\sigma)\,\text{d}\sigma \right),\;\phi(\sigma) = \int_0^\sigma \kappa(\tau)\,\text{d}\tau. \tag{1} $$ The goal is to determine whether $\gamma$ is closed and simple. In principle, parametrization $(1)$ suffices to check whether $\gamma$ is closed and simple; however, the nested integrals makes this cumbersome in practice. In my case, I have a family of functions $\kappa(s)$ as periodic orbits of a given dynamical system, and I would like to select those $\kappa$-orbits that give rise to a simple closed curve $\gamma$. A priori, one could consider the total curvature $K = \int_0^{2\pi} \kappa(s)\,\text{d}s$. For a closed curve $\gamma$, the condition $K = 2\pi$ is necessary to avoid self-intersections. However, this condition is unfortunately not sufficient. It is straightforward to construct an example where a homotopy within a curve family for which $K=2\pi$ induces self-intersection: Are there results from differential geometry that I can use here, is my only option to check the injectivity and periodicity of the explicit parametrization $(1)$? [Related question for algebraic, non-closed curves: https://mathoverflow.net/questions/170320/conditions-for-a-parametric-curve-to-avoid-self-intersection] REPLY [3 votes]: The Four-vertex theorem dictates that the curvature function needs to have at least two local minima and maxima. The converse of the four-vertex theorem tells us that there's at least one parametrization, such that the resulting curve is simple and closed. It basically turns the curve into something closely resembling an ellipse. For arbitrary curvature functions satisfying the Four-vertex condition and a fixed parametrization, I don't believe it is possible to answer the question.<|endoftext|> TITLE: Factor states on C*-algebras QUESTION [12 upvotes]: Which C$^*$-algebras admit factor states for which the von Neumann algebra it generates in the corresponding GNS representation is a type III$_1$ factor? For example, do all purely infinite algebras admit such a state? Even more generally: what about a type III$_\lambda$ factor for $\lambda\in(0,1]$? REPLY [8 votes]: By the main result of this paper of Odile Maréchal, which was generalizing James Glimm's famous paper, the following extreme dichotomy holds. If $A$ is any separable C*-algebra, precisely one of the following statements holds. $A$ is of type I: for every representation $\pi$ of $A$, we have that $\pi(A)''$ is a von Neumann algebra of type I. For every infinite injective factor $M$, there exists a representation $\pi$ of $A$ such that $\pi(A)'' \cong M$. In particular, every separable C*-algebra that is not of type I, admits states $\omega_\lambda$ such that $\pi_{\omega_\lambda}(A)''$ is the unique injective factor of type III$_\lambda$, with $\lambda \in (0,1]$.<|endoftext|> TITLE: Reference for a Grünwald–Letnikov-type definition of the $n$-th derivative of a function QUESTION [5 upvotes]: Let $U\subset\mathbb R$ be an open set. Let $n\in\mathbb N$ and suppose that $f\in\mathcal C^n(U)$, i.e. that $f$ is $n$-times continuously differentiable on $U$. The $n$-th derivative of $f$, denoted by $f^{(n)}$, then satisfies, for all $x\in U$, $$f^{(n)}(x)=\lim_{h\to 0}\frac{\sum_{k=0}^n f(x+ k h) \binom nk (-1)^{n-k}}{h^n}.$$ I proved this fact here and the result is closely related to the Grünwald–Letnikov derivative. My question. Suppose that $f: U \to\mathbb R$ is any function such that $$\lim_{h\to 0}\frac{\sum_{k=0}^n f(x+ k h) \binom nk (-1)^{n-k}}{h^n}$$ exists for all $x\in U$. Does it follow that $f$ is $n$ times differentiable? If yes, is there an easy proof or a reference that proves this? REPLY [6 votes]: First, let's write out what your expression requires when $n = 2$: $$ \lim_{h \to 0} \frac{f(x) + f(x + 2h) - 2 f(x+h)}{h^2} $$ is required to exist for all $x \in U$. Let $U = (-1,1)$. Take $f(x) = |x|$. When $x \neq 0$, there exists a sufficiently small interval $I_x$ around $x$ such that $f|_{I_x}$ is $C^2$, and clearly the expression evaluates to $0$. When $x = 0$, you have that for any $h$: $$ \frac{f(0) + f(2h) - 2 f(h)}{h^2} = 0 $$ and hence the limit exists and equals 0. The absolute value function is clearly not twice differentiable at 0. A little bit more analysis: Let $D_{h} f(x)$ be the difference quotient of the function $f$ $$ D_h f(x) = \frac1h (f(x+h) - f(x)) $$ The expression you wrote down is $$ \underbrace{D_h D_h D_h \cdots D_h}_{n \text{ times}} f(x) $$ $n$-times differentiability requires the limit $$ \lim_{h_n \to 0} \lim_{h_{n-1}\to 0} \cdots \lim_{h_1\to 0} D_{h_n} \cdots D_{h_1} f(x) $$ to exist. Your condition only requires the limit to exist along the diagonal, so can be quite far from enough. REPLY [4 votes]: $\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}$Let $$f(x):=\sum_{n\in\Z}2^{-n}K\Big(\frac{x-2^{-n}}{2^{-n-2}}\Big) =\sum_{n\in\Z}2^{-n}K\Big(\frac{x-2^{-n}}{2^{-n-2}}\Big)1(x\in2^{-n}[3/4,5/4]),$$ where $K$ is a smooth function such that $K(0)=1$ and $K(x)=0$ if $|x|\ge1$. Note that the intervals $2^{-n}[3/4,5/4]$ are disjoint for distinct integers $n$. So, $f$ is smooth on $\R\setminus\{0\}$, $f(0)=0$, and $f(2x)=2f(x)$ for all real $x$. So, for $n=2$, $$\lim_{h\to 0}\frac{\sum_{k=0}^n f(x+k h)\binom nk (-1)^{n-k}}{h^n}$$ exists for all $x\in\R$, but $f$ is not even differentiable at $0$, because $f(2^{-k})=2^{-k}$ and $f(\frac34\,2^{-k})=0$ for all natural $k$. REPLY [4 votes]: The problem is that the existence of the limits only sees how well the function is approximated but it does not see the derivatives. A simple example is something like $f(x)=\exp(-1/x^2)\sin(\exp(1/x^4))$: This is extremely small near zero and the limit for $x=0$ in your condition is $0$. At other points, the limit exists because the function is smooth there. However, the derivatives explode at $0$.<|endoftext|> TITLE: Categorical semantics of universe levels in dependent type theory QUESTION [6 upvotes]: I know that locally closed cartesian categories provide categorical semantics for type theories with dependent products. What kind of categories model type theories with infinite universe hierarchies (cumulative or non-cumulative)? REPLY [3 votes]: A universe in a category with display maps is a specified display map $\tilde U \to U$ (cf. Section 5.5 of Taylor's thesis Recursive Domains, Indexed Category Theory and Polymorphism, or Section 9.6 of Taylor's Practical Foundations of Mathematics; the notation in this answer follows Voevodsky's A C-system defined by a universe in a category). Recall that objects represent contexts, and display maps represent projections from extended contexts, and so this display map can be interpreted as a term $X : U, x : X \vdash X : U$. We can extend a context $\Gamma$ by a $U$-small type $A$ (that is, a type in the universe $U$) by taking a pullback along the universe: The morphism $\Gamma, a : A \to \tilde U$ is viewed as picking out the type $A$ and the term $a : A$. Conversely, any display map $\Gamma, a : A \to \Gamma$ that fits into a pullback square along the universe may be seen as an extension by a $U$-small type in this way. We can therefore define a hierarchy of (non-cumulative) universes to be a family of display maps $\{ \tilde U_i \to U_i \}_{i \in \mathbb N}$ such that each $U_i$ is $U_{i + 1}$-small. One may additionally want to impose axioms on these universes asserting that they are closed under the type constructors appropriately. Modulo coherence issues, a locally cartesian-closed category is a display map category, and so dependent type theories with dependent sums, dependent products, extensional identity types and universes are modelled by locally cartesian-closed categories with universes. (Mike Shulman points this out in the comments, but it's surprisingly difficult to find a definition of universe in this setting, so I thought it would be helpful to spell out.)<|endoftext|> TITLE: Recovering an abelian category from the Ext of its simple objects QUESTION [14 upvotes]: Let $C$ be an abelian category, assume for simplicity that $C$ is enriched over $Vect_k$ (vector spaces over $k$) for some fixed field $k$. Suppose also that $C$ is both Artinian and Noetherian, so that for any object $X$ there is a sequence of objects $0=X_0 \hookrightarrow \ldots \hookrightarrow X_n = X$ with $X_i/X_{i-1}$ simple. Finally suppose that $C$ has enough injective/projective objects so that $\operatorname{Ext}_C$ can be defined. Given $C$, we build a new category $S$, enriched over graded $k$-vector spaces, in the following way: The objects of $S$ are the simple objects of $C$ If $X,Y\in Ob(S)$ then $\operatorname {Hom}_S(X,Y) = \bigoplus_{n\geq0}\operatorname{Ext}^n_C(X,Y)$ Compositions of morphisms are defined using the natural maps $\operatorname{Ext}^n_C(X,Y)\otimes \operatorname{Ext}^m_C(Y,Z)\to\operatorname{Ext}^{n+m}_C(X,Z)$ My question is: Can we recover $C$ from $S$ (say up to equivalence)? Assuming the answer is "yes", I guess that there is an analogue for when $C$ is only enriched over $Ab$, maybe if we redefine $S$ so that $\operatorname{Hom}_S(X,Y)=\operatorname{Hom}_{D(C)}(X,Y)$ or something REPLY [18 votes]: Here's a counterexample that appears in nature. Fix a prime $p$ and a field $k$ of characteristic $p$, and let $G=C_{p^{n}}$ be a cyclic group of order $p^{n}$ (where $n\geq1$ if $p$ is odd, and $n\geq2$ if $p=2$). Then the category $\operatorname{mod}kG$ of finitely generated $kG$-modules has only one simple module: the trivial module $k$. As graded $k$-algebras $$\operatorname{Ext}^{\ast}_{kG}(k,k)=H^{\ast}(G,k)\cong k[s,t]/(s^{2}),$$ with $s$ and $t$ in degree $1$ and $2$ respectively, independent of $n$. But $\operatorname{mod}kG$ determines $n$, as $p^n$ is the length of an indecomposable projective module. So, for example, $\operatorname{mod}\mathbb{F}_{2}C_{4}$ and $\operatorname{mod}\mathbb{F}_{2}C_{8}$ are not equivalent, but have the same $\operatorname{Ext}$-algebra. Alternatively, this generalizes to looking at $\operatorname{mod}k[x]/(x^{m})$ for any field $k$ and varying values of $m\geq3$. So a slightly smaller counterexample is $\operatorname{mod}k[x]/(x^{3})$ and $\operatorname{mod}k[x]/(x^{4})$.<|endoftext|> TITLE: On the connection between sums of prime numbers and distribution of prime numbers QUESTION [6 upvotes]: As an amateur mathematician, I have always been fascinated by the magic of prime numbers, and their apparently random distribution. I was utterly amazed when I found the following connection between sums of prime numbers and its distribution: Theorem 1 Let us define the prime counting function up to a given natural number $n$ as $\pi(n)$, and the sum of consecutive prime numbers up to the integer part of the square root of a given natural number $n$ as $S(n)=\sum_{p\leq\sqrt{n}}p$. Then, we have that $S(n)\sim\pi(n)$. Proof By partial summation $$S(n)=(\left\lfloor \sqrt{n}\right\rfloor \pi(\sqrt{n}))-\sum_{m=2}^{\left\lfloor \sqrt{n}\right\rfloor -1}\pi(m)$$ Where $\left\lfloor \sqrt{n}\right\rfloor$ denotes the integer part of $\sqrt{n}$. By the Prime Number Theorem with error term, there exists a constant $C$ such that $$\Big|\pi(x)-\frac{x}{\log x}\Big|\le C\frac{x}{\log^{2}x}\qquad\text{for }x\ge2$$ Therefore, substituting $\pi(\sqrt{n})$ and $\pi(m)$ by the application of the Prime Number Theorem, we get that $$S(n)=(\left\lfloor \sqrt{n}\right\rfloor \frac{\sqrt{n}}{\log(\sqrt{n})})-\sum_{m=2}^{\left\lfloor \sqrt{n}\right\rfloor -1}\frac{m}{\log(m)}+O\left(\frac{n}{\log^{2}(\sqrt{n})}\right)$$ Applying Riemman Sums theory, we get that $$\sum_{m=2}^{\left\lfloor \sqrt{n}\right\rfloor -1}\frac{m}{\log(m)}=\int_{2}^{\left\lfloor \sqrt{n}\right\rfloor }\frac{x}{\log\left(x\right)}dx+O\left(\frac{n}{\log^{2}(\sqrt{n})}\right)$$ Solving the integral by partial integration, we have that $$\int_{2}^{\left\lfloor\sqrt{n}\right\rfloor}\frac{x}{\log\left(x\right)}dx=$$ $$=\frac{n}{2\ln\left(\left\lfloor\sqrt{n}\right\rfloor \right)}+O\left(\frac{n}{\log^{2}(\sqrt{n})}\right)$$ It is easy to see that $$\frac{n}{2\log\left(\left\lfloor \sqrt{n}\right\rfloor \right)}\sim\frac{n}{\log n}$$ Thus, $$\sum_{m=2}^{\left\lfloor \sqrt{n}\right\rfloor -1}\frac{m}{\log(m)}\sim\frac{n}{\log\left(n\right)}+O\left(\frac{n}{\log^{2}(\sqrt{n})}\right)$$ It can be seen that $$\left\lfloor \sqrt{n}\right\rfloor \frac{\sqrt{n}}{\log(\sqrt{n})}\sim\frac{n}{\log\left(\sqrt{n}\right)}=\frac{n}{\frac{1}{2}\log\left(n\right)}=\frac{2n}{\log\left(n\right)}$$ Substituting, we have that $$S(n)\sim\frac{2n}{\log\left(n\right)}-\frac{n}{\log\left(n\right)}+O\left(\frac{n}{\log^{2}(\sqrt{n})}\right)$$ As $$\frac{2n}{\log\left(n\right)}-\frac{n}{\log\left(n\right)}=\frac{n}{\log\left(n\right)}$$ Thus, $$S(n)\sim\frac{n}{\log\left(n\right)}$$ And subsequently, as by the Prime Number Theorem $\pi(n)\sim\frac{n}{\log(n)}$ it can be stated that $S(n)\sim\pi(n)$ After reaching this nice result, I wondered if it could happen that $S(n)=\pi(n)$, and found that the answer was YES. In fact, as I found that there were many cases such that $S(n)=\pi(n)$, I conjectured that this equality occured infinitely often. And thanks to GH from MO, the conjecture was proven here for $n$ being prime, and thus we can state that Theorem 2 $S(n)=\pi(n)$ infinitely often As the prime counting function up to some composite number equals the prime counting function up to the inmediate prior prime number, considering the set $M=\{m_{1},m_{2},...,m_{k}\}$ as the set of values of $n$ such that $\pi\left(n\right)=S\left(n\right)$ and $n$ is some prime number, if $\pi\left(m_{k}\right)=S\left(m_{k}\right)$, then, as $\pi\left(m_{k}\right)=\pi\left(m_{k}+1\right)=\pi\left(m_{k}+2\right)=...=\pi\left(m_{k+1}-1\right)$, it follows that all the composite numbers between $m_{k}$ and $m_{k+1}$ are intersection points. A pending question to be solved is that he first value of $n$ with a concrete $\pi\left(n\right)=S\left(n\right)$ seems to be always a prime number. This conjecture assumes the truth of the following Conjecture. It does not exist any squared prime number $p_{k}^{2}$ such that $\pi\left(p_{k}^{2}\right)=S\left(p_{k}^{2}\right)$ except of $p_{1}=2$. That is, $\pi\left(p_{k}^{2}\right)\neq\sum_{i=1}^{k}p_{i}$ The conjecture has been tested and found to be true for the first thousands of primes. As if $\pi\left(p_{k}^{2}\right)=S\left(p_{k}^{2}\right)$ then necessarily $\pi\left(p_{n}\right)-S\left(p_{n}\right)=p_k$, where $p_n$ is the prime number inmediately lesser than $p_{k}^2$, one way to prove the conjecture would be to improve the result proved by GH from MO $$\pi(x)-S(x)=\Omega_\pm(x^\frac{1}{2})$$ to $$\pi(x)-S(x)\leq{x^\frac{1}{2}}$$ for all $x\in\mathbb{P}$, or at least for those prime numbers inmediately lesser than $p_{k}^{2}$. Now, many further questions regarding the connection between sums of prime numbers and its distribution remain unanswered, and honestly I do not really know how to deepen more on them. Do the two theorems exposed imply that the number of primes between $p_{n}^{2}$ and $p_{n+1}^{2}$ , on average, do not differ much from $p_{n+1}$? Does this connection add any new insight on the distribution of prime numbers, and it is worthy to be studied and exploited further to know more about the distribution of prime numbers? Is the result $$\pi(x)-S(x)\leq{x^\frac{1}{2}}$$ for all $x\in\mathbb{P}$, or at least for those prime numbers $p_n$ inmediately lesser than $p_{k}^{2}$ provable? Does it exist some counterexample? I would appreciate your comments and ideas. Thanks in advance! REPLY [4 votes]: The post that you quoted contains a proof that $$\pi(x)-S(x)=\Omega_\pm(x^c)\quad\text{for any}\quad c<3/4.$$ Equivalently, for any $c<3/4$ and any $C>0$, both $\pi(x)-S(x)\leq Cx^c$ and $\pi(x)-S(x)\geq -Cx^c$ are false.<|endoftext|> TITLE: Real analyticity of continuous function via restriction to analytic curves QUESTION [6 upvotes]: Suppose $X\subset \mathbb R^n$ is an irreducible real analytic sub-variety (i.e. the set of solutions of a system $f_1=\ldots=f_k=0$ with $f_i$ analytic) Let $x\in X$ be a point and let $F: X\to \mathbb R^1$ be a continuous function defined on $X$ in a neighbourhood of $x$. I want to understand whether $F$ is real analytic on $X$. The question is whether the following would be sufficient to know. Property. Suppose that for any real analytic map $\varphi: (-1,1)\to X$ sending $0$ to $x$ the composition $F\circ \varphi$ is analytic on $(-1,1)$. Question. Does it follow from the property that $F$ is real analytic in a neighbourhood of $x$? I am interested both in positive statements in this direction (possibly strengthening the condition of the Property) and in counterexamples. REPLY [8 votes]: No. Take $X = \mathbb R^2$ and $F(x,y) = \frac{x^3}{x^2+y^2}$. Then $F$ is real-analytic everywhere but $(0,0)$. Moreover, at $(0,0)$, any curve passing through $(0,0)$ must have coordinates two analytic functions $x,y$ vanishing to orders $a,b$, in which case $x^2+y^2$ vanishes to order $2\min(a,b)$ and $x^3$ vanishes to order $3a > 2 \min(a,b)$ so the ratio is a well-defined analytic function. But $F$ is not analytic at $(0,0)$. A similar trick can be used to construct worse functions, like the irrational $F(x,y) = \frac{ x^5}{ \sqrt{ (x^2+y^2) (x^2 + 2y^2)}} $ and, by summing terms of this form, functions that fail to be analytic at many points.<|endoftext|> TITLE: Paths $tg_1+(1-t)g_0$ in the moduli space of Riemann surfaces QUESTION [7 upvotes]: Suppose $S$ is a smooth compact oriented surface without boundary. Let $g_0$ and $g_1$ be two smooth Riemannian metrics on $S$. Consider the interpolating path of metrics $g_t=g_1t+g_0(1-t)$. Recall that a Riemannian metric $g$ on an oriented surface defines a unique (integrable) almost complex structure $J$ satisfying $J(g)=g$, $J^2=-1$. So we get a path $\gamma: [0,1]\to M$ in the moduli space of Riemann surfaces. Question. Is it true that $\gamma$ is a real analytic path in $M$? If so, how can I convince myself in this? (the statement strikes me as counter-intuitive...) (We recall that the moduli space of Riemann surfaces (of fixed genus) has a natural real analytic structure (for example given by Fenchel–Nielsen coordinates)) PS. The comment of abx below suddenly makes this statement much more plausible for me. Indeed if we look at the path of $J_t$, then at each point $x\in S$, ${J_t}_x$ depends analytically on $t$ (by obvious linear algebra). But still, how to go from here to saying that the path in the moduli space is real analytic? REPLY [4 votes]: Answering your PS: as you point out, the complex structure $J_t$ is given in each coordinate chart by a matrix which depends real-analytically on $t$. Now you can find a neighborhood $U$ of $[0,1]$ in $\mathbb{C}$ and extend $J_t$ as a holomorphic function of $t\in U$; by analytic continuation it will still be a complex structure. Covering $S$ by finitely many coordinate charts and taking the intersection of the corresponding $U$ you get a holomorphic family of complex structures $J_t$ on $S$, for $t$ in a certain neighborhood $V$ of $[0,1]$ in $\mathbb{C}$. This gives a holomorphic map $V\rightarrow M$ which extends your path $\gamma$, hence $\gamma$ is real-analytic. Edit: To see that the classifying map $V\rightarrow M$ is holomorphic, define an almost complex structure $\mathscr{J}$ on $V\times S$ as follows. For $(t,x)$ in $V\times S$, $\mathscr{J}_{(t,x)}$ is the endomorphism $(I(t),J_{t}(x))$ of $T_{(t,x)}(V\times S)=T_t(V)\oplus T_x(S)$, where $I$ is the standard complex structure on $\Bbb{C}$. Integrability should not be difficult to check, since the action on $T(V)$ and on $T(S)$ are separated. Thus we have a complex structure on $X:=V\times S$, and the projection $X\rightarrow V$ is holomorphic. So this gives a holomorphic family of Riemann surfaces, hence a holomorphic map to the moduli space.<|endoftext|> TITLE: Are the trace relations among matrices generated by cyclic permutations? QUESTION [10 upvotes]: Let $X_1,\dots,X_n$ be non commutative variables such that $\operatorname{tr} f(X_1,\dots,X_n) = 0$ whenever the $X_i$ are specialized to square matrices in $M_r(k)$ for any $r \geq 1$. Does this imply that $f$ is in the ideal generated by cyclic permutations: $g_1\dots g_k - g_2\dots g_k g_1$ for any polynomials $g_i$ in the $X_i$ and $k \geq 2$? (And if I have missed any obvious relations, is the statement true up to adding in those relations to the ideal?) REPLY [2 votes]: For simplicity, let us assume that $K=\mathbb{C}$ (though, these results apply to any field of characteristic zero). We will use capital letters like $X,Y,Z$ to denote matrices while lower case letters like $x,y,z$ shall denote variables. We will use the fact that the trace of matrices produces an inner product on $M_{n}(\mathbb{C})$ defined by $\langle A,B\rangle=\operatorname{Tr}(AB^{*})$. If $F$ is a field, then let $F\langle x_{1},\dots,x_{n}\rangle$ denote the ring of non-commutative polynomials over $F$ in the variables $x_{1},\dots,x_{n}$. We will now go over a few lemmas. Lemma: Suppose that $f,g\in F\langle x_{1},\dots,x_{n}\rangle$. Then $f=g$ if and only if whenever $r\geq 1$ and $X_{1},\dots,X_{n}\in M_{r}(F)$, we have $f(X_{1},\dots,X_{n})=g(X_{1},\dots,X_{n})$. Proof: Suppose that $f\neq g$, and $u>\text{Deg}(f)+\text{Deg}(g)$. Let $V$ be a finite dimensional vector space over $F$ with linearly independent set $$(e_{i_{1},\dots,i_{k}}|0\leq k\leq u,i_{1},\dots,i_{k}\in\{1,\dots,n\})$$ You can let $X_{1},\dots,X_{n}:V\rightarrow V$ be linear maps such that $X_{i}(e_{i_{1},\dots,i_{k}})=e_{i,i_{1},\dots,i_{k}}$ whenever $k TITLE: Log-concavity of matroids: characterization of equality? QUESTION [14 upvotes]: Let $M$ be a (loopless) matroid of rank $r$. The characteristic polynomial $\chi_M(x)$ is defined by $\chi_M(x)=\sum_{F \in \mathcal{L}(M)}\mu(\hat{0},F) \cdot x^{\mathrm{rk}(F)}$, where $ \mathcal{L}(M)$ is the lattice of flats of $M$ and $\mu$ its Möbius function. It is known that the signs of the characteristic polynomial alternate, and the so-called Whitney numbers of the 1st kind $\omega_i$ are defined by $\chi_M(x) = \sum_{i=0}^{r} (-1)^i \omega_i x^i$. In Hodge theory for combinatorial geometries, affirming a long-standing conjecture of Rota and others, Adiprasito-Huh-Katz (AHK) showed that these $\omega_i$ form a log-concave sequence, i.e., $\omega_i^2 \geq \omega_{i-1}\omega_{i+1}$. Actually, they proved something stronger. It is known that $(1-x)$ is always a factor of $\chi_M(x)$, and the reduced Whitney numbers $\overline{\omega}_i$ are thus defined by $\chi_M(x)/(1-x)=\sum_{i=0}^{r-1} (-1)^i \overline{\omega}_i x^i$. AHK showed that in fact the reduced Whitney numbers form a log-concave sequence: $\overline{\omega}_i^2 \geq \overline{\omega}_{i-1}\overline{\omega}_{i+1}$ (which implies log-concavity of the $\omega_i$ by a straightforward argument). In fact, an easy reduction using the truncation of the matroid $M$ shows that it is enough to prove log-concavity of the $\overline{\omega}_i$ in the "last-spot": i.e., that $\overline{\omega}_{r-2}^2 \geq \overline{\omega}_{r-3}\cdot\overline{\omega}_{r-1}$. To prove this, AHK use the Chow ring $A(M) = A^0(M)\oplus A^1(M)\oplus \cdots\oplus A^{r-1}(M)$ of the matroid $M$: they show that $\overline{\omega}_k=\langle \alpha^{r-1-k}\beta^{k}\rangle$ for certain linear elements $\alpha,\beta\in A^1(M)$ of the Chow ring, where $\langle \cdot \rangle\colon A^{r-1}(M)\to \mathbb{R}$ is the canonical degree map isomorphism; and they deduce $\overline{\omega}_{r-2}^2 \geq \overline{\omega}_{r-3}\cdot\overline{\omega}_{r-1}$ from the so-called Kähler package for $A(M)$, in particular, the Hodge-Riemann relations, which imply that the relevant $2\times 2$ determinant is nonpositive. Question: From the work of AHK (or elsewhere) is it possible to deduce when (i.e. for which matroids) we have an equality $\overline{\omega}_{r-2}^2 = \overline{\omega}_{r-3}\cdot\overline{\omega}_{r-1}$? (In other words, when the determinant of the $2\times 2$ HR relations matrix is $0$ rather than negative?) REPLY [6 votes]: I have nothing to add to Hunter's answer to your main question, but I thought it might be helpful to comment more generally on where the difficulty lies in extracting such information from a Kähler package. Let us adopt the general setting in this review of Huh. Let $A(X)=\bigoplus_{q=0}^d A^q(X)$ be a graded algebra with Kähler cone $\mathrm{K}(X)\subset A^1(X)$, satisfying Poincare duality, hard Lefschetz (HL), and Hodge-Riemann (HR). A simple consequence of (HR) is that for any $\mathrm{L}_0,\ldots,\mathrm{L}_{d-2}\in\mathrm{K}(X)$ and $\eta\in A^1(X)$, we have $$(\eta \mathrm{L}_0\cdots \mathrm{L}_{d-2})^2\ge (\eta^2\mathrm{L}_1\cdots \mathrm{L}_{d-2})\, (\mathrm{L}_0^2\mathrm{L}_1\cdots \mathrm{L}_{d-2}).$$This is what gives rise to log-concavity in various applications. Moreover, (HL) implies that in this inequality, equality holds if and only if $\eta$ is proportional to $\mathrm{L}_0$. There are various ways to see this. For example, if one has equality in the above inequality, then the quadratic polynomial $$q(t) = ((\eta+t\eta') \mathrm{L}_0\cdots \mathrm{L}_{d-2})^2-((\eta+t\eta')^2\mathrm{L}_1\cdots \mathrm{L}_{d-2})\, (\mathrm{L}_0^2\mathrm{L}_1\cdots \mathrm{L}_{d-2})$$satisfies $q(0)=0$ and $q(t)\ge 0$ for all $t$ and $\eta'\in A^1(X)$. Computing $q'(0)=0$ shows that a linear combination of $\eta$ and $\mathrm{L}_0$ lies in the kernel of the map $\mathrm{L}_1\cdots \mathrm{L}_{d-2}:A^1(X)\to A^{d-1}(X)$. But (HL) states this map is a bijection, so its kernel is trivial, and thus $\eta$ and $\mathrm{L}_0$ are proportional. The difficulty is that in many applications, the $\mathrm{L}_i$ of interest do not lie in $\mathrm{K}(X)$, but can only be approximated by elements of $\mathrm{K}(X)$. This is the case in AHK (see, e.g., the last section of this Notices paper), in the Khovanskii-Teissier inequality when one deals with nef classes rather than ample classes, etc. In such cases, the inequality still follows by approximation, but all information on the equality cases is lost. While no nontrivial equality cases turn out to appear in the particular case of AHK (as per Hunter's answer), many new equality cases can appear for the limiting inequalities in other situations. As far as I know, the only structures of this kind where the equality cases of the limiting inequalities are fully understood (in the sense that there is a complete geometric/combinatorial characterization) are mixed discriminants and mixed volumes of convex polytopes. The last section of the latter paper contains some speculative remarks on what such results might look like more generally.<|endoftext|> TITLE: Is $\operatorname{Fun}^\text{small}(\operatorname{Fun}(C,\mathsf{Set}),\mathsf{Set})$ total when $C$ is small? QUESTION [7 upvotes]: Let $\mathcal C$ be a small category. Then it is known that the category $\operatorname{Fun}^\text{small}(\operatorname{Fun}(\mathcal C,\mathsf{Set}),\mathsf{Set})$ of functors $\operatorname{Fun}(\mathcal C,\mathsf{Set}) \to \mathsf{Set}$ which are small colimits of representable functors is cocomplete and (less obviously) complete. For instance, this is shown in ABLR - A classification of accessible categories, and also in Day and Lack - Limits of small functors. Question: Is this category total? Is it cototal? (Totality is a property which is implied by local presentability and which implies both completeness and cocompleteness. I am fairly certain that the category in question is not locally presentable unless $\mathcal C$ is empty, but if it were total, it would be “almost as good” as being locally presentable.) One sufficient criterion for totality is for a category to be cocomplete, be epi-complete, and have a generator. So it would be useful to know along the way whether this category has a generator, is cowellpowered, etc. To simpify notation, write $P(\mathcal C) = \operatorname{Fun}^\text{small}(\mathcal C^{op},Set)$ and $L(\mathcal C) = P(\mathcal C^{op})^{op}$. In these terms, the question asks, for $\mathcal C$ small, whether $P(L(\mathcal C))$ is total. REPLY [2 votes]: I think one can see that even when $\mathcal C = 1$, the category $P(L(1)) = Acc(Set)$ of accessible functors $Set \to Set$ is not total or cototal, or even compact or cocompact in the sense of Isbell. In other words, there exist continuous functors $\Phi: Acc(Set)^{op} \to Set, \Psi: Acc(Set) \to Set$ which are not representable: Proof of Non-cototality / cocompactness: For sufficiently large regular cardinals $\kappa$, choose some $F_\kappa \in Acc(Set)$ such that $F_\kappa(X) = 1$ for $|X|<\kappa$, but $|F_\kappa(X)| > 1$ for $|X| = \kappa$. For instance, we may take $F_\kappa(X) = X^\kappa / X^{<\kappa}$ to be the set of $\kappa$_sequences in $X$ modulo those which are eventually constant. Then let $\Phi: Acc(Set)^{op} \to Set$ be defined by $\Phi(G) = \prod_\kappa Nat(G,F_\kappa)$. This functor is well-defined because if $G$ is $\lambda$-accessible, then $Nat(G,F_\kappa) = 1$ for $\kappa \geq \lambda$, so that the indicated product is small. Moreover, as a product of continuous functors, this functor is continuous. However, this functor is not representable. If it were, then by testing with the objects $G = Hom(X,-) \in Acc(Set)$, we would see that the representing functor is $\prod_\kappa F_\kappa: Set \to Set$. But this functor is not $\lambda$-accessible for any $\lambda$, since it has as a retract the functor $\prod_{\kappa \geq \lambda} F_\kappa$, which is clearly not $\lambda$-accessible. Proof of Non-totality / compactness: As before, set $F_\kappa(X) = X^\kappa / X^{<\kappa}$ for each regular $\kappa$, and let $\Psi(G) = \prod_{\kappa,G(\ast)} Nat(F_\kappa,G)$ to be the large fiber product of the representables on these functors, with the fiber product taken over the functor $ev_\ast$ which evaluates at a point. Note that if $G$ is $\lambda$-accessible, then for $\kappa \geq \lambda$ we have that $G(\kappa) = \varinjlim_{\alpha < \kappa} G(\alpha) \to \varprojlim_{\alpha < \kappa} G(\alpha)$ is injective. It follows that $Nat(F_\kappa, G) \to G(\ast)$ is likewise injective for $\kappa \geq \lambda$. Therefore $\Psi(G)$ is always a small set. Moreover, as a product of representable functors, $\Psi$ is continuous. Further, the unique map $\kappa \to \ast$ induces a map $ev_\ast \to \Psi$, which is a section of the canonical map in the other direction. Using this map, we also see that for each regular $\lambda$, the representable on $F_\lambda$ (which is not $\kappa$-accessible for $\kappa < \lambda$) is a retract of $\Psi$. Thus $\Psi$ is not accessible, and therefore not corepresentable.<|endoftext|> TITLE: Nowhere negative polynomials form a semialgebraic set QUESTION [5 upvotes]: Let $P_{d, n}$ be the space of polynomial maps $\mathbb{R}^n\to \mathbb{R}$ of degree at most $d$. Is the subset $S\subset P_{d, n}$ of nowhere negative polynomials semialgebraic? REPLY [10 votes]: As I said in the comments this is very well known: $S$ is the complement of the projection of the semialgebraic set $\{(f,a)\in P_{d,n}\times\mathbb{R}^n:f(a)<0\}$, hence semialgebraic by the Tarski-Seidenberg theorem. REPLY [5 votes]: The answer is yes. Recall first that the sums-of-squares polynomials of degree at most $D$ and in $n$ variables form a spectrahedron, and in particular a semialgebraic set. Now if $f \in P_{d,n}$ is a positive¹ polynomial, then by Hilbert's 17th problem and Artin's solution there are sums-of-squares polynomials $g$ and $h$ such that $f = \frac{g}{h}$. It is possible to bound the relevant degrees of $g$ and $h$ in terms of $n$ and $d$. (I'm not sure what the best currently known degree bounds are, but the linked 2014 paper gives a tower of five exponentials!) Thus the $f \in P_{d,n}$ are characterized in terms of the equations $hf = g$ and $h \neq 0$ for sums-of-squares polynomials $g$ and $h$ with a given number of unknown coefficients. Hence $P_{d,n}$ is the projection of a semialgebraic set, and by Tarski's theorem therefore semialgebraic itself. ¹ "Positive polynomial" is the standard term for nowhere negative polynomial.<|endoftext|> TITLE: First use of term "Hilbert's Nullstellensatz" QUESTION [33 upvotes]: The post below first appeared on hsm.stackexchange over a week ago and has received no answers there yet, so by now I think it is okay to ask it here. This year (2021) marks the 100th anniversary of Emmy Noether's 1921 paper in which she introduced Noetherian rings and proved the primary ideal decomposition for them. The original version of her paper is here and an English translation is on the arXiv here. She of course does not call rings "Noetherian" but instead refers to rings satisfying "the finiteness condition" (die Endlichkeitsbedingung). When referring to results of Hilbert that we'd call Hilbert's basis theorem and Hilbert's Nullstellensatz at the start of Section 10 of the paper, she writes "Hilbert’s Module Basis Theorem" (Hilbertschen Theorem von der Modulbasis) and "a famous theorem of Hilbert’s" (... eines bekannten Hilbertschen Satzes). In two footnotes on this page she also refers to "Hilbert's theorem" instead of the Nullstellensatz. The Nullstellensatz was published by Hilbert in 1893 here (see the theorem starting on the bottom of p. 320), and that was nearly 30 years earlier. I am surprised that Noether refers to that result simply as "Hilbert's Theorem". Thus my question: What is the earliest known reference to Hilbert's theorem as the Nullstellensatz or Hilbert's Nullstellensatz (in German, presumably)? The answer is no later than 1927, when a paper by van der Waerden here refers to Hilbert's Nullstellensatz on the top of p. 202. This is the earliest of two papers mentioned in Zentralblatt that has "Nullstellensatz" in a review. The other earliest paper in Zentralblatt is by Grete Hermann from 1926 here, which refers to Hentzelt's Nullstellensatz in the 2nd and 3rd paragraphs of the third page (Hentzelt's Nullstellensatz also appears in van der Waerden's paper on p. 203). I didn't find her mentioning Hilbert's Nullstellensatz. Hermann's paper is in English translation here -- see the first column of the second page. The paper by Hentzelt is from 1923 here (it was edited by Noether since Hentzelt went missing in Belgium in 1914 during WW1), and I did not find the term "Hilbert's Nullstellensatz" used in it. For what it's worth, the earliest article in Zentralblatt with Nullstellensatz in a title is the paper by Rabinowitsch here (1930) in which he used his trick to prove the Nullstellensatz. That is also the earliest paper in MathSciNet's records with Nullstellensatz appearing anywhere. It's not from a real MathSciNet review since it predates the birth of MathSciNet (well, the birth of Mathematical Reviews) in 1940. The earliest real review in MathSciNet mentioning the Nullstellensatz is Zariski's paper from 1947 with his new proof of the Nullstellensatz by "Zariski's lemma". REPLY [15 votes]: Below is the Dutch paper mentioned by Francois Ziegler; This paper did indeed appear before the 1927 paper mentioned in the OP, but Van der Waerden does refer to that forthcoming publication in a footnote; I translate: This theorem is a special case of the "Nulpuntenstelling" of HILBERT$^{5})$ $^{5})$ Hilbert assumes that $\Omega$ is the field of complex numbers; his proof actually holds more generally. For a different proof see my forthcoming paper in the Mathematische Annalen entitled "Zur Nullstellentheorie der Polynomideale". The Mathematische Annalen paper was submitted 14 August 1925, while the paper below was read at the Netherlands Academy on 28 November 1925, which explains why Van der Waerden could refer to former as a an earlier work. B.L. Van der Waerden, Een algebraies kriterium voor de oplosbaarheid van een stelsel homogene vergelijkingen, Verslag van de Gewone Vergadering van de Afdeling Natuurkunde van de Koninklijke Akademie van Wetenschappen 34, 1123-1130 (1925) [source] -- only accessible online from a US IP number. In response to a request in a comment: Van der Waerden wrote again on this topic in 1927, Neue Begründung der Eliminations– und Resultantentheorie, Nieuw Archief Wiskunde 15, 302–320 (1927) [source]<|endoftext|> TITLE: How to motivate constructible sheaves QUESTION [12 upvotes]: I'm writing some notes for some students which just finished a first course in scheme theory. There I would like to talk about constructible sheaves, but I found it hard to give a compelling motivation for these objects. (And I don't like to give a definition without at least trying to explain why this is a nice thing to consider.) Since the students mostly never saw what a stratification is, perhaps it is best to begin with local systems and then focus on the fact that the derived pushforward of a local system is not necessarily a local system, but constructible sheaves are closed under the usual six functors. I would like to know what the community here thinks of such approach and if there are other possible motivations. REPLY [19 votes]: Even if you're only interested in say cohomology with coefficients in the constant sheaf, working with constructible sheaves gives you extra flexibility and is more amenable to inductive proofs. Here is a basic theorem in the topology of algebraic varieties one of whose proofs could serve as a motivation. I discussed it in some MO answer, maybe I'll link it later, and I learned this from Lazarsfeld's book "Positivity in Algebraic Geometry" (volume 1). Theorem. Let $X\subseteq \mathbf{C}^n$ be an affine algebraic variety, i.e. a closed subset cut out by polynomial equations. Then $H^i(X, \mathbf{Z}) = 0$ for $i>\dim X$. Here $\dim X$ is the algebraic dimension, can be defined in various ways, see any textbook on algebraic geometry. The "dimension" (in whatever sense) of $X$ as a topological space is thus $2\dim X$. The above result is really special to affine varieties: the projective space $\mathbf{C}P^n$ has dimension $n$ and nonzero cohomology in even degrees $i\leq 2n$. One proof of the above result uses Morse theory (a careful analysis of $X$ by means of an auxiliary function $\mathbf{C}^n\to [0,\infty)$ such as $\sum |z_i|^2$), and shows a bit more: $X$ is homotopy equivalent to a CW complex with cells in dimensions $\leq n$ (the Andreotti-Frankel theorem). This is not the proof I'd like to mention. The proof which generalizes well to other contexts, e.g. to $\ell$-adic cohomology, due to Michael Artin, proceeds by induction on $d=\dim X$. By the Noether normalization lemma, there exists a finite morphism $$ f\colon X\to \mathbf{C}^d $$ ("finite" here is equivalent to "proper with finite fibers"). How is the cohomology of $X$ related to cohomology of $\mathbf{C}^d$? You really need sheaf cohomology to answer that. In this case (because the map is finite!) we obtain isomorphisms $$ H^i(X, \mathbf{Z}) \simeq H^i(\mathbf{C}^d, f_*\mathbf{Z}). $$ The sheaf $f_*\mathbf{Z}$ on the right hand side is no longer the constant sheaf, but can be shown to be a constructible sheaf. (Over a Zariski dense open subset of $\mathbf{C}^n$, it will be a locally constant sheaf of rank $e$ where $e$ is the degree of the finite map.) Therefore the theorem will follow from a more general statement below. Theorem. Let $F$ be a sheaf on $\mathbf{C}^d$, constructible with respect to a Zariski stratification. Then $H^i(\mathbf{C}^d, F)=0$ for $i>d$. Now we are able to proceed by induction on $d$: we take a linear projection $$ p\colon \mathbf{C}^d \to \mathbf{C}^{d-1}. $$ If $p$ is generic, then the cohomology of $F$ fits inside a long exact sequence (a form of the Leray spectral sequence): $$ \cdots \to H^i(\mathbf{C}^{d-1}, p_* F)\to H^i(\mathbf{C}^d, F) \to H^{i-1}(\mathbf{C}^{d-1}, R^1 p_* F)\to \cdots $$ Here $R^1 p_* F$ is the first higher push-forward, and (again for $p$ generic) the sheaves $p_* F = R^0 p_* f$ and $R^1 p_* F$ will have stalks which compute the cohomology of the fibers: $$ (R^i p_* F)_y = H^i(p^{-1}(y), F). $$ (I am telling this backwards: one shows the above formula for all $i$, and since we know the theorem for $d=1$, we know that $R^i p_* F = 0$ for $i>1$, and then we get the long exact sequence.) One can show that $p_* F$ and $R^1 p_* F$ are again constructible with respect to an algebraic stratification, and we proceed by induction. (The case $d=1$ still has to be done by hand.)<|endoftext|> TITLE: Normal numbers, Liouville function, and the Riemann Hypothesis QUESTION [11 upvotes]: This is a question about whether or not some number $\lambda^*$ is normal in base 2. More specifically, I am wondering if $\lambda^*$ is not normal. Proving it is normal would be next to impossible, and more difficult than proving the Riemann Hypothesis (RH), because it is known that its normality would imply RH. But this is not the purpose of my question. Even if $\lambda^*$ is not normal, it is conjectured that its binary digits behave in a way that is random enough, to make RH true. My question (see below) is essentially a probability theory question, and maybe not a difficult one. The number $\lambda^*$ is defined as follows. Let $\Omega(k)$ be the number of prime factors of $k$, including repetition. For instance $\Omega(2^5\cdot 11^7\cdot 19^1)=5+7+1=13$. Then $\lambda(k)=(-1)^{\Omega(k)}$ is known as the Liouville function, see here and here (these Wikipedia entries also show you the connection to the Riemann zeta function $\zeta$). Note that $\lambda$ is a multiplicative function, with $\lambda(p)=-1$ if $p$ is prime. Then $$\lambda^*=\frac{1}{2}\sum_{k=1}^\infty \frac{1+\lambda(k)}{2^k}=\frac{1}{2}\Big(1+\sum_{k=1}^\infty \frac{\lambda(k)}{2^k}\Big).$$ The binary digits of $\lambda^*$ are the numbers $\lambda(k)$ except that $-1$ is replaced by zero. It is known that $\lambda^*$ is irrational (see update at the bottom of this post; initially I claimed transcendental, but this was an error due to misreading some paper) the proportions of digits equal to $0$ or $1$ are both 50%, the sequence $\lambda(k)$ is conjectured to be ergodic, see here. (thanks to Will Sawin for clarifying this one is a conjecture, not a theorem) So $\lambda^*$ is simply normal (too weak to imply RH), but not necessarily normal (much stronger than needed to imply RH, if it was normal). Almost all irrational numbers (except a set of Lebesgue measure zero) are normal, but a proof that $\pi$ or $e$ (or $\lambda^*$ for that matter) is normal, still remains elusive, and it is probably more difficult to prove than RH. I suspect (and this is the purpose of my question, see below) that $\lambda^*$ is not normal, a fact much easier to prove. A normal number in base $2$ is a number with binary digits behaving like an infinite sequence of Bernoulli trials. In particular, successive digits appear to be independent, but because we are dealing with a deterministic sequence, this is akin to asymptotic independence of the joint empirical distribution. Note that the first $n$ binary digits of $\pi$ do not fully appear random (there are some little discrepancies). The same is true for $\lambda^*$. By contrast, the number $e$ is better behaved. But eventually, as $n$ increases, we are getting closer to pure randomness. One weird fact about $\lambda^*$ is that $\sum_{k=1}^n \lambda(k)$ is always negative until $n=906150257$ (see here) but then again unexpected things like that also happen with otherwise perfect random walks, see here.) My question Is it possible that $\lambda^*$ is not normal? It passes all the statistical tests I tried, yet what bothers me is this. Let's replace $\lambda(k)$ by a random variable $X_k$ defined as follows: $X_k=-1$ with probability $p$ if $k$ is prime, and $X_k=1$ with probability $1-p$ if $k$ is prime. In addition $X_{nm}=X_n\cdot X_m$. It is easy to show that the sequence is fully defined for all $k$'s, and generally, $X_{nm}$, $X_n$, $X_m$ are not independent (nor even asymptotically) and thus can not lead to normality in case of a deterministic sequence. Or is there something wrong with this argument? Implications for RH Even if $\lambda^*$ is not normal, there are no negative implications for RH, but this is worth a little discussion. Even if there are weak dependencies among the $\lambda(k)$'s, the law of the iterated logarithm might still apply to that sequence, see example here. An heuristic proof of RH, published in AMS (here) only assumes that the law of the iterated logarithm applies to the $\lambda(k)$'s, not that $\lambda^*$ is normal. The proof uses the Moebius function $\mu$ instead of $\lambda$, but arguments are otherwise identical (I think). But even the law of the iterated logarithm is more than we need. A weaker property (see below) is all that is needed to prove RH. $$\lim_{n \to \infty}\frac{\lambda\left(1 \right)+ \lambda\left(2 \right) + \dots + \lambda\left(n \right)}{n^{\frac{1}{2}+\epsilon}} =0,$$ for any $\epsilon >0$. This fact dates back to Liouville's thesis in 1899, see here. The law of the iterated logarithm, assuming it applies to the sequence $\lambda(k)$, would be this: $$\lim\sup_{n\rightarrow\infty} \frac{\Big|\sum_{k=1}^n \lambda(k)\Big|}{\sqrt{n\log\log n}} = C,$$ where $0 TITLE: continued fraction for logarithmic integral QUESTION [6 upvotes]: Does the logarithmic integral function $\operatorname{li}(x)$ have the continued fraction expansion $$\operatorname{li}(x) = \cfrac{x}{\log x -1 -{}} \ \cfrac{1}{\log x -3 -{}} \ \cfrac{4}{\log x -5 -{}}\ \cfrac{9}{\log x - 7-{}}\ \cfrac{16}{\log x - 9 -{}} \ \cfrac{25}{\log x - 11 -{}} \ \cdots$$ for $x > 1$? If so, is there a proof or a reference that proves it? This question is not answered by any reference I can find because standard results in the literature will verify the identity (appropriately interpreted) for complex values of $x$ excluding $x > 1$. For $x > 1$, I do not even know if the given continued fraction converges. It might help to note that the $n$th convergent of the continued fraction above is equal to $-\sum_{k = 1}^n \frac{1}{kL_k(x)L_{k-1}(x)}$, where $L_k(x)$ denotes the $k$th Laguerre polynomial (at least at values of $x$ that are not roots of any Laguerre polynomial). Thus, the question is equivalent to the following: does one have $\operatorname{li}(x) = -\sum_{k = 1}^\infty \frac{1}{kL_k(x)L_{k-1}(x)}$ for all real $x > 1$ that are not roots of any Laguerre polynomial? It is well known that the exponential integral $E_1(z)$ has the continued fraction expansion $$E_1(z) = \cfrac{e^{-z}}{z+1 -{}} \ \cfrac{1}{z+3 -{}} \ \cfrac{4}{z+5 -{}}\ \cfrac{9}{z+7-{}}\ \cfrac{16}{z+9 -{}} \ \cfrac{25}{z+11 -{}} \ \cdots, \quad z \in \mathbb{C} \setminus (-\infty,0].$$ A third equivalent formulation of the question is the following: For $z \in (-\infty,0]$, does the continued fraction above converge to $-\operatorname{li}(e^{-z}) = E_1(z)+\pi i$? EDIT: In all the references I can find, including the ones given in the proposed answer by Alexey Ustinov, the domain for which the given expansions hold exclude the domains I inquired about in my question. The domain $x> 1$ of $\operatorname{li}(x)$ is the domain number theorists care most about, and it would be nice if it had the proposed continued fraction expansion on that domain. FINAL EDIT: I now think it's more likely that the continued fraction diverges for $x > 1$, but I don't know how to prove this. REPLY [4 votes]: You can find this expansion in the book Lorentzen L. & Waadeland H. Continued fractions with applications North-Holland Publishing Co., 1992 (formula (4.3.10)). As I understand, this document is a more recent version of the appendix of this book. Here the desired formula has the number (3.3.10). However, the validity of the expansion for $\operatorname{li}(x)$ does not include the domain $x > 1$ that you seek.<|endoftext|> TITLE: What are the current research directions in the geometric theory of dynamical systems? QUESTION [6 upvotes]: By geometric theory of dynamical systems, I mean the kind found in the book by Palis, or papers like this one. In other words, dynamics on manifolds, but not specifically hyperbolic dynamical systems. What are some recommended papers, survey articles, lecture notes, or books to read to explore this topic further? I really like this flavour of dynamics and would like to know what the modern research directions/questions are. Thanks in advance! REPLY [4 votes]: I think the book by Bonatti, Diaz and Viana: "Dynamics beyond uniform hyperbolicity' can give you a nice overview of one possible point of view. https://link.springer.com/book/10.1007/b138174 The book by Katok-Hasselblatt and their Handbook contains a lot of other points of view. REPLY [3 votes]: Perhaps not the type of dynamical system in which you are interested, but an interesting example: Schwartz, Richard Evan. "The Farthest Point Map on the Regular Octahedron." Experimental Mathematics (2021): 1-12. Preliminary version: arXiv abs. The farthest point map associates to each point $p$ on the surface the set of points $\mathcal{F}_p$ that are furthest from $p$, with distance measured by shortest paths (geodesics segments). Of special interest are the points $p$ for which $\mathcal{F}_p$ is a single point. Even on the regular octahedron, the dynamics are quite intricate, but calculable.<|endoftext|> TITLE: Anti-concentration of Gaussian when conditioning on event QUESTION [5 upvotes]: Let $v$ be a given vector with $\|v\|_{\Sigma^{-1}} \leq 1$, where $\Sigma$ is a positive semi-definite matrix and $\|v\|_{\Sigma^{-1}} = \sqrt{v^\top\Sigma v}$. Meanwhile, let $u$ be a random vector drawn from $N(0,\Sigma^{-1})$. We know that for any given vector $\phi$, it holds that $$ P(\phi^\top u > \phi^\top v) > c $$ where $c$ is a positive absolute constant that depends on $v$. Question: Does a similar anti-concentration property still hold when we additionally condition on the event $ \mathcal{E} = \{u \in \mathcal{C}\} $, where $\mathcal{C}$ is a given set such that $v \in \mathcal{C}$ and moreover $v$ is in the interior of $\mathcal{C}$? In other words, does it hold that $$ P(\phi^\top u > \phi^\top v \,|\, \mathcal{E})> c', $$ where $c'$ is a positive absolute constant that depends on $v$ and $\mathcal{C}$? One particular example of $\mathcal{C}$ that I am interested in is $$ \mathcal{C} = \{u: \| u + w \| \leq 1 \}, $$ where $w$ is a given vector such that $\| v + w \| \leq 1-\delta$ with $\delta > 0$ (e.g., $\delta = 0.01$). REPLY [2 votes]: $\newcommand{\si}{\sigma}\newcommand{\Si}{\Sigma}\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}$The answer is: in general, no -- even for convex $\mathcal C$. Indeed, let $C:=\mathcal C=(-\infty,1)\times\R$, $v=(0,0)$, $f:=\phi=(1,0)$, and $\Si=\begin{pmatrix}\si^2&0\\0&1\end{pmatrix}$, with $\si\to\infty$. Then $$P(u\in C)\ge P\big(u\in(-\infty,0)\times\R\big)=1/2$$ and \begin{equation*} P(u\in C,f\cdot u>f\cdot v)=P(1>f\cdot u>0)\to0, \end{equation*} so that $P(f\cdot u>f\cdot v\,|\,u\in C)\to0$. This example is written for dimension $2$, but can be easily modified for any dimension $\ge1$. On a positive note, let us also show that the answer becomes yes if $C$ is assumed to be bounded (as in your "particular example") and, of course, $f\ne0$. Indeed, then there are real $\ep>0$ and $R>\ep$ (depending only on $C$ and $v$) such that $B_v(\ep)\subseteq C\subseteq B_v(R)$, where $B_v(r)$ is the open ball of radius $r$ centered at $v$. Replacing $u,v,C$ by $\Si^{1/2}u,\Si^{1/2}v,\Si^{1/2}C$, respectively, we see that it is enough to to prove the following: \begin{equation*} \frac{P(u\in E_\ep,f\cdot u\ge f\cdot v)}{P(u\in E_R)}\ge c_{\ep,R,n} \tag{1} \end{equation*} for some $c_{\ep,R,n}>0$ depending only on $\ep$, $R$, and $n:=\dim C$, where $u=(u_1,\dots,u_n)\sim N(0,I_n)$, $v=(v_1,\dots,v_n)$, $\|v\|\le1$, $\|\cdot\|$ is the Euclidean norm on $\R^n$, \begin{equation*} E_r:=E_{v,\Si,r}:=\{x=(x_1,\dots,x_n)\in\R^n\colon\sum_1^n(x_i-v_i)^2/\si_i^20$ without loss of generality $v_i\ge0$ and hence $P(s>u_i-v_i>0)\le P(-s0$ \begin{equation*} P(|u_i-v_i|0$. Now (1) follows from (2) and (3).<|endoftext|> TITLE: Self-homeomorphism of Stone-Čech boundary with an isolated fixed point QUESTION [5 upvotes]: $\DeclareMathOperator\bso{\beta^*\!\omega}\DeclareMathOperator\Homeo{Homeo}$Let $\bso$ be the complement of the countable discrete space $\omega$ in its Stone-Čech compactification $\beta\omega$ (some authors denote it $\omega^*$). Question. Assume ZFC+CH. Does there exist a self-homeomorphism of $\bso$ with an isolated fixed point (i.e., a fixed point that is not a limit of other fixed points)? Bonus question: if yes: it it true that for every $x\in\bso$ there exist a self-homeomorphism of $\bso$ with $x$ an isolated fixed point? Other bonus question: what about asking the self-homeomorphism to have order 2? Contextual remarks: Say that a self-homeomorphism of $\bso$ is smooth if it is induced by a bijection between two subsets of $\omega$ with finite complement. If $f\in\Homeo(\bso)$ is smooth then the set of fixed points of $f$ is clopen, hence has no isolated point. Shelah in the early 1980s showed the existence of models of ZFC in which every self-homeomorphism of $\bso$ is smooth (and in particular in which $\#(\Homeo(\bso))=\mathfrak{c}$). Hence in such models the above question has a negative answer. Rudin proved in the 1950s that under ZFC+CH, $\#(\Homeo(\bso))=2^\mathfrak{c}$ and its action on $\bso$ is not transitive. If there's a self-homeomorphism with an isolated fixed point $x$, it is easy to modify it to get another one with unique fixed point $x$. REPLY [5 votes]: As an answer to the bonus question: no, see K. P. Hart and J. Vermeer. Fixed-point sets of autohomeomorphisms of compact F-spaces, Proceedings of the American Mathematical Society, 123 (1995), 311–314: every fixed-point set of an autohomeomorphism of $\omega^*$ is a $P$-set. The same paper also answers the other bonus question: every $P$-set is the fixed-point set of an involution. Apply this to a $P$-point.<|endoftext|> TITLE: Alternatives to the law of the excluded middle QUESTION [32 upvotes]: As a sentential logic, intuitionistic logic plus the law of the excluded middle gives classical logic. Is there a logical law that is consistent with intuitionistic logic but inconsistent with classical logic? REPLY [21 votes]: In first-order logic, the sentence $$\neg\forall x,y(\neg\neg x=y \to x=y)$$ is consistent with intuitionist logic but not with classical logic. One might call this "the fuzziness of identity". In synthetic differential geometry, as axiomatized in Models for smooth infinitesimal analysis by Moerdijk and Reyes, this is actually a theorem about the real numbers.<|endoftext|> TITLE: Computing the invariants of ball quotient surfaces QUESTION [5 upvotes]: The two-dimensional complex unit ball $B$ has group of biholomorphic automorphisms $PU(2,1)$. If $Γ$ is an arithmetic subgroup of $PU(2,1)$, the quotient $Γ\text{\\}B$ is an orbifold. Taking its minimal resolution gives a complex manifold $X$. Question: How to compute the Hodge numbers of $X$, given the group $Γ$ (in matrix form or generators-relation form)? REPLY [7 votes]: I'm assuming that you know "where" in the commensurability class your lattice is. By this, I mean you perhaps have $\Gamma$ as a subgroup of some principal arithmetic lattice $\Lambda$ of known index, e.g., as a subgroup of $\mathrm{PU}(h, \mathcal{O}_k)$ where $h$ is a hermitian form on $k^3$ for an appropriate number field $k$. You then know the orbifold Euler characteristic of $X = \Gamma \backslash \mathrm{B}$, using Prasad's formula for the orbifold Euler characteristic of $Y = \Lambda \backslash \mathrm{B}$. To start, you need to know the conjugacy classes of finite subgroups of $\Gamma$. This can be a bit subtle. You cannot, for example, read these all off of a presentation. You could, for example, have a computer find a fundamental domain for the action of $\Gamma$ on $\mathrm{B}$ (if you have matrix generators), then find the stabilizers of the lower-dimensional faces. Alternately, you can sometimes figure out the torsion in $\Lambda$ by hand, but you still need to convert these elements to words in $\Gamma$ if you're just working with a presentation. Long story short, if you don't already know $X$ as a normal variety (along with the orbifold locus), this step can be a gigantic pain. Assuming we now have the conjugacy classes of finite subgroups, you can figure out all the singularities of $X$ (as a normal variety). In essence, you care about the subgroups with an isolated fixed point for their action on $\mathrm{B}$. These are all classical singularities from subgroups of $\mathrm{U}(2)$, so you're now ready to resolve them to get the smooth resolution $\overline{X}$. Also, from these finite subgroups, you can figure out the topological Euler characteristic of $X$ from the orbifold Euler characteristic (this is, for instance, explained in the book of Barthel, Hirzebruch, and Hofer), and hence you can find $c_2(\overline{X})$. There is an additional subtle point here: you need to know the genus of each curve in the orbifold locus. These are arithmetic Fuchsian subgroups of $\Gamma$ associated with stabilizers of certain sub-balls of $\mathrm{B}$, so you need to figure these out as well - this is possible, but again quite annoying. We now figure out the irregularity $q(\overline{X}) = h^{1,0}(\overline{X})$. Take the quotient of $\Gamma$ by its subgroups generated by complex reflections, you get a presentation for $\pi_1(X)$. You can then abelianize to find $H_1(X, \mathbb{Z})$. From all the standard relations between Hodge numbers and characteristic numbers, all we need now is the holomorphic Euler characteristic $\chi(\mathcal{O}_{\overline{X}})$. This you can deduce from Hirzebruch proportionality. To be a bit careful, let $Z \to X$ be a cover by a smooth ball quotient (e.g., you can find an explicit congruence subgroup of $\Gamma$ that is torsion-free), and we then have $$ c_1^2(Z) = 3 c_2(Z) = 9 \chi(\mathcal{O_Z}), $$ and we know $c_2(Z)$ from the orbifold Euler characteristic of $Y$ and $[\Lambda : \pi_1(Z)]$. Now use properties of the holomorphic Euler characteristic under finite maps to figure out $\chi(\mathcal{O}_{\overline{X}})$. We now have $p_g(\overline{X}) = h^{2,0}(\overline{X})$, and can get $h^{1,1}(\overline{X})$ from $c_2(\overline{X})$. You could really argue that everything I've described comes from finding an explicit Galois cover $Z \to Y$ by a smooth ball quotient with finite Galois group $F$. You know $q(Z)$ from abelianizing $\pi_1(Z) \le \Gamma$, then Hirzebruch proportionality gives you the other Hodge numbers of $Z$. Now it's all about computing invariants for resolutions of quotient surfaces. Knowing the number field $k$, one can find a completely explicit $F$ by taking $\pi_1(Z) < \Lambda$ to be a torsion-free congruence subgroup (and only only needs to avoid certain explicitly computable congruence subgroups to ensure torsion-free). You could probably go for the canonical divisor instead of the holomorphic Euler characteristic, but I think that's more work. Also, when $X$ has cusps, there is additional fuss about resolving the cusp singularities, but the effect on the characteristic numbers can be found in the literature (Holzapfel's book, work of Hirzebruch, Feustel, ...). While this all sounds impossible in practice, people have done this for small Picard modular groups (i.e., where $k$ is an imaginary quadratic field). However, the problem does get out of hand pretty quickly.<|endoftext|> TITLE: $\pi_{2n-1}(\operatorname{SO}(2n))$ element represents the tangent bundle $TS^{2n}$, not torsion and indivisible for $n>1$? QUESTION [5 upvotes]: Question: Is the element $\alpha$ in $\pi_{2n-1}(\operatorname{SO}(2n))$ representing the tangent bundle $TS^{2n}$ of the sphere $S^{2n}$ indivisible and not torsion? My understanding so far — An $\operatorname{SO}(2n)$ bundle over $S^{2n}$ corresponds to an element in $\pi_{2n}\operatorname{BSO}(2n) =\pi_{2n-1}\operatorname{SO}(2n)$. Not torsion: There does not exist any integer $m > 0$ such that $m\alpha$ is a trivial element. Indivisible: There does not exist any integer $k > 1$ and any element $\beta$ in $\pi_{2n-1}\operatorname{SO}(2n)$ such that $\alpha=k\beta$. Ref: Mimura, Toda: Topology of Lie groups. Chapter IV Corollary 6.14. p.s. I am asking $n>1$ so if you vote down for $n=1$, you may reconsider your vote... Lol REPLY [5 votes]: For $n=1$, the answer to your question is negative, as explained by Gregory Arone in the comments. In the cases $n\neq 1,2,4$, there is the following easy argument: The long exact sequence of the fibration $S^{2n-1}\to BO(2n-1)\to BO(2n)$ induces an exact sequence $$\pi_{2n}(BSO(2n))\xrightarrow{e} \mathbb{Z}\xrightarrow{[TS^{2n-1}]}\pi_{2n-1}(BSO(2n-1)),$$ where $e$ is the evaluation of the Euler class. As the order of $[TS^{2n-1}]\in \pi_{2n-1}(BSO(2n-1))$ is $2$ unless $n=1,2,4$, the image of $e$ is $2\mathbb{Z}$. As $\chi(S^{2n})=2$, we have $e([TS^{2n}])=2$, so $[TS^{2n}]$ must be nontorsion and indivisible as long as $n\neq 1,2,4$. Let me also mention the following convenient description of $\pi_{2n}(BO(2n))$ (this can be used to settle the cases $n=2,4$): Combining the Euler class with the inclusion $SO(2n)\subset SO$, we have a morphism $$ (e,s)\colon \pi_{2n}(BO(2n))\rightarrow \mathbb{Z}\oplus \pi_{2n}(BO)=\mathbb{Z}\oplus\begin{cases} \mathbb{Z} & \text{ if }n\equiv 0,2\text{ (mod) } 4\\ \mathbb{Z}/2 & \text{ if }n\equiv 1\text{ (mod) } 4\\ 0 & \text{ if }n\equiv 3\text{ (mod) } 4 \end{cases}. $$ From Kervaire's Some nonstable homotopy groups of Lie groups, one can deduce that as long as $n\neq 1$ this morphism is injective with image $$ \begin{cases} 2\mathbb{Z}\oplus \mathbb{Z} & \text{ if }n\equiv 0\text{ (mod) } 4,n\neq 2,4\\ \{(k,l)\in\mathbb{Z}^2|k+l\text{ even} \} & \text{ if }n=2,4\\ 2\mathbb{Z}\oplus\mathbb{Z}/2 & \text{ if }n\equiv 1\text{ (mod) } 4\\ 2\mathbb{Z}\oplus 0 & \text{ if }n\equiv 3\text{ (mod) } 4. \end{cases} $$ As $S^{2n}$ is stably parallelisable, the image of $[TS^{2n}]$ under this morphism is $(2,0)$, so this class is indivisible and nontorsion.<|endoftext|> TITLE: Conjecture on minimum size of graph QUESTION [10 upvotes]: Given a graph $G(V,E)$, let $\chi(G)$ be its chromatic number, and $\chi_1(G)$ its 1-improper chromatic number (meaning that each node can have at most 1 neighbor with the same color; or another way of looking at this is that you are allowed to remove any matching from $G$). It is fairly easy to prove that a graph that satisfies $\chi_1=\chi$ has at least $2\chi-1$ vertices (a proof by induction exists). However, determining the minimum number of edges in order for the equality to hold seems more difficult. Quick drawings suggest the number of edges must be at least $2(\chi-1)^2$, but I cannot manage to prove it. Any suggestions? Note: it is easy to see that the number of edges must be larger than $\chi(\chi-1)$. Indeed, extremal theory tells us the number of edges in a graph is always larger than (or equals) $\chi(\chi-1)/2$, but if we can remove any matching, it has to be larger than (or equal to) $\chi(\chi-1)$. Here is a possible MIP formulation for the 1-improper chromatic number $\chi_1(G)$: Variables $y_c \in \{0,1\}$, takes value $1$ if color $c\in K=\{1,...,n\}$ is used $x_{vc}\in \{0,1\}$, takes value $1$ if color $c \in K$ is assigned to node $v \in V$ $\delta_{uv}\in \{0,1\}$, takes value $1$ if vertices $u$ and $v$ share the same color, $(u,v)\in E$ Objective Function $$ \min \; \sum_{c \in K} y_c $$ Constraints One color per node: $$ \sum_{c \in K} x_{vc} = 1 \quad \forall v \in V $$ If vertex $v$ takes color $c$, $y_c$ is activated: $$ x_{vc} \le y_c \quad \forall v \in V, \forall c \in K $$ If endpoints of an edge $(u,v)$ share the same color, $\delta_{uv}$ is activated: $$ x_{uc}+x_{vc} \le 1 + \delta_{uv}\quad \forall (u,v) \in E, \forall c \in K $$ At most one conflict per node: $$ \sum_{u| (u,v)\in E} \delta_{uv} \le 1 \quad \forall v \in V $$ Using data from findstat.org, here is a compilation of results for a few graphs. The conjecture holds for all of the graphs for which the data is available on findstat.org. Disclaimer: This question has been posted here (math.stackexchange) 5 years ago, and has not been answered, so I am trying another community. However, someone attempted to make a proof, and although the proof is not correct, it may inspire. REPLY [2 votes]: Let us prove that any graph with $\chi_1(G)>n$ has at least $2n^2$ edges (with no assumptions on $\chi(G)$). This provides a sharp estimate (and the method also shows how to construct an optimal graph). Lemma. Assume that the maximal degree in $G$ does not exceed $2k-1$. Then $\chi_1(G)\leq k$. Proof. Consider a coloring in $k$ colors with the smallest number of monocolor edges. Assume that a vertex $v$ has two neighbors of its color $c$; then there is a color $c’$ appearing among the neighbors of $v$ at most once. Recoloring $v$ with $c’$ decreases the number of monocolor edges. This contradiction proves the Lemma. Back to our statement. Induction on $n$. The base case $n=0$ is trivial. For the step, arguing indirectly, assume that $G$ has less than $2n^2$ edges but $\chi_1(G)>n$. Find a vertex $v_1$ with $d(v_1)\geq 2n$ (otherwise apply the Lemma). Next, in $G-v_1$ find a vertex $v_2$ of degree at least $2n-2$ (otherwise, color $G-v_1$ in $n-1$ colors $1$-improperly and color $v_1$ with the $n$th color). Then $G-v_1-v_2$ has less than.$2n^2-2n-(2n-2)=2(n-1)^2$ edges, so $\chi_1(G-v_1-v_2)\leq n-1$ by the inductive hypothesis. It remains to color $v_1$ and $v_2$ with the $n$th color to get a $1$-improper coloring of $G$. Remark. In the same manner, one can obtain an estimate for $e(G)$ in terms of $\chi_m(G)$.<|endoftext|> TITLE: Simple-looking problem with integrals QUESTION [9 upvotes]: Let $f: [0,\infty) \rightarrow \mathbb{R}$ be a continuous function such that $f(0) = 0$. Is it true that if the integral $$ \int_0^{\pi/2} \sin(\theta) f(\lambda \sin(\theta)) \, d\theta $$ is zero for every $\lambda > 0$, then $f$ is identically zero? It's rather obviously true if $f$ is a polynomial and I'm hoping it is true in general, which is perhaps why I'm stuck. Edit. I came across this problem in two different, but related contexts. I'll describe the easier one: given a positive continuous function $F : \mathbb{R} \rightarrow \mathbb{R}$, the functions $\kappa_\lambda(\theta) := F(\lambda\cos(\theta))$, with $\lambda > 0$, are all curvature functions of plane ovals evaluated at the point of the curve where $(\cos(\theta),\sin(\theta))$ is the normal vector if and only if $$ \int_0^{2\pi} e^{-i\theta} \kappa_\lambda(\theta) \, d\theta = \int_0^{2\pi} e^{-i\theta} F(\lambda\cos(\theta)) \, d\theta = 0 $$ for all $\lambda > 0$. If $F$ is even, this is always the case, but does it have to be even? Well, after you decompose $F$ into even and odd parts and play around with this you come to the problem posed above and so nicely solved by Fedor Petrov and Mateusz Kw'asnicki below. Their solution readily implies that A continuous function $F : \mathbb{R} \rightarrow \mathbb{R}$ is even if and only if $$ \int_0^{2\pi} e^{-i\theta} F(\lambda\cos(\theta)) \, d\theta = 0 $$ for all $\lambda > 0$. In other words, all the ovals will be centrally symmetric. REPLY [12 votes]: As suggested by Fedor Petrov, we write $$ g(x) = f(\sqrt x) , $$ and we substitute $\lambda \sin\theta = \sqrt{x}$ and $t = \lambda^2$. This leads to $$ \begin{aligned} 0 & = 2 \lambda \int_0^{\pi/2} f(\lambda \sin \theta) \sin \theta \, d\theta \\ & = \int_0^{\lambda^2} 2 f(\sqrt x) \sqrt{x} \, \frac{1}{2 \sqrt{x (\lambda^2 - x)}} \, dx \\ & = \int_0^t \frac{g(x)}{\sqrt{t - x}} \, dx . \end{aligned} $$ Now, this is the "half-integral" of $g$: the convolution of $g$ with $x^{-1/2}$ is the fractional Riemann–Liouville integral of $g$, up to a constant factor. Adding another half-integral leads to the usual integral: by Fubini, $$ \begin{aligned} 0 & = \int_0^s \biggl(\int_0^t \frac{g(x)}{\sqrt{t - x}} dx\biggr) \frac{1}{\sqrt{s - t}} dt \\ & = \int_0^s g(x) \biggl(\int_x^s \frac{1}{\sqrt{t - x}} \frac{1}{\sqrt{s - t}} dt \biggr) dx . \end{aligned} $$ The inner integral is just $\pi$: substituting $t = x + (s - x) u$, we obtain $$ \int_x^s \frac{1}{\sqrt{t - x}} \frac{1}{\sqrt{s - t}} dt = \int_0^1 \frac{1}{(s - x) \sqrt{u (1 - u)}} \, (s - x) \, du = \pi .$$ It follows that $$ 0 = \pi \int_0^s g(x) dx , $$ which clearly implies that $g$ is zero almost everywhere. Edit: ...and since $g$ is continuous, almost everywhere yields everywhere (as pointed out by Fedor Petrov in a comment below).<|endoftext|> TITLE: Special nilpotent elements QUESTION [9 upvotes]: Let $R$ be a (noncommutative, associative) ring. Set $N_2:=\{x\in R : x^2=0\}$, the set of nilpotent elements of degree $2$ (also called the square-zero elements). If $x,y\in R$ satisfy $xy=0$, then $yx\in N_2$, but not every element in $N_2$ arises in this way. (See the example below.) Question: Has the set $F:=\{yx : xy=0\}$ been studied before? Are there characterizations of these "special square-zero" elements? Note that every element in $F$ is a commutator (if $xy=0$, then $yx=yx-xy=[y,x]$.) Thus, a precise question would be if $F=N_2\cap[R,R]$? Example: Consider $R=\left\{ \begin{pmatrix} a & b \\ 0 & a\end{pmatrix} : a,b\in\mathbb{R} \right\}$. Then $z=\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$ belongs to $N_2$ but not to $F$, since every commutator in $R$ is zero. REPLY [4 votes]: The answer to your precise question is no: it is not always the case that $F = N_2\cap [R,R]$. A nice way to see this is by fixing a field $k$, and constructing the universal example of a $k$-algebra equipped with a square-zero commutator. That is, $R = k\langle a,b\rangle/((ab - ba)^2)$. In that $k$-algebra, the element $ab - ba$ is certainly a square-zero commutator. The question is whether that element is equal to $yx$ for some pair of elements $x,y\in R$ such that $xy=0$. You can rule the existence of such a pair $x,y\in R$ by grading $R$ so that $a,b$ are each in degree $1$. Then the relation $(ab - ba)^2$ in $R$ is homogeneous of degree $4$. So if you have $yx = ab- ba$, then: $x(ab - ba) = xyx = 0$, so $x$ must have no nonzero terms of degree $<2$, and $(ab - ba)y = yxy = 0$, so $y$ must have no nonzero terms of degree $<2$. But $ab-ba$ is homogeneous of degree $2$, so there's no way to multiply $x$ and $y$, each with no terms of degree $<2$, to yield $ab-ba$. As an aside, in the case where $k = \mathbb{F}_2$, a certain $k$-algebra quotient of $k\langle a,b\rangle/((ab - ba)^2)$ arises quite naturally in topology: the $k$-algebra $$ k\langle a,b\rangle/((ab - ba)^2,a^2, aba - b^2)$$ is isomorphic to the subalgebra of the Steenrod algebra generated by the Steenrod squares $Sq^1$ and $Sq^2$. The element $ab - ba = Sq^1 Sq^2 - Sq^2 Sq^1$ is one of the famous Milnor primitives, usually denoted $Q_1$. That element is also an example of a square-zero commutator which is not in your set $F$. Thanks for asking an interesting question, which I enjoyed thinking about.<|endoftext|> TITLE: On the coefficients that appear in finite groups of matrices with integer entries QUESTION [9 upvotes]: Let $n$ be a positive integer and $G$ be a finite group of $n\times n$ matrices with integer coefficients, i.e. $G\subset\operatorname{GL}_n(\mathbb{Z})$. It is known that for sufficiently large $n$, the maximum order of such a group is $2^nn!$ by Feit (although relying on an unpublished manuscript of Weisfeiler I believe) and it is attained. I am interested in the coefficients that appear in the matrices of such a group. Specifically, I would like to know how many distinct numbers can appear. Clearly it cannot be more than $2^nn!n^2$ if all coefficients of all matrices are distinct but I have the feeling that such a bound cannot be reached. By considering the companion matrix of a cyclotomic polynomial, I think it is possible to produce roughly $n$ or maybe even $n^2$ distinct coefficients. On the other hand, a group like the symmetric group can be encoded with $0,1$ matrices and the group that attains the $2^nn!$ upper bound only uses $-1,0,1$. My questions are: is it possible to derive a much better bound on the number of distinct coefficients, for example polynomial in $n$ ? or is there an example family of finite groups whose number of distinct coefficients is exponential in $n$ ? I have the feeling that representation theory could help but I am not well-versed in the theory of finite groups. REPLY [8 votes]: It is possible to get a number of distinct coefficients exponential in $n$. Here is an example. Let $$B = \begin{bmatrix} 1 & -1 & -1 & -1 & \cdots & x_1 \\ 0 & 1 & 0 & 0 & & x_2 \\ 0 & 0 & 1 & 0 & \cdots & x_3 \\ 0 & 0 & 0 & 1 & & \vdots \\ & \vdots & & & \ddots & x_{n-1} \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$ We have $$B^{-1} = \begin{bmatrix} 1 & 1 & 1 & 1 & \cdots & -S \\ 0 & 1 & 0 & 0 & & -x_2 \\ 0 & 0 & 1 & 0 & \cdots & -x_3 \\ 0 & 0 & 0 & 1 & & \vdots \\ & \vdots & & & \ddots & -x_{n-1} \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$ where $S=x_1+x_2+\cdots+x_{n-1}$. We consider the conjugate by $B$ of the group of diagonal $\pm 1$ matrices, of size $2^n$. Let $X$ be a diagonal matrix with $X_{i,i}=\varepsilon_i \in \{\pm 1\}$ and suppose $\varepsilon_n=1$. Then we can compute the upper right entry of $B^{-1}XB$ as $$-S + \sum_{i=1}^{n-1} \varepsilon_i x_i$$ If we take, for example, $x_i=2^i$, then these sums are all different for the $2^{n-1}$ possible choices of $X$, so at least $2^{n-1}$ different coefficients appear in the matrices of the group.<|endoftext|> TITLE: Yoneda map for a composition of a representable functor and an arbitrary functor QUESTION [5 upvotes]: Let $\mathcal{C}$ and $\mathcal{D}$ be categories and let $T : \mathcal{C} \rightarrow \mathcal{D}$ be a functor. Suppose that $F : \mathcal{D}^\mathrm{op} \rightarrow \mathrm{Set}$ is a functor. (So in the older language I am more used to, $F$ is a contravariant functor.) Let $Y \in \mathcal{D}$ be an object. Under what conditions on $T$ and $Y$ is there a bijection $$\mathrm{Nat}\Bigl(\mathrm{Hom}_\mathcal{D}\bigl(T(-), Y\bigr), F\circ T\Bigr) \stackrel{\mathrm{Yon}}{\longrightarrow} F\bigl(Y\bigr)$$ with the properties of the Yoneda embedding? The rest of my question is background and motivation. I'll first show that there is a unique way to define part of $\mathrm{Yon}$ when $Y$ is $T(B)$ for some object $B \in \mathcal{C}$. Suppose that $\alpha$ is a natural transformation from the contravariant functor $\mathrm{Hom}_\mathcal{D}\bigl(T(-), T(B) \bigr) : \mathcal{C} \rightarrow \mathrm{Set}$ to the contravariant functor $F \circ T : \mathcal{C} \rightarrow \mathrm{Set}$. (Again these become ordinary functors replacing $\mathcal{C}$ with $\mathcal{C}^\mathrm{op}$.) Apply $$\alpha_{B} : \mathrm{Hom}_{\mathcal{D}}\bigl(T(B),T(B) \bigr) \rightarrow F\bigl(T(B)\bigr)$$ to $\mathrm{id}_{T(B)}$ to get $t = \alpha_B(\mathrm{id}_{T(B)}) \in F(TB) = F(Y)$. Then, as in the usual proof of the contravariant version of Yoneda's Lemma, a chase around a commutative square shows that if $f : A \rightarrow B$ is any morphism in $\mathcal{C}$, and so $Tf : T(A) \rightarrow T(B)$ is a morphism in $\mathcal{D}$, then $\alpha_A : \mathrm{Hom}_\mathcal{D}\bigl(T(A),T(B)\bigr) \rightarrow F\bigl(T(A)\bigr)$ satisfies $$\alpha_A(Tf) = (FTf) \alpha_B(\mathrm{id}_{T(B)}) = (FTf)(t) \in F\bigl(T(A)\bigr).$$ This determines $\alpha_A$ on those morphisms of the form $Tf$, for an arbitrary object $A$ of $\mathcal{C}$, and so when $T$ is full, we have the required bijection. Restated in terms of $t$, my question asks for a necessary and sufficient condition on $T : \mathcal{C} \rightarrow \mathcal{D}$ and $Y \in \mathcal{D}$ for $t \in F(Y)$ to determine $\alpha$. Example 1. Let $\mathcal{C} = \mathcal{D}$ be the category with one object $\star$ and $\mathrm{Mor}(\star,\star) = \mathcal{S}$ where $\mathcal{S} = \langle \sigma, \tau \rangle$ is the group of permutations of $\{1,2,3\}$, generated by $\sigma = (1,2,3)$ and $\tau = (1,2)$. Since $Y = \star = T(\star)$, the requirement on $Y$ for part of $\mathrm{Yon}$ to be defined is obviously satisfied. Define $T : \mathcal{C} \rightarrow \mathcal{C}$ by abelianization with embedding $\langle 1, \tau \rangle$, so $$T(\mathrm{id}_\star) = T(\sigma) = T(\sigma^2) = \mathrm{id}_\star, \quad T(\tau) = T(\sigma\tau) = T(\sigma^2 \tau) = \tau.$$ The partial definition of $\alpha_\star$ requires $\alpha_\star(Tf) = t(Tf)$ where $t = \alpha_\star(\mathrm{id}_\star)$, for all $f \in \mathcal{S}$. There are only two possibilities for $Tf$. Taking $f \in \{\mathrm{id}_\star, \sigma, \sigma^2 \}$ gives nothing, whereas taking $f \in \{\tau, \sigma\tau, \sigma^2\tau\}$ gives $\alpha_\star(\tau) = t\tau$. Thus $t$ determines $\alpha_\star(\tau)$, but not, for instance, $\alpha_\star(\sigma)$. Example 2. (Edited, since although this was the original motivation, I realised later it doesn't exactly fit the setup above.) Let $\mathcal{C}$ be the category of representations of the algebraic group $\mathrm{GL}_d(\mathbb{C})$ and let $\mathcal{D}$ be the category of bimodules with left $\mathrm{GL}_d(\mathbb{C})$ action and right $S_r$ action. Let $T : \mathcal{C} \rightarrow \mathcal{D}$ be the functor defined by $T(U) = U^{\otimes r}$, where the tensor product is regarded as a representation of $S_r$ acting on the right by place permutation on tensors. Let $F$ be the representable functor $$\mathrm{Hom}_{\mathcal{D}}\bigl( -, \mathrm{Sp}^\mu \bigr),$$ taking values in $\mathcal{C}$, not $\mathrm{Set}$ as above. (There is a $\mathrm{GL}_d(\mathbb{C})$ action on the hom-set because the $\mathrm{GL}_d(\mathbb{C})$ action on each $T(U)$ commutes with $S_r$; the action on $\mathrm{Sp}^\mu$ must be defined somehow to make this module an object in $\mathcal{D}$, so take the trivial action.) By Schur–Weyl duality, $$U^{\otimes r} \cong \bigoplus_\nu \mathrm{\Delta}^\nu(U) \boxtimes \mathrm{Sp}^\lambda,$$ where $\Delta^\nu$ is the Schur functor for $\nu$. Hence by Schur's Lemma, $F\bigl(T(U)\bigr) \cong \Delta^\mu(U)$, naturally in $U$. Taking $Y = \mathrm{Sp}^\lambda$, we have, $$\mathrm{Nat}\Bigl(\mathrm{Hom}_\mathcal{D}\bigl((-)^{\otimes r}, \mathrm{Sp}^\lambda\bigr), \mathrm{Hom}_\mathcal{D}\bigl((-)^{\otimes r}, \mathrm{Sp}^\mu\bigr)) \cong \mathrm{Nat}(\Delta^\lambda, \Delta^\mu) $$ and by further applications of Schur's Lemma, $$ \mathrm{Nat}(\Delta^\lambda, \Delta^\mu) \cong \begin{cases} \mathbb{C} & \text{if $\lambda=\mu$} \\ 0 & \text{otherwise} \end{cases} \cong \mathrm{Hom}_{\mathbb{C}S_r}(\mathrm{Sp}^\lambda, \mathrm{Sp}^\mu) \cong F(\mathrm{Sp}^\lambda)$$ so every natural transformation comes from an element of $F(\mathrm{Sp}^\lambda) = F(Y)$, as in the Yoneda embedding, but now into the category $\mathcal{C}$ of representations of $\mathrm{GL}_d(\mathbb{C})$, rather than $\mathrm{Set}$. REPLY [7 votes]: This property of the functor $T$ is called being a dense functor, or "densely generating functor". The notion was first introduced by Isbell in the case where $T$ is fully faithful under the name "adequate subcategory", but that usage has by now disappeared, and "dense" is preferred. It has been occasionally rediscovered -- for example, in the $\infty$-categorical context, Lurie calls such a functor "strongly generating", but even there the term "dense" is now preferred, especially because "strong generator" means something else classically. One of the equivalent ways to say that a functor $T: C \to D$ is dense is to say that the induced functor $T^\ast: \mathrm{Set}^{D^{op}} \to \mathrm{Set}^{C^{op}}$ defined by $F \mapsto F \circ T^{op}$ on categories of presheaves is fully faithful. The condition as you've stated it says that $T^\ast$ is fully faithful on the homset from the representable $\mathrm{Hom}_D(-,Y)$ to the functor $F$, but this is equivalent because every presheaf is a colimit of representables such as $\mathrm{Hom}_D(-,Y)$.<|endoftext|> TITLE: Cohomology of BG, G non-connected Lie group, and spectral sequence relating to classifying space of connected component of the identity QUESTION [6 upvotes]: Suppose $G$ is a Lie group, with $\pi_0(G)$ not necessarily finite, but might as well assume $G_0$, the connected component of the identity, is compact. In the case that $\pi_0(G)$ is finite, then we know that there is an injection $H^*(BG,\mathbb{Q})\to H^*(BG_0,\mathbb{Q})$, and this can apparently be seen via a spectral sequence argument, using the fact that the rational cohomology of $B\pi_0(G)$ is concentrated in degree zero. So this is some kind of Leray–Serre spectral sequence argument on either $\pi_0(G)\to BG_0\to BG$ or $BG_0\to BG\to B\pi_0(G)$ (and I suspect the latter), probably using the degeneration and some kind of "edge homomorphism is injective" argument. I suspect that in the case that we know something strong about the rational cohomology of $B\pi_0(G)$, then we might be able to say something in the case where $\pi_0(G)$ is not finite. Unfortunately my spectral sequence knowledge is limited, and I can't find a treatment of spectral sequences that seems general enough to deal with this setup in general (namely non-simply-connected base, and possibly non-connected fibre, plus non-finiteness issues, depending on which fibration is used). Is my intuition correct, that $H^*(B\pi_0(G),\mathbb{Q}) = H^0(B\pi_0(G),\mathbb{Q})$ can let us conclude something about how the cohomology of $BG$ relates to that of $BG_0$? Also, what would be a good reference that covers a general-enough version of the relevant spectral sequence? REPLY [8 votes]: Think about the case where $\pi_0(G)=\mathbb{Z}$, so $B(\pi_0(G))=S^1$, so we have a fibre bundle $BG_0\to BG\to S^1$. In this case $G$ is always a semidirect product formed using an automorphism $\alpha$ of $G_0$. By considering the preimages of the the complements of two points in $S^1$, we can express $BG$ as $U\cup V$, where $U$ and $V$ are each open and homotopy equivalent to $BG_0$, and $U\cap V$ is homotopy equivalent to $BG_0\amalg BG_0$. This gives a Mayer-Vietoris sequence. If we set up the identifications carefully we find that $\alpha^*$ appears in one of the Mayer-Vietoris maps, and we deduce that there is a short exact sequence $C^{*-1}\to H^*(BG)\to K^*$, where $K^*$ and $C^*$ are respectively the kernel and cokernel of $\alpha^*-1\colon H^*(BG_0)\to H^*(BG_0)$. All this is valid integrally as well as rationally. For more general $\pi_0(G)$, we have a spectral sequence $$ H^p(\pi_0(G);H^q(BG_0)) \Longrightarrow H^{p+q}(BG). $$ Note that the $E_2$ term involves group cohomology of the group $\pi_0(G)$ with coefficients in the module $H^*(BG_0)$, which typically has nontrivial action of $\pi_0(G)$. This is the same as the cohomology of the space $B\pi_0(G)$ with coefficients in a local system that typically has nontrivial twisting. In the case $\pi_0(G)=\mathbb{Z}$ the only group cohomology groups are $E_2^{0q}=K^q$ and $E_2^{1q}=C^q$. The differentials are like $d_r\colon E_r^{pq}\to E_r^{p+r,q-r+1}$ so there is no room for them to be nonzero. Thus $E_\infty=E_2$ and we recover the picture in the first paragraph above. On the other hand, if $\pi_0(G)$ is finite and we use rational coefficients then $H^p(\pi_0(G);H^q(BG_0))=0$ for $p>0$ and the spectral sequence collapses to an isomorphism $H^n(BG)=H^n(BG_0)^{\pi_0(G)}$. Note also that $G$ could just be $G_0\times\Gamma$ for an arbitrary discrete group $\Gamma=\pi_0(G)$. In this case $BG=BG_0\times B\Gamma$, and the Kan-Thurston Theorem tells us that $H^*(B\Gamma)$ can be essentially anything.<|endoftext|> TITLE: Orthogonal Hamiltonian cycles in (n x n x n) grids QUESTION [8 upvotes]: Let $C_n$ be a cubical $n \times n \times n$ subset of the integer lattice, so consisting of $n^3$ vertices. I am interested in special Hamiltonian cycles in $C_n$, special in the sense that (a) each edge of the cycle has unit length, and (b) no two adjacent edges are collinear---the path turns $90^\circ$ at every vertex. Call these orthogonal Hamiltonian cycles. For $C_2$, this is straightforward:          I believe that $C_3$ has no orthogonal Hamiltonian cycle. However, I don't have a clean proof, just an unconvincing exhaustive case analysis. Q1. Prove (or disprove) that $C_3$ has no orthogonal Hamiltonian cycle. For $C_4$, the below shows an orthogonal Hamiltonian cycle (colored on the right to reveal its structure). This is known as a Moore curve, a space-filling curve. The next level Moore curve, however, has collinear edges, as do Hilbert curves. Q2. For which $n$ does $C_n$ have an orthogonal Hamiltonian cycle? It is natural to hope that $8$ copies of the $C_4$ solution could be stitched $(2 \times 2 \times 2)$-together to form an ortho-Ham cycle for $C_8$, but I couldn't see my way through the complications. REPLY [8 votes]: In the recent book "Bicycle or Unicycle?" by Velleman and Wagon, this is problem #16 "Wiggle Room." (Actually, there it's generalized to computing the maximum length path, with a Hamiltonian path in the next-best case, and the best case being a Hamiltonian cycle.) It's known that there is a Hamiltonian cycle for all even $n$, and a nice parity argument that there isn't even a path for $n=3$, and evidence for conjecture that there is a path (but not necessarily a cycle) for all $n\geq 4$.<|endoftext|> TITLE: How to find equations of a sub-Riemannian problem QUESTION [7 upvotes]: I am working on sub-Riemannian geometry and try to understand what are the tools to find the equations of a sub-Riemannian problem. Here is an example: Let us consider the system defined by a lagrangian: \begin{equation} L=\frac{1}{2}m(\overset{\cdot}{x}^2+\overset{\cdot}{y}^2)+\frac{1}{2}I\overset{\cdot}{\theta}^2 \end{equation} And take the following non-holonomic constraint \begin{equation} \mathscr{D}=span\left\{\cos\theta\frac{\partial}{\partial x}+\sin\theta\frac{\partial}{\partial y},\frac{\partial}{\partial\theta}\right\} \end{equation} So we have, \begin{equation} \mathscr{D}^\perp=span\{\omega\} \end{equation} with $\omega=-\sin\theta dx+\cos\theta dy$ If I understand clearly the sub-Riemannian problem is to minimize \begin{equation} \int_0^TLdt \end{equation} under the constaint $\dot{z}(t)\in\mathscr{D}(z(t))$. Now, how do I find these equations? So far I am not asking how to solve these equations but just to understand how to find them in the first place. Thanks in advance for your help. Edit: I quite understand your answer, thank you for the clear explanation. I am though not really familiar with canonical coordinates. I agree with the expression of the Hamiltonian. Now, I am right if I say that the Hamilton's equations are given by: \begin{equation} \begin{cases} \overset{\cdot}{x}(t)&=\frac{\partial H}{\partial p_x}=p_x\cos^2\theta +p_y\cos\theta\cdot\sin\theta \\ \overset{\cdot}{y}(t)&=\frac{\partial H}{\partial p_y}=p_y\sin^2\theta +p_x\cos\theta\cdot\sin\theta \\ \overset{\cdot}{\theta}(t)&=\frac{\partial H}{\partial p_\theta}=p_\theta\\ \overset{\cdot}{p_x}(t)&=-\frac{\partial H}{\partial x}= 0 \\ \overset{\cdot}{p_y}(t)&=-\frac{\partial H}{\partial y}= 0 \\ \overset{\cdot}{p_\theta}(t)&=-\frac{\partial H}{\partial \theta}= (p_x^2-p_y^2)\sin(\theta)cos(\theta)+p_xp_y(\sin^2(\theta)-\cos^2(\theta)) \\ \end{cases} \end{equation} I'm not sure to understand how to interpret them. Are these the equations of the sub-Riemannian problem? REPLY [7 votes]: You can reformulate your problem in the language of geometric control (where your dynamical system is sometimes called Dubin's car). Let $$ X_1=\frac{1}{\sqrt{m}}(\cos\theta\partial_x + \sin\theta\partial_y),\qquad X_2=\frac{1}{\sqrt{I}}\partial_\theta$$ be a basis of your distribution. Horizontal curves are trajectories $z :[0,T]\to M$ (here $M=\mathbb{R}^2\times S^1$), that are tangent to the distribution, that is there exist $u_1,u_2:[0,T]\to \mathbb{R}$ (let us forget about regularity issues), such that $$ \dot z(t) = u_1(t) X_1|_{\gamma(t)} + u_2(t)X_2|_{\gamma(t)}, \qquad t\in [0,1]. $$ You look for horizontal trajectories between fixed endpoints, that minimize the cost functional: $$ \int L(\gamma(t))dt = \frac{1}{2}\int_0^T \left(u_1(t)^2 + u_2(t)^2\right)dt. $$ (this should clarify the normalization of $X_1,X_2$). This is a very simple optimal control problem. An application of the Pontryagin Maximum Principle yields first order necessary conditions for solutions of this problem. In this very simple all solutions correspond to projections on $M$ of integral trajectories of an Hamiltonian system on $T^*M$, with Hamiltonian given by $$ H(\lambda) = \frac{1}{2}\sum_{i=1}^2 \lambda(X_i)^2, \qquad \lambda \in T^*M. $$ In canonical coordinates $(x,y,\theta,p_x,p_y,p_\theta)$ on $T^*M$, your Hamiltonian reads explicitly $$ H=\frac{1}{2}(p_x \cos\theta + p_y \sin\theta)^2 + \frac{1}{2}p_\theta^2. $$ Then your "geodesic equations" correspond to the Hamilton's equations for $H$. A good reference to this general kind of optimal control problems can be found in the following book (chapter 12, Theorem 12.10, case (b) is vacuous in your case, so only case (a) remains): Agrachev, Andrei A.; Sachkov, Yuri L., Control theory from the geometric viewpoint., Encyclopaedia of Mathematical Sciences 87. Control Theory and Optimization II. Berlin: Springer (ISBN 3-540-21019-9/hbk). xiv, 412 p. (2004). ZBL1062.93001. For a specifically sub-Riemannian reference, see Chapter 4.3 in Agrachev, Andrei; Barilari, Davide; Boscain, Ugo, A comprehensive introduction to sub-Riemannian geometry. From the Hamiltonian viewpoint., ZBL07073879. The references proposed by Robert Bryant are also very good, but the two aforementioned one are closer (I think) to the spirit of your question.<|endoftext|> TITLE: Is every variety an image of a smooth variety? QUESTION [6 upvotes]: Let $X$ be a finite type scheme over a field $k$. Is it true that there exists a surjective morphism $f : Y \rightarrow X$, where $Y$ is smooth over $k$? In other words, is every such scheme a quotient of a smooth scheme over $k$? REPLY [8 votes]: Presumably, you want to say that $X$ is reduced, since $Y$ will be reduced and any map from a reduced scheme lands in $X^{\mathrm{red}}$. Once you've said that, this follows from De Jong's alterations theorem, appearing in A.J. De Jong, Smoothness, semi-stability and alterations, Publications Mathématiques de l'IHÉS, Tome 83 (1996), pp. 51–93, doi:10.1007/BF02698644, Numdam.<|endoftext|> TITLE: Comparing notions of $A_{\infty}$ homotopy (in char 0): Markl's definition versus "Sullivan homotopy" QUESTION [7 upvotes]: Suppose that we have $A_{\infty}$ algebras $A,B$ (over a field of characteristic $0$), with $A_{\infty}$ maps $f,g: A \rightarrow B$. In the paper https://arxiv.org/abs/math/0401007 (top of page 4, item $(6)$) Markl defines a notion of $A_{\infty}$ homotopy, which can analogously be used to define a homotopy between maps $f$ and $g$. At least over characteristic $0$, there is a competing definition of a homotopy, defined as an $A_{\infty}$ morphism $H: A \rightarrow B \otimes \Omega^{\bullet}_{[0,1]}$ such that $H|_{t = 0} = f$ and $H|_{t =1} = g$. Supposedly, in this context, the two definitions are equivalent. How can do I go from Markl's definition to the other definition? Thank you for your time! REPLY [3 votes]: $\newcommand{\dd}{\mathrm d}$Markls notion is the same as an $A_\infty$-morphism from $A$ to $B\otimes C^\bullet_{[0,1]}$, where the second factor are the cellular cochains on the interval with (non-commutative!) dga structure defined via the Alexander-Whitney map. The two notions are then related by producing $A_\infty$-morphisms between this dga and $\Omega^\bullet_{[0,1]}$. At least the $A_\infty$-morphism from forms to singular cochains can be written down explicitly via so-called iterated integrals, compare this question I asked a while ago. There are still a lot of combinatorics involved, eg in finding the inverse $A_\infty$-morphism and taking tensor products of $A_\infty$-algebras. Two recent articles which discuss generalisations of this question are Mazuir's Higher algebra of A∞ and ΩBAs-algebras in Morse theory II and Robert-Nicoud and Vallette's Higher Lie Theory.<|endoftext|> TITLE: Which representations of $\mathfrak{sl}(2)$ are homomorphic images of the tensor product of finitely many copies of $\mathbb{C}^2$? QUESTION [9 upvotes]: My questions may turn out to be related to Schur functors. If $\mathfrak{g}$ is a complex semisimple Lie algebra and $\lambda$ is the highest weight of an irreducible representation $V$ of $\mathfrak{g}$, I am interested in restricting the representation to an $\mathfrak{sl}(2)$ subalgebra of $\mathfrak{g}$. For example, if $\mathfrak{g} = \mathfrak{sl}(n)$, then I am interested in $V = \mathbb{C}^n$ being the standard representation. If $\mathfrak{g} = \mathfrak{so(n)}$, I am interested in $V = \mathbb{C}^n$, i.e. in the vector representation. Similarly if $\mathfrak{g} = Sp(m, \mathbb{C})$, I am interested in $V = \mathbb{C}^{2m}$ being its standard representation. I am also interested in the $5$ exceptional cases and their fundamental representations. As for the $\mathfrak{sl}(2)$ subalgebra of $\mathfrak{g}$, I am mostly interested in "the" so-called "regular" $\mathfrak{sl}(2)$ subalgebra (it is unique up to conjugation in $\mathfrak{g}$), which was studied for instance by Kostant and others. I can now formulate my questions. From now on, consider the representation $V$, but view it as a representation of a fixed $\mathfrak{sl}(2)$ subalgebra of $\mathfrak{g}$ (for the sake of this post, one may for instance assume it to be a regular $\mathfrak{sl}(2)$ subalgebra). Question 1: Does there always exist a positive integer $m$, and a $\mathfrak{sl}(2)$ intertwining (meaning the corresponding notion at the Lie algebra level) linear map $$ \mathbb{C}^2 \otimes \cdots \otimes \mathbb{C}^2 \to V, $$ which is onto, where the domain is the tensor product of $m$ copies of $\mathbb{C}^2$? Question 2: Assuming the answer to question 1 is yes, can one write down such a linear map using weights and roots? For example, what would $m$ be in terms of Lie-theoretic data, etc? Note: I did edit my post quite a bit, so some of the comments may not make sense now to a new reader, because of my edits. The new reader should keep this in mind. However, these are now the precise questions that I would like to see answered. I apologize for the little mess that I have made, with my latest edits. I think though that my questions are more focused now, in this version. REPLY [8 votes]: $\def\CC{\mathbb{C}}\def\ZZ{\mathbb{Z}}\def\sl{\mathfrak{sl}}$In the comments to hm2020's answer, the OP explains that they want a surjection from $(\mathbb{C}^2)^{\otimes m}$, not from a direct sum $\bigoplus_i (\mathbb{C}^2)^{\otimes m_i}$. In that case, there is one obstruction: The weights of the representation must be either all even or all odd, and then the answer is yes. Notation: Let $h = \left[ \begin{smallmatrix} 1&0 \\0&-1 \end{smallmatrix} \right]$. Recall that the eigenvalues of $h$ on any (finite dimensional) $\sl(2)$ representation are integers. We write $V_i$ for the $i$-eigenspace of $h$ on a representation $V$, and write $\chi_V = \sum_i \dim V_i q^i$ for a formal variable $q$. We call $\chi_V$ the character of $V$. We put $R = \mathbb{C}^2$ with the obvious $\sl(2)$ action, and put $S_k:=\operatorname{Sym}^k R$ (so $R = S_1$). It is well known that the finite dimensional representation theory of $\sl(2)$ is semisimple, and that the simple representations are the $S_k$ with character $q^k + q^{k-2} + \dotsb + q^{-k}$. Meanwhile, the character of $R^{\otimes m}$ is $(q+q^{-1})^m$. We thus see that $\chi_{R^{\otimes m}}$ is either an even or odd Laurent polynomial in $q$, according to the parity of $m$, and that $\chi_{S_k}$ is either even or odd according to the parity of $k$. So $S_k$ can only be a summand of $R^{\otimes m}$ if $m \equiv k \bmod 2$. Thus, a representation $\bigoplus S_{k_i}$ is only a summand of $R^{\otimes m}$ if all the $k_i$ are congruent modulo $2$. This is also a sufficient criterion. The multiplicity of $S_k$ in $R^{\otimes (k+2 \ell)}$ is $\binom{k+2\ell}{\ell} - \binom{k+2\ell}{\ell-1}$, which goes to infinity as $\ell$ goes to infinity. So, for any fixed list of $k_i$ which are all congruent to a fixed value $k$ modulo $2$, the multiplicities of $S_{k_i}$ in $R^{\otimes k+2 \ell}$ will get arbitrarily large. Thus, any finite dimensional representation obeying this partiy condition will eventually be a summand of $R^{\otimes m}$. I find this more intuitive with the Lie group $SL(2)$ instead of the Lie algebra: $\chi_V(q)$ is the trace of the Lie group element $\left[ \begin{smallmatrix} q & 0 \\ 0 & q^{-1} \\ \end{smallmatrix} \right]$, and the parity condition is that $\left[ \begin{smallmatrix} -1 & 0 \\ 0 & -1 \\ \end{smallmatrix} \right]$ must act either $1$ or by $-1$.<|endoftext|> TITLE: Does ZF + BPI alone prove the equivalence between "Baire theorem for compact Hausdorff spaces" and "Rasiowa-Sikorski Lemma for Forcing Posets"? QUESTION [8 upvotes]: Rasiowa-Sikorski Lemma (for forcing posets)is the statement: For any p.o. $\mathbb{P}$ (i.e. $\mathbb{P}$ is a reflexive transitive relation) and for any countable family of dense subsets of $\mathbb{P}$ there is a generic filter which intersects all dense subsets of the countable family. It is well-known that this statement is equivalent to the Baire Category Theorem for Complete Metric Spaces - and thus it is also equivalent to the Principle of Dependent Choices. A masters student of mine has found in the literature the following statement: "Rasiowa-Sikorski Lemma is equivalent to the Baire Category Theorem for Compact Hausdorff Spaces, modulo the Boolean Prime Ideal Theorem". We understood this as the assertion that the theory ZF + BPI alone is able to prove the equivalence between the Baire Category Theorem for Compact Hausdorff Spaces and the Rasiowa-Sikorski Lemma. Well, I asked my student to verify such claim, and at first glance I suggested him to follow the results 3.1 to 3.4 of Chapter II of Kunen's book, where there are proofs for some equivalences of Martin's Axiom at $\kappa$, MA($\kappa$): the idea was to discard the hypothesis "c.c.c." and adapt the reasoning, arguing for $\kappa = \omega$. It turns out that it was not a good suggestion, because in 3.1 a kind of Downward-Lowenheim-Skolem argument is done, to show that it is equivalent to work with a restricted form of the forcing axiom, considering only partial orders of bounded cardinality. However, such argument seems to require the Axiom of Choice, or some part of it other than BPI. Does any of you know if it is indeed possible to prove the equivalence between "Baire Category Theorem for Compact Hausdorff Spaces" and "Rasiowa-Sikorski Lemma for forcing posets" from ZF + BPI alone ? Any suggestions or references would be appreciated. REPLY [6 votes]: The key observations are that BPI is equivalent to the Stone representation theorem for Boolean algebras, and that for the Rasiowa–Sikorski lemma we can focus on [complete] Boolean algebras, since they are forcing equivalent (so we can restrict the generality of partial orders). Now, one implication is a consequence of ZF. Since RS is equivalent to Dependent Choice, which itself is equivalent to the downward Löwenheim–Skolem theorem, one can just use that argument. Alternatively, if $X$ is a compact Hausdorff space and $D_n$ are dense open sets, and without loss of generality $D_{n+1}\subseteq D_n$. take $U$ to be a non-empty open set, and consider the forcing whose conditions are sequences $(x_i,W_i)_{i TITLE: Cohomology and base change without Noetherian assumption QUESTION [6 upvotes]: In the "The Rising Sea" by Vakil one can find the base change theorem for proper morphisms over a locally Noetherian base (28.1.6). He later indicates (28.2.M) how one could exchange the locally Noetherian condition by finitely presented using a result of Grothendieck. And indeed, it does not seem too hard to show this. I am wondering if this version of the theorem is written down anywhere else in the literature (with or without a proof), as I seem unable to find it. I see the stronger version getting applied quite often (most recently in Olsson's book "Algebraic Spaces and Stacks" proof of lemma 8.4.6) but always without a proper reference (Olsson references Hartshorne, who only shows the statement for projective morphisms). I could imagine that there is a formulation for algebraic spaces or stacks such that the theorem for schemes is just a special case. The closest thing I could find was theorem 1.9 / 1.10 in "Compatifying the Picard Scheme" by Altman and Kleiman, but this also does not seem to imply the base change theorem as stated by Vakil. It just seems really odd to me that such a prominent theorem can not be found in the Stacks Project or any other reputable source (without implying that Vakil is not reputable...) Edit: To add to the list below: In Conrad, Brian, Grothendieck duality and base change, Lecture Notes in Mathematics. 1750. Berlin: Springer. x, 296 p. (2000). ZBL0992.14001. in Chapter 5.1 one can find an argument how to remove the Noetherian condition. But he only does so for a slightly weaker statement. REPLY [4 votes]: There is a fairly general version of base change for schemes in Lipman's "yellow book": Lipman, Joseph; Hashimoto, Mitsuyasu: Foundations of Grothendieck duality for diagrams of schemes. Lecture Notes in Mathematics, 1960. Springer-Verlag, Berlin, 2009. Also available at: https://www.math.purdue.edu/~jlipman/Duality.pdf Concretely, Theorem (3.10.3) establishes the base-change theorem for an independent square of quasi-compact and quasi-separated maps of quasi-separated schemes. The proof does not use noetherian schemes.<|endoftext|> TITLE: Is the function $k(g,h) = \frac{1}{1+\lvert gh^{-1}\rvert}$ positive definite? QUESTION [6 upvotes]: Let $G$ be a finite group, $S \subset G$ a generating set, closed under taking inverses, and $\lvert\cdot\rvert$ the word length with respect to this set $S$. Question. Is the function $k(g,h) = \frac{1}{1+\lvert gh^{-1}\rvert}$ positive definite, for $g,h \in G$? A positive answer would allow every finite group $G$ to be embedded as a metric space in Euclidean space with $d(g,h) = \sqrt{2-2k(g,h)}$, and this could be useful in Geometric Group Theory. Motivation: This matrix $k(g,h)$ plays a role in a group theoretic reformulation of the Lagarias inequality: What properties characterize the function $L(x) = x+\exp(x) \log(x)$? Also asked on MSE, since I was not sure if it is research relevant: https://math.stackexchange.com/questions/4126491/is-the-function-kg-h-frac11gh-1-positive-definite REPLY [6 votes]: [NB. Addendum below gives a counterexample to OP's question.] Computationally, this appears to be true for $S_n$, $2 \le n \le 6$. The MATLAB code I used to check this is below (using the case $n = 4$ to eyeball the Cayley graph). By Cayley's theorem and appropriately tweaking the code (essentially, the arrays gen and G), you should be able to check other small groups with this as well (this is just exploiting the permutation representation, albeit very expensively). As an aside, it also appears that the kernel $\exp(-\beta|gh^{-1}|)$ is positive definite on $S_n$ for all $\beta > 0$ using adjacent transpositions, but the story is more delicate for all transpositions (see links in reference). n = 6; % Generating set of S_n: adjacent tranpositions gen = ones(n-1,1)*(1:n); gen(1:n:(n-1)^2) = 2:n; gen(n:n:((n-1)*n)) = 1:(n-1); % S_n itself G = perms(1:n); % Indices of generating set genInd = find(ismember(G,gen,'rows')); % Represent symmetric group using permutation matrices P = cell(size(G,1),1); for j = 1:numel(P) P{j} = zeros(n); for k = 1:n P{j}(k,G(j,k)) = 1; end end % Multiplication table in terms of group indices mul = nan(size(G,1)); for j = 1:numel(P) for k = 1:numel(P) [prod_jk,~] = find(P{j}*P{k}); mul(j,k) = find(ismember(G,prod_jk','rows')); end end % Get indices of inverses from this [ver,~] = find(mul==find(ismember(G,1:n,'rows'))); % don't overload inv % Division table div = mul(:,ver); % Adjacency matrix of Cayley graph A = ismember(div,genInd); % Cayley graph (Bruhat for this particular case) nodenames = join([repmat("x",[size(G,1),1]),string(G)],''); cayleyGraph = digraph(A,nodenames); % Distances equal word lengths d = distances(cayleyGraph); % Form kernel ker = 1./(1+d); % Inspect spectrum visually to check positive definiteness spec = eig(ker)' % Optional: plot Cayley graph for assurance % figure; % plot(cayleyGraph); ADDENDUM: For $S_5$, using all transpositions shows this is false. The key bit to change in the above code is to put this generating set in after G is defined: % All transpositions transInd = find(sum(G~=ones(size(G,1),1)*(1:n),2)==2); gen = G(transInd,:); You can verify that gen obeys the hypotheses by checking that transInd and ver(transInd) have the same indices. In this case, spec shows an eigenvalue of -0.0333.<|endoftext|> TITLE: Reference request: recurrence relation for Catalan numbers QUESTION [5 upvotes]: I would like to know if the following recurrence relation for Catalan numbers (see mathoverflow.net/questions/191524 and also math.stackexchange.com/questions/2113830) has appeared in a paper or a book, so that I can cite it. $$C_n = 1 + \sum_{k=1}^{\left\lceil\frac{n}{2} \right\rceil } (-1)^{k+1} \binom{n-k}{k} C_{n-k}$$ where $C_n$ is the $n$-th Catalan number. Fixed error: added + 1 to the right hand side REPLY [7 votes]: T. Koshy, Catalan Numbers with Applications (Oxford, 2009), page 322, proves a very similar identity: $$C_n=\sum_{k=1}^{\left\lfloor\frac{n+1}{2}\right\rfloor}(-1)^{k-1} \binom{n-k+1}{k} C_{n-k}$$ $$\Leftrightarrow \sum_{k=1}^{\left\lfloor\frac{n+1}{2}\right\rfloor} (-1)^{k-1} \binom{n-k+1}{k} C_{n-k}=0.$$ (I did not check the book, the identity is quoted in this paper, equation 4.) I also note that Mathematica knows the identity in the OP, in the form $$\sum_{k=0}^{\left\lceil\frac{n}{2} \right\rceil } (-1)^{k+1} \binom{n-k}{k} C_{n-k}=-1,$$ (Floor or ceiling in the summation limits does not matter, the sum may also be extended to infinity, since the binomial coefficients vanish.) In response to a question in the comment, I record the q-analog of Koshy's identity, derived by George Andrews. (The sum over $r$ may be terminated at $\left\lfloor\frac{n+1}{2}\right\rfloor$.)<|endoftext|> TITLE: Can the product of a 3-dimensional lens space with a circle be diffeomorphic to another such product when the lens spaces aren't diffeomorphic? QUESTION [19 upvotes]: This is a question that I need to answer in order to resolve an issue for my dissertation and I am looking for a reference. Here is the precise statement of the question. Suppose we have two three-dimensional lens spaces $L(n;r)$ and $L(n;s)$ which are homotopy equivalent but not diffeomorphic. Can their products with a circle , $L(n;r) \times S^1$ and $L(n;s) \times S^1$, be diffeomorphic? Note that the condition that the lens spaces be homotopy equivalent is necessary because one can lift a diffeomorphism of $L(n;r) \times S^1$ and $L(n;s) \times S^1$ to a homotopy equivalence from $L(n;r) \times \mathbb{R}$ to $L(n;s)\times \mathbb{R}$. Then use the fact that these deformation retract to $L(n;r)$ and $L(n;s)$, respectively. REPLY [28 votes]: The answer is no. If $L$, $L^\prime$ are 3-dimensional lens spaces and $S^1\times L$ is diffeomorphic to $S^1\times L^\prime$, then the covering space of $S^1\times L$ corresponding to the torsion subgroup defines an h-cobordism between $L$ and $L^\prime$ (we have embeddings of L and L′ in the covering space with disjoint images, and the images bound an h-cobordsim). It is an application of Atiyah-Singer fixed point theorem (with contributions by Bott and Milnor), that h-cobordant lens spaces are diffeomorphic. One reference is p.479 in "A Lefschetz Fixed Point Formula for Elliptic Complexes: II. Applications" by Atiyah and Bott. Various related results and generalizations are discussed in "Toral and exponential stabilization for homotopy spherical spaceforms" by Kwasik and Schultz. Both papers can be easily found online, I think.<|endoftext|> TITLE: Assuming decidable equality but not LEM in HoTT QUESTION [5 upvotes]: The law of excluded middle in homotopy type theory is a term of $$\prod_{A:\mathcal{U}}\Big(\mathrm{isProp}(A)\to(A+\neg A)\Big).$$What if we assume a term of$$\prod_{A:\mathcal{U}}\Big(\mathrm{isSet}(A)\to\prod_{x,y:A}\big((x=y)+\neg (x=y)\big)\Big)$$instead? Is this a strictly weaker axiom than LEM? Is it useful? Are there any philosophical reasons to accept this axiom but not LEM? EDIT: If we only assume a term of$$\prod_{A:\mathcal{U}_0}\Big(\mathrm{isSet}(A)\to\prod_{x,y:A}\big((x=y)+\neg (x=y)\big)\Big)$$does that change much? REPLY [4 votes]: In addition to Simon Henry's proof you can also use suspensions to show that decidable equality implies the law of excluded middle. Given a proposition $A : \mathrm{hProp}$ you can show using an encode-decode argument that the suspension $\Sigma A$ is a set and that the equality $N = S$ is equivalent to $A$. So decidable equality for every set of the form $\Sigma A$ implies the law of excluded middle.<|endoftext|> TITLE: Is $xz+1 $ a proper divisor of $a_3z^3+a_2z^2+a_1z+1$ finitely often? QUESTION [5 upvotes]: Given a polynomial $P=a_3z^3+a_2z^2+a_1z+1, z >0$ with non-negative integer coefficients $a_1, a_2, a_3\ne 0$, it appears if $P$ is not factorizable then there are finitely many positive integers $x, z$ such that $xz+1 \mid P(z)$, $xz+1 a^2+2a$. However the proof for the general case doesn't follow directly from the proof for the case $P=a_3z^3+1$. Also for a particular triple $(a_1, a_2, a_3)$, what's the minimum value of $z$ such that $xz+1$ is not a proper divisor of $P(z)$ for all $x>1$? My thoughts: If $P$ is factorizable then we can find integers $b_1, b_2, b_3$ such that $a_3z^3+a_2z^2+a_1z+1=(b_1z+1)(b_2z^2+b_3z+1)$. Expanding and comparing coefficients we get $b_1+b_3=a_1$, $b_1b_3+b_2=a_2$, and $b_1b_2=a_3$. Since $P$ is assumed non-factorizable, we will have to use this result somewhere in the proof. REPLY [8 votes]: The conjecture is true. That is, if the integral cubic polynomial $$P(Z)=a_3 Z^3+a_2 Z^2+a_1 Z+1$$ is irreducible in $\mathbb{Z}[Z]$ (hence also in $\mathbb{Q}[Z]$ by Gauss's lemma), then there are only finitely many positive integer solutions of the equation $$(xz+1)(yz+1)=P(z).$$ 1. First we consider the case when $x\mid a_3$ or $y\mid a_3$. By symmetry, it suffices to deal with the case $x\mid a_3$. We fix $x$ for this section. By long division, we get an integral quadratic polynomial $Q\in\mathbb{Z}[Z]$ and a nonzero integer $r\in\mathbb{Z}$ such that $$a_3^2 P(Z)=(xZ+1)Q(Z)+r.$$ If $xz+1\mid P(z)$, then $xz+1\mid r$, hence there are finitely many possibilities for $z$ (and also for $y$). 2. Now we consider the case when $x\nmid a_3$ and $y\nmid a_3$. We rewrite the original equation as $$tz=x+y-a_1\qquad\text{where}\qquad t:=a_3z+a_2-xy.$$ Here $t$ is an integer. If $t\leq 0$, then $x+y\leq a_1$, which leads to finitely many triples $(x,y,z)$. So let us focus on the case $t>0$. We use an identity inspired by the OP's earlier post: \begin{align*} (tx-a_3)(ty-a_3)&=t^2 xy-a_3 t(x+y)+a_3^2\\ &=t^2(a_3z+a_2-t)-a_3 t(tz+a_1)+a_3^2\\ &=-t^3+a_2 t^2-a_1 a_3 t+a_3^2. \end{align*} We conclude that $t\leq 3\max(|a_1|,|a_2|,|a_3|)$, for otherwise the LHS is positive, while the RHS is negative. Moreover, the factors on the LHS are nonzero integers by $x\nmid a_3$ and $y\nmid a_3$. So there are finitely many possibilities for the factors on the LHS (namely they are integral divisors of the finitely many possible values of the RHS), hence also for the triple $(x,y,z)$.<|endoftext|> TITLE: A continuous map with zero-dimensional fibres QUESTION [6 upvotes]: Let $f: X\to Y$ be a continuous surjective map between compact metric spaces. Suppose the fibre $f^{-1}(y)$ has zero topological dimension for each $y\in Y$. Then by Hurewicz dimension lowering theorem, one has $${\rm dim}(X)\le {\rm dim}(Y)+\sup_{y\in Y} {\rm dim}(f^{-1}(y))={\rm dim}(Y).$$ I would like whether there is a direct/simple proof of ${\rm dim}(X)\le {\rm dim}(Y)$ without using Hurewicz dimension lowering theorem. Thanks. REPLY [2 votes]: Yes, there is such a proof. A reference is Theorem 6.4.11 from Pears Dimension Theory of General Spaces, which is a great reference for dimension theory in general even though it's not as well known as other books on the topic. The other good news is that, assuming a small lemma which might even be a definition for you if you are only working with metrizable spaces, the proof is short and clear enough to be completely written here, let's start with said lemma (which is Corollary 3.4.4 in the aforementioned book and I will not prove it here). Lemma: Let $X$ be a normal space with $\dim X\leq n$ and let $\{G_i\}_{i\in I}$ be a locally finite open cover of $X$. Then there exist and open cover $\{H_i\}_{i\in I}$ of $X$ such that the order of $\{H_i\}_{i\in I}$ is at most $n$ and $H_i\subseteq G_i$ for all $i$. With this lemma available we can prove the result you want (I'm copying the statement and the proof from the book by Pears in full generality but you can ignore most point-set technicalities since you are interested in compact metric spaces). Theorem: Let $X$ be a normal space, let $Y$ be a paracompact Hausdorff space and let $f\colon X\to Y$ be a continuous closed surjection such that $\dim f^{-1}(y)=0$ if $y\in Y$. Then $\dim X\leq\dim Y$. Proof: We shall show that if $\dim Y\leq n$ then $\dim X\leq n$. Let $$\{U_1,\ldots U_k\}$$ be an open covering of $X$. If $y\in Y$, then since $Y$ is a $T_1$-space, $f^{-1}(y)$ is a closed subspace of $X$ and thus is a normal space. Since $$\dim f^{-1}(y)=0,$$ there exist a disjoint closed covering $\{F_{1y},\ldots, F_{ky}\}$ of $f^{-1}(y)$ such that $F_{iy}\subseteq U_i$ for each $i$. Since $X$ is a normal space and the sets $F_{iy}$ are closed in $X$, there exist disjoint open sets $G_{iy}$ such that $$F_{iy}\subseteq G_{iy}\subseteq U_i,\text{ for } i=1,\ldots,k.$$ Since $f^{-1}(y)\subseteq \bigcup_{i=1}^k G_{iy}$ and $f$ is a closed mapping, there exist an open neighbourhood $W_y$ of $y$ in $Y$ such that $f^{-1}(W_y)\subseteq\bigcup_{i=1}^k G_{iy}$. Since $Y$ is a paracompact normal space such that $\dim Y\leq n$, it follows from Corollary 3.4.4. that there exists an open covering $\{V_y\}_{y\in Y}$ of $Y$ of order not exceeding $n$ such that $V_y\subseteq W_y$ for each $y$. If $y\in Y$ and $i=1,\ldots,k$ let $H_{iy}=G_{iy}\cap f^{-1}(V_y)$. Then $\{H_{iy}\}_{y\in Y,i=1,\ldots k}$is an open covering of $X$ of order not exceeding $n$ which is a refinement of $\{U_1,\ldots, U_k\}$.Thus $\dim X\leq n$. $\square$<|endoftext|> TITLE: Existence of translation-invariant basis on $C_c(\mathbb R)$ QUESTION [17 upvotes]: Consider the space $C_c(\mathbb R)$ of complex-valued continuous functions of compact support. This is a vector space over $\mathbb C$, and I am not considering any topology, so the question is algebraic. A set $A \subseteq C_c(\mathbb R)$ is called shift-invariant if for all $f \in A$ and $t \in \mathbb R$ we also have $T_t f \in A$, where $T_tf(x)=f(x-t)$. I have the following question: Question Does there exist a shift-invariant basis $B$ for $C_c(\mathbb R)$? It is a standard application of Zorn's lemma that there exists a subset $A \subseteq C_c(\mathbb R)$ which is a maximal linearly independent shift-invariant subset, but I do not think that $A$ is always a basis. Such a set $A$ has the property that for all $g \notin \mbox{Span}(A)$ the elements $\{ T_t g : t \in \mathbb R \}$ are linearly independent and none belong to $\mbox{Span}(A)$, but unfortunately this does not seem enough to show that $A \cup \{ T_t g : t \in \mathbb R \}$ is linearly independent. REPLY [11 votes]: No, there is no shift-invariant basis of $V=C_c(\mathbb R).$ I'll use the formulation in YCor's answer, so we need to show that $V$ is not a free $B$-module where $B=\mathbb C[T^r:r\in \mathbb R],$ with $T^r$ acting as the translation $T_r.$ Let $f(x)=\max(0,1-|x|).$ Suppose there is a $B$-module basis $\{v_i\}$ for $V.$ Then we could write $f$ as a finite sum $\sum P_i v_i$ with $P_i\in B.$ By integrating both sides, we have $P_i(1)\neq 0$ for some $i$ (we can evaluate elements of $B$ at $T=1$ meaningfully - just sum the coefficients). If we define $f_n(x)=f(2^nx)$ then $f_n=\tfrac12(T^{-2^{-n}/4}+T^{2^{-n}/4})^2 f_{n+1}.$ This means $f,$ and hence $P_i,$ are divisible by $1+T^{2^{-n}}$ for all $n\geq 1.$ This is a contradiction: Lemma. Assume $P\in B$ is divisible by $1+T^{2^{-n}}$ for all $n\geq 1.$ Then $P(1)= 0.$ Proof: Write $P$ as a finite sum $\sum r a_rT_r.$ Consider a coset $C\in \mathbb R/\mathbb Z[1/2].$ Each polynomial $\sum_{r\in C} a_rT^r$ must still be divisible by $1+T^{2^{-n}}$ for all $n\geq 1.$ So we can reduce to the case where only one of these cosets has non-zero coefficients $a_r.$ This means that $P$ can be written in the form $\sum_m a_m T^{m2^{-n}+r}$ for some integer $n$ and some real $r,$ and where $m$ ranges over a finite set of integers. By applying a shift we can furthermore assume $m$ ranges over non-negative integers, and $r=0.$ We have then reduced to the case where $P$ is a polynomial in $T^{2^{-n}}.$ By injectivity of the unit circle $S^1,$ we can pick a group homomorphism $\phi:\mathbb R\to S^1$ sending $2^{-n-1}$ to $-1.$ This gives a ring automorphism $\sigma$ of $B$ taking $T^r$ to $\phi(r)T^r.$ By assumption $P$ is divisible by $1+T^{2^{-n-1}},$ so $P=\sigma(P)$ is also divisible by $1-T^{2^{-n-1}},$ giving $P(1)=0.$<|endoftext|> TITLE: When do two topoi have the same cohomology of constant sheaves QUESTION [7 upvotes]: Recently, I have some questions for some generalizations from algebraic topology. I learn some homotopy theory in algebraic topology. We know that, if two spaces are homotopy, then they have same cohomology groups for constant sheaves. I want to know if there are similar theory for topoi, such as homotopy theory for topoi? Thanks for your answers REPLY [6 votes]: There is a notion of the étale homotopy type of a (Grothendieck) topos, going back to Artin and Mazur (I think). However, in classic "French" fashion they turned a theorem (in one setting) into a definition (in a more general setting): roughly speaking, two toposes have the same étale homotopy type if they have the same étale fundamental group and the same cohomology for all (constant sheaf) coefficients. So perhaps this isn't what you're looking for, but let me continue. Let $\mathcal{E}$ be a topos. Artin and Mazur define the étale homotopy type of $\mathcal{E}$ under the assumption that $\mathcal{E}$ is locally connected, which for the purpose of this discussion means the unique colimit- and finite-limit-preserving functor $\Delta : \textbf{Set} \to \mathcal{E}$ has a left adjoint $\pi : \mathcal{E} \to \textbf{Set}$. Let $\textbf{Hc} (\mathcal{E})$ be the category of hypercovers of the terminal object $1$ in $\mathcal{E}$. (The objects are simplicial objects $K$ in $\mathcal{E}$ such that the unique morphism $K \to 1$ is a local trivial Kan fibration and the morphisms are simplicial homotopy classes of morphisms.) Let $\mathcal{K} = \operatorname{Ho} \textbf{sSet}$ be the category of simplicial sets localised with respect to weak homotopy equivalences. The étale homotopy type of $\mathcal{E}$, as defined by Artin and Mazur, is the following pro-object in $\mathcal{K}$: $$\Pi (\mathcal{E}) = \varprojlim_{K : \textbf{Hc} (\mathcal{E})} \pi (K)$$ Here, $\pi (K)$ denotes the simplicial set obtained by applying $\pi : \mathcal{E} \to \textbf{Set}$ degreewise to the simplicial object $K$. (The reason for using simplicial homotopy classes of morphisms in the definition of $\textbf{Hc} (\mathcal{E})$ is so that we get a cofiltered category. This category is usually not small, but it is coinitially small, so we do indeed get a pro-object.) Observe that, for a Kan complex $X$, $$\textbf{Pro} (\mathcal{K}) (\Pi (\mathcal{E}), X) \cong \varinjlim_{K : \textbf{Hc} (\mathcal{E})^\textrm{op}} \mathcal{K} (\pi (K), X) \cong \varinjlim_{K : \textbf{Hc} (\mathcal{E})^\textrm{op}} \pi_0 [\textbf{s} \mathcal{E}] (K, \Delta (X))$$ where $\pi_0 [\textbf{s} \mathcal{E}]$ is the category of simplicial objects in $\mathcal{E}$ modulo simplicial homotopy. The simplicial analogue of Verdier's hypercovering theorem states that the RHS computes $\pi_0 (\textbf{R} \Gamma (\Delta (X)))$, which you might interpret as analogous to $H_0$ of the derived global sections of a chain complex of sheaves. This is literally the case when $X$ is a simplicial abelian group, so you can extract the classical Verdier hypercovering theorem from this. Put it another way, the pro-object $\Pi (\mathcal{E})$ encodes enough information to determine the cohomology of constant sheaves on $\mathcal{E}$, which is what you wanted – but as I said in the first paragraph, in some sense the definition was constructed to make this true. Now let me discuss the notion of shape, which has been mentioned in the comments. The category $\mathcal{K}$ is not a pleasant category to work with, and $\textbf{Pro} (\mathcal{K})$ is even less pleasant. The restriction to locally connected $\mathcal{E}$ is also somewhat unsatisfying. In the years since Artin–Mazur new technologies have been developed for abstract homotopy theory and using this we obtain an improved version of the étale homotopy type. First, we must obtain an $\infty$-topos from $\mathcal{E}$. Take the category of simplicial objects in $\mathcal{E}$ and localise (in the $(\infty, 1)$-categorical sense now) with respect to local weak homotopy equivalences to obtain an $\infty$-topos $\tilde{\mathcal{E}}$. Let $\mathcal{S}$ be the $(\infty, 1)$-category of $\infty$-groupoids. The global sections functor $\Gamma : \tilde{\mathcal{E}} \to \mathcal{S}$ has a left adjoint $\Delta : \mathcal{S} \to \tilde{\mathcal{E}}$, and as before we are interested in the composite functor $X \mapsto \pi_0 (\Gamma (\Delta (X)))$. Since $\Gamma$ is an accessible right adjoint and $\Delta$ is a left adjoint that preserves finite limits, the composite $\Gamma \Delta$ is an accessible functor that preserves finite limits. The $(\infty, 1)$-category of accessible functors $\mathcal{S} \to \mathcal{S}$ that preserve finite limits is equivalent to the $(\infty, 1)$-category of pro-objects in $\mathcal{S}$, so $\Gamma \Delta$ corresponds to some pro-object $\Pi (\tilde{\mathcal{E}})$ – the shape of $\tilde{\mathcal{E}}$ is this object. (Given an inverse diagram $T$ in $\mathcal{S}$, the functor $\varinjlim \mathcal{S} (T, -) : \mathcal{S} \to \mathcal{S}$ is an accessible functor that preserves finite limits; the fact is that all accessible functors $\mathcal{S} \to \mathcal{S}$ that preserve finite limits arise in this way.) The advantage of the formulation of shape is that it is immediate how the shape encodes the cohomology of constant sheaves, but again this seems to be a trick – a different one from before, but in some sense there is still nothing deep going on. One might even say there is even less depth here because even hypercovers have disappeared from view. But perhaps this answers your question about whether there is a homotopy theory of toposes.<|endoftext|> TITLE: How to generalize the various vector calculus theorems to distributions? QUESTION [16 upvotes]: Here is a list of vector calculus identities; in the proof of these identities, we all assume that these functions are $^$ in an open set, and we usually use these identities to calculate integrals, however, sometimes the function may has singularities in the domain of integration (for example $\frac{1}{r}$ has 0 as a singularity), and the div of such a function contains a Dirac delta function (e.g. $\mathop{\mathrm{div}}\left(\frac{}{||^3}\right)=4\pi ()$ ), and in this case we must regard the div of this function as a distribution (a linear functional on the vector space of all smooth real-valued functions with compact support). My question is the following: do the identities of vector calculus still hold? And if so, what is the meaning of curl, div in this case, do they mean distribution? REPLY [13 votes]: This is an addition to Dirk's answer. It is of course impossible to give an all-embracing answer to your question—there are so many identities. But the general answer is a resounding YES. The proofs are usually very simple, using a density argument—distributions can be approximated by smooth functions in their natural topologies so one can take limits of the classical versions for smooth functions. By the way this means that one can avoid some of the complications involved in using minimal smoothness assumptions. There are, of course, some caveats (already hinted at above). If products are involved there can be difficulties. However, all is not lost—avoid the common fallacy that the fact that one cannot always multiply distributions means that one can NEVER multiply them. Usually one can! There are many situations where one can a rigorous and elementary definition—for starters the product of a smooth scalar function and distributional vector field (or vice versa) is well-defined and the usual rules (product formula, etc.) hold. The question of integration (e.g., in the Stokes theorem). Distributions, say on the line, always have primitives but not necessarily definite integrals. Again this doesn’t mean that they NEVER have—again they “usually do in practice” and have the sort of properties you would expect from the classical case. These concepts were developed decades ago using elementary methods (no functional analysis or duality theory for locally convex spaces. Not that I have anything against the latter—they just don’t usually occur in the toolbox of many mathematicians who want to use distributions). I could go on and on about this but I will give two further remarks: Evaluation at points: it is not true that every distribution has a value at every point but again this does NOT mean that one can NEVER evaluate a distribution at a point. Again this concept was investigated in detail (in the 50’s and 60’s of the last century) using elementary methods and most distributions of practical relevance can be evaluated at most points. Composition with functions. Again symbols like $T\circ f$, i.e., the substitution of a function in a distribution, are not always well-defined. However, there are general situations where it can be rigorously defined and then it has the familiar properties. The theory of distributions was developed by Sobolev and Schwartz, the latter using the language and techniques of duality for locally convex spaces. A number of mathematicians were then motivated to develop the theory from a more direct and elementary standpoint (i.e., at the level of an advanced calculus, or in germanophone countries “Analysis”, course). However, most texts use the Schwartz approach and the latter seems to have slipped into oblivion. All these approaches are on record and easily available online (some names: Mikusinski, Heinz König, Sebastião e Silva, Sikorski). I would be happy to supply more details.<|endoftext|> TITLE: How to read the paper of Arthur on trace formula on general reductive groups QUESTION [7 upvotes]: My question is about the correct order to read the papers by Arthur on trace formula. Arthur's papers are perfectly well-written, but maybe a little too hard for me to go through easily. I would like to start with unrefined trace formula, but there are bunches of papers already. There is a paper on trace formula for rank-1 groups (1974), and there are two papers on trace formula for general reductive groups (1978 & 1980). Also there is a paper in Corvallis Proceedings (1979), and a notes on trace formula (2005) which has become the standard introductory notes on this subject. I am seeking for any advice on how to read these references. I wish to at least go through some details to feel how the formula is proved. Now I have read the notes by Whitehouse on trace formula for $\operatorname{GL}(2)$ (which covers a great many details about computation), and read a small part of Arthur's paper on rank-1 groups (1974). My question is that whether it is better to take 1974 paper first to gain more intuition, or instead start with the more general setting of reductive groups (like 1978 & 1980 papers), where maybe things are actually more clarified. Any recommendation of other references is also sincerely welcomed. I am not aware of any other introductory references on trace formula on general reductive groups. Thank you very much for reading this lengthy question! I am totally new to this community, so you are welcomed to point out any mistake. REPLY [6 votes]: One of the best places to learn about trace formula, other than David Whitehouse's wonderful notes, are the notes by Erez Lapid. https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.494.5118&rep=rep1&type=pdf Arthur's trace formula relies on Langlands's work on Eisenstein series and Spectral decomposition from 1962-64. This stuff is extremely difficult for general reductive groups. It would be near insanity to directly go to Arthur's papers or notes without mastering the spectral decomposition of automorphic forms on GL(2) and the Selberg trace formula. It is not wise to learn trace formula without learning about spectral decomposition. For the general case, Borel's notes are reasonable beginning. http://homepages.math.uic.edu/~rtakloo/papers/borel/borel3.pdf My advisor Paul Garrett's book is, of course, my favorite place. They treat several examples, but do not do any higher rank case completely. https://www-users.cse.umn.edu/~garrett/m/v/current_version.pdf<|endoftext|> TITLE: Group extensions with non-abelian kernel QUESTION [9 upvotes]: If $N$ is a normal subgroup of $G$ then there is a coupling: that is, a representation of $G/N$ in $\operatorname{Out}(N)$. In that case, the extensions of $N$ by $G/N$ affording the same coupling are classified by the elements of $H^2(G/N, \,Z(N))$. Given a coupling there may be no extension corresponding: the obstructions are non-zero elements of $H^3(G/N, \, Z(N))$. Question. Does anyone know a good reference for the functioriality of this theory wrt morphisms of group extensions and in particular to the application to the study of forming Malcev completions of nilpotent normal subgroups in a group? REPLY [4 votes]: There is an account of the general theory of group extensions with non-abelian kernel in Gruenberg's Springer Lecture Notes 143, Cohomological Topics in Group Theory: here he refers to his own paper 'A New Treatment of Group Extensions, Math. Zeit. 102, 1967, 340--350. It's possible that Gruenberg was a pioneer in this field as well as taking the novel approach using what we now call the 'Gruenberg resolution'. I have not explored Schreier's paper from 1926: it would be surprising if this included more than a fragment of what Gruenberg knew but I am ready to be corrected. Eilenberg and Mac Lane seem to confine themselves largely to the case of abelian kernel. More modern treatments include D.J.S.Robinson's in 'A Course in the Theory of Groups' Graduate Texts in Mathematics 80, Spriner, and Robinson's treatment follows that of Gruenberg moderately closely. The idea of applying this theory to the behaviour of groups under Mal'cev completion of their Fitting subgroups must have been widely imagined and it would be interesting to know if there is any account of this specific direction.<|endoftext|> TITLE: Existence of a translation-invariant basis of $\ell^2$ QUESTION [13 upvotes]: This question is heavily inspired by this other one, but is meant to be a hopefully more accessible variant of it (and I think slightly more natural). I give four equivalent formulations of the same question in the hope of inspiring different ways to view it: Let $\ell^2(\mathbb{Z};\mathbb{C})$ be the $\mathbb{C}$-vector space of square-summable $\mathbb{Z}$-indexed complex-valued sequences. Does there exist a basis $B$ of $\ell^2(\mathbb{Z};\mathbb{C})$ as a $\mathbb{C}$-vector space (not as a Hilbert space: we completely ignore topology here) which is translation-invariant in the sense that if $e \in B$ then any shift $T^r(e) : k \mapsto e(k-r)$ of $e$ is also in $B$ for any $r\in\mathbb{Z}$? Is $\ell^2(\mathbb{Z};\mathbb{C})$ a free $\mathbb{C}[T^{\pm 1}]$-module, where $\mathbb{C}[T^{\pm 1}]$ is the ring of Laurent polynomials with complex coefficients, and $T$ acts on $\ell^2(\mathbb{Z};\mathbb{C})$ by shift ($T^r(e) : k \mapsto e(k-r)$)? Let $L^2(\mathbb{R}/\mathbb{Z};\mathbb{C})$ be the $\mathbb{C}$-vector space of square-integrable complex-valued functions on $\mathbb{R}/\mathbb{Z}$ modulo equality a.e. Does there exist a basis $B$ of $L^2(\mathbb{R}/\mathbb{Z};\mathbb{C})$ as a $\mathbb{C}$-vector space (ignoring topology) which is stable under multiplication by $\exp(\pm 2i\pi\theta)$ (in the sense that if $f\in B$ then $\theta \mapsto \exp(2i r \pi\theta) \thinspace f(\theta)$ also belongs to $B$ for any $r\in\mathbb{Z}$)? Is $L^2(\mathbb{R}/\mathbb{Z};\mathbb{C})$ a free $\mathbb{C}[T^{\pm 1}]$-module, where now $\mathbb{C}[T^{\pm 1}]$ is seen as the ring of trigonometric polynomials ($\mathbb{C}$-linear combinations of $\theta \mapsto \exp(2i k\pi\theta)$)? (The equivalence of (1) and (2) is proved analogously to YCor's answer in the aforementioned question: the fact that $\ell^2(\mathbb{Z};\mathbb{C})$ is a torsion-free $\mathbb{C}[T^{\pm 1}]$-module follows from the fact that no linear recurrence sequence is square-integrable. The equivalence of (1) and (3) or (2) and (4) is by Fourier transform.) A few remarks: $\ell^2(\mathbb{Z};\mathbb{C})$ (or, of course, $L^2(\mathbb{R}/\mathbb{Z};\mathbb{C})$, since they are the same) is a flat $\mathbb{C}[T^{\pm 1}]$-module since it is torsion-free and $\mathbb{C}[T^{\pm 1}]$ is a PID (hence a Dedekind domain). Moreover, because of a theorem of Bass (“Big Projective Modules are Free”, Ill. J. Math. 7 (1963), 24–31: theorem 4.3), to see that it is free it is enough to see that it is projective. (I don't think any of this helps, but it is still worth pointing out.) As I note in a partial answer to the other question, $\mathbb{C}[T^{\pm 1}]^I$ is not free for any infinite set $I$. I suspect it can probably be deduced from this that $\mathbb{C}^{\mathbb{Z}}$ (viꝫ., all complex-valued sequences) does not admit a translation-invariant vector-space basis (the module reformulation of this would be that it is a direct sum of cyclic modules: obviously, here, it is not free since it is not even torsion-free). At the other extreme, $\mathbb{C}^{(\mathbb{Z})}$ (viꝫ., complex-valued sequences with finite support) is trivially a free $\mathbb{C}[T^{\pm 1}]$-module (with rank $1$). So $\ell^2(\mathbb{Z};\mathbb{C})$ strikes me as something natural to ask “in between”. But if someone is capable of answering question (1) for other natural intermediate spaces between $\mathbb{C}^{(\mathbb{Z})}$ and $\mathbb{C}^{\mathbb{Z}}$, like $\ell^1(\mathbb{Z};\mathbb{C})$ or $c_0(\mathbb{Z};\mathbb{C})$, this also counts as an answer to this question! An vaguely analogous situation that may be noted is that $\mathbb{Z}^{\mathbb{N}}$ is not a free $\mathbb{Z}$-module (i.e., abelian group), $\mathbb{Z}^{(\mathbb{N})}$ trivially is one, and in between them, the set of bounded $\mathbb{Z}$-valued sequences is free (a theorem of Specker). REPLY [3 votes]: Inspired by Harry West's answer to the question about continuous functions with compact support, I can now answer my own version thusly: $\ell^2(\mathbb{Z};\mathbb{C})$ has no translation-stable $\mathbb{C}$-vector space basis. To show this, I will use form (4) of the question and prove that $L^2(\mathbb{R}/\mathbb{Z};\mathbb{C})$ is not free as a module over the ring $\mathbb{C}[T^{\pm1}]$ of trigonometric polynomials (where $T$ is now $\theta \mapsto \exp(2i\pi\theta)$). Indeed, let $f \in \mathbb{C}[T^{\pm1}]$ be a non-constant trigonometric polynomial which does not vanish on $\mathbb{R}/\mathbb{Z}$, e.g., $T + 3 + T^{-1}$ (that is, $3 + 2\cos(2\pi\theta)$). Now, crucially, $\frac{1}{f}$ is a well-defined continuous function $\mathbb{R}/\mathbb{Z} \to \mathbb{C}$ (in particular, it is bounded). So, for any $g \in L^2(\mathbb{R}/\mathbb{Z};\mathbb{C})$, we have $\frac{g}{f} \in L^2(\mathbb{R}/\mathbb{Z};\mathbb{C})$; in particular, $\frac{1}{f^n} \in L^2(\mathbb{R}/\mathbb{Z};\mathbb{C})$ for any $n\in\mathbb{N}$. Now if $L^2(\mathbb{R}/\mathbb{Z};\mathbb{C})$ had a $\mathbb{C}[T^{\pm1}]$-basis, expressing $\frac{1}{f^n}$ on this basis and multiplying by $f^n$, we see that the coefficients of $1$ on the basis must be multiple of $f^n$ for all $n$. But zero is the only element of $\mathbb{C}[T^{\pm1}]$ which is divisible by $f^n$ for all $n$, because $\mathbb{C}[T^{\pm1}]$ is a principal ideal domain (as a localization of $\mathbb{C}[T]$ which is one), and $f$ is not invertible (the invertible elements of $\mathbb{C}[T^{\pm1}]$ are the $c T^r$ for $r\in\mathbb{Z}$, and $f$ is not of this form). This is a contradiction.<|endoftext|> TITLE: Question about a proof in Berthelot's crystalline book QUESTION [5 upvotes]: Below is an excerpt from Berthelot's book on crystalline cohomology. I don't understand the last sentence, namely why it follows that $\sigma\circ \varepsilon$ is an isomorphism. For what it's worth, $P^1$ is the sheaf of principal parts and $E$ is an $\mathcal O _X$-module. I can elaborate on what $\sigma,\varepsilon,\tau$ are, but perhaps I'm just missing some basic algebra... We have an endomorphism of a module which becomes the identity modulo a square zero ideal. Why, in this case, is it an isomorphism? REPLY [10 votes]: Suppose you have a commutative ring $R$, a square zero ideal $I\subset R$, a $R$-module $M$ and an endomorphism $u$ of $M$ which is the identity modulo $IM$. Then $v:= 1_M-u$ maps $M$ to $IM$, hence $v^2=0$. Therefore $u=1_M-v$ is invertible — its inverse is $1_M+v$.<|endoftext|> TITLE: Does the strict henselization satisfy Going-Up? QUESTION [7 upvotes]: $\DeclareMathOperator\sh{sh}$This question is cross-posted from Math.SE where it has gone unanswered for a week -- perhaps it is harder than I guessed. My question is this: Let $A$ be a local commutative, unital ring and let $A^{\sh}$ be a strict henselization of $A$. Does the structure map $A\rightarrow A^{\sh}$ satisfy Going-Up? (If the answer is no, are there broad natural conditions on $A$ that would make it yes, such as being noetherian, or excellent?) The Going-Up property does not appear in standard lists of properties of the strict henselization such as those at the Stacks Project and in EGA IV 18.8.12 and 18.8.13. I have some vague reasons to think the strict henselization "should" satisfy Going-Up, and some vague reasons to think it's plausible that it does. But one of these reasons would also seem to apply to the completion, which does not always satisfy Going-Up, see here. In any case, here they are: The most concrete reason I have is that the strict henselization is a colimit of a directed system of étale $A$-algebras, which themselves are localizations of finite $A$-algebras, with the directed system strung together in a way that guarantees every prime of $A$ (has a lift that) survives all the localizations. Finite $A$-algebras do satisfy Going-Up; the localization step can interfere with this by destroying primes, but the structure of the directed system guarantees that (i) $A$'s maximal $\mathfrak{m}$ lifts to the maximal of $A^{\sh}$, so that $A\rightarrow A^{\sh}$ is faithfully flat, thus all primes of $A$ do have a lift in $A^{\sh}$, and (ii) all lifts of any of $A$'s primes in any intermediate $A'$, that are not in the kernel of $A'\rightarrow k^{sep}$, will eventually be destroyed. This does not add up to a proof, of course (or I wouldn't be asking the question), but it does mean that if $A\rightarrow A^{\sh}$ fails to have Going-Up, it does so in kind of a funny way: it seems to me it would have to look like a chain $\mathfrak{p}\subset\mathfrak{q}$ in $A$, and a prime $\mathfrak{P}$ of $A^{\sh}$ lifting $\mathfrak{p}$, and some intermediate étale algebra $A'=B_h$ with $B$ finite over $A$ and $h\in B$, such that none of the lifts of $\mathfrak{q}$ in $B$ containing the pullback of $\mathfrak{P}$ in $B$ (which exist, because $A\rightarrow B$ satisfies going up) are contained in the kernel of $B\rightarrow B_h \rightarrow k^{sep}$, even though this kernel does contain at least one prime that pulls back to $\mathfrak{q}$, and any such prime does contain a prime that pulls back to $\mathfrak{p}$ (this is all known by pulling back to $B$ a chain in $A^{\sh}$ meeting these requirements, which must exist because $A\rightarrow A^{\sh}$ is faithfully flat and thus satisfies Lying-Over and Going-Down). Does this really happen? On a more metaphorical and handwavy level, I think of $A^{\sh}$ as a kind of "universal cover of a very small neighborhood of (the closed point of) $\operatorname{Spec}A$." If I reason by analogy with covering space theory in topology, Going-Up translates to the claim that if I have a flag of irreducible subvarieties all going through the same point, and I have a lift of the highest-dimension of these to the universal cover, then I can extend it to a lift of the whole flag. This is true in covering space theory. (But the imprecision of this reasoning is made clear by considering that it's not obvious why something similar wouldn't also apply in the completion, where Going-Up does not always hold.) On a perhaps even less precise level, there do exist theorems about the strict henselization that seem to hew to the motto, "chains of primes are respected by strict henselization"; for example, $A$ is universally catenary iff $A^{\sh}$ is universally catenary (EGA IV 18.8.17). This is not true for the completion, which is always universally catenary (at least in the noetherian case), whether or not $A$ is. I'm looking forward to your thoughts. Thanks in advance. REPLY [7 votes]: Here is a counterexample. Let $S$ be a smooth surface (irreducible). Let $C_1, C_2 \subset S$ be two disjoint smooth curves in $S$ which happen to be isomorphic. Let $X$ be the result of glueing $C_1$ and $C_2$ by this isomorphism. The singular locus of $X$ is a curve $C$ which is mapped onto isomorphically by $C_1$ and $C_2$ via the morphism $S \to X$. On the other hand, let $Y$ be the result of glueing two copies $S_1, S_2$ of $S$ where $C_1 \subset S_1$ is glued to $C_2 \subset S_2$ via the given isomorphism of $C_1$ with $C_2$. Again the singular locus of $Y$ is a curve $C' \subset Y$ which is mapped onto isomorphically by $C_1 \subset S_1$ and $C_2 \subset S_2$ via the morphism $S_1 \amalg S_2 \to Y$. We may and do think of $S_1$ and $S_2$ as closed subschemes of $Y$ (in fact $S_1$ and $S_2$ are the irreducible components of $Y$). There is a morphism $$ f : Y \longrightarrow X $$ compatible with the obvious morphism $S_1 \amalg S_2 \to S$ and the morphisms $S \to X$ and $S_1 \amalg S_2 \to Y$ mentioned above. For any closed point $c' \in C'$ corresponding to $c = f(c') \in C$ the morphism $f$ is etale at $c'$. OK, now we are going to choose a curve $D \subset X$ which is smooth, meets $C \subset X$ exactly at $c$. Just take a general smooth curve on $S$ passing through one of the two points above $c$ and take the image. As primes in $\mathcal{O}_{X, c}$ we will use $(0)$ and the prime ideal cutting out $D$. We can do this because $X$ and $D$ are irreducible. Then $f^{-1}(D)$ will have two irreducible components $D_1$ and $D_2$ with $D_1 \subset S_1$ and $D_2 \subset S_2$. But only one of these will pass through $c'$ by our choice of D (to see this I suggest drawing a picture). Say $c' \in D_1$. Then we pick our prime ideal in $\mathcal{O}_{Y, c'}$ to be the prime ideal $\mathfrak P$ cutting out $S_2$ which does indeed lie over $(0) \subset \mathcal{O}_{X, c}$ as $S_2 \to X$ is dominant. But there is no prime containing $\mathfrak P$ in $\mathcal{O}_{Y, c'}$ lying over the prime ideal cutting out $D$ because $D_2$ does not pass through $c'$.<|endoftext|> TITLE: The integral closure $\overline{\mathbb{Z}}$ and the group $\overline{\mathbb{Z}}^{\times}$ QUESTION [5 upvotes]: Let $\mathbb{Q}$ be the field of rational numbers, and let $\overline{\mathbb{Q}}$ be its algebraic closure. Assume $\overline{\mathbb{Z}}$ is the integral closure of $\mathbb{Z}$ in $\overline{\mathbb{Q}}$, my question is that, what can we say about the integral domain $\overline{\mathbb{Z}}$? Explicitly: (1) Is it noetherian? (Hardly I think) (2) How do we describe the nonzero prime ideals of $\overline{\mathbb{Z}}$ ? Are they maximal ? What are the residue fields? (I guess the residues are algebraic closure of finite field) (3) Is it factorial? (4) What is $\overline{\mathbb{Z}}^{\times}$ as an abstract group? For (4), Obviously it contains all roots of unity, and as an abstract group it is isomorphic to $\mathbb{Q}/\mathbb{Z}$. Quotient it out, it becomes uniquely divisible, and hence an $\mathbb{Q}$-vector space. Can we find an explicit basis of it? All these questions are clear for the integer ring $\mathcal{O}_K$ of a number field $K$, from algebraic number theory. And it is obvious that $\overline{\mathbb{Z}}$ is the direct limit of the integer ring of number fields, i.e $\overline{\mathbb{Z}}=\varinjlim_{K}\mathcal{O}_K$. But I am not quite familiar with the property of direct limit of rings. Can anyone give me some detail? Additionally, one can ask the following question: Consider $\overline{\mathbb{Q}}^{\times}$. It can be split into (not canonically) a product of a torsion part $\overline{\mathbb{Q}}^{\times}_{tor}$ and a free part $\overline{\mathbb{Q}}^{\times}/\overline{\mathbb{Q}}^{\times}_{tor}$ . The torsion part are exactly roots of unity, and is $\mathbb{Q}/\mathbb{Z}$. The quotient part becomes a $\mathbb{Q}$ vector space, too. What can we say about it? REPLY [6 votes]: I think we can describe $P$ a bit more, using Dirichlet’s unit theorem. Since $\overline{\mathbf{Z}} = \varinjlim_{[K:\mathbf{Q}] < \infty} \mathcal{O}_K$, the same is true for the units. Now Dirichlet tells us that taking logs of the archimedean valuations $v \mid \infty$ of $K$ gives us an embedding: $$\mathcal{O}_K^\times/\mu_\infty(K) \hookrightarrow \mathcal{L}_K := (\oplus_{v \mid \infty} \mathbf{R}_v)_0$$ Here, the subscript $(\underline{ })_0$ means the subspace where the coordinates sum to zero. Moreover, this induces an isomorphism after extending scalars to $\mathbf{R}$. The image is a discrete subgroup, so it must be a $\mathbf{Z}$-lattice with rank equal to the real dimension of the space on the right, which is $1$ fewer than the number of archimedean valuations. If $L/K$ is a finite extension, the inclusion $\mathcal{O}_K^\times \hookrightarrow \mathcal{O}_L^\times$ corresponds to the diagonal embedding of $\mathbf{R}_v$ into $\oplus_{v’\mid v} \mathbf{R}_{v’}$ where the $v’$ are the archimedean valuations of $L$ which extend $v$. Now, direct limits of abelian groups are exact, so taking limits of both sides gives us an embedding $$P = \overline{\mathbf{Z}}^\times/(\mu_\infty) \hookrightarrow \overline{\mathcal{L}} := \varinjlim_K \mathcal{L}_K \subseteq \prod_{v \mid \infty} \mathbf{R}_v$$ where now the $v$ range through the set of archimedean valuations on $\overline{\mathbf{Q}}$, i.e. the set of conjugate pairs of complex embeddings. The space $\overline{\mathcal{L}}$ is the set of vectors in the infinite product which are fixed by an open subgroup of the absolute Galois group (which acts transitively on the factors by permutation, with the stabilizer of a given $v$ given by the associated complex conjugation) which moreover have “trace” $0$, in the sense that averaging over the finite Galois orbit gives $0$. A spanning set is given by the countable set $\{e_{K, v_K} - e_{K, w_K}\}$ where $K$ is a number field, $v_K, w_K$ are archimedean places of $K$, and $e_{K, v_K}$ is the vector with $1$’s in every component $\mathbf{R}_v$ such that $v$ restricts to $v_K$ and $0$’s elsewhere. Certainly this set is not linearly independent, but I can’t think of a natural basis (if anyone else can, please share in the comments!). Since direct limits of abelian groups moreover commute with tensor products, we know that the induced map $P \otimes_{\mathbf{Z}} \mathbf{R} = P \otimes_{\mathbf{Q}} \mathbf{R} \rightarrow \overline{\mathcal{L}}$ is an isomorphism. Thus, the $\mathbf{Q}$-vector space $P$ is a $\mathbf{Q}$-“lattice” inside $\overline{\mathcal{L}}$.<|endoftext|> TITLE: Uniqueness of stationary measures for $(G,\mu)$ boundaries QUESTION [7 upvotes]: Let $G$ be a countable group acting minimally by homeomorphisms on a compact Hausdorff space $X$ and $\mu$ be a probability measure on $G$ whose support generates $G$ as a semigroup. Let $\nu$ is a $\mu$-stationary measure on $X$ such that $(X,\nu)$ is a $(G,\mu)$ boundary, ie for $\mu^{\mathbb{N}}$ almost every trajectory $g_n$ the pushforwards $g_1...g_n \nu$ weakly converge to a dirac measure. Does it follow that $\nu$ is the unique $\mu$-stationary Borel probability measure on $X$? This is the case in many geometric situations, for instance when the action is on the Gromov boundary of a hyperbolic group. In general I suspect the answer is false but cannot think of a counterexample... REPLY [3 votes]: A way to force a counter example is by taking $G$ to be an amenable group and choose $\mu$ such that the Furstenberg-Poisson boundary of $(G,\mu)$ is non-trivial. Now take $X$ to be a compact model of this Furstenberg-Poisson boundary. Then, by amenability, $X$ admits an invariant measure in addition to the $\nu$, thus at least two different stationary measures.<|endoftext|> TITLE: The double cover in the classical limit of $U_q(\mathfrak{sl}_2)$ QUESTION [9 upvotes]: I am trying to learn about Drinfeld–Jimbo quantum groups and I am having trouble with the classical limit of $U_q(\mathfrak{sl}_2)$. When properly expressed the limit makes sense as $q\to 1$ — see for example this question. But we get $U(\mathfrak{sl}_2) \otimes \mathbb{Z}/2$. This fact goes on to create many problems, such as the effective doubling of the number of representations, half of which are generally regarded as "uninteresting". It seems to me that finding a presentation of $U_q(\mathfrak{sl}_2)$ would be of great benefit. However I can't seem to find any discussion of this anywhere. Can anybody explain what is going on here? Is the double cover somehow a 'necessary consequence' of quantization? Is there some reason why people felt it best to stick with the double cover approach? REPLY [2 votes]: Well, i am not sure if this is what the OP is looking for but here is an heuristic method for computing the limit, avoiding the use of another algebra defined at $q=1$ and thus bypassing the "double cover problem" (mentioned in the OP): Let us use the standard presentations in terms of generators and relations, $U\big(sl(2)\big)$ is: $\ \ \ [H,X]=2X$, $[H,Y]=-2Y$, $[X,Y]=H$ whereas $U_q\big(sl(2)\big)$ is: $\ \ KK^{-1}=K^{-1}K=1$, $KE=q^2EK$, $KF=q^{-2}FK$, $[E,F]=\frac{K-K^{-1}}{q-q^{-1}}$ The problem is that we cannot just plug-in $q=1$ in the above relations because the limit is indeterminate. In order to work around this problem (essentially following Drinfeld's original approach), let us use the "change of variables": $$ q=e^{h/2} \ \ \ \ K=q^H=e^{hH/2} $$ These imply: $$\frac{dK}{dh}=\frac{HK}{2}, \ \ \ \ \ \ \ \ \ \ \lim_{h\to 0}K=1$$ Now $$ [K,E]=KE-EK=(q^2-1)EK=(e^h-1)EK $$ Differentiating the last relation wrt to $h$ gives: $$ \frac{1}{2}[HK,E]=e^hEK+(e^h-1)E\frac{h}{2}K $$ and finally take the limit of the above while $h\to 0$ to get: $$[H,E]=2E$$ With the correspondence $E\leftrightarrow X$ and $F\leftrightarrow Y$ (the $[K,F]$ relation is treated in a similar manner) these give the first two relations of $U\big(sl(2)\big)$. Now the third relation is written as $$ [E,F]=\frac{K-K^{-1}}{q-q^{-1}}=\frac{e^{hH/2}-e^{-hH/2}}{e^{h/2}-e^{-h/2}} $$ Take the limit of both sides at $h\to 0$, using Del' Hospital in the rhs: $$ [E,F]=\lim_{h\to 0}\frac{\frac{HK}{2}-(-\frac{HK}{2})}{\frac{1}{2}e^{h/2}-(-\frac{1}{2}e^{-h/2})}=H $$ which gives the third relation describing the multiplication of the $U\big(sl(2)\big)$ algebra as the $q\to 1$ limit of the multiplication of the $U_q\big(sl(2)\big)$ algebra. Finally, the coalgebra structure (comultiplication, counity) and the antipode limits can be handled in a similar manner to conclude that the $q\to 1$ limit of the Quantum group $U_q\big(sl(2)\big)$ is the Hopf algebra $U\big(sl(2)\big)$. Edit: Regarding your last question: Is there some reason why people felt it best to stick with the double cover approach? i would say that (contrary to the remark in the OP: This fact goes on to create many problems, such as the effective doubling of the number of representations ...) this "doubling" of the representations is the reason behind the double-cover approach in the sense that this provides a more direct analogy between the representations of $U_q\big(sl(2)\big)$ and the representations of $U\big(sl(2)\big)$: The representation theory of $U_q$ includes two classes of highest weight modules, parameterized by the positive integers and the sign $\epsilon =\pm 1$, unlike the rep theory of $U$ which includes a single class of highest weight modules parameterized by the positive integers. (This happens genererally for $q$ being either a root of unity or not, although for the root of unity case things are a little more subtle i.e. there are other classes of reps as well, an upper bound in their dim etc; but this is irrelevant to the rest). So, when taking the $q\to 1$ limit of $U_q$, its $\epsilon =1$ reps give the usual $U$-highest weight modules while its $\epsilon =-1$ reps correspond to the "double" class of reps of $U(\mathfrak{sl}_2) \otimes \mathbb{Z}/2$ (since the last algebra can be viewed as an extension of $U$ by adjoining a single, central generator satisfying $g^2=1$).<|endoftext|> TITLE: Does there exist a countable metric space which is Lipschitz universal for all countable metric spaces? QUESTION [7 upvotes]: Is there a countable metric space $U$ such that any countable metric space is bi-Lipschitz equivalent to a subset of $U$? How about $c_{00}(\mathbb{Q})$ where $\mathbb{Q}$ is the rational numbers? Thanks! REPLY [5 votes]: The affirmative answer to this problem follows from Lemma. For any countable dense subsets $X,Y$ in the half-line $\mathbb R_+=[0,+\infty)$ there exists a $C^2$-smooth function $f:\mathbb R_+\to\mathbb R_+$ such that $\bullet$ $f(X)\subseteq Y\cup\{0\}$; $\bullet$ $f(0)=0$; $\bullet$ $10$; $\bullet$ $f''(x)<0$ for all $x>0$. Proof. Such a function $f$ can be found by a standard back-and-forth argument. $\quad\square$ Now take any countable metric space $(X,d)$ and consider the countable subset $d(X\times X)$ of $\mathbb R_+$. By the above lemma, there exists a function $f:\mathbb R_+\to\mathbb R_+$ such that $f(d(X\times X))\subseteq \mathbb Q$, $f(0)=0$, $1< f'(x)< 2$ and $f''(x)<0$ for all $x>0$. These properties of the function $f$ imply that $$f(x+y)\le f(x)+f(y)\quad\mbox{and}\quad x TITLE: Possible contemporary improvement to bounded gaps between primes? QUESTION [19 upvotes]: In his summary of his book Bounded gaps between primes: the epic breakthroughs of the early 21st century, Kevin Broughan writes Which brings me to my final remark: where to next in the bounded gaps saga? As hinted before, the structure of narrow admissible tuples related to the structure of multiple divisors of Maynard/Tao, and variations of the perturbation structure of Polymath8b, and of the polynomial basis used in the optimization step, could assist progress to the next target. Based on “jumping champions” results, this should be 210. But who knows! [emphasis added] We have a few years of computer and also presumably analytic development since Polymath8, are we able to push the bound down to 210 with current technology? What would it take, if not just someone with enough motivation to sit down and do it/spend the computational cycles? Are we still a long way from getting to below 200? These are of course arbitrary numbers, in the grand scheme of things, since true progress down towards the bound of 12 enabled by the Elliott–Halberstam conjecture requires absolutely new ideas that seem too far from current knowledge. But there should be satisfaction in making some progress, even if not fundamentally substantial, similar to the recent case-by-case progress by Booker and Sutherland on outstanding sum-of-three-cubes cases below 1000. REPLY [20 votes]: I think that there is indeed some possibility to lower the bound, and this is something I've looked at seriously a few times. I spent a semester (in 2019) with the Computational Number Theory Group here at BYU trying to do it, without success. Let me outline some of the difficulties we found. Issue 1. The main technique used in establishing the current bound is optimizing a quotient of integrals. This integral expression was first developed by Maynard and Tao (independently). The optimized integral quotient, in turn, can be approximated very well by finding the largest eigenvalue of a product of certain matrices, $M_1 M_2^{-1}$, related to those integral expressions. To get better approximations, one needs to increase the size of the matrices. One of the roadblocks in improving knowledge of numerics is that the sizes of these matrices can be too large to store reasonably. I've spent some time optimizing code in Mathematica, storing the matrices as sparse association lists. One still runs into storage issues very easily, even in small dimensions. I'm happy to share this code with anyone who is interested. (Feel free to email me. Disclaimer: There are no comments explaining the code.) That said, the real bottleneck is the next issue. Issue 2. Generating the sparse entries that give rise to the eigenvalue problem takes time. To get to the current bound of $246$ in the Polymath 8b project we needed to slightly generalize the integral quotient mentioned above to an "epsilon-enlarged" region. This adds some complexity to the computations used to form the matrices. This epsilon-enlarged region is still amenable to quick computation, but the bound of $246$ might be the best one can get here. The polymath project did go beyond the epsilon-enlarged region, using "vanishing marginals", and this led (after a huge amount of effort) to the prime gap bound of $6$ (under a Generalized Elliot-Halberstam conjecture). Unfortunately, these vanishing marginal conditions do not seem to be amenable to quick computation in medium-sized dimensions (or even small dimensions). In the polymath 8b paper, there are different versions of the epsilon-enlargement idea, and perhaps one of these can immediately give an improved upper bound, but I've not been able to make that work. To sum up: I believe that there is some small amount of improvement possible to the current bound, and this could easily be done if someone were to improve the epsilon-enlarged region in a way that is amenable to quick computations.<|endoftext|> TITLE: Extending contents QUESTION [6 upvotes]: Suppose $\mu$ is a finitely additive measure on $X$ (aka “content”) with $\mu(X) < \infty$, defined on an algebra of sets $\mathcal A$. Let $$\mu^*(Y) = \inf \{ \mu(E) : E \in \mathcal A \wedge E \supseteq Y \}.$$ Question: Is it true that for all $Y \subseteq X$, there is an extension $\nu$ of $\mu$, where $\nu$ is a content defined at $Y$, such that $\nu(Y) = \mu^*(Y)$? If so, can this be proven without the axiom of choice? REPLY [6 votes]: I believe part of the question is answered in the following paper: Łoś, J.; Marczewski, E. Extensions of measure. Fund. Math. 36 (1949), 267–276. Let $c$ be any number between (inclusive) the inner and outer measure of $Y$. Then there is an extension $\nu$ which assigns $c$ to $Y$, with the domain of $\nu$ being the algebra of sets generated by the set $Y$ together with the sets in the domain of $\mu$. ($\mu$ is the measure to be extended.) I am not certain about the role (if any) of the axiom choice. Roger Purves REPLY [6 votes]: $\let\bez\smallsetminus\let\sdif\vartriangle$Such as extension exists, and it can be constructed explicitly without using the axiom of choice. Let $\mathcal B$ denote the algebra generated by $\mathcal A\cup\{Y\}$, i.e., $$\mathcal B=\bigl\{(Z_0\cap Y)\cup(Z_1\bez Y):Z_0,Z_1\in\mathcal A\bigr\}.$$ Put $$\mathcal F=\{E\in\mathcal A:E\supseteq Y\}.$$ Note that $\mathcal F$ is a filter of $\mathcal A$. For any $Z\in\mathcal A$, we define $$\begin{align} \mu_0(Z)&=\inf\{\mu(Z\cap E):E\in\mathcal F\},\\ \mu_1(Z)&=\sup\{\mu(Z\bez E):E\in\mathcal F\}. \end{align}$$ Since $\mu$ is monotone and $\mathcal F$ is a filter, we have $$\mu_0(Z)=\inf\{\mu(Z\cap E):E\in\mathcal F,E\subseteq E_0\}$$ for any $E_0\in\mathcal F$. In particular, $$Z\cap Y=Z'\cap Y\implies\mu_0(Z)=\mu_0(Z'),\tag1$$ because $E_0=X\bez(Z\sdif Z')\in\mathcal F$, and $Z\cap E=Z'\cap E$ for any $E\subseteq E_0$. Also, $$Z\bez Y=Z'\bez Y\implies\mu_1(Z)=\mu_1(Z')\tag2$$ as then $Z\bez E=Z'\bez E$ for all $E\in\mathcal F$. Thus, we can define $\nu\colon\mathcal B\to\mathbb R_{\ge0}$ by $$\nu\bigl((Z_0\cap Y)\cup(Z_1\bez Y)\bigr)=\mu_0(Z_0)+\mu_1(Z_1).\tag3$$ This is independent of the choice of $Z_0$ and $Z_1$ by (1) and (2). Clearly, $$\nu(Y)=\mu_0(X)+\mu_1(\varnothing)=\mu^*(Y).\tag4$$ For any $Z\in\mathcal A$, we have $$\mu_1(Z)=\sup\bigl\{\mu(Z)-\mu(Z\cap E):E\in\mathcal F\bigr\}=\mu(Z)-\mu_0(Z),$$ thus $$\nu(Z)=\mu(Z).\tag5$$ It remains to prove that $\nu$ is additive, which follows from the properties $$\begin{align} Z\cap Z'\cap Y=\varnothing&\implies\mu_0(Z\cup Z')=\mu_0(Z)+\mu_0(Z'),\tag6\\ Z\cap Z'\subseteq Y&\implies\mu_1(Z\cup Z')=\mu_1(Z)+\mu_1(Z').\tag7 \end{align}$$ For (6), put $E_0=X\bez(Z\cap Z')\in\mathcal F$; we have $$\begin{align} \mu_0(Z)+\mu_0(Z')&=\inf\{\mu(Z\cap E):E\in\mathcal F\}+\inf\{\mu(Z'\cap E):E\in\mathcal F\}\\ &=\inf\{\mu(Z\cap E)+\mu(Z'\cap E'):E,E'\in\mathcal F\}\\ &=\inf\{\mu(Z\cap E)+\mu(Z'\cap E):E\in\mathcal F,E\subseteq E_0\}\\ &=\inf\{\mu((Z\cup Z')\cap E):E\in\mathcal F,E\subseteq E_0\}=\mu_0(Z\cup Z'), \end{align}$$ where the third equality follows from $\mathcal F$ being a filter and the monotonicity of $\mu$. The equation (7) is proved in the same way, except we do not have to bother with $E_0$. We have constructed a content $\nu\supseteq\mu$ such that $$\nu(Y)=\mu^*(Y).$$ Using the same construction with $X\bez Y$ in place of $Y$, we obtain a content $\nu'\supseteq\mu$ such that $$\nu'(Y)=\mu_*(Y)=\sup\{\mu(E):E\in\mathcal A,E\subseteq Y\}.$$ Then, for any $y$ such that $\mu_*(Y)\le y\le\mu^*(Y)$, $$t\nu(Y)+(1-t)\nu'(Y)$$ is a content that gives $Y$ value $y$, where $t=(y-\mu_*(Y))/(\mu^*(Y)-\mu_*(Y))\in[0,1]$.<|endoftext|> TITLE: General solution to an ultrahyperbolic PDE QUESTION [5 upvotes]: $\DeclareMathOperator\SO{SO}$The following PDE defined on $\mathbb{R}^2$ $$\frac{\partial}{\partial x}\frac{\partial}{\partial y}f(x,y) = 0,$$ has solution $$f(x,y) = g(x) + h(y),$$ where $g,h : \mathbb{R} \to \mathbb{R}$ are arbitrary (nice enough) functions. I believe this to be a general solution to the equation, for any choice of boundary conditions. I don't have a proof of this and I'm happy to be shown otherwise if that's the case. I want to solve a similar looking PDE for $(x,y) \in \mathbb{R}^{n}\times\mathbb{R}^{n},$ which is given by $$\frac{\partial}{\partial x}\cdot\frac{\partial}{\partial y}f(x,y) = 0,$$ where the dot product is with the Euclidean metric on $\mathbb{R}^{n}$. There is an $\SO(n)$ symmetry under $x \mapsto M x, y \mapsto M^T y$. I can see that there are still solutions of the form $f_1(x,y) = g(x) + h(y)$ where $g,h : \mathbb{R}^n \to \mathbb{R}$ , but that these are no longer the only solutions. I've come up with something quite general that solves the equation;$$f_2(x,y) = \int d^na\,\left( g^x(x \wedge a) \,h^y(y\cdot a) + g^y(y \wedge a) \,h^x(x\cdot a)\right) + f_1(x,y),$$ where $g^x,g^y:\mathbb{R}^n\times\mathbb{R}^n\to \mathbb{R}$ and $h^x,h^y:\mathbb{R} \to \mathbb{R}$ are arbitrary (sufficiently nice) functions, which can depend on $a$. I added on $f_1(x,y)$ explicility, although I think it may be possible to fiddle the form of the integral and the choice of $g^x,g^y,h^x,h^y$ to include solutions of the form $f_1(x,y)$ without adding them on separately. One feature that I like about this solution is that it respects the $\SO(n)$ symmetry of the original PDE. The integral over the vector $a$ is also natural, in the context of the physical problem where the PDE comes from. Is anyone able to find a general solution to this equation, in a form which respects the $\SO(n)$ symmetry, possibly with some (hopefully quite general) restriction on the boundary conditions? Is my solution general for some choice of boundary conditions? REPLY [2 votes]: This is not an answer to the main question as the one offered by Robert Bryant, but it is a comment to the statement pertaining the two-variable boundary value problem for the ultrahyperbolic equation: I believe this to be a general solution to the equation, for any choice of boundary conditions. I don't have a proof of this and I'm happy to be shown otherwise if that's the case. The Dirichlet problem $$ \begin{cases} \dfrac{\partial}{\partial x}\dfrac{\partial}{\partial y} f(x,y)=f_{xy}(x,y)=0 & x,y\in[0,1]\\ f(x,\alpha(x))=\varphi_1(x) & x\in [0,1]\\ f(x,\beta(x))=\varphi_2(x) & x\in [0,1] \end{cases}\label{1}\tag{1} $$ where $\alpha, \beta:[0,1] \to [0,1]$ are appropriate strictly monotonic functions, is overdetermined (this, according to [1] pp. 6, was proved by Mauro Picone in [3]). Problem \eqref{1} was thoroughly studied by Gaetano Fichera: in [2] he found necessary and sufficient conditions (compatibility conditions) on the Dirichlet datum $(\varphi_1,\varphi_2)$ for it to be solvable with $f\in C^1([0,1]^2)$. In [1] he drops these compatibility conditions in order to accurately describe the singularities developed by the general solution $$ f(x,y) =g(x)+ h(y),\label{2}\tag{2} $$ by describing the kind of singularities developed by $g(x)$ and $h(y)$ as $x,y\to 0^+$. Therefore, it is probably right that \eqref{2} is the general solution to \eqref{1}, even when the boundary data are such that the solution behaves singularly at some point of the domain where the problem is posed. References [1] Gaetano Fichera, "Studio delle singolarità della soluzione di un problema di Dirichlet per l'equazione $u_{xy} = 0$" (Italian), Atti della Accademia Nazionale dei Lincei, Classe di Scienze Fisiche, Matematiche e Naturali. Rendiconti Serie 8 50 (1971), fasc. n.1, pp. 6-17, MR0310434, Zbl 0231.35046. [2] Gaetano Fichera, "Su un problema di Dirichlet per l'equazione $u_{xy}=0$". (Italian) Atti dell'Accademia delle Scienze di Torino, Classe di Scienze Fisiche Matematiche e Naturali 105 (1971), pp. 355–366, MR298228. [3] Mauro Picone, "Sulle equazioni alle derivate parziali del second’ordine del tipo iperbolico in due variabili indipendenti" (Italian), Rendiconti del Circolo Matematico di Palermo 30, pp. 349-376 (1910), DOI: 10.1007/BF03014882, JFM 41.0424.04.<|endoftext|> TITLE: The Fibonacci sequence modulo $5^n$ QUESTION [11 upvotes]: Let $(F_k)_{k=0}^\infty$ be the classical Fibonacci sequence, defined by the recursive formula $F_{k+1}=F_k+F_{k-1}$ where $F_0=0$ and $F_1=1$. For every $n\in\mathbb N$ let $\pi(n)$ be the smallest positive number such that $$\begin{cases} F_{\pi(n)}=F_0 \mod n,\\ F_{\pi(n)+1}= F_1\mod n. \end{cases}$$ The number $\pi(n)$ is called the $n$-th Pisano period. It is known that $\pi(5^n)=4\cdot 5^n$ for every $n\in\mathbb N$. Definition. A number $n\in\mathbb N$ is called Fibonacci uniform if $n$ divides $\pi(n)$ and for every $a\in \{1,\dots,n\}$ the set $\{k\in\{1,\dots,\pi(n)\}:F_k=a\mod n\}$ has cardinality $\pi(n)/n$. Example. The number $n=5$ is Fibonacci uniform which is witnessed by the first $20=\pi(5)$ Fibonacci numbers modulo 5: $$\mbox{0 1 1 2 3 0 3 3 1 4 0 4 4 3 2 0 2 2 4 1.}$$ On the other hand, the number $n=6$ is not Fibonacci uniform since among the first $24=\pi(6)$ Fibonacci numbers (modulo 6) $$\mbox{0 1 1 2 3 5 2 1 3 4 1 5 0 5 5 4 3 1 4 5 3 2 5 1}$$ 0 appears 2 times; 1: 6 times; 2: 3 times; 3: 4 times; 4: 3 times; 5: 6 times. It can be shown that each Fibonacci uniform number is a power of 5. Computer calculations show that for $n\le 10$ the power $5^n$ is indeed Fibonacci uniform. This suggests the following Conjecture. For every $n\in\mathbb N$ the number $5^n$ is Fibonacci uniform, which means that for every $a\in\{1,\dots,5^n\}$ the set $\{k\in\{1,\dots,4\cdot 5^n\}:F_k=a\mod 5^n\}$ contains exactly 4 numbers. Now the question how to prove this conjecture. REPLY [4 votes]: Fedor Petrov gave a satisfactory answer, but here is another one, based on a linear algebraic reasoning. The following is a general fact, which proof is left as an exercise. Lemma: Let $C$ be a $\mathbb{Z}$-valued matrix and fix a prime $p\geq 5$. If $C^2=0$ over $\mathbb{F}_p$ then for every natural $m$, $(I+C)^m=I+mC$ over the ring $\mathbb{Z}/p^{\nu_p(m)+1}$. Here $\mathbb{F}_p$ denotes the field with $p$ elements and $\nu_p(m)$ denotes the $p$-adic valuation of $m$. My goal is to explain how the following claim, which clearly implies the OP's conjecture, is implied by the above lemma. Cliam: For every $i=1,\ldots 4$ and every natural $n$, the map $\pi:\mathbb{Z}\to \mathbb{Z}/5^n\mathbb{Z}$ forms a bijection when restricted to the set $S=\{F_{4k+i}\mid k=0,\dots,5^n-1\}$. Recall that the Fibonacci sequence could be computed by $ \begin{bmatrix} F_{k+1} \\ F_{k} \end{bmatrix} = A^k \begin{bmatrix} 1 \\ 0 \end{bmatrix} $, where $A=\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$. Note that over $\mathbb{F}_5$, 3 is the unique eigenvalue of the matrix $A$, and its eigenspace $E$ is the span of $\begin{bmatrix} 3 \\ 1 \end{bmatrix}$. The "moral" reason for the period 4 appearing in the claim is that 3 is a 4th root of unity in $\mathbb{F}_5$. Thus the $\mathbb{Z}$-valued matrix $B=A^4$, becomes unipotent over $\mathbb{F}_5$: 1 is its unique eigenvalue, with eigenspace $E$. We get that over $\mathbb{F}_5$, $E$ is both the image and the kernel of $C=B-I$ and in particular, $C^2=0$. We can now prove the claim. Clearly it is enough to show that $\pi$ is injective on $S$: for every $0\leq k TITLE: Is there a "minimal" Whitney stratification of a complex hypersurface? QUESTION [8 upvotes]: Let $X\subset \mathbb C^n$ be a complex hypersurface (given by $F=0$ where $F$ is a polynomial). It is known then that $X$ admits a Whitney stratification. This is a decomposition of $X$ into smooth submanifolds (strata) that have some adjacency properties (Whitney conditions a and b). Question. Does $X$ have a minimal stratification, i.e. such a stratification that for any other Whitney stratification of $X$ the strata of the minimal one are unions of the strata of the other one? At least maybe this is known for varieties with certain type of singularities? REPLY [7 votes]: The answer is yes for any reduced equidimensional analytic space. This is the proposition 3.2 (and remark after the proof) page 479 of Variétés polaires II by Bernard Teissier.<|endoftext|> TITLE: Every half-dimensional subspace of a symplectic vector space has a Lagrangian complement QUESTION [6 upvotes]: Let $(V, \omega)$ be a finite-dimensional real symplectic vector space, i.e. $\omega : V \times V \to \mathbb{R}$ is a non-degenerate skew-symmetric bilinear map. A linear subspace $L \subset V$ is called Lagrangian if $L = L^\perp$, where $L^\perp = \{v \in V : \omega(v, L) = \{0\}\}$. Let $U \subset V$ be a linear subspace such that $$\dim U = \frac{1}{2} \dim V.$$ Question. Why is there always a Lagrangian subspace $L \subset V$ such that $V = U \oplus L$? This is easy if $U$ is also Lagrangian, and there are plenty of sources explaining this (we can set $L = I(U)$ where $I$ is a compatible complex structure). But I didn't find any reference explaining this more general fact, although I have seen it used in some research papers. It is used, for instance, for the existence of local Lagrangian bisections in symplectic groupoids. Any hints on the proof would be appreciated. REPLY [8 votes]: If $U$ is non-degenerate, then choose a symplectic isomorphism $f : {-U} \to U^\perp$, where $-U$ is $U$ equipped with the opposite $-\langle\cdot, \cdot\rangle$ of the original symplectic pairing, and put $L = \{f(u) - u \mathrel: u \in U\}$. (I think that this is closely related to your idea of choosing a compatible complex structure, but I'm trying to avoid assuming anything about the ground field.) If $U$ is not non-degenerate (!), then choose a non-$0$ vector $x \in U \cap U^\perp$, and then a vector $y \in V \setminus x^\perp$. By induction, there is a totally isotropic complement $\tilde M/x^{\perp\perp}$ to $U/x^{\perp\perp}$ in $x^\perp/x^{\perp\perp}$. Put $L = (\tilde M \cap y^\perp) + y^{\perp\perp}$.<|endoftext|> TITLE: Is there a higher categorical structure which models the (higher) conjugation actions of a group acting on itself? QUESTION [5 upvotes]: Let $G$ be a group, and consider the action of $G$ on itself by conjugation. If we think of $G$ as a one object category, then the conjugation action can be realised as automorphisms of this category, and we may build the associated $2$ category with one object, with additional $2$ morphisms given by elements of $G$, acting as conjugation. The question is whether one can extend this construction to take into account that the $2$ morphisms also have a notion of "equivalence", whether they are conjugate when viewed as elements of $G$. These relations should be witnessed by elements of $G$, expressing when two $2$ morphisms are conjugate, which then have more relations witnessed by elements of $G$ (by conjugation on these witnesses), and so on. Ideally this whole package would respect the underlying group structure in some sense, since we are considering automorphisms of all the data in the previous stage to obtain the next stage. One can give silly ways of describing this heap of data, so as a test/benchmark, one could ask for some general categorical object, which is built only out $G$ as a one object category, which observes "categorically" the following fact about finite groups: For $G$ finite, with $p\in \mathbb{Z}$ coprime to $|G|$, then for any $x,y\in G$, there exists $g,h\in G$ with $(xy)^p=gx^p g^{-1}h y^p h^{-1}$. REPLY [3 votes]: Following up a bit on Fernando's comment, you look to be trying to define things a bit like the tensor square of a group (acting on itself). Look at the work by Ronnie Brown, et al, Some computations of non-abelian tensor products of groups, J. Algebra, 111, (1987), 177 – 202. This would give you a crossed square. There are variants using an exterior product, which may also be of interest (see work by Graham Ellis) As Fernando mentions there are $cat^n$-groups / crossed n-cubes of groups that might encode higher commutator data but I think you would be needing more than just a single group, for instance a group with a family of normal subgroups. There is a separate response, which says that if G is thought of as a one object groupoid, then there is the functor category $G^G$, which encodes the conjugation. It has an associated crossed module which is exactly the $G\to Aut(G)$ one that was mentioned by Fernando.<|endoftext|> TITLE: An explicit equation of the canonical morphism $X_1(N) \to X_0(N)$ QUESTION [6 upvotes]: I know there are some research about explicit equations for affine models in $\mathbb{A}^2$ of many modular curves over $\mathbb{Q}$, for example of $X_i(N), X(N)$ (where $i = 0, 1, 2$) for small $N$. These equations are very useful in order to study elliptic curves. (Indeed there are so many papers which use them in crucial point.) But are there similar researches about explicit equations for the canonical morphism $X_1(N) \to X_0(N) : [E, P] \mapsto [E, \left< P \right>]$? REPLY [6 votes]: It is possible to find equations for $\pi : X_1(N) \to X_0(N)$ using work of Yifan Yang. In the article Defining equations of modular curves, he explains an algorithm to obtain equations for the modular curves $X_1(N)$ and $X_0(N)$. Let me explain how to use his results to get an equation for $\pi$. Yang shows that there exist two modular units $F,G$ on $X_1(N)$ such that $F$ and $G$ are regular away from the cusp at infinity, and the orders of the poles of $F$ and $G$ at this cusp are relatively prime. Then $F$ and $G$ generate the function field of $X_1(N)$ and using the $q$-expansions of $F$ and $G$, it is easy linear algebra to compute an equation $P(F,G)=0$ of $X_1(N)$. If $F$ has a pole of order $m$ and $G$ has a pole of order $n$, then the equation has degree $n$ in $F$, and $m$ in $G$. The same can be done for $X_0(N)$, giving an equation $Q(X,Y)=0$. Now the question is to express the modular units $X$ and $Y$ in terms of $F$ and $G$. Since Yang's construction uses explicit modular units, we know the $q$-expansions of $X,Y$ and $F,G$. Every rational function on $X_1(N)$ is of the form $\sum_{j=0}^{m-1} R_j(F) G^j$, where the $R_j$ are rational functions. In the case of $X$, if we know that the $R_j$ have degree $\leq d$, then we can simply use the $q$-expansions to find a linear relation of the form \begin{equation*} (*) \qquad \sum_{i=0}^d a_i F^i X = \sum_{i=0}^d \sum_{j=0}^{m-1} b_{i,j} F^i G^j. \end{equation*} with $a_i, b_{i,j} \in \mathbb{Q}$, and deduce $X$ in terms of $F$ and $G$ (and proceeding similarly for $Y$). As a variant, we may also search for a relation of the form $A(F,G)X=B(F,G)$. Once the computer detects a relation $A(F,G)X=B(F,G)$ as above, then we should certify it. To this end, it suffices to show that the function $R=A(F,G)X-B(F,G)$ is regular at all cusps of $X_1(N)$ above $\infty \in X_0(N)$, and is zero at $\infty \in X_1(N)$. Indeed, the function $R$ is already regular away from these cusps. The cusps of $X_1(N)$ above $\infty \in X_0(N)$ are acted on simply transitively by the diamond automorphisms $\langle \gamma \rangle : X_1(N) \to X_1(N)$ with $\gamma \in \Gamma_0(N)/\pm \Gamma_1(N)$, the last group being isomorphic to $(\mathbb{Z}/N\mathbb{Z})^\times / \pm 1$. Checking regularity at these cusps can be done using the transformation formulas for modular units (Proposition 2 in Yang's article). An alternative way is to bound the orders of the poles of $X$ at these cusps. In fact these orders are equal since $X$ comes from $X_0(N)$, so we already know them. Then we just need to check that the order of vanishing of $R$ at $\infty \in X_1(N)$ is greater than the sum of the orders of the other possible poles. This can be done by computing the $q$-expansion to enough accuracy. It would be nice to give an a priori bound on the degrees of the $R_j$ (that is, a bound on $d$). It is not difficult to find a polynomial $H$ in $F$ such that $HX$ is regular away from infinity. The idea is to cancel the poles of $X$ at the cusps $\langle \gamma \rangle \infty \in X_1(N) \backslash \{\infty\}$, by multiplying by suitable powers of $F-F(\langle \gamma \rangle \infty)$. Yet is not clear to me that the resulting function $HX$ is a polynomial in $F,G$, because the curve $P(F,G)=0$ may be singular, so that its ring of regular functions may be a strict subring in the normalisation $\mathcal{O}(X_1(N) \backslash \infty)$. However, searching for a relation as in $(*)$ should work in practice, by increasing the value of $d$ progressively if necessary. What this answer doesn't address is whether the pairs $(F,G)$ and $(X,Y)$ can be chosen consistently, that is, in such a way that the resulting equation for $X_1(N) \to X_0(N)$ is as simple as possible. In fact, once $(F,G)$ has been found, there is a standard choice, namely $X=\sum_\gamma F | \gamma$ and $Y=\sum_\gamma G | \gamma$, where the average is over the diamond automorphisms. These are modular funtions on $X_0(N)$ and as Yang explains, the functions $X$ and $Y$ satisfy the assumptions of his theorem, so they generate the function field of $X_0(N)$ and we can compute the resulting model of $X_0(N)$. Then the method above gives an equation for the morphism $X_1(N) \to X_0(N)$. I don't know whether this choice of $(X,Y)$ produces simpler equations. With the same technique, we can also express $F|\gamma$ and $G|\gamma$ in terms of $F$ and $G$ (recall that the $q$-expansions of $F|\gamma$ and $G|\gamma$ can be obtained using the transformation formulas for modular units). As a bonus, this gives equations for the diamond automorphisms $\langle \gamma \rangle : X_1(N) \to X_1(N)$. Note that if $(\mathbb{Z}/N\mathbb{Z})^\times / \pm 1$ is cyclic, generated by $\gamma_0$, then we only need to compute $F|\gamma_0$ and $G|\gamma_0$ in terms of $F$ and $G$. I haven't done experiments with this method, but since Yang's functions and equations are pretty simple, we may hope the same for the equation of the morphism $X_1(N) \to X_0(N)$.<|endoftext|> TITLE: What alternatives are there to the binomial poset theory of generating function families? QUESTION [13 upvotes]: A natural question in combinatorics is, why are certain families of generating functions combinatorially useful, like $\Sigma_n a_n x^n$ and $\Sigma_na_n\frac{x^n}{n!}$, why are other families are not, like $\Sigma_na_n\frac{x^n}{n^2+1}$? Doubilet, Rota, and Stanley proposed in this 1972 paper that $\Sigma a_n\frac{x^n}{B_n}$ is combinatorially useful if and only if $B_n$ is the factorial function of some binomial poset. They justify this on the basis of an isomorphism between the reduced incidence algebra of a binomial poset and the ring of formal power series (viewed as an algebra). But my question is, what alternate theories are there concerning when a family of generating function is and is not combinatorially useful? This 1978 paper by Richard Stanley contains the following statement: Two abstract theories of generating functions have been formulated to try to solve this problem - the Doubilet-Rota-Stanley theory of “reduced incidence algebras”, and the Bender-Goldman theory of “prefabs” (cf. also the “dissect” theory of M. Henley which combines features of both the preceding theories). But I’m not familiar with either the prefab theory or the dissect theory, so can anyone tell me what alternative account they give for which families of generating functions are combinatorially meaningful? And also have additional theories been developed since Stanley wrote this in 1978? For instance, does Joyal’s theory of combinatorial species address which families of generating functions are combinatorially meaningful? REPLY [4 votes]: This is just a comment, but too long to fit in the 600 character limit. It would be interesting for someone to compile a list of all functions $B(n)$ for which some combinatorial use has been found for a generating function $\sum a_n\frac{x^n}{B(n)}$. Moreover, for which of these functions $B(n)$ can an example of such a generating function be given that can be explained in a natural way by some existing theory (binomial posets, prefabs, dissects, species, $\dots$)? For example, what about $B(n)=(q^n-1)(q^n-q)\cdots(q^n-q^{n-1})$, which occurs in the enumeration of linear transformations over $\mathbb{F}_q$ (EC1, second ed., Section 1.10)? This function $B(n)$ is the factorial function of a binomial poset (see EC1, second ed., Example 3.18.3(c) and last sentence of Example 3.18.3(e)), but can this poset actually be used to obtain some of the generating functions in EC1, Section 1.10?<|endoftext|> TITLE: Smoothness of distance function to a compact set QUESTION [11 upvotes]: Fix a non-empty compact subset $K\subseteq \mathbb{R}^n$ and let $d_K(x):=\min_{z \in K} \,\|z-x\|$ be the map sending any $x\in \mathbb{R}^n$ to its distance from $K$. Suppose that: $K$ is regular : it has a non-empty interior $\overset{\circ}{K}$, and the closure of $\overset{\circ}{K}$ is $K$; in particular $K$ has co-dimension $0$. $K$ has a $C^{k+1}$ boundary. $K$ is convex. Then: Question 1: Is there some exponent $k+1k$ we have $d_K^p\in C^k(\mathbb{R}^n)$. Proof. Let $\nu:\partial K\to\mathbb{R}^n$ be the unit outer notrmal. If $\partial K$ is represented (locally) as a graph of $\varphi\in C^k$, then $\nu$ can be represented in terms of $D\varphi$ and hence $\nu\in C^{k-1}$. Consider the function $$ \Phi:\partial K\times\mathbb{R}\to\mathbb{R}^n, \quad \Phi(x,t)=x+\nu(x)t. $$ Clearly $\Phi\in C^{k-1}$. It follows from the computation in Lemma 14.16 in [1] that the Jacobian $J_\Phi$ of $\Phi$ can be expressed in terms of the principal curvatures and in fact $$ J_\Phi(x,t)>0 \quad \text{for all} \quad (x,t)\in \partial K\times [0,\infty). $$ Hence $\Phi$ is a diffeomorphism in a neighborhood of $\partial K\times\{0\}$. However, the normal lines never intersect outside $K$, so it follows that $\Phi$ is actually a diffeomorphism in an open set $U$ that contains $\partial K\times [0,\infty)$. Clearly $V=\Phi(U)$ is an open subset of $\mathbb{R}^n$ that contains $\mathbb{R}^n\setminus\overset{\circ}{K}$. Let $\hat{d}_K$ be the signed distance to $\partial K$ which is positive and equals $d_K$ in $\mathbb{R}^n\setminus K$ and is negative in $\overset{\circ}{K}$. For $y\in V$, let $\pi(y)\in\partial K$ be the unique closest point so $$ y=\pi(y)+\nu(\pi(y))\hat{d}_K(y), \quad \text{i.e.,} \quad y=\Phi(\pi(y),\hat{d}_K(y)), $$ and hence $(\pi(y),\hat{d}_K(y))=\Phi^{-1}(y)$ which proves that $\hat{d}_K\in C^{k-1}(V)$. This is smaller regularity than we wanted, but a nice trick allows us to show that actually $\hat{d}_K\in C^{k}(V)$. Indeed, $\nabla \hat{d}_K$ points in directional in which the function growths fastest which is $\nu(\pi(y))$. Since the distance growths linearly in that direction, we have that $\nabla \hat{d}_K(y)=\nu(\pi(y))\in C^{k-1}$, and hence $\hat{d}_K\in C^k(V)$. Now $$ d_K=\begin{cases} \hat{d}_K & \text{in } \mathbb{R}^n\setminus\overset{\circ}{K}\\ 0 & \text{in } K. \end{cases} $$ For that reason reason we lose regularity of the function at the boundary. However, the function $d_K^p$, where $p>k$ has all partial derivatives of order up to $k$ equal zero on the boundary of $K$ and it follows that $d_K^p\in C^k$. [1] Gilbarg, D. Trudinger, N. S. Elliptic partial differential equations of second order. Reprint of the 1998 edition. Classics in Mathematics. Springer-Verlag, Berlin, 2001.<|endoftext|> TITLE: Is every 1-million-connected graph rigid in 3D? QUESTION [20 upvotes]: It is an old result that every $6$-connected graph is rigid in $\mathbb{R}^2$: Lovász, László, and Yechiam Yemini. "On generic rigidity in the plane." SIAM Journal on Algebraic Discrete Methods 3, no. 1 (1982): 91-98. DOI. It is natural to hope that sufficiently high connectivity implies rigidity in $\mathbb{R}^3$. Q. Is it known that there is some $k$ such that any $k$-connected graph is generically rigid in $\mathbb{R}^3$? Informally, $G$ is rigid if the distances between vertices connected in $G$ determine all the distances between vertices not connected in $G$. More formally, $G$ is generically rigid in $\mathbb{R}^d$ if every generic representation in $\mathbb{R}^d$ is infinitesimally rigid—no infinitesimal length-preserving velocities (if they exist) can be extended. An embedded graph representation (a framework) is generic if the coordinates of its configuration do not satisfy any non-trivial algebraic equation with rational coefficients. There have been recent advances in 3D rigidity and I am unclear on the current status of this question Q. REPLY [11 votes]: I think this is still an open problem, but recent work of Clinch, Jackson, and Tanigawa (almost) shows every $12$-connected graph is generically rigid in $\mathbb{R}^3$. In that paper, they prove that $12$-connectivity is sufficient to force rigidity in the $C_2^1$-cofactor matroid (see the paper for precise definitions). In an earlier paper, the same authors showed that the $C_2^1$-cofactor matroid is the unique maximal abstract $3$-rigidity matroid. A long-standing conjecture in rigidity theory is that the unique maximal abstract $3$-rigidity matroid is in fact the generic $3$-dimensional rigidity matroid. If you believe this conjecture, then the answer to your quesiton is yes, with one million replaced with $12$. Acknowledgement. This answer is entirely due to Katie Clinch.<|endoftext|> TITLE: Set of primes $p$ such that $\mathrm{Hom}(A, \mathbb{F}_p)=\emptyset$ QUESTION [10 upvotes]: For which sets of primes $P$ is there a finite type $\mathbb{Z}$-algebra $A$ such that$$p\in P\iff\mathrm{Hom}(A, \mathbb{F}_p)=\emptyset?$$Do all the finite $P$ arise this way? $A=\mathbb{Z}/n$ works for the cofinite $P$. REPLY [10 votes]: The answer of RP_ is correct, so my main contribution is cleaning it up, providing more detail, and uniformising the different cases. Definition. Let $\Omega$ be the set of prime numbers. For subsets $S, T \subseteq \Omega$, write $S \sim T$ if the symmetric difference $S \mathbin\triangle T = (S \setminus T) \cup (T \setminus S)$ is finite; note that this is an equivalence relation. For a finite type $\mathbf Z$-scheme $X$, write $S_X$ for the set of primes such that $X(\mathbf F_p) \neq \varnothing$. If $X = \operatorname{Spec} A$ is affine, write $S_A$ for $S_X$. Finally, if $f \in \mathbf Z[x]$ is a polynomial, set $S_f = S_{\mathbf Z[x]/(f)}$, as in RP_'s answer. Define subsets of the power set $\mathcal P(\Omega)$ by \begin{align*} \mathcal P_{\text{sch}}(\Omega) &:= \{S_X \mathrel| X \in \mathbf{Sch}_{\mathbf Z}^{\text{f.t.}}\},\\ \mathcal P_{\text{aff}}(\Omega) &:= \{S_A \mathrel| A \in \mathbf{Alg}_{\mathbf Z}^{\text{f.t.}}\},\\ \mathcal P_{\text{poly}}(\Omega) &:= \{S_f \mathrel| f \in \mathbf Z[x]\},\\ \mathcal P_{\text{monic}}(\Omega) &:= \{S_f \mathrel| f \in \mathbf Z[x] \text{ monic}\}. \end{align*} In the main proposition below, we will show that the first three agree, and the fourth one as well if we only consider subsets of $\Omega$ up to $\sim$. Example. The set of primes congruent to $1$ modulo $4$ is $S_{4x^2+1}$, since $-1$ is a square modulo a prime $p$ if and only if $p = 2$ or $p \equiv 1 \pmod 4$. In general, the Chebotarev density theorem says that elements of $\mathcal P_{\text{poly}}(\Omega)$ have rational density. For example, if $\mathbf Z[x]/(f) \cong \mathcal O_K$ is the ring of integers in a finite Galois extension $\mathbf Q \subseteq K$ of degree $n$, then $S_f$ has density $\tfrac{1}{n}$. Remark. Note that $S_{X \times Y} = S_X \cap S_Y$ and $S_{X \amalg Y} = S_X \cup S_Y$, and more generally $S_{X \cup Y} = S_X \cup S_Y$ if $X,Y \subseteq Z$ are subschemes (not necessarily disjoint). This shows that $\mathcal P_{\text{sch}}(\Omega)$ and $\mathcal P_{\text{aff}}(\Omega)$ are closed under finite unions and intersections. This also gives $S_{fg} = S_f \cup S_g$, since $\operatorname{Spec} \mathbf Z[x]/(fg)$ is the union $\operatorname{Spec} \mathbf Z[x]/(f) \cup \operatorname{Spec} \mathbf Z[x]/(g)$ (the intersection between the two components need not be empty, but it doesn't matter), so $\mathcal P_{\text{poly}}(\Omega)$ is closed under unions. In addition, the Corollary to Lemma 2 below shows that it is also closed under intersection. Lemma 1. If $S \in \mathcal P_{\text{poly}}(\Omega)$ and $T \sim S$, then $T \in \mathcal P_{\text{poly}}(\Omega)$. Proof. By assumption, there exists $f \in \mathbf Z[x]$ such that $S = S_f$. It suffices to show that if $p \in \Omega$, then $S \cup \{p\}$ and $S \setminus \{p\}$ are in $\mathcal P_{\text{poly}}(\Omega)$. For $S_f \cup \{p\}$, we may use $S_{pf}$. If $f = 0$, then $S_{px-1} = \Omega\setminus \{p\} = S \setminus \{p\}$, showing that $S \setminus \{p\} \in \mathcal P_{\text{poly}}(\Omega)$. If $f \neq 0$, choose $a \in \mathbf Z$ with $f(a) \neq 0$, and let $r = v_p(f(a)) \in \mathbf Z_{\geq 0}$. After replacing $f(x)$ by $f(x-a)$, we may assume $a = 0$. Then $g(x) = \tfrac{f(p^{r+1}x)}{p^r}$ has a solution modulo a prime $q \neq p$ if and only if $f$ does, since $p^{r+1}$ is invertible modulo $q$. This gives an integer polynomial whose terms are all divisible by $p$ except the constant term, so $g$ has no zeroes modulo $p$. Therefore, $S_g = S \setminus \{p\}$, showing that $S \setminus \{p\} \in \mathcal P_{\text{poly}}(\Omega)$. $\square$ Lemma 2. Let $f, g \in \mathbf Z[x]$. Then there exists a monic polynomial $h \in \mathbf Z[x]$ such that $S_f \cap S_g \sim S_h$. Proof. Write $A = \mathbf Z[x]/(f)$ and $B = \mathbf Z[x]/(g)$, and set $C = A \otimes B$, so $S_f \cap S_g = S_C$. Then $(C \otimes \mathbf Q)^{\text{red}}$ is a finite product of fields, so can be written as $\prod_{i=1}^r \mathbf Q[x]/(h_i)$ for monic polynomials $h_i \in \mathbf Z[x]$. Then $C' = \prod_{i=1}^r\operatorname{Spec} \mathbf Z[x]/(h_i)$ differs from $(\operatorname{Spec} C)^{\text{red}}$ in finitely many closed fibres above $\operatorname{Spec} \mathbf Z$, so away from the corresponding primes we have $S_{C'} = S_C$. Setting $h = h_1 \dotsm h_r$ gives the result, since $S_{C'} = \bigcup_i S_{h_i} = S_h$. $\square$ Corollary. The set $\mathcal P_{\text{poly}}(\Omega)$ is closed under (finite) intersections. Proof. If $S, T \in \mathcal P_{\text{poly}}(\Omega)$, then there exists $U \in \mathcal P_{\text{monic}}(\Omega)$ with $U \sim S \cap T$ by Lemma 2. Then Lemma 1 gives $S \cap T \in \mathcal P_{\text{poly}}(\Omega)$. $\square$ The main claim is the following: Proposition. Let $S \subseteq \Omega$ be a set of primes. Then the following are equivalent: $S \in \mathcal P_{\text{sch}}(\Omega)$; $S \in \mathcal P_{\text{aff}}(\Omega)$; $S \in \mathcal P_{\text{poly}}(\Omega)$; $S \sim T$ for some $T \in \mathcal P_{\text{monic}}(\Omega)$. That is, there exists a monic polynomial $f \in \mathbf Z[x]$ such that $S \mathbin\triangle S_f$ is finite. Note that this is not quite the Boolean lattice generated by the sets $S_f$, as $\mathcal P_{\text{aff}}(\Omega)$ is not closed under complements (e.g. it doesn't contain the set of primes congruent to $3$ modulo $4$, right?). Proof of Proposition. Note that all sets contain singletons and complements of singletons, and are closed under finite unions and intersections (for intersections in $(4)$ and $(3)$, use Lemma 2 and its Corollary above). Implications $(3) \Rightarrow (2) \Rightarrow (1)$ are clear, and breaking up an arbitrary finite type $\mathbf Z$-scheme into locally closed affine subschemes shows $(1) \Rightarrow (2)$. Note that Lemma 2 implies $(3) \Rightarrow (4)$, but we don't need this. The converse follows from Lemma 1. It remains to show $(2) \Rightarrow (4)$, where we may assume $A$ is an integral domain. Let $K$ be the algebraic closure of $\mathbf Q$ in $\operatorname{Frac}(A)$. Then $A \otimes \mathbf Q$ is a geometrically integral $K$-algebra [Tags 020I, 037P, and 054Q]. There is a finite set of primes $T$ of $K$ such that $A[1/T]$ is a flat $\mathcal O_{K,T}$-algebra with geometrically integral fibres [EGA IV$_3$, Prop. 8.9.4 and Thm. 9.7.7]. The Lang–Weil bound (or more precise versions coming from the Weil conjectures as proven by Deligne) then gives $$\lvert A(\kappa(\mathfrak p))\rvert \geq \lvert\kappa(\mathfrak p)\rvert^d - c \lvert\kappa(\mathfrak p)\rvert^{d-\tfrac{1}{2}}$$ for all prime ideals $\mathfrak p \subseteq \mathcal O_{K,T}$ and some $c > 0$ that does not depend on $\mathfrak p$. In particular, for all but finitely many primes $\mathfrak p \subseteq \mathcal O_{K,T}$, we get $A(\kappa(\mathfrak p)) \neq \varnothing$. Therefore, for all but finitely primes $p \in \Omega$, we get $A(\mathbf F_p) \neq \varnothing$ if and only if $\mathcal O_{K,T}$ has a prime with residue field $\mathbf F_p$, i.e. if and only if $\mathcal O_{K,T}(\mathbf F_p) \neq \varnothing$. In other words, $S_A \sim S_{\mathcal O_{K,T}}$. Thus we may replace $A$ by $\mathcal O_{K,T}$, and now we proceed as in Lemma 2: if $f \in \mathbf Z[x]$ is a monic polynomial such that $K \cong \mathbf Q[x]/(f)$, then $A$ and $\mathbf Z[x]/(f)$ differ in finitely many closed fibres, so $S_A \sim S_f$. $\square$<|endoftext|> TITLE: $\pi_0^{\mathbb{A}^1}(\mathbb{A}^m-0/\mu_n)\simeq\mathbb{G}_m/\mathbb{G}_m^n$ QUESTION [6 upvotes]: I have been reading Asok and Doran - $\mathbb A^1$-homotopy groups, excision, and solvable quotients. In the Example 2.17, page 1155, there is a claim that for $m>1$ and $n\geq 1$, $\pi_0^{\mathbb{A}^1}(\mathbb{A}^m-0/\mu_n)\simeq\mathbb{G}_m/\mathbb{G}_m^n$. Let me explain the notation quickly: $\mu_n$ is the group scheme of $n^\text{th}$ roots of unity acting on $\mathbb{A}^m-0$ by diagonal action. So $\mathbb{A}^m-0/\mu_n$ is a smooth scheme. $\mathbb{G}_m/\mathbb{G}_m^n$ is the cokernel of the $n^\text{th}$-power map $\mathbb{G}_m\xrightarrow{(-)^n} \mathbb{G}_m$ in the category of Nisnevich sheaves of abelian groups. My question is how to see the isomorphism $\pi_0^{\mathbb{A}^1}(\mathbb{A}^m-0/\mu_n)\simeq\mathbb{G}_m/\mathbb{G}_m^n$? Comments are most welcome! REPLY [3 votes]: Over a field of characteristic coprime to $n$, the classifying space ${\rm B}_{\rm et}\mu_n$ is $\mathbb{A}^1$-local. In this situation, any $\mu_n$-covering $\widetilde{X}\to X$ is an $\mathbb{A}^1$-covering space and provides a classifying morphism $X\to {\rm B}_{\rm et}\mu_n$. From the Kummer sequence $\mu_n\to \mathbb{G}_{\rm m}\to \mathbb{G}_{\rm m}$ we get a fiber sequence ${\rm B}_{\rm et}\mu_n\to {\rm B}\mathbb{G}_{\rm m}\to {\rm B}\mathbb{G}_{\rm m}$. This implies that the sheaf of $\mathbb{A}^1$-connected components is $\mathbb{G}_{\rm m}/\mathbb{G}_{\rm m}^n$ (the Nisnevich sheaf quotient) which we can alternatively write as the Nisnevich sheafification of ${\rm H}^1_{\rm et}(-,\mu_n)$. (This is in the Morel-Voevodsky paper.) Now turn to $\mathbb{A}^m\setminus\{0\}$. The $\mu_n$-covering $\mathbb{A}^m\setminus\{0\}\to\mathbb{A}^m\setminus\{0\}/\mu_n$ is classified by a map $\mathbb{A}^m\setminus\{0\}/\mu_n\to {\rm B}_{\rm et}\mu_n$ and the more precise claim is that the induced map on $\pi_0^{\mathbb{A}^1}$ is an isomorphism of Nisnevich sheaves. For surjectivity, we consider an embedding $\mathbb{G}_{\rm m}\to\mathbb{A}^m\setminus\{0\}$ as $i$-th component. This is a $\mu_n$-equivariant map if we take the componentwise action on $\mathbb{A}^m\setminus\{0\}$. Therefore, we get an induced map on quotient schemes $\mathbb{G}_{\rm m}\to \mathbb{A}^m\setminus\{0\}/\mu_n$. On sheaves of $\mathbb{A}^1$-connected components we get induced maps $$ \mathbb{G}_{\rm m}\to \pi_0^{\mathbb{A}^1}\mathbb{A}^m\setminus\{0\}/\mu_n\to \mathbb{G}_{\rm m}/\mathbb{G}_{\rm m}^n $$ such that the composition is the natural quotient map. This proves the surjectivity. We check injectivity locally in the Nisnevich topology. For that, let $R$ be a henselian local ring. A morphism ${\rm Spec}R\to \mathbb{A}^m\setminus\{0\}/\mu_n$ is represented by a tuple $(x_1,\dots,x_m)$ with $x_i\in R$ such that at least one of the $x_i$ is not in the maximal ideal of $R$. Two such tuples $(x_1,\dots,x_m)$ and $(y_1,\dots,y_m)$ give the same morphism if the tuples $(x_1^n,\dots,x_m^n)$ and $(y_1^n,\dots,y_m^n)$ agree. Now we want to show that any map ${\rm Spec} R\to \mathbb{A}^m\setminus\{0\}/\mu_n$ (corresponding to a tuple $(x_1,\dots,x_m)$) factors through a coordinate embedding $\mathbb{G}_{\rm m}\to\mathbb{A}^m\setminus\{0\}/\mu_n$ used before. One entry $x_i$ is not contained in the maximal ideal, assume it's the first. Then $(x_1,t\cdot x_2,\dots,t\cdot x_m)$ provides a naive $\mathbb{A}^1$-homotopy from the given tuple to $(x_1,0,\dots,0)$. So up to $\mathbb{A}^1$-homotopy, any map ${\rm Spec} R\to \mathbb{A}^m\setminus\{0\}/\mu_n$ factors through a coordinate embedding $\mathbb{G}_{\rm m}\to \mathbb{A}^m\setminus\{0\}$. If such a homotopy class maps to the base point of $\mathbb{G}_{\rm m}/\mathbb{G}_{\rm m}^n$ then by the identifications above, it must lift through the $\mu_n$-covering $\mathbb{G}_{\rm m}\xrightarrow{n} \mathbb{G}_{\rm m}$. But since we have a morphism from the covering $\mathbb{G}_{\rm m}\to \mathbb{G}_{\rm m}$ to the covering $\mathbb{A}^m\setminus\{0\}\to \mathbb{A}^m\setminus\{0\}/\mu_n$, the map ${\rm Spec}R\to \mathbb{A}^m\setminus\{0\}/\mu_n$ lifts (up to $\mathbb{A}^1$-homotopy) through a map ${\rm Spec}R\to \mathbb{A}^m\setminus\{0\}$. But the target there is $\mathbb{A}^1$-connected, meaning that the original map ${\rm Spec} R\to \mathbb{A}^m\setminus\{0\}/\mu_n$ must be null-homotopic, showing injectivity.<|endoftext|> TITLE: Smooth morphism of smooth varieties with fibres isomorphic to an affine space QUESTION [7 upvotes]: Let $X$ and $Y$ be smooth varieties over the field of complex numbers $\bf C$ (smooth integral separated schemes of finite type over $\bf C$). Let $$f\colon X\to Y$$ be a surjective morphism such that for any closed point $y\in Y$, the schematic fibre $f^{-1}(y)\subset X$ is isomorphic to the affine space ${\Bbb A}_{\bf C}^{n(y)}$. Moreover, assume that the morphism $f$ is smooth (which is equivalent to the assumption that $n(y)$ is the constant function $n(y)=n$, where $n=\dim X-\dim Y$). Consider the real $C^\infty$-manifolds $X^\infty=X({\bf C})$ and $Y^\infty=Y({\bf C})$ and the induced $C^\infty$-map $$f^\infty\colon X^\infty\to Y^\infty.$$ Since $f$ is smooth, the map $f^\infty$ is a submersion, that is, for any $x\in X^\infty$, the differential $$d_x f\colon T_x(X)\to T_{f(x)}Y$$ is surjective. Moreover, each fibre of $f^\infty$ is diffeomorphic to ${\bf R}^{2n}$. By Corollary 31 of G. Meigniez, Submersions, fibrations and bundles, Trans. Amer. Math. Soc. 354 (2002), no. 9, 3771-3787, the map $f^\infty$ is a locally trivial fibre bundle of $C^\infty$-manifolds, that is, for any $y\in Y^\infty$ there exists an open neighborhood ${\mathcal U}_y$ of $y$ in $Y^\infty$ such that $f^{-1}({\mathcal U}_y)\simeq {\bf R}^{2n}\times {\mathcal U}_y$, where $\simeq$ denotes a $C^\infty$-diffeomorphism compatible with the projections onto ${\mathcal U}_y$. Question 1. Does it follow that the morphism $f$ is a locally trivial fibre bundle in the étale topology, that is, for any closed point $y\in Y$ there exists an étale open neighborhood $ U_y\to Y$ of $y$ such that $$X\times_Y U_y\simeq {\Bbb A}_{\bf C}^n\times_{\bf C} U_y\,,$$ where $\simeq$ denotes an isomorphism of $\bf C$-varieties compatible with the projections onto $U_y$ ? Question 2. Is $f$ a locally trivial fibre bundle in the flat topology? Consider the complex analytic manifolds $X^{\rm an}=X({\bf C})$, $Y^{\rm an}=Y({\bf C})$ and the induced complex analytic morphism $$f^{\rm an}\colon X^{\rm an}\to Y^{\rm an}.$$ Question 3. Is $f^{\rm an}\colon X^{\rm an}\to Y^{\rm an}$ a locally trivial fibre bundle of complex analytic manifolds, that is, for any $y\in Y^{\rm an}$ there exists an open neighborhood ${\mathcal U}_y$ of $y$ in $Y^{\rm an}$ such that $(f^{\rm an})^{-1}({\mathcal U}_y)\simeq {\bf C}^n\times {\mathcal U}_y$, where $\simeq$ denotes an analytic isomorphism compatible with the projections onto ${\mathcal U}_y$ ? REPLY [3 votes]: Regarding Question 1, it seems to be an open problem, known as a variant of Dolgachev–Weisfeiler Conjecture. The article $\mathbb{A}^2$-fibrations between affine spaces are $\mathbb{A}^2$-trivial (A. Dubouloz) shows that an $\mathbb{A}^2$-fibration $f\colon X\to S$ is étale-locally trivial if and only if $\Omega^1_{X/S}$ is a pullback of a locally-free sheaf $\mathcal{E}$ on $S$. Similar questions are also mentioned in Vénéreau polynomials and related fiber bundles (S. Kaliman, M. Zaidenberg), page 276. Perhaps some experts can answer this question in greater detail.<|endoftext|> TITLE: Is it possible for the reduction modulo $p$ of an non-commutative semisimple algebra to be commutative? QUESTION [8 upvotes]: Suppose that $I, X_1, \ldots, X_{d-1}$ are $n \times n$ matrices with integer entries whose $\mathbb{Z}$-span is a subalgebra of $\mathrm{Mat}_n(\mathbb{Z})$. Suppose that, thought of as a subalgebra of $\mathrm{Mat}_n(\mathbb{C})$, this algebra is semisimple and non-commutative. Thus, by Wedderburn's Theorem, it is isomorphic to a direct product of complete matrix algebras $\mathrm{Mat}_r(\mathbb{C})$, with $r \ge 2$ for at least one factor. It is possible that there exists a prime $p$ and a field $K$ of characteristic $p$ such that, regarding $I, X_1, \ldots, X_{d-1}$ as elements of $\mathrm{Mat}_n(K)$ by reduction modulo $p$, the subalgebra of $\mathrm{Mat}_n(K)$ spanned by $I, X_1, \ldots, X_{d-1}$ is commutative, and still of dimension $d$? As a follow-up, in my specific setup, the matrices are the orbital matrices for the action of a finite group $G$ on a set $\Omega$: the orbital matrix with $1$ in position $(\alpha,\beta)$ has $1$s exactly in the positions $(\alpha g, \beta g)$, for $g \in G$. It is known that these matrices span the centralizer algebra of the permutation module (over $\mathbb{Z}$ or any field). Moreover, the algebra is commutative in characteristic zero if and only if the associated permutation character $\pi(g) = |\mathrm{Fix}_\Omega(g)|$ is multiplicity-free. Is it possible that in this situation the centralizer algebra is non commutative in characteristic zero, but commutative after reduction modulo a prime? REPLY [4 votes]: Example 5.10 of Towers, Matthew, Endomorphism algebras of transitive permutation modules for $p$-groups., Arch. Math. 92, No. 3, 215-227 (2009) (whose author you might know) gives a positive answer to the second question. The group action involved is of $$G=\langle x,y,x\mid x^4, y^4, z^4, [x,y]=z\in Z(G)\rangle$$ (which has order $64$) acting on the cosets of the Klein four subgroup generated by $x^2$ and $y^2$,<|endoftext|> TITLE: Uncountable counterexamples in algebra QUESTION [32 upvotes]: In functional analysis, there are many examples of things that "go wrong" in the nonseparable setting. For instance, my favorite version of the spectral theorem only works for operators on a separable Hilbert space. Within C${}^*$-algebra there are many examples of nice results that require separability. (Dixmier's problem: is every prime C${}^*$-algebra primitive? Yes for separable C*-algebras, no in general.) I wondered whether there is a similar phenomenon in pure algebra. Are there good examples of results that hold for countable groups, countable dimensional vector spaces, etc., but fail in general? One example I know about is Whitehead's problem, which has a positive solution for countable abelian groups, but is independent of ZFC in general. REPLY [9 votes]: Since you mentioned Whitehead's problem here is another interesting independence example. For a group $G$ define its dual $G^\ast$ to be $\mathrm{Hom}(G,\Bbb Z)$. Like in the vector spaces case we get a canonical evaluation homomorphism $j\colon G\to G^{\ast\ast}$ given by $g\mapsto(f\mapsto f(g))$ and we call a group reflexive if $j$ is an isomorphism. Now let $G$ be free abelian, must it be reflexive? The answer is provably positive for all "small" free abelian groups in $\mathsf{ZFC}$, in fact it is positive for all free abelian groups iff there is no measurable cardinal.<|endoftext|> TITLE: Golden ratio as a property of conic section (is it known?) QUESTION [5 upvotes]: I am looking for a proof of a discovery as follows: Let $ABC$ be arbitrary triangle and $(\Omega)$ be an arbitrary circumconic of $ABC$ let $A'B'C'$ is its tangential triangle of $ABC$ respect to $(\Omega)$. Let $BB'$ meet $AC$ at $D$ and $CC'$ meet $AB$ at $E$, let $DE$ meet the circumconic at $F$. A line through $F$ and parallel to $B'C'$ meets $AB$, $AC$ at $H$, $G$ (see Figure) then: $$\frac{HG}{GF}=\frac{\sqrt{5}+1}{2}.$$ REPLY [4 votes]: Consider generalizing the last line to Let $I$ be a point on $B'C'$, and let $FI$ meet $AB$, $AC$ at $H$, $G$. Then the cross ratio of $H,F,G,I$ is the negative golden ratio: $$\frac{HG\cdot FI}{FG\cdot HI}=-\frac{1+\sqrt{5}}{2} = -\phi$$ The original version is the limit of the general version as $I$ goes to infinity. But the general version is projectively invariant, so we can prove it using coordinates where $A$ is at vertical infinity, the conic is $y=x^2$, and the tangent at $A$ is the line at infinity. Then the first coordinates are simple: \begin{align} A &= \infty(0,1)\\ B &= (b,\ b^2)\\ C &= (c,\ c^2)\\ A' &= ((b+c)/2,\ bc)\\ B' &= \infty(1,2c)\\ C' &= \infty(1,2b)\\ D &= (c,\ (b-c)^2+c^2)\\ E &= (b,\ (b-c)^2+b^2)\\ F &= ((1-\phi)b+\phi c,\ ((1-\phi)b+\phi c)^2)\\ \end{align} and we can stop computing coordinates there. The graphic below shows the case $b=-1$, $c=2$. (There is also another solution for $F$, not mentioned in the original question, which leads to switching $\phi$ and $1-\phi$ in the above and the below.) We want to evaluate the cross ratio $(HG\cdot FI)/(FG \cdot HI)$. Since $I$ is a point at infinity, the $FI$ and $HI$ terms will cancel. Then we can compute the directed quotient $HG/FG$ from the $x-$coordinates, which are $c$ and $b$ for $G$ and $H$. This gives us the desired result: $$\frac{HG\cdot FI}{FG \cdot HI}=\frac{HG}{FG}=\frac{b-c}{(b-c)(1-\phi)}=-\phi$$<|endoftext|> TITLE: Quotient groups obtained by quotienting $G^n$ by $G^{n-1}$ QUESTION [5 upvotes]: Notation: For a group $G$, we write $G^n$ to denote the $n$-fold direct product of $G$ with itself. Problem set up: Consider, for a finite group $G$, and $n > 1$, the set $Q(G)_n$ of all isomorphism classes of groups of the form $\frac{G^{n}}{H_{n-1}}$, for any normal subgroup $H_{n-1}$ of $G^n$ isomorphic to $G^{n-1}$. Note that $Q(G)_1 \subset Q(G)_2 \subset \cdots$, and so for finite groups $G$, the sequence stabilises at some $N$ - i.e. we have $Q(G)_n =Q(G)_{n+1}$ for all $n \geq N$. We call the sequence $Q(G)_1, \cdots, Q(G)_N$ the quotient series of $G$. We recall that by the classification theorem for finite abelian groups, we can write any finite abelian group as $\bigoplus_{i = 0}^M \mathbb Z_{p_{i}^{k_i}}$ for primes $p_i$ and positive integers $k_i$. Question: Let $G = \bigoplus_{i = 0}^M \mathbb Z_{p_{i}^{k_i}}$ be a finite abelian group. What is the quotient series of $G$? REPLY [10 votes]: I'll abbreviate $\mathbb{Z}/p^k \mathbb{Z}$ to $L_k$. Let $G = \bigoplus_{i=1}^M L_i^{a_i}$. I claim that $\bigoplus_{i=1}^N L_i^{b_i}$ is in the quotient series of $G$ if and only if the Littlewood-Richardson coefficient $$c_{(1^a_1),\ \ (2^{a_2}),\ \ \dotsc,\ \ (M^{a_M})}^{(1^{b_1} 2^{b_2} \cdots N^{b_N})}$$ is nonzero. Note the parentheses and commas: We have $M$ distinct partitions on the bottom and only one partition on the top. You can decide whether or not this is an explicit enough answer to be useful to you. Proof. In general, let $\alpha$, $\beta$ and $\gamma$ be three partitions. Then there is a short exact sequence $$0 \to \bigoplus L_{\alpha_i} \to \bigoplus L_{\beta_i} \to \bigoplus L_{\gamma_i} \to 0$$ if and only if $c_{\alpha \gamma}^{\beta}$ is nonzero. This is the same as asking that there be a semistandard Young tableaux of shape $\beta/\alpha$ whose rectification is a given tableaux of shape $\gamma$. See, for example, Fulton's survey Eigenvalues, invariant factors, highest weights, and Schubert calculus. The short exact sequences of abelian groups are in Section 2; the tableaux are briefly mentioned in Section 9 and can be found in more detail in many places, such as Stanley's Enumerative Combinatorics II or Fulton's Young Tableaux. In your situation, you want to know when we have $c_{T \alpha,\ \gamma}^{(T+1) \alpha}$ nonzero for $T$ large. So we are looking for tableaux of shape $(T+1)\alpha/T \alpha$ whose rectification is a given tableaux of shape $\gamma$. But, once $T$ is large enough, $(T+1)\alpha/T \alpha$ is just a union of disconnected rectangles. Specifically, if $\alpha = 1^{a_1} 2^{a_2} \cdots M^{a_M}$, then the connected components of $(T+1)\alpha/T \alpha$ are rectangles of shape $k \times a_k$, for $1 \leq k \leq M$. Standard properties of Young tableaux then give the answer above.<|endoftext|> TITLE: What are those 'others' in 'the classification table of algebraic three-folds' QUESTION [5 upvotes]: In the Wikipedia page for 'Kodaira dimension', there is 'the classification table of algebraic three-folds' in the section 'Any dimension'. What are those 'others' in the very bottom of this table? Are there some classifications for these 'others'? Are there any examples of these 'others' that are not rational or birational to Fano threefolds? REPLY [4 votes]: Here is a different example. Let $S$ be an Enriques surface. Then $X = S \times \mathbb{CP}^1$ is uniruled, hence has Kodaira dimension $-\infty$. But $b_{1}(X) = 0$, hence the irregularity satisfies $q(X) = h^{0,1}(X)=0$. Then we can can conclude that $X$ is not rational or birational to any Fano since for example $\pi_{1}(X) = \mathbb{Z}_{2}$.<|endoftext|> TITLE: Dependent choices (DC) in ${\bf HOD}(\mathbb{R},X)$, where $X$ is a set of reals QUESTION [5 upvotes]: In Turing invariant sets and the perfect set property, Math. Log. Quart. 66 (2020), Hamel, Horowitz and Shelah, the authors work in ZF + DC. They claim that DC can be dispensed with, asserting: if $V \models {\rm ZF}\: + $ “all Turing invariant sets have the perfect set property” and $X \in V$ is a set of reals, then ${\bf HOD}(\mathbb{R},X) \models {\rm ZF} + {\rm DC}\: +$ “all Turing invariant sets have the perfect set property”. Question: Why would ${\bf HOD}(\mathbb{R},X) \models {\rm DC}$, here? For context: the paper proves: If all Turing invariant sets have the perfect set property, then all sets of reals have the perfect set property. It's available on arXiv: https://arxiv.org/abs/1912.12558 REPLY [5 votes]: Consider the "singular Solovay-style model" of John Truss from Truss, John, Models of set theory containing many perfect sets, Ann. Math. Logic 7, 197-219 (1974). ZBL0302.02024. In that model the following holds: $V=L(\Bbb R)$, and therefore for any set of reals, $X$, the equalities $\mathrm{HOD}(\Bbb R,X)=L(\Bbb R)=V$ hold. $\sf DC$ fails, since $\omega_1$ is singular. Every set of reals, Turing invariant or not, has the perfect set property. If I understand their claim correctly, this is a counterexample.<|endoftext|> TITLE: A corollary of the non-existence of positive scalar curvature QUESTION [5 upvotes]: I've been done some work with scalar curvature and managed to give a simple proof for the following result: Let $M$ be a closed manifold which do not admit a metric of positive scalar curvature. Then for any Riemannian metric $g$ on $M$ it holds that $$\int_M \mathrm{scal}_g \leq 0.$$ In particular, the above condition is necessary and sufficient in order that a manifold does not admit a metric of positive scalar curvature. Since I am quite new in this area of positive scalar curvature specifically, I would like to know how known is this result and, also, if anyone can provide some references. EDIT: perhaps the correct statement is under the hypothesis of $M$ being a non-enlargeable spin manifold. REPLY [14 votes]: I am suspicious of your result. The three torus $\mathbb{T}^3$ is well-known to not admit any metric of positive scalar curvature. Let $g_0$ be the flat metric on $\mathbb{T}^3$. Given a positive function $u > 0$, consider the metric $g = u^{4} g_0$. Then we have the identity $$ - 8 \Delta_0 u = S_g u^5 \tag{1} $$ where $S_g$ is the scalar curvature of the metric $g$. Note that the volume form of $g$ is $\mathrm{dvol}_g = u^6~ \mathrm{dvol}_0$. Multiply both sides of (1) by $u ~\mathrm{dvol}_0$ we find $$ - 8 \Delta_0 u \cdot u ~\mathrm{dvol}_0 = S_g ~\mathrm{dvol}_g$$ Integrating both sides, for any non-constant $u$ you have that the left hand side is manifestly positive. But your "result" would imply that the right hand side must be non-positive. More generally: Let $g_0$ be any constant scalar curvature ($S_0$) metric, on an $n$-dimensional manifold $M$ with $n > 2$. Let $g_u = u^{4/(n-2)} g_0$, where $u$ is a positive function. Then we have that the scalar curvature $S_u$ of $g_u$ satisfies $$ - \gamma \Delta_0 u + S_0 u = S_u u^{2n/(n-2)} u^{-1} $$ where $\gamma = 4(n-1)/(n-2)$. Using again that $\mathrm{dvol}_u = u^{2n/(n-2)} ~\mathrm{dvol}_0$ we find that $$ \int (- \gamma \Delta_0 + S_0)u \cdot u ~\mathrm{dvol}_0 = \int S_u ~\mathrm{dvol}_u $$ Using that $\Delta_0^{-1}$ is compact, the operator $(-\gamma \Delta_0 + S_0)$ has arbitrarily large and positive eigenvalues. This shows that you can always find a metric conformal to $g_0$ with positive Einstein-Hilbert integral. (Remark: that $g_0$ has constant scalar curvature is inessential; it just makes the description of the spectrum of $-\gamma \Delta_0 + S_0$ easier to state. You can do the same argument with any metric; or you can bring out the big guns and use Yamabe to first transform the metric to one with constant scalar curvature.) When $n = 2$, your result is true by Gauss-Bonnet.<|endoftext|> TITLE: Differentiation of functions over graphs QUESTION [7 upvotes]: In short: There are various ways to define differentiation over a graph. I am trying to get the big picture, like a more complete and structured bestiary. Definitions. Let $G=(V,E)$ be a directed weighted graph; we denote by $\omega(u,v)$ the weight of edge $(u,v)$, with $\omega(u,v) = 0$ if $(u,v) \not\in E$. We consider undirected and unweighted graphs as special cases. For any $v$ in $V$, let $d(v)$ denote the (weighted) (out-)degree of $v$: $d(v) = \sum_{u\in V} \omega(v,u)$. Let $f$ be a function with domain $V$, which we call a graph function: $f$ associates a value to each vertex. Equivalently, a graph function may be seen as a vector $\mathbf{x}$ with $\mathbf{x}_v = f(v)$ for all $v\in V$. Let $\mathbf{A}$ denote the adjacency matrix of $G$: $\mathbf{A}_{uv} = \omega(u,v)$ for all $u$ and $v$ in $V$. Let $\mathbf{D}$ denote the diagonal matrix such that $\mathbf{D}_{vv} = d(v)$. There are several ways to define a (discrete) derivative $f'$ of $f$ over $G$ or, equivalently, the derived vector $\mathbf{y}$ associated to $\mathbf{x}$ (with $\mathbf{y}_v = f'(v)$). Shift+difference approach. A natural approach consists in saying that differentiating a graph function is just shifting its values and making their component-wise difference: $f'(v) = f(v) - \overline{f}(v)$ where $\overline{f}$ denotes a shifted $f$. Then, defining a derivative boils down to defining a shift. Several shift operators may be defined, for instance: $\overline{f}(v) = \sum_u \omega(u,v)\cdot f(u)$: the shifted value at $v$ is the (weighted) sum of the value at its neighbors. Equivalently, $\mathbf{\overline{x}} = \mathbf{A}\cdot \mathbf{x}$ and so this shift is called the adjacency shift. $\overline{f}(v) = \sum_u \omega(u,v)\cdot \frac{f(u)}{d(u)}$: the shifted value at $v$ is the (weighted) sum of the value at its neighbors divided by their (weighted) (outgoing) degree. Equivalently, $\mathbf{\overline{x}} = \mathbf{R}\cdot \mathbf{x}$, where $\mathbf{R}= \mathbf{D}^{-1}\cdot \mathbf{A}$ denotes the (weighted) random walk transition matrix of $G$, and so this shift is called the random walk shift. Direct matrix approach. Another approach directly defines matrix operators. The most classical probably is the graph Laplacian $\mathbf{L} = \mathbf{D}-\mathbf{A}$. Then, $\mathbf{y} = \mathbf{L}\cdot \mathbf{x}$ means that $f'(v) = \sum_u \omega(u,v)\cdot (f(v)-f(u))$: the differentiated value at $v$ is the (weighted) sum of the differences between the value at $v$ and the one at its neighbors. Several variants of this Laplacian operator exist. In particular, the random walk Laplacian defined as $\mathbf{D}^{-1}\cdot \mathbf{L}$ is nothing but the random walk differentiation, based on the random walk shift above. Questions and hints. All these differentiation definitions are used in the literature, as well as others. What other differentiation operators do you know? Which are your favorite ones? Why? Is there any meaningful classification of these operators? Which criteria are the most relevant? The classification may be property-oriented. For instance, some shifts preserve the global sum, others the global energy. The classification may rely on the operator form, like for instance the class of generalized Laplacians (for all $u\ne v$: $\mathbf{Q}_{uv}<0$ if $(u,v)\in E$, $\mathbf{Q}_{uv}=0$ otherwise; and $\mathbf{Q}_{vv}$ equal to any number). Similarly, one may distinguish operators having a matrix expression from operators having a shift+difference expression. Some may have both kinds of expression, and others none. For instance: the adjacency operator and the random walk operator above have both a matrix and a shift+difference expression. the Laplacian operator above is defined by a matrix expression but it does not seem to have a shift+difference expression, since it leads to $f'(v) = d(v)\cdot f(v) - \sum_u \omega(u,v)\cdot f(u)$. if instead we define $\mathbf{L'}=(\mathbf{D}^{-1}\cdot \mathbf{L})^\top$, then $\mathbf{y} = \mathbf{L'}\cdot \mathbf{x}$ leads to $f'(v) = f(v) - \sum_u \omega(v,u)\cdot \frac{f(u)}{d(v)}$; this operator is close to the Laplacian one, but it has a shift+difference expression. I am personally most interested by operators having a shift+difference expression and wonder if there are contexts where other kinds of differentiation operators make more sense. REPLY [2 votes]: This is just the tip of the iceberg so to speak, but let me add to your bestiary of discrete derivatives the David-Eynard $\nabla$-operator as defined in https://arxiv.org/pdf/1307.3123.pdf. Briefly, the picture is this: Begin with an abstract triangulation $T$, in other words an abstract graph with vertex and edge sets $\mathrm{V}(T)$ and $\mathrm{E}(T)$ together with an face set $\mathrm{F}(T)$ consisting of vertex triples ("triangles"). The triangulation is usually assumed to be locally finite in the sense each vertex $\mathrm{v} \in \mathrm{V}(T)$ participates as an end-point in only finitely many edges (and consequently finitely many triangles). Now let us assume we may equip $T$ with an polygonal embedding into the plane, namely an injective, complex-valued function $z: \mathrm{V}(T) \longrightarrow \Bbb{C}$. We interpret the embedding as mapping edges to straight line segments and we require (!) that these line segements are pairwise non-crossing, i.e. the interior of any line segment does not intersect the closure of any other line segment. Clearly each abstract triangle $\mathrm{f} = \{ \mathrm{u}, \mathrm{v}, \mathrm{w}\}$ is mapped to an actual planar triangle $z(\mathrm{f})$ with vertices $\{z(\mathrm{u}), z(\mathrm{v}), z(\mathrm{w})\}$ under this set-up. We'll orient an abstract triangle $\mathrm{f} = \{ \mathrm{u}, \mathrm{v}, \mathrm{w}\}$, i.e. read it as an ordered triple $\mathrm{f}^\circlearrowleft = ( z(\mathrm{u}), z(\mathrm{v}), z(\mathrm{w}) )$ up to a cyclic shirt, according to whether the area of $z(\mathrm{f})$ is calculated by the determinant \begin{equation} A(\mathrm{f}) = {1 \over {4 \mathrm{i}}} \, \det \, \begin{pmatrix} 1 & 1 & 1 \\ \overline{z}(\mathrm{u}) &\overline{z}(\mathrm{v}) &\overline{z}(\mathrm{w}) \\ z(\mathrm{u}) &z(\mathrm{v}) &z(\mathrm{w}) \end{pmatrix} \end{equation} with the correct sign. In other words orientation is determined by reading triangles in the plane counter-clockwise. Now given any complex-valued function $\phi: \mathrm{V}(T) \longrightarrow \Bbb{C}$ we may defined its discrete derivative $\nabla \phi: \mathrm{F}(T) \longrightarrow \Bbb{C}$ by \begin{equation} \nabla \phi(\mathrm{f}) \ := \ {1 \over {4 \mathrm{i} A(\mathrm{f})}} \, \det \, \begin{pmatrix} 1 & 1 & 1 \\ \overline{z}(\mathrm{u}) &\overline{z}(\mathrm{v}) &\overline{z}(\mathrm{w}) \\ \phi(\mathrm{u}) &\phi(\mathrm{v}) &\phi(\mathrm{w}) \end{pmatrix} \end{equation} where $\mathrm{f}^\circlearrowleft = ( z(\mathrm{u}), z(\mathrm{v}), z(\mathrm{w}))$. One can similiarly define a complex conjugate $\overline{\nabla}$-operator by making the substitution $z(\mathrm{p}) \mapsto \overline{z}(\mathrm{p})$ for each vertex $\mathrm{p} \in \mathrm{f}$ occuring the formula for the $\nabla$-operator. Given a smooth, complex-valued function $g: \Bbb{C} \longrightarrow \Bbb{C}$ we may restrict it to vertices $\mathrm{v} \in \mathrm{V}(T)$ by setting \begin{equation} g(\mathrm{v}) \ := \ g(z(\mathrm{v})) \end{equation} If we do this then clearly \begin{equation} \begin{array}{ll} \nabla \, g = 1 &\nabla \, \overline{g} = 0 \\ \overline{\nabla} \, g = 0 &\overline{\nabla} \, \overline{g} = 1 \end{array} \end{equation} where $g(w) = w$ for $w \in \Bbb{C}$. So $\nabla$ and $\overline{\nabla}$ agree with the continuous Wirtinger derivatives $\partial_w$ and $\overline{\partial}_w$ respectively up to linear functions in the global coordinates $w, \bar{w}$. If the polygonal embedding of $T$ has the added property that the circumcenter of each embedded triangle $z(\mathrm{f})$ is contained in its closure $\overline{z(\mathrm{f})}$ then on can define a postive semi-definite Laplace operator $\Delta: \Bbb{C}^{\mathrm{V}(T)} \longrightarrow \Bbb{C}^{\mathrm{V}(T)}$ by \begin{equation} \Delta = 4 \, {\frak{Re}} \Big[ \overline{\nabla}^{\scriptscriptstyle \top} A \, \nabla \Big] \end{equation} where $\Bbb{C}^{\mathrm{V}(T)}$ and $\Bbb{C}^{\mathrm{V}(F)}$ are the vector spaces of all complex-valued functions $\phi: \mathrm{V}(T) \longrightarrow \Bbb{C}$ and $\psi: \mathrm{V}(F) \longrightarrow \Bbb{C}$ respectively and where $A: \Bbb{C}^{\mathrm{V}(F)} \longrightarrow \Bbb{C}^{\mathrm{V}(F)}$ is the area operator defined by $A \phi (\mathrm{f}) := A(\mathrm{f}) \, \phi(\mathrm{f})$ for $\phi \in \Bbb{C}^{\mathrm{V}(F)}$ and $\mathrm{f} \in \mathrm{F}(T)$. One can show that $\Delta$ satisfies an adjunction formula (with respect to an appropriately defined inner product on $\Bbb{C}^{\mathrm{V}(T)}$) which justifies the view that $\Delta$ is the correct analogue of the Beltrami-Laplace operator in the polygonal setting (with the condition about circumcenters added). ines.<|endoftext|> TITLE: Nonseparable counterexamples in analysis QUESTION [21 upvotes]: When asking for uncountable counterexamples in algebra I noted that in functional analysis there are many examples of things that “go wrong” in the nonseparable setting. But most of the examples I'm thinking of come from $\mathrm C^*$-algebras (equivalent formulations of being type I; anything having to do with direct integral decompositions; primeness versus primitivity, etc.). Question. Are there also good examples from Banach space theory, or from harmonic analysis? Or from analysis in general? (Or even other examples from $\mathrm C^*$-algebra that I might not know about?) REPLY [3 votes]: For AF C*-algebras, there are a number of differences between separable and nonseparable algebras, and something weird happens at $\aleph_2$, too. Recall that an AF (C*)-algebra is the norm-closure of a directed limit of finite-dimensional C*-algebras (the norm is uniquely determined), equivalently, for every $\epsilon > 0$, any finite set of elements can be approximated to within $\epsilon$ by elements of a finite-dimensional algebra. George Elliott proved that for separable AF algebras (the usual kind), ordered pointed K$_0$ is a complete invariant (for isomorphism), and Effros Handelman and Shen determined exactly which partially ordered abelian groups can so arise; these are known as dimension groups (partially ordered abelian gps satisfying Riesz interpolation and unperforation) When the locally semisimple subalgebra has first uncountable dimension (which corresponds to the smallest nonseparability condition for the C*-algebra), it was already known (from the 40s, in a Russian paper) that ordered pointed K$_0$ is not complete. However, it was shown (by Goodearl and Handelman) that every first uncountable dimension group could arise as the ordered K$_0$ of an AF algebra with first uncountable dimensional locally semisimple algebra. However, when the dimension group is of cardinality $\aleph_2$, it was shown (by a well-known mathematician whose name escapes me at the moment), that there exist dimension groups of this cardinality which cannot be realized by a suitable direct limit, hence cannot be realized by corresponding AF algebras.<|endoftext|> TITLE: Finding a volume form on a fibre of a submersion between oriented manifolds QUESTION [5 upvotes]: Let $f:X\to Y$ be a submersion between orientable smooth manifolds of respective dimensions $n,p$ and let $j:M=f^{-1}(y)\hookrightarrow X$ denote the inclusion of some fibre of $f$. My naïve (I am not a differential geometer) question is: what data do we need in order to define a volume form on $M$, given the existence of volume forms on $X$ and $Y$ ? In the case where $Y=\mathbb R^p$ and $f=(f_1,\cdots,f_p)$, I think that if for some $\omega \in \Omega^{n-p}(X)$ the form $df_1\wedge\cdots\wedge df_p\wedge\omega\in \Omega^n(X)$ is a volume form (i.e. vanishes nowhere), then the form $j^* \omega\in\Omega^{n-p}(M)$ is a volume form for $M$. In the even more particular case where $X=\mathbb R^n$ and $p=1$ we may take for $\omega$ the $(n-1)$-form $$\omega=\sum_{i=1}^{n} (-1)^{i-1}\partial_i f \wedge dx_1\cdots\wedge \widehat{ dx_i}\wedge\cdots \wedge dx_n \in \Omega^{n-1}(\mathbb R^n)$$ and obtain the required volume form $j^* \omega\in\Omega^{n-p}(M)$. I once long ago read that last beautiful formula somewhere and the remembrance of that thing past motivates my present question, which is to find generalizations of that formula. REPLY [2 votes]: Take a unit tangent p-vector $a$ based at $y \in Y$ (you need the volume form on $Y$ to say what a unit p-vector is). At each point $x$ of the fiber $f^{-1}(y)$ consider the contraction of your volume form on $X$ with any p-vector $b$ in $\Lambda^p(T_xX)$ that projects down to $a$ by the linear map $D_xf$. Note that the pullback of this contracted form to the fiber (you only evaluate it on $n-p$ vectors tangent to the fiber) does not depend on the choice of the p-vectors that project to $a$. This is basically the pushforward construction for forms, which also works for densities.<|endoftext|> TITLE: Reference request for Linton's theorems on equational theories QUESTION [13 upvotes]: This is a reference request for the following "well-known" theorems in category theory: There is an equivalence of categories between finitary monads on $\mathbf{Set}$ and finitary Lawvere theories (i.e. single-sorted finite product theories). There is an equivalence of categories between monads on $\mathbf{Set}$ and infinitary Lawvere theories (i.e. singled-sorted product theories). The category of models of a finitary/infinitary Lawvere theory is equivalent to the category of algebras of the corresponding monad. I am not looking for proofs (I have written them up). I am looking for references which actually give proofs, so that I can cite them in a paper (instead of writing up the proof in the paper). Ideally, they should be classical references. I have scrolled through lots of papers which mention the theorems, and most of the time one (or several) of Linton's papers (let's give them letters) are cited: E. Some aspects of equational theories, Proceedings of the Conference on Categorical Algebra, Springer, 1966 F. An outline of functorial semantics, Seminar on triples and categorical homology theory, Springer, 1969 A. Applied functorial semantics, Seminar on triples and categorical homology theory, Springer, 1969 Sometimes, also the book "Toposes, Triples and Theories" by Barr & Wells is cited. In E Linton only mentions 2) in the end of section 6, without proof. In fact, Linton proves a characterization theorem of the concrete categories of models of varietal theories (his name for infinitary Lawvere theories), via a version of the first isomorphism theorem, and he only mentions in passing that a combination with Beck's monadicity theorem yields the equivalence to monads (but this detour is actually not necessary to get the equivalence; so this is not the simplest proof anyway). I do not understand much what is going on in F (I find it very hard to read, also because of the typesetting), but it seems to deal with a much more general situation, and therefore I don't see where 1) or 2) is proven either. This is funny because both in the introduction of the Lecture Notes and in the introduction of A it is claimed that Linton proves 2) in F. Can perhaps someone help me to "decipher" F and explain where 2) is actually proved? The only result which looks similar is Lemma 10.2, but its proof is omitted... Maybe 3) follows from Theorem 9.3, but I don't see how. The paper A focusses on monadicity criterions, and just mentions 2) in the introduction. I could not find a proof in the book by Barr-Wells either. They talk about the history of these theorems in section 4.5 and attribute 1) and 2) to Linton's E and F. I have found references with proofs of more general versions of 1), for example in the enriched case (Nishizawa, Power, Lawvere theories enriched over a general base), but it is probably awkward to cite such a paper for a classical result. I haven't found a published proof of 2) so far. The only thing which comes very close to 2) and 3) is the nlab article on algebraic theories: https://ncatlab.org/nlab/show/algebraic+theory, but the proof is sketchy. REPLY [2 votes]: A detailed, self-contained and beginner-friendly proof of both theorems can now be found in my new paper on limit sketches in the appendix.<|endoftext|> TITLE: Is pure mathematics useful outside of mathematics itself? QUESTION [33 upvotes]: From time to time Mathoverflow allows soft questions because they are arguably best answered by active mathematicians and they can benefit other mathematicians/PhD students/math undergraduates. I think this is such a question. I'm a mathematics student planning to enroll in a good math PhD program this Fall. I have always been extremely disciplined in math and my goal has always been to pursue a math PhD. However, I've had the opportunity to work in computer science, and this has caused some doubts about the significance of my future work in mathematics. I imagine such doubts are nonunique to myself and that the best place to ask is here, from people who've been through a PhD themselves, who are wiser, and who may possibly have had these same thoughts. (I hope it is clear I am asking this out of good nature and that this is not dismissed as a cynical thing to ask.) My main question: Is pure mathematics useful, specifically, outside of mathematics itself? Instead of giving a definition of "useful," perhaps I can share some doubts I have about the significance of pure mathematics research. It seems to me that in all honesty, pure mathematics does not immediately benefit the population at large in a direct and obvious way. At best benefits are usually theoretical (e.g., "These methods could..."). I think that very, very few people actually read and care about the average published pure mathematics paper. I think it's because math papers are hard and it's not clear that they are interesting or useful to math as a whole or to the future of humanity. There are very obvious exceptions, for example, for papers like Fermat's Last Theorem, which are arguably achievements for humanity. But most papers are objectively not of this level of significance and may not always contribute to major problems. It seems that the only reason we, as a population, care about mathematics, is because of the "cool" open problems which are simple to understand but difficult to prove. But this account for only a very small portion of active and successful mathematical work (since math papers don't always try to solve such problems because they're very hard). So doesn't this imply that my work as a future research mathematician is actually not useful for the future of humanity? It seems that pure mathematics was originally created to solve practical and interesting problems, and that as we turned to use abstraction as a tool to solve things (because abstraction is a very useful problem solving tool), we have arrived many years later to nested layers of subproblems of subproblems, whose depth is so deep that such problems of these areas are hard to understand and are not obviously useful for the world or for anything outside of that area of mathematics itself. It seems that mathematics is a science that studies itself, and so at a certain point, it does not have an immediate practical use outside of itself. I can't be the only math person to have every had these thoughts. As a hardcore pure math person it almost feels like a sin to have such doubts (not literally of course). I would very much like to be wrong, to learn from anyone's objections, and to do my PhD as I planned (although I obviously can't enroll with these doubts and will just continue working in CS). This leads to my secondary questions: Have any mathematicians ever had these thoughts? How did they reconcile these thoughts with their career choice? REPLY [4 votes]: It's kind of funny that when I was a young silly undergrad my view on Math was completely opposite to yours: in my immature eyes the best Math to get involved with would be the most impractical one! I wasn't aware at the time of the Steve Jobs quote about making the dent in the Universe, but that was basically the idea. To go somewhere where no man has gone before. To study the Forms, not the Shadows. To discover something beautiful outside the mundane realm of practical applications. Looking back, I was wrong. The most fruitful pure Math is not perfectly pure; it grows around applications like a pearl grows around a spec of dust. This applies to the most pure abstract things as well. People try to solve algebraic equations, that leads to algebraic varieties, and that leads to schemes. You don't start with schemes. Maybe that's the reason that pure Math tends to find applications eventually, sometimes decades after its development. Maybe that happens because pure Math ultimately grew from the real World, maybe that's just magic, but even the most impractical abstract subjects somehow find their applications.<|endoftext|> TITLE: Variety having infinite and perfect $\pi_1$ QUESTION [8 upvotes]: Question. Does there exist a smooth complex projective variety with infinite and perfect fundamental group? A group $G$ is perfect if its Abelianisation $G/[G,\, G]$ is the trivial group. REPLY [8 votes]: The simplest example I know of such a group is given by the presentation $$ \langle a, b, c| a^2, b^3, c^7, abc\rangle. $$ This group $G$ is obviously perfect and (less obviously) is isomorphic to the fundamental group of a smooth complex-projective variety. The reason for the latter is that $G$ acts holomorphically, properly and cocompactly on the unit disk and contains a finite index torsion-free subgroup. With this in mind, one uses the following lemma a form of which one can find in the book "Kahler groups": Lemma. Suppose that $G$ is a group containing a torsion-free normal subgroup $H$ of finite index. Suppose that $G$ is acting holomorphically, properly on a simply-connected complex manifold $X$ such that $X/H$ is biholomorphic to a smooth complex-projective variety. Then $G$ is isomorphic to the fundamental group of a smooth complex projective variety. Proof. There exists a simply-connected projective variety $Z$ with a free holomorphic action of $G/H$. Now, form the fiber product $$ Y=(X\times Z)/G $$ where the action of $G$ on $X$ is as above and $G$ acts on $Z$ via the action of $G/H$. Then $\pi_1(Y)\cong G$ and $Y$ is biholomorphic to a smooth projective variety, since it is the quotient of the projective variety $W=(X/H) \times Z$ by a free holomorphic action of a finite group, namely $G/H$.<|endoftext|> TITLE: Low dimensional noncommutative non-cocommutative Hopf algebras QUESTION [5 upvotes]: Sweedler's Hopf algebra (see here) is the lowest dimesnional ($4$-dimensional) Hopf algebra that is noncommutative and non-cocommutative. What are the next examples? Are there noncommutative, noncocommutative Hopf algebras of dimension $6,8,9,10$? Edit: Looking at the Stefan's paper (as suggested by John) I found that If $A$ is Hopf algebra of prime dimension $p$ over an algebraically closed field of characteristic $0$, then $A$ is the group algebra of the cyclic group $C_p$ of order $p$. Thus we see that prime order gives a commutative example. From this i removed $5$ and $7$ from the question. Edit: I have added non-cocommutative - since as explained below, these also come from groups, via the group algebra $k(G)$, and are dual to the ring constriction $k^G$. REPLY [7 votes]: By standard results (in fin dim, over an alg closed field of zero char), all cocommutative HAs are group algebras (for some finite group), all commutative HAs are duals of group HAs (for some finite group) (see for example: About the classification of commutative and of cocommutative, fin. dim. Hopf algebras) More generally, if we consider fd hopf algebras, over a field of char zero then, by the Larson-Radford theorem, semicimplicity and cosemisimplicity are equivalent notions and they both result either from commutativity or cocommutativity. Consequently, since you are looking to construct noncommutative, noncocommutative HAs, going beyond the semisimple/cosemisimple case would guarantee that the non-(co)semisimple HAs will also be non-(co)commutative. So i guess a natural class of examples would be pointed HAs. The Taft algebras constitute a standard class of non-commutative, non-cocommutative examples. Sweddler's hopf algebra (mentioned in the OP) is a special case of them. There are various results on the classification of f.d. hopf algebras which have been proved during the last couple of decades (with still lots of open problems) that can be of use for the purposes of the OP. Assuming that we are speaking about an algebraically closed field, of zero char, one should take into account both (co)semisimple and non-(co)semisimple cases: First note that all prime dimensions should be excluded, since by Zhu's theorem (also mentioned in the OP), these are all group hopf algebras of cyclic groups of prime order. Hence, these are commutative, cocommutative, semisimple and cosemisimple. $dim=6$: a Hopf algebra of dimension 6 is either commutative or cocommutative. These are all semisimple. $dim=8$: There are $7$ group algebras (up to isomorphism) and their duals (which are either comm or cocom). There is a unique (up to isomorphisms) semisimple Hopf algebra of dimension 8 that is neither commutative nor cocommutative. This is Kac's hopf algebra. See: Semisimple Hopf algebras of dimension 6, 8, A. Masuoka, Israel J. of Math. v. 92, p. 361–373, (1995). It interesting to note that dim $8$ is the smallest dimension for which (co)semisimplicity does not imply (co)commutativity. The non-semisimple case for dim 8 is more interesting though: It includes $6$ more isomorphism classes of HAs. These are either pointed or their duals are pointed. See: A Survey of Hopf Algebras of Low Dimension, M. Beattie, Acta Applicandae Mathematicae, v. 108, p.19–31, (2009) for their explicit description. $dim=9$: All semisimple HAs of dim $p^2$, $p=2,3$, are trivial (i.e. group algebras or their duals). In the non-semisimple case we have $p-1$ (pointed) Taft algebras. $dim=10$: the semisimple cases are trivial, i.e. we have only group algebras and their duals. This is a general result for $pq$-dim HAs: Semisimple Hopf Algebras of Dimension $pq$ Are Trivial, J. of Algebra, v. 210, 2, 1998, p. 664-669. Τhere are no non-semisimple HAs of dim $10$. See also the table included in p. 23 of Classifying Hopf algebras of a given dimension, arXiv:1206.6529v2 [math.QA] and the relevant literature included there. Following it, you can extend the above classification for higher dimensional HAs. Edit: A particularly interesting article is Hopf Algebras of Low Dimension, Stefan, Journal of Algebra 211, 343 361 (1999). There, the author classifies all isomorphism types of Hopf algebras of dimension $\leq 11$ over an algebraically closed field of characteristic 0.<|endoftext|> TITLE: Can Matsumoto's theorem for the symmetric group be proved using a monovariant? QUESTION [5 upvotes]: This is a question that can be asked for any Coxeter group, but for the sake of simplicity I will restrict myself to symmetric groups. Recall the main definitions: Let $n$ be a nonnegative integer. The symmetric group $S_n$ consists of the permutations of the set $\left[n\right] := \left\{1,2,\ldots,n\right\}$. For each $i \in \left[n-1\right]$, we let $s_i$ be the $i$-th simple transposition; this is the permutation in $S_n$ that swaps $i$ with $i+1$ while leaving all other elements of $\left[n\right]$ unchanged. It is well-known that the symmetric group $S_n$ can be presented as the group with generators $s_1, s_2, \ldots, s_{n-1}$ and relations $s_i^2 = 1$ for all $i \in \left[n-1\right]$, $s_i s_j = s_j s_i$ for all $i,j \in \left[n-1\right]$ satisfying $\left|i-j\right| > 1$, $s_i s_{i+1} s_i = s_{i+1} s_i s_{i+1}$ for all $i \in \left[n-2\right]$. This is known as the Coxeter-Moore presentation of $S_n$. Let us take a look at the combinatorics of this presentation. If $w \in S_n$, then a Coxeter word for $w$ shall mean a tuple $\left(i_1, i_2, \ldots, i_k\right) \in \left[n-1\right]^k$ satisfying $w = s_{i_1} s_{i_2} \ldots s_{i_k}$; a reduced word for $w$ shall mean a Coxeter word for $w$ that has the smallest length among all Coxeter words for $w$. For example, the cycle $\left(1,2,3\right)$ has reduced word $s_1 s_2$ and a (non-reduced) Coxeter word $s_3 s_1 s_3 s_2$. (The typical permutation has many reduced words and infinitely many Coxeter words.) Given a reduced word $\mathbf{i} = \left(i_1, i_2, \ldots, i_k\right)$ of $w$, we can obtain other reduced words of $w$ by the following transformations: We can pick two adjacent entries $i_u$ and $i_{u+1}$ of $\mathbf{i}$ that satisfy $\left|i_u - i_{u+1}\right| > 1$, and swap them. This is called a commutation move; for example, we can use such a move to transform $\left(1,2,3,1,2\right)$ into $\left(1,2,1,3,2\right)$. We can pick three adjacent entries $i_u$, $i_{u+1}$ and $i_{u+2}$ of $\mathbf{i}$ that satisfy $i_u = i_{u+2} = i_{u+1} \pm 1$, and replace them by $i_{u+1}$, $i_u$ and $i_{u+1}$, respectively. This is called a braid move; for example, we can use such a move to transform $\left(1,2,1,3,2\right)$ into $\left(2,1,2,3,2\right)$, and we can use another such move to transform this result further into $\left(2,1,3,2,3\right)$. Theorem (Matsumoto's theorem for the symmetric group). Let $w \in S_n$. Then, any two reduced words of $w$ can be transformed into one another by a sequence of commutation moves and braid moves. This theorem is often illustrated by drawing the graph whose vertices are the reduced words of a given $w \in S_n$, with two vertices $\mathbf{i}$ and $\mathbf{j}$ being joined by an edge if the reduced word $\mathbf{j}$ can be obtained from $\mathbf{i}$ by a single commutation move or braid move. Page 6 of Yuval Roichman's SLC67 slides shows such a graph. An elementary proof of Matsumoto's theorem appears, e.g., in the LLPT notes (Chapter SYM, Proposition (2.6)). Most texts on Coxeter groups prove it as well, in one or the other (usually more general) form. Some (I believe) derive it from the PBW property of the Hecke algebra. However, these proofs (to my knowledge) are rather tricky, and appear to rely on massaging the reduced words until they either begin or end with the same letter. (Not sure about the Hecke-based proof, but the proof in the LLPT notes definitely works this way.) I am a bit surprised that the seemingly more natural monovariant approach is never used: assigning a number (or some other kind of object in a totally ordered set) to each reduced word (for example, we could assign $\sum_{j=1}^k j^2 i_j$ to the reduced word $\left(i_1, i_2, \ldots, i_k\right)$); then showing that if $\mathbf{i}$ is any reduced word, then we can apply a braid or commutation move to $\mathbf{i}$ that decreases this number, unless $\mathbf{i}$ has some specific property that characterizes it uniquely (there are several known choices of "canonical" reduced word for a permutation $w$ -- it could be one of them, or a new one); then concluding by the monovariance principle (as the set of reduced expressions of $w$ is finite). (Intuitively, this sort of reasoning is the first thing one might try. After all, one can think of transforming reduced words as pulling a string around the surface of a permutahedron, while keeping the ends of the string fixed on two vertices; the permutahedron being convex, there should be a way to pull it "all the way down" without it getting stuck. The monovariant would then be responsible for defining the meaning of "down".) Alternatively, and to some extent equivalently, I'm surprised to have never seen a diamond lemma argument for the theorem. Question: Do such proofs of Matsumoto's theorem exist? Or are there some obstructions to them? REPLY [3 votes]: Adriano Garsia gives a proof of essentially this form in his book "The Saga of Reduced Factorizations of Elements of the Symmetric Group". See Theorem 1.1.2 for details. In particular, Garsia's argument does not rely on the exchange lemma. Basically, the idea is to show the minimal value $i$ occuring in a reduced word can be made to occur once. Then separating the reduced word into two, $(\textbf{a},i,\textbf{b})$ we can do the same to $\textbf{a},\textbf{b}$ to obtain $(\dots,i+1,\dots,i,\dots, i+1,\dots)$ and so on. Eventually this procedure results in a canonical form. If the first step can be interpreted using a monovariant argument, the whole proof can be viewed through that lense.<|endoftext|> TITLE: Can we construct a filtered chain complex from a spectral sequence? QUESTION [5 upvotes]: Suppose $\{(E_r,d_r)\}$ for $r>0$ forms a spectral sequence over a field $\mathbb{F}$, i.e. for any $r$, $(E_r,d_r)$ is a chain complex over $\mathbb{F}$ and $E_{r+1}=H(E_r,d_r)$. For simplicity, suppose $E_N=E_{N+1}=\cdots=E_{\infty}$ for some fixed number $N$. I have the following questions: Can we construct a filtered chain complex $(C,d)$ such that the induced spectral sequence is $\{(E_r,d_r)\}$? Can we determine $(C,d)$ up to some equivalence (like filtered quasi-isomorphism)? Can we relax the assumption to obtain the similar results? Are there any good references? Indeed, the spectral sequence in my mind is from the unrolled spectral sequence in https://ncatlab.org/nlab/show/conditional+convergence There are many similar questions, which gives some potential counterexamples when we replace $\mathbb{F}$ by $\mathbb{Z}$ or without the assumption about the convergence at some finite page $E_N$. Complete invariant of filtered chain complexes under chain homotopy equivalence Functoriality of filtered spectral sequences isomorphic spectral sequences => quasi-isomorphic filtered chain complexes? REPLY [3 votes]: Yes, since you're working over a field, you can decompose your spectral sequence into a big direct sum of a) permanent cycles, and b) for each r>1, chains supporting a d_r, and the cycle hit by that d_r. It is straightforward to represent each such summand by a simple filtered chain complex, so do so, and then take a direct sum. The resulting filtered chain complex has its spectral sequence isomorphic to the one you began with. I don't know a written reference for this, but it is an old idea, and it is not difficult to write out the details. I think this question is addressed by the more general results in "Model category structures and spectral sequences" by Cirici, Santander, Livernet, and Whitehouse: https://arxiv.org/pdf/1805.00374.pdf If I recall correctly, you can replace your ground field with any commutative ring without changing the answer to #1. But I am only going from memory here! See answers to #1 and #2. You mentioned unrolled exact couples. Let me mention that question #2 becomes much, much more subtle if you work with exact couples rather than filtered chain complexes, but at least question #1 still has the same answer (for basically the same reasons) in that setting.<|endoftext|> TITLE: Existence of harmonic maps onto the $n$-sphere QUESTION [7 upvotes]: Let $(M^n,g)$ be a closed smooth Riemannian $n$-manifold with positive scalar curvature (or positive Ricci curvature) and $(S^n, g_{st})$ be the standard round $n$-sphere. Whether there exists a non-zero degree harmonic map $f$ from $M^n$ onto $S^n$, $f:M^n\to S^n$? REPLY [5 votes]: A simple example where the answer is 'no' is when $M=\mathbb{RP}^2$ (with, say, the standard metric of Gauss curvature $K\equiv1$, though, in dimension $2$, only the conformal structure on $M$ matters in the definition of harmonic map). There is no non-constant harmonic map $f:\mathbb{RP}^2\to S^2$ (when $S^2$ given the standard metric with $K\equiv1$). In particular, there is not one that has nonzero degree. The reason is that such a map would lift to $\tilde f:S^2\to S^2$ as a non-constant harmonic map, and it is well-known that such a map would have to be either holomorphic or anti-holomorphic when $S^2$ is regarded as $\mathbb{CP}^1$, i.e., the Riemann sphere. Since $\mathbb{RP}^2$ is $\mathbb{CP}^1=S^2$ divided by the anti-holomorphic involution $[z,w]\to [-\bar w,\bar z]$, it would follow that the holomorphic mapping $\tilde f$ would have to satisfy $\tilde f\bigl([z,w]\bigr) =\tilde f\bigl([-\bar w,\bar z]\bigr)$, which would be impossible unless $\tilde f$ were constant.<|endoftext|> TITLE: Partitions of the real line into Borel subsets QUESTION [17 upvotes]: Problem 1. Is it true that for every cardinal $\kappa\le\mathfrak c$ there exists a partition $(B_\alpha)_{\alpha\in\kappa}$ of the real line into $\kappa$ pairwise disjoint non-empty Borel subsets? Remark. The answer to this problem is affirmative if $\mathfrak c\le \aleph_2$. If the answer to Problem 1 is negative, then what about Problem 2. Let $\mathcal P$ be a partition of the real line into Borel subsets. Is $|\mathcal P|=\mathfrak c$ or $|\mathcal P|\le\aleph_1$ in ZFC? REPLY [14 votes]: The answer to both problems is no! If the Cohen forcing is used to add lots of reals to a countable transitive model of GCH, then in the resulting extension, any partition of the real line into Borel sets has size $\leq \aleph_1$ or $\mathfrak{c}$. (But the continuum can be anything with uncountable cofinality.) This is a theorem of Arnie Miller, Theorem 3.7 in A. W. Miller, "Infinite combinatorics and definability," Annals of Pure and Applied Mathematical Logic 41 (1989), pp. 179-203. On the other hand, it is consistent with any permissible value of $\mathfrak{c}$ that there is a partition of the real line into $\kappa$ closed sets, for every uncountable $\kappa \leq \mathfrak{c}$. This is a joint theorem of myself and Arnie Miller, Theorem 3.11 in W. Brian and A. W. Miller, "Partitions of $2^\omega$ and completely ultrametrizable spaces," Topology and Its Applications 184 (2015), pp. 61-71 (link).<|endoftext|> TITLE: 2-REA PA degrees QUESTION [7 upvotes]: Remember that an n-REA set is a set of the form $A_0 \oplus A_1 \oplus \cdots \oplus A_n$ with $A_n$ relatively r.e. in $A_m, m TITLE: Are we able to estimate the fraction of the domain where $\cos (ax)+2\cos (b x)$ with $\frac ab \notin\mathbb{Q}$ is positive? QUESTION [8 upvotes]: We know that the two functions $\{\;\cos (ax),\;2\cos (b x)\;\}$ where $\frac ab \notin \mathbb{Q}$ are independently positive (and negative) over $\frac 12$ of the domain. Is it possible to estimate the fraction of the domain in which $\;\cos (ax)+2\cos (b x) \;$ is positive (in this case, the function is not periodic)? In other words, can we estimate $$ \lim_{m\to\infty} \frac 1m \int_{0}^{m} \Big(\cos (ax)+2\cos (b x) >0\Big) \,dx?$$ Any hints and comments are greatly appreciated. REPLY [11 votes]: Since $a,b$ are incommensurable, $(ax,bx)$ is asymptotically equidistributed in the torus $({\bf R} / 2\pi{\bf Z})^2$. [One proof is via a continuous version of Weyl's equidistribution criterion: for any integers $r,s$ with $(r,s) \neq (0,0)$ we have $$ \frac1m \int_0^m e^{i(rax+sbx)}\, dx = O_{r,s}(1/m) \to 0 $$ as $m \to \infty$.] Therefore $\{x > 0 \mid \cos ax + 2 \cos bx > 0 \}$ has the same density in the positive reals as $\{ (\theta,\phi) \in ({\bf R} / 2\pi{\bf Z})^2 \mid \cos \theta + 2 \cos \phi > 0\}$, which is $1/2$ by symmetry. REPLY [3 votes]: Let $r$ be an irrational real number. For real $x>0$, let $U_x$ be a random variable (r.v.) uniformly distributed on the interval $[0,x]$, and then let $$C_x:=\cos rU_x+2\cos U_x.$$ Then the problem can be restated as follows: Is it true that \begin{equation*} P(C_x>0)\to1/2\,\text{?} \tag{1} \end{equation*} Everywhere here, the limits are taken for $x\to\infty$. The answer to this question is yes. Indeed, for each $(k,n)\in\{0,1,\dots\}^2$, \begin{align*} &E\cos^k rU_x\,\cos^n U_x \\ &=2^{-k-n}\,\frac1x\,\int_0^x du\,(e^{iru}+e^{-iru})^k \,(e^{iu}+e^{-iu})^n \\ &=2^{-k-n}\sum_{p=0}^k\sum_{q=0}^n\binom kp\binom nq \frac1x\,\int_0^x du\, \exp\{iu[(2p-k)r+2q-n]\} \\ &\to2^{-k-n}\sum_{p=0}^k\sum_{q=0}^n\binom kp\binom nq 1(2p=k,2q=n), \end{align*} since $r$ is irrational. So, \begin{align*} &E\cos^k rU_x\,\cos^n U_x\to m_k m_n, \end{align*} where \begin{equation*} m_k:=1(k\text{ is even})2^{-k}\binom k{k/2}=E\cos^k U_{2\pi}. \end{equation*} So, by dominated convergence, for the joint characteristic function (c.f.) $f_{r,x}$ of the pair $(\cos rU_x,\cos U_x)$ of r.v's and all real $s,t$ we have \begin{align*} f_{r,x}(s,t)&=E\exp\{i(s\cos rU_x+t\cos U_x)\} \\ &=\sum_{n=0}^\infty \frac{i^n}{n!}\,E(s\cos rU_x+t\cos U_x)^n \\ &=\sum_{n=0}^\infty \frac{i^n}{n!}\, \sum_{k=0}^n\binom nk s^k t^{n-k}E\cos^k rU_x\,\cos^{n-k} U_x \\ &\to\sum_{n=0}^\infty \frac{i^n}{n!}\, \sum_{k=0}^n\binom nk s^k t^{n-k}m_k m_{n-k} \\ &=h(s)h(t), \end{align*} where \begin{equation*} h(s):=\sum_{k=0}^\infty \frac{i^k s^k m_k}{k!}=E\exp\{is\cos U_{2\pi}\}, \end{equation*} so that $h$ is the c.f. of the (symmetric absolutely continuous) r.v. $\cos U_{2\pi}$. So, the pair $(\cos rU_x,\cos U_x)$ of r.v's converges in distribution to a pair $(A,B)$ of independent copies of the r.v. $\cos U_{2\pi}$. So, for any real $b$, the r.v. $\cos rU_x+b\cos U_x$ converges in distribution to the symmetric absolutely continuous r.v. $A+bB$. Thus, (1) follows.<|endoftext|> TITLE: An introductory text on expanders QUESTION [9 upvotes]: I am looking for a book that covers expander graphs rigorously. Preferably a book aimed at beginners. REPLY [8 votes]: One should mention Lubotzky's lovely book "Discrete groups, expanding graphs and invariant measures" (1994), I think the first book written on the subject, which shows how the problem of constructing expander families connects to lots of other interesting mathematics. My already-mentioned book with Giuliana Davidoff and Peter Sarnak was based on a master class I taught on the subject of Ramanujan graphs, itself based on a set of unpublished notes by Giuliana and Peter. Our aim is to present one family of expanders: the graphs of Lubotzky-Phillips-Sarnak (also constructed independently by Margulis). The prerequisites are those of a 2nd year master class: linear algebra, a standard course in algebra (groups, rings and fields), measure theory, some combinatorics. The book "Expander families and Cayley graphs - a beginner's guide", by Mike Krebs and Anthony Shaheen (2011), starts lower in terms of prerequisites and has lots of drawings, tons of exercises, as well as students projects. Depending what kind of beginner you are, it may be more suitable...<|endoftext|> TITLE: Kullback–Leibler chains QUESTION [5 upvotes]: The following question was asked and then deleted by the post author: Let $P$ and $Q$ be two probability distributions defined over the same space, with $KL(P \parallel Q) < \infty$. For $\epsilon > 0$, is it possible to choose an $n \in \mathbb{N}$ and a sequence of distributions $R_1, \dots, R_n$ such that $KL(P \parallel R_1) + \dots + KL(R_n \parallel Q) < \epsilon$? Due to the strict convexity of the KL divergence in either one of its arguments (fixing the other), for each $i$ we have (assuming $R_i \neq R_{i+1}$ and $\lambda \in (0, 1)$) $$KL(R_i \parallel \lambda R_i + (1-\lambda)R_{i+1}) + KL(\lambda R_i + (1-\lambda)R_{i+1} \parallel R_{i+1}) < (1 - \lambda) KL(R_i \parallel R_{i+1}) + \lambda KL(R_i \parallel R_{i+1}) = KL(R_i \parallel R_{i+1}).$$ Therefore it would seem we can always just insert $\lambda R_i + (1-\lambda)R_{i+1}$ into the sequence of distributions, between $i$ and $i+1$, and push down the sum of the KL divergences. But maybe there are diminishing returns from repeating this procedure and there is some limit to how low you can go. This question may be of interest for some users and is therefore being revived here, accompanied with a (positive) answer. REPLY [3 votes]: $\newcommand{\ep}{\epsilon}\newcommand{\de}{\delta}$The Kullback--Leibler (KL) divergence may be defined by the formula \begin{equation*} D(P\parallel Q):=KL(P\parallel Q):=\int p\ln\frac pq=\int q\,g(p/q), \tag{0} \end{equation*} where $P$ and $Q$ are probability measures on a measurable space; $p$ and $q$ are, respectively, densities of $P$ and $Q$ with respect to a measure $\mu$ such that $P$ and $Q$ are absolutely continuous with respect to $\mu$; $\int f:=\int f\,d\mu$; and $$g(u):=u\ln u$$ for $u\in(0,\infty)$, with $g(0):=0$ and $g(\infty):=\infty$. Here we are using the standard conventions $a/0:=\infty$ for $a>0$ and $0\times\text{anything}=\text{anything}\times0:=0$. For $\mu$, one can always take e.g. $P+Q$. It is easy to see and very well known that we always have $D(P\parallel Q)\in[0,\infty]$. Here, it is given that \begin{equation*} c:=D(P\parallel Q)<\infty. \end{equation*} Without loss of generality $c>0$ (otherwise, there is nothing to prove). Take any $\ep\in(0,3c/2]$, so that \begin{equation*} \de:=\ep/(3c)\in(0,1/2]. \end{equation*} Take now any natural $n\ge2$ and for $j\in[n]:=\{1,\dots,n\}$ let \begin{equation*} R_j:=P_{t_j}, \end{equation*} where \begin{equation*} P_t:=(1-t)P+tQ \end{equation*} and \begin{equation*} t_j:=\de+\frac{j-1}{n-1}\,(1-2\de), \end{equation*} so that $t_1=\de\le1-\de=t_n$, $P_0=P$, $R_1=P_\de$, $R_n=P_{1-\de}$, and $P_1=Q$. In view of (0), $D(P\parallel Q)$ is convex in $P$ (because the function $g$ is convex) and in $Q$ (because $\ln\frac pq$ is convex in $q$). So, for all $t\in[0,1]$ \begin{equation*} D(P\parallel P_t)\le(1-t)D(P\parallel P_0)+tD(P\parallel P_1)=tc \end{equation*} and \begin{equation*} D(P_t\parallel Q)\le(1-t)D(P_0\parallel Q)+tD(P_1\parallel Q)=(1-t)c. \end{equation*} So, \begin{equation*} D(P\parallel R_1)\le\ep/3,\quad D(R_n\parallel Q)\le\ep/3. \end{equation*} To bound $D(R_j\parallel R_{j+1})$ for $j\in[n-1]$, we will use Lemma 1: For any $s$ and $t$ in $[\de,1-\de]$, \begin{equation} D(P_s\parallel P_t)\le\frac{2(s-t)^2}\de. \end{equation} This lemma will be proved at the end of the answer. At this point, just note that, by Lemma 1, \begin{equation} D(R_j\parallel R_{j+1})\le\frac{2(1-2\de)^2}{(n-1)^2\de} \end{equation} for $j\in[n-1]$, whence \begin{equation} D(P\parallel R_1)+D(R_1\parallel R_2)+\dots+D(R_{n-1}\parallel R_n)+D(R_n\parallel Q) \\ \le\ep/3+\frac{2(1-2\de)^2}{(n-1)\de}+\ep/3<\ep, \end{equation} as desired, if $n$ is taken to be large enough. It remains to provide Proof of Lemma 1: Note that $g(u)\le u-1+(u-1)^2$ for all real $u>0$. So, letting $p_t:=(1-t)p+tq$ for $t\in[0,1]$, so that $p_t$ is the density of $P_t$, we have \begin{align*} D(P_s\parallel P_t)&=\int p_t g(p_s/p_t) \\ &\le\int p_t \Big[\frac{p_s}{p_t}-1+\Big(\frac{p_s}{p_t}-1\Big)^2\Big] \\ &\le\int p_t \Big(\frac{p_s}{p_t}-1\Big)^2 \\ &=\int \frac{(p_s-p_t)^2}{p_t} \\ &=(s-t)^2\int \frac{(p-q)^2}{(1-t)p+tq} \\ &\le\frac{(s-t)^2}\de\,\int \frac{(p-q)^2}{p+q} \\ &\le\frac{(s-t)^2}\de\,\int(p+q)=\frac{2(s-t)^2}\de. \quad\Box \end{align*}<|endoftext|> TITLE: Commutator problem vs conjugacy/word problem QUESTION [12 upvotes]: For a finitely presented group $G$, generated by a finite set $A$, the commutator problem is the decision problem: given a word $w$ over the alphabet $A \cup A^{-1}$, can one decide if $w$ is a commutator, i.e. whether there exist words $x, y$ such that $w = [x, y]$ in $G$. Here $[x, y] = x^{-1}y^{-1}xy$ is the commutator. The commutator problem was solved for free groups by Wicks in 1962. In 1981, Comerford & Edmunds [1] asked whether decidability of the commutator problem for $G$ implies decidability of the conjugacy problem, or even the word problem, for $G$. Has there been any recent progress on this question since then, or any results in a similar direction? [1] Comerford, Leo P. jun.; Edmunds, Charles C., Quadratic equations over free groups and free products, J. Algebra 68, 276-297 (1981). ZBL0526.20024. REPLY [14 votes]: Denis Osin [Osin, Denis, Small cancellations over relatively hyperbolic groups and embedding theorems, Ann. Math. (2) 172, No. 1, 1-39 (2010). ZBL1203.20031.] proved that every torsion-free countable group can be embedded into a $2$-generated group with exactly two conjugacy classes of elements, i.e., any two non-trivial elements will be conjugate. So, let $H$ be a finitely generated torsion-free countable group with unsolvable word problem. Let's embed $H$ into a $2$-generated group $G$ with $2$ conjugacy classes using Osin's theorem. Then every element in $G$ is a commutator and $G$ has unsolvable word problem. Thus the commutator problem in $G$ is solvable but neither the word problem nor the conjugacy problem are decidable.<|endoftext|> TITLE: Lower convex envelope of polynomial functions QUESTION [6 upvotes]: Let $P\in{\mathbb R}[X]$ be a polynomial and $[a,b]$ be a bounded interval. Of course, the graph of $P$ is an algebraic set. I am interested in the lower convex envelope $\bar P$ of $P|_{[a,b]}$. It seems to me that its graph is semialgebraic. However I am not at all familiar with real algebraic geometry. Is there a clear justification ? At least, the graph of $\bar P$ is the union of arcs of the graph of $P$, together with bi-tangents. The latter are obtained by an elimination in a system of two algebraic equations. Same question when replacing $P$ by $\min(P,Q)|_{[a,b]}$ where $P$ and $Q$ are two polynomials. Perhaps my question looks obvious for specialists, in which case I should be happy with a proper reference. REPLY [2 votes]: The answer is yes, but not in the most exciting way. The idea is to write the envelope in the first-order logic and then kill the problem with quantifier elimination for the theory of real-closed fields. For example, we can describe the envelope as the set of points $(z, w)$ such that for all $x, y\in [a, b]$ we have $(z, w)$ is above or on the segment between $(x, P(x))$ and $(y, P(y))$ but for any $t > 0$ there are $x, y$ such that this is not true for $(z, w-t)$. One can write it out as a first-order formula of quantifier degree 6 or so. Then, running the elimination algorithm and then reducing the result to a disjunctive normal form we get the expression as a semi-algebraic set (small caveat here is that one should treat $a, b$ and the coefficients of $P$ as free parameters and the resulting semi-algebraic set of course depends on these parameters). Note that one can also bound the number of pieces of this semi-algebraic set in terms of the degree of $P$ only (something like $2^{2^{O(deg P)}}$ or whatever the complexity of the quantifier elimination algorithm). Of course, this strategy can be easily modifed to two or more polynomials and many other situations.<|endoftext|> TITLE: PL-embeddings of balls into PL-manifolds QUESTION [8 upvotes]: Let $B$ be $k$-dimensional PL-ball and let $M$ be a connected $n$-dimensional PL-manifold, let's say without boundary. Furthermore let $f,g\colon B\to M$ be two PL-embeddings. If $k=n$, then the disc theorem (see Theorem 3.34 in Rourke-Sanderson) says that, after possibly composing $f$ by a reflection, $f$ and $g$ are ambiently isotopic. Now assume that $k TITLE: Functoriality of Thurston's norm QUESTION [5 upvotes]: Let $M$ be a manifold of dimension $3$ and let $N$ be an embedded submanifold of $M$ (also of dimension $3$). Then, both second homologies $H_2(M)$ and $H_2(N,\delta N)$ are equipped with a norm (technically a semi-norm if we don't assume more properties, but not relevant here), called Thurston's norm. I am wondering whether there is any relation between the two normed vector spaces $H_2(M)$ and $H_2(N,\delta N)$. More precisely, if $i \colon N \to M$ is the embedding, is there some induced map $i_* \colon H_2(N, \delta N) \to H_2(M)$ of normed vector spaces, possibly making the construction functorial. The main issue for me here is that if $x \in H_2(N,\delta N)$ is represented by a surface (possibly not connected) with boundaries, I don't see a canonical way to extend this surface to a closed surface in $M$. REPLY [6 votes]: For submanifolds that are link complements, Ken Baker and I recently investigated this question in our paper "Dehn Filling and the Thurston Norm" available at arXiv:1608.02443 The basic idea (as in so many 3-manifold papers!) is to play surfaces of different boundary slopes against each other. It's most natural to think of this in terms of comparing Dehn surgeries, but one can also think of it in terms of using a surface with non-meridional boundary on the link to say something about the situation when the ambient 3-manifold has a class with smaller than expected Thurston norm compared to the corresponding (relative) class in the submanifold.<|endoftext|> TITLE: How special are homogeneous spaces? QUESTION [8 upvotes]: Let $M$ be a smooth finite dimensional manifold, how restrictive is it to require $M$ to admit a smooth action by a finite dimensional Lie group $G$? Related questions/approaches: Of course we need $\mathrm{dim}(G) \geq \mathrm{dim}(M)$, are there any results relating the minimal dimension of a Lie group acting transitively to that of $M$, perhaps in special cases? Going in the other direction, any criteria which easily allow to say that a given smooth manifold does not admit a transitive group action? EDIT: in the first question I meant to write transitive smooth group action. REPLY [10 votes]: I suppose you want the action to be transitive as your title suggests. In this case, a classical theorem of Mostow (in 1950 for surfaces, in 2005 in general) says that for a compact homogeneous space $M = G/H$ the Euler characteristics is non-negative. Mostow, G.D.: The extensibility of local Lie groups of transformations and groups on surfaces. Ann. Math. (2) 52, 606–636 (1950) Mostow, G.D.: A structure theorem for homogeneous spaces. Geom. Dedic. 114, 87–102 (2005)<|endoftext|> TITLE: Have any proposals been advanced for the analytic continuation of the divisor function? QUESTION [11 upvotes]: While I was working on the evaluation of a certain series, the following limit came up: \begin{align} \lim_{n \to 1} \frac{d(n)-1}{n(n-1)} &= \lim_{n \to 1} \frac{d'(n)}{2n-1} \\ &= d'(1) .\end{align} Here, I used l'Hôpitals rule, and $d(\cdot)$ denotes the divisor function. In order to compute the derivative of a function, one must know how it is defined on the real numbers. Unfortunately, I have not found any proposals that described the notion of the analytic continuation for the divisor function so far. It seems that it is defined on $\mathbb{Z}$ only. I did find identities for $\sigma_{\alpha} (x)$ for any $\alpha \in \mathbb{C}$ and $x \in \mathbb{Z}_{\geq 1}$. However, this is not what I am looking for. I seek to find a suitable expression for $\sigma_{\alpha} (x)$ when $\alpha = 0$, and real or complex arguments $x$. I believe it may be possible to find such an extension of the divisor function, in part because such extensions have been found for the lowercase prime omega function. This arithmetic function is related to the divisor function. The continuation of the function is as follows: $$ \DeclareMathOperator{\sinc}{sinc} \omega(z) = \log_{2} \Bigg{(} \sum_{x=1}^{\lceil Re(z) \rceil} \sinc \Bigg{(} \prod_{y=1}^{\lceil Re(z) \rceil + 1 } (x^{2} + x - yz) \Bigg{)} \Bigg{)} ,$$ where $\sinc(\cdot)$ is the normalized sinc function. Question: have any proposals been advanced for the analytic continuation of the divisor function, thereby extending the domain to $\mathbb{R}$ or $\mathbb{C}$ ? REPLY [2 votes]: Since the formula I posted in my previous answer only converges as $N\to\infty$ at $x\in\mathbb{Z}_{\ne 0}$, I decided to post another formula which I believe converges as $N\to\infty$ for $x\in\mathbb{C}$. The previous answer I posted is consistent with the fact that zero has an infinite number of divisors, but this answer assumes the definition $\sigma_0(0)=0$. Consider the following definitions of the analytic extension $\tilde{\sigma}_0(x)$ of $\sigma_0(n)$ and it's first order derivative $\tilde{\sigma}_0'(x)$ which are based on partial evaluations of real analytic formulas for $\tilde{f_{\sigma_0}}'(x)=\sum\limits_n \sigma_0(n)\,\delta(x-n)$ and $\tilde{f_{\sigma_0}}''(x)=\sum\limits_n \sigma_0(n)\,\delta'(x-n)$ (see this answer I posted to a related question on Math StackExchange). $$\tilde{\sigma}_0(x)=\underset{N\to\infty}{\text{lim}}\left(\frac{1}{2}\sum\limits_{n=1}^N\left(-\frac{\sin(2 \pi x)}{\pi x}+\frac{1}{n}\sum\limits_{k=1}^n\left(\cos\left(\frac{2 \pi k x}{n}\right)+\cos\left(\frac{2 \pi (k-1) x}{n}\right)\right)\right)\right)\tag{1}$$ $$\tilde{\sigma}_0'(x)=\underset{N\to\infty}{\text{lim}}\left(\frac{1}{2}\sum\limits_{n=1}^N\left(\frac{\sin(2 \pi x)-2 \pi x \cos(2 \pi x)}{\pi x^2}-\frac{2 \pi}{n^2}\sum\limits_{k=1}^n\left(k \sin\left(\frac{2 \pi k x}{n}\right)+(k-1) \sin\left(\frac{2 \pi (k-1) x}{n}\right)\right)\right)\right)\tag{2}$$ Figure (1) below illustrates formula (1) for $\tilde{\sigma}_0(x)$ evaluated at $N=100$ in blue where the red discrete evaluation points illustrate $\sigma_0(x)$ for $x=n\in\mathbb{Z}$. Figure (1): Illustration of formula (1) for $\tilde{\sigma}_0(x)$ The following table illustrates formula (2) for $\tilde{\sigma}_0'(x)$ evaluated at $x=1$ seems to converge to approximately $-3.35$ as $N\to\infty$. $$\begin{array}{cc} N & \tilde{\sigma}_0'(1) \\ 100 & -3.31334 \\ 1000 & -3.34278 \\ 10000 & -3.34574 \\ \end{array}$$<|endoftext|> TITLE: Mittag-Leffler function QUESTION [6 upvotes]: Let the Mittaq-Leffler function be defined by the expression $$ E_{\mu,\nu}(z) = \sum_{k=0}^{\infty} \frac{z^k}{\Gamma(k\mu+\nu)}\quad \text{$\mu>0$ and $\nu\in \mathbb R$}$$ Now let $n\in \mathbb N$, $\alpha \in (\frac{n-1}{n},1)$. My question is whether the Mittag-Leffler function $E_{n\alpha,1}(t)$ is non vanishing for all $t<0$ or not. REPLY [4 votes]: The answer is no. E.g., for $n=2$ and $\alpha=19/20$, we get the following: The idea of this example is based on the identity $E_2(-z^2)=E_{2,1}(-z^2)=\cos z$ and the continuity of $E_{a,1}(t)$ in $a$. More counterexamples can be obtained from the asymptotics $$E_{a,1}(t)\sim-\frac1{\Gamma(1-a)t}=-\frac{\Gamma(1+a)}{\pi at}\,\sin\pi a$$ as $t\to-\infty$ for $a\in\bigcup_{n=1}^\infty(n-1,n)$ and the equality $E_{a,1}(0)=1$.<|endoftext|> TITLE: Can the intersection of a unitary and an irreducibly represented injective $C^*$-algebra be $\{0\}$? QUESTION [7 upvotes]: Let $\mathcal{A}$ be an injective $C^*$-algebra irreducibly acting on a Hilbert space $\mathcal{H}$, and let $\phi$ be a completely positive idempotent from $\mathbb{B}(\mathcal{H})$ onto $\mathcal{A}$, and let $u\in\mathbb{B}(\mathcal{H})$ be a unitary. Question: Can we conclude that $\phi(u)\ne0$? If necessary, we can assume that $u$ normalizes $\mathcal{A}$, that is, $u\mathcal{A}u^*=\mathcal{A}$. The point is the irreducibility, since without it $\phi(u)$ can easily be $0$. The question appears to be easy, but I have been struggling for months, and I have already asked five experts. REPLY [6 votes]: No. Take any irreducibly represented simple and injective C*-algebra $A\subset B(H)$ and an outer automorphism $\theta$ of period $2$ (which is easy to find when $A$ is the hyperfinite II_1 factor). Case 1: The representation is covariant, i.e., $\exists u\in{\mathcal U}(H)$ such that $uau^*=\theta(a)$ for $a\in A$. We may assume $u^2=1$ by irreducibility. Since $\theta$ is outer, $A\rtimes_\theta({\mathbb Z}/2{\mathbb Z})$ is simple by Kishimoto's theorem. Hence, $\mathrm{C}^*(A,u)\cong A\rtimes_\theta({\mathbb Z}/2{\mathbb Z})$ and there is a conditional expectation $\phi\colon\mathrm{C}^*(A,u)\ni a+bu\mapsto a\in A$, $a,b\in A$. The map $\phi$ extends on $B(H)$. Case 2: The representation is not covariant, in which case there is no nonzero $x\in B(H)$ such that $xa=\theta(a)x$ for all $a\in A$, by Schur's lemma. Consider $A_1:=A\rtimes_\theta({\mathbb Z}/2{\mathbb Z})=\mathrm{C}^*(A,v)$ represented on $H\oplus H$, where $a\in A$ acts by $a\oplus\theta(a)$ and the switching unitary operator $v(\xi\oplus\eta)=\eta\oplus\xi$ implements $\theta$. Now, $A_1$ is injective, irreducibly represented on $H\oplus H$, and the dual automorphism $\hat{\theta}\colon a+bv\mapsto a-bv$ is outer and implemented by the unitary operator $u:=1\oplus-1$ on $H\oplus H$. Hence it reduces to Case 1.<|endoftext|> TITLE: Swapping non-commuting generators in Coxeter group QUESTION [6 upvotes]: Let $a$ and $b$ be two generators in a Coxeter group which do not commute. Is it possible for $ab$ to be equal to a product of generators where all instances of $b$ come before all instances of $a$? I've tried coming up with invariants that are preserved after applying the conditions of a Coxeter group to a string of generators, but the fact that one can insert either $aa$ or $bb$ at any point in the string has made this complicated. Particularly, the more general claim that a string with all $a$s before $b$s can't be turned into one with all $b$s before $a$s turns out to be false, since for instance $aab=baa$. REPLY [12 votes]: The answer is no. The Deletion Condition says that any expression in the generators of a Coxeter group contains a reduced expression for the same element as a subexpression. Since $a$ and $b$ don't commute, the unique reduced expression for $ab$ is $ab$, which must therefore occur as a subexpression of any longer expression that evaluates to $ab$, i.e. any such expression must contain an $a$ before a $b$.<|endoftext|> TITLE: How similar are the c.e. degrees and the CEA(Cohen) degrees? QUESTION [6 upvotes]: Given reals $A,B,X$, let $A\le_{T/X}B$ iff $A\oplus X\le_TB\oplus X$. For each real $X$ we can define a version of the c.e. degrees over $X$: we look at the preorder on $X$-c.e. reals given by $\le_{T/X}$ (or, if preferred, take the corresponding partial order). Call this $\mathcal{R}_X$; equivalently, $\mathcal{R}_X$ could be given by the reals which are CEA $X$. My question is the following: Suppose $G$ is "sufficiently" Cohen generic. Is $\mathcal{R}_G\cong\mathcal{R}$? Here $\mathcal{R}$ is the usual c.e. degrees. Incidentally, it's not hard to show that if $G,H$ are "sufficiently" Cohen generic then $\mathcal{R}_G\cong\mathcal{R}_H$. This is false. I vaguely recall a result (due to Shore?) that the answer is no, but I can't track it down. If the answer is no, I'm curious how similar they are nonetheless. For example, do $\mathcal{R}_G$ and $\mathcal{R}$ have the same $\Pi_3$ theories? REPLY [7 votes]: I am reading ≅ as meaning "elementarily equivalent." There are several questions here. I believe that some of them are accessible by known methods. The first is that if $X$ is not (low-level)arithmetically definable, then $\mathcal{R}_X$ is not elementarily equivalent to $\mathcal{R}$. The proof should go by relativising the methods of Nies, André ; Shore, Richard A. ; Slaman, Theodore A. Interpretability and definability in the recursively enumerable degrees. Proc. London Math. Soc. (3) 77 (1998), no. 2, 241--291. They show that there is a way to interpret the structure of first order arithmetic in the theory of $\mathcal{R}.$ Relativizing, the method should show that there is an interpretation of the structure of first order arithmetic with a predicate for $X$ in $\mathcal{R}_X$. The elementary difference with $\mathcal{R}$ would be of the form, there is an interpretation of the structure of first order arithmetic with a unary predicate $U$ such that $U$ is not $\Sigma_k$, where $k$ is sufficiently large that it cannot be so represented in $\mathcal{R}$. It should suffice that $k=10$. You might look at Shore's paper, Shore, Richard A. Degree structures: local and global investigations. Bull. Symbolic Logic 12 (2006), no. 3, 369--389, for a survey of known results and methods. I agree that it would be interesting to know the exact complexity of the optimal elementary difference. That question also touches on the old and open problem of deciding the two-quantifier theory of $\mathcal{R}$. It would be very interesting if working above $X$ a generic, or a random, or anything at, made it possible to decide the two-quantifier theory of $\mathcal{R}_X$.<|endoftext|> TITLE: Elementary embeddings and replacement QUESTION [10 upvotes]: Let $\alpha\not= 0$ be such that for every $\beta<\alpha$ there is $\beta<\gamma<\alpha$, where $V_\gamma$ is an elementary substructure of $V_\alpha$. In other words, $V_\alpha$ is a limit of its $V_\beta$ elementary substructures. Then it is a simple result that $V_\alpha$ models replacement. My question: Let $\alpha\not= 0$ be such that for every $\beta<\alpha$ there is $\beta<\gamma<\alpha$ and an elementary embedding from $V_\gamma$ to $V_\alpha$. Does it follow that $V_\alpha$ models replacement? REPLY [11 votes]: If $\alpha$ is a limit of $2^\alpha$-supercompact cardinals, then by the Magidor characterization of supercompactness, for each $2^\alpha$-supercompact cardinal $\kappa < \alpha$, for some $\gamma < \kappa$, there is an elementary embedding $j : V_{\gamma}\to V_\alpha$ with critical point arbitrarily large below $\kappa$. Thus for each $\beta < \gamma$, there is an elementary embedding $j : V_{\gamma}\to V_\alpha$ such that $\gamma$ is between $\beta$ and the next $\alpha$-supercompact cardinal, which yields the property you asked about. But if $\alpha$ is the least cardinal that is a limit of $\alpha$-supercompact cardinals, then $V_\alpha$ does not model replacement: from the perspective of $V_\alpha$, there are $\omega$-many supercompact cardinals that are cofinal in the ordinals. The reason is that the $2^{\alpha}$-supercompacts of $V$ are precisely the supercompacts of $V_\alpha$. Clearly the forwards implication holds, but conversely if $\delta$ is supercompact in $V_\alpha$, then $\delta$ is supercompact up to a cardinal $\kappa < \alpha$ that is $2^\alpha$-supercompact, and as a consequence, $\delta$ itself is $2^\alpha$-supercompact. The optimal hypothesis is the existence of an ordinal $\alpha$ that is a limit of $\alpha$-Magidor supercompact cardinals, where a cardinal $\kappa$ is $\alpha$-Magidor supercompact if for some $\gamma < \kappa$, there is an elementary embedding $j : V_\gamma\to V_\alpha$ such that $\kappa = j(\text{crit}(j))$. Let $\alpha$ be the least ordinal as in your question. Let $\beta$ be the supremum of the $\alpha$-Magidor supercompact cardinals. If $\beta < \alpha$, then fix an elementary embedding $j : V_\gamma\to V_\alpha$ with $\beta < \gamma < \alpha$, and note that $j(\text{crit}(j)) \leq \beta < \gamma$ and hence $j$ witnesses that $\text{crit}(j)$ is huge, contrary to the minimality of $\alpha$. So $\alpha = \beta$ and hence $\alpha$ is a limit of $\alpha$-Magidor supercompact cardinals.<|endoftext|> TITLE: Uses for Sharkovskii's theorem QUESTION [6 upvotes]: Sharkovskii's theorem is a fascinating result and ready to stand on its own. But is it also used somewhere? Are there other theorems that rely on it or somehow use the Sharkovskii ordering? REPLY [2 votes]: Quoting from Thomas D Rogers, Remarks on Sharkovsky's Theorem, Rocky Mountain J. Math. 15 (1985) 565-569: In regard to applications of Sharkovsky's theorem (difference equation models) it is interesting that the result is sturdy to perturbations in $f$. Block [5] shows that if $f$ has a point of period $n$, then there is a neighborhood $N$ of $f$ in $C(I)$ such that for all $g$ in $N$ and all positive integers $k$, with $k$ left of $n$ in the Sharkovsky order, $g$ has a point of period $k$. Kloeden [10] and Butler and Pianigiani [6] have related results. [5] is L. Block, Stability of periodic orbits in theorem of Sharkovsky, Proc. Amer. Math. Soc. 81 (1981), 333-336. [10] is P. E. Kloeden, Chaotic difference equations are dense, Bull. Austral. Math. Soc. 15 (1976), 371-379. [6] is G. J. Butler and G. Pianigiani, Periodic points and chaotic functions in the unit interval, Bull. Austral. Math. Soc. 18 (1978), 255-265.<|endoftext|> TITLE: On the comparison map $MU^\bullet(X)\otimes_{MU^\bullet(pt)}E^\bullet(pt)\to E^\bullet(X)$ for complex oriented multiplicative cohomology theories QUESTION [7 upvotes]: Whatever complex oriented multiplicative cohomology theories are, they come with two basic properties (among many others): i) a complex oriented multiplicative cohomology theory is a contravariant functor form (nice) topological spaces to (graded) rings; ii) there is an initial object $MU^\bullet$. Properties i) and ii) alone imply that for a (nice) topological space $X$ we have two distinguished morphisms of rings, $$ E^\bullet(pt)\to E^\bullet(X) $$ induced by the terminal morphism $X\to pt$, and $$ MU^\bullet(X)\to E^\bullet(X) $$ induced by the fact $MU^\bullet$ is initial. By naturality we have the commutative diagram of rings $\require{AMScd}$ \begin{CD} MU^\bullet(pt) @>>> MU^\bullet(X)\\ @V V V @VV V\\ E^\bullet(pt) @>>> E^\bullet(X) \end{CD} Then, by the universal property of the tensor product of rings, we have a distinguished morphism of rings $$ MU^\bullet(X)\otimes_{MU^\bullet(pt)}E^\bullet(pt)\to E^\bullet(X). $$ What I'm unable to understand is whether this is always an isomorphism or, if it is not, which are examples where it is not an isomorphism. Clearly, if $E^\bullet$ is Landweber exact, the above comparison map is easily seen to be an isomorphism. Yet, as far I understand it, Landweber exactness is more concerned with determining whether $MU^\bullet(X)\otimes_{MU^\bullet(pt)}R$ defines a muliplicative cohomology theory given a ring morphism $MU^\bullet(pt)\to R$ than into investigating what happens when one a priori knows $R$ is $E^\bullet(pt)$ and the morphism $MU^\bullet(pt)\to E^\bullet(pt)$ the universal one. Moreover, Landweber exact functor theorem gives a sufficient condition for $MU^\bullet(pt)\to R$ to induce a multiplicative complex oriented cohomology theory and not a necessary one. In other words, my question is: the comparison morphism $$ MU^\bullet(X)\otimes_{MU^\bullet(pt)}E^\bullet(pt)\to E^\bullet(X). $$ is always defined, and there are "nice" cohomology theories, for which this is an isomorphism. Nice cohomology theories include Landweber exact theories. But what are nice theories? Are they all? If not, what is an example of a not nice theory? REPLY [6 votes]: Firstly, even in the Landweber exact case your comparison map need not be an isomorphism unless $X$ is a finite complex. It is more often the case that $E^*(X)$ agrees with the completed tensor product $MU^*(X)\widehat{\otimes}_{MU^*}E^*$. However, I am still not sure whether that works for all $X$; exactness properties of completed tensor products are quite delicate. On the other hand, the Landweber exactness criterion is actually equivalent to $MU^*(-)\otimes_{MU^*}R^*$ being a cohomology theory on finite complexes. One direction is Landweber's theorem, the other follows easily from the existence of generalized Moore spectra $S/(v_0^{i_0},\dotsc,v_n^{i_n})$ (which is a highly nontrivial theorem, of course).<|endoftext|> TITLE: Absolute values of non-unit algebraic integers at all infinite places QUESTION [9 upvotes]: Let $K$ be a number field and $O_K$ its ring of integers. Given any non-unit element $\alpha\in O_K$, does there exist a unit $u\in O^\times_K$ such that $$ |\sigma(u \alpha)| >1 \quad \text{ for all } \sigma \in \mathrm{Hom}(K, \mathbb{C}). $$ I do not find a counter-example; and I can not see how to use Minkowski theory to find such an element in the ideal lattice $\alpha O_K \subset \prod_{v\mid \infty} K_v$. REPLY [15 votes]: No. Let $K=\mathbb Q(\sqrt{6})$ and let $\alpha=2+\sqrt{6}\in O_K$. All units of $O_K$ are given by $u=\pm v^n$, where $v=5+2\sqrt{6}$ is the fundamental unit. The key here is that $\alpha$ has relatively small norm while the fundamental unit $v$ is relatively large. Consider $u\alpha=\pm v^n\alpha$. If $n<0$, then $|u\alpha|=|v|^n|\alpha|\leq|5-2\sqrt{6}|\cdot |2+\sqrt{6}|<1$. If $n\geq 0$, then $|u\alpha|=|v|^n|\alpha|\geq|2+\sqrt{6}|>2$. Since $\alpha$, and hence also $u\alpha$, has absolute norm $2$, this implies that the conjugate of $u\alpha$ has absolute value smaller than $1$.<|endoftext|> TITLE: Number of subgroups of a $p$-group of index $p^k$ QUESTION [5 upvotes]: Let $p$ be a prime, let $n$ and $k$ be positive integers and let $G$ be a group of order $p^n$. Further, let $a_{p^k}$ denote the number of subgroups of $G$ of index $p^k$. If $a_{p^k}$ is greater than 1 and not congruent to $p+1$ modulo $p^2$ -- does it follow that $p = 2$ and $G$ is either a dihedral group, a quasidihedral group or a generalized quaternion group? REPLY [10 votes]: It seems that the $p>2$ part of this was proved in Kulakoff, A., Über die Anzahl der eigentlichen Untergruppen und der Elemente von gegebener Ordnung in $p$-Gruppen., Math. Ann. 104, 778-793 (1931). ZBL57.0146.03. and the $p=2$ part in Easterfield, T. E., An extension of a theorem of Kulakoff, Proc. Camb. Philos. Soc. 34, 316-320 (1938). ZBL0019.10802.<|endoftext|> TITLE: Branching rule for Specht modules over Kazhdan-Lusztig basis QUESTION [8 upvotes]: Suppose $\lambda\vdash n$ is a partition and $S^\lambda$ is the associated irreducible representation of $S_n$. As we know from the branching rule, we have an isomorphism of $S_{n-1}$ modules $$S^\lambda\cong\bigoplus_{\mu}S^{\mu}$$ where $\mu$ are the partitions obtained by deleting an outer corner from $\lambda$. In particular, I am concerned with the case when $S^\lambda$ is spanned by basis elements $\{B_Q\mid Q\in\text{SYT}(\lambda)\}$ of the Kazhdan-Lusztig basis (with the relevant left-action of the generators). Can the isomorphism from the branching rule be realised in such a way that preserves the KL-basis in some form? That is, if $\phi$ is a realisation of the isomorphism, then $$\phi\cdot B_Q=\sum_{\mu}\sum_{P\in\text{SYT}(\mu)}z_pB_P$$ for some constants $z_P$. What can we say about these constants $z_P$? For example, when $\lambda$ is a rectangular partition (and so $S^\lambda\cong S^\mu$), then the isomorphism can be realised as $B_Q\mapsto B_{d(Q)}$, where $d(Q)$ is the tableau obtained by deleting the $n$-box from $Q$. I am unsure as to the theory for arbitrary partitions. Any help would be appreciated! REPLY [4 votes]: It turns out that this is indeed true. The relevant proofs are given in Chapter 3 of this extended abstract.<|endoftext|> TITLE: Reconstruction of coalgebras QUESTION [5 upvotes]: In the paper Reconstruction of hidden symmetries of Bodo Pareigis in the subsection "3.1 Reconstruction of coalgebras" there is the following proposition (3.3.). Let $\mathcal{C}$ be a braided monoidal category, $\mathcal{A}$ a $\mathcal{C}$-monoidal category and $\mathcal{B}$ a $\mathcal{C}$-category such as $\omega : \mathcal{B}\longrightarrow\mathcal{A}$ a $\mathcal{C}$-functor. If the functor \begin{equation*} \mathrm{Nat}_{\mathcal{C}}(\omega,\omega\otimes-) : \mathcal{A}\longrightarrow\mathrm{Set} \end{equation*} is representable by an object $C$ (also denoted as $\mathrm{coend}_{\mathcal{C}}(\omega)$), then $C$ is a coalgebra in $\mathcal{A}$. In the proof of this proposition the author just says, that the comultiplication $\Delta$ und counit $\epsilon$ are uniquely definied by $(1_{\omega}\otimes\Delta)\circ\delta = (\delta\otimes 1_{C})\circ\delta$ and $(1_{\omega}\otimes\epsilon)\circ\delta = \rho_{\omega}^{-1}$, where $\delta : \omega\longrightarrow\omega\otimes C$ is the universal arrow of the representation , but he doesn't prove the coassociativity and the counit axioms. I tried it myself and in fact it is straightforward to show the coassiciativity and that $\epsilon$ is a left counit. For example the diagram $\require{AMScd}$ \begin{CD} \omega @>{\delta}>> \omega\otimes C\\ @VV{\delta}V @VV{1_{\omega}\otimes\Delta}V \\ \omega\otimes C @>{\delta\otimes 1_{C}}>> \omega\otimes C\otimes C \\ @V{1_{\omega}\otimes\lambda_{C}^{-1}}V{\rho_{\omega}^{-1}\otimes 1_{C}}V @VV{1_{\omega}\otimes\epsilon\otimes 1_{C}}V \\ \omega\otimes 1\otimes C @= \omega\otimes 1\otimes C \end{CD} shows that $\epsilon$ is a left counit. However I can not prove that $\epsilon$ is also a right counit, i.g. $\rho_{C}^{-1} = (1_{C}\otimes\epsilon)\circ\Delta$. Does anyone know how to do that? Or maybe the author means by a coalgebra something more general, a coalgebra just with a left counit? REPLY [2 votes]: The diagram $\require{AMScd}$ \begin{CD} \omega @>{\delta}>> \omega\otimes C\\ @VV{\delta}V @VV{1_{\omega}\otimes\Delta}V \\ \omega\otimes C @>{\delta\otimes 1_{C}}>> \omega\otimes C\otimes C \\ @V{1_{\omega}\otimes\epsilon}VV @VV{1_{\omega\otimes C}\otimes\epsilon}V \\ \omega\otimes1 @>\delta\otimes1_1>> \omega\otimes C\otimes1 \end{CD} commutes and moreover the left vertical composite is $\rho_\omega^{-1}$ by requirement. (The lower square commutes since it is the tensor product of two commuting squares, $$ \begin{CD} \omega@>\delta>> \omega\otimes C\\ @| @| \\ \omega@>\delta>> \omega\otimes C \end{CD}\quad\bigotimes\quad \begin{CD} C @= C \\ @VV\epsilon V @VV\epsilon V \\ 1 @= 1 \end{CD}) $$<|endoftext|> TITLE: Geometric intuition behind Garside's paper? QUESTION [8 upvotes]: I apologize in advance for a somewhat wishy-washy question. I just read the paper "The Braid Group and Other Groups" by F. A. Garside in which he solves the conjugacy problem for the braid group and also obtains a standard form for braids that yields a solution to the word problem and computes the center. The paper is straightforward to follow line by line, however the last section "Other groups" makes me feel like I am completely missing the geometric insight that Garside had while writing the paper - in particular he says that all of the results given above hold for a certain class of more general groups to give the solution to the word/conjugacy problems. He also notes that the diagram that he has drawn for the element $\Delta \in B_4$ is the 2-skeleton of the truncated octahedron and the are similar higher-dimensional polytopes for the other elements $\Delta \in B_n$. For example, corresponding to the truncated cuboctahedron, Garside tells us that for the group generated by $a_1, a_2, a_3$ subject to the relations $a_1a_2a_1a_2 = a_2 a_1a_2a_1, a_2a_3a_2= a_3a_2a_3, a_1a_3=a_3a_1$, if we consider the element $\Delta = (a_1a_2a_3)^3$, then the results proved in Garside's paper, all suitably modified, still hold. How is this group related to that polytope? Can someone guide me with the geometric rehydration process for this paper? For concreteness, this might involve relating some of the results in the paper to constructions with a polytope. REPLY [2 votes]: If you add the relations that the standard generators have order two, then instead of getting the braid group on $n$ strands, you get the symmetric group on $n$ symbols. The generator that corresponded to swapping the $i$th and $(i+1)$st strands maps to the transposition $(i,i+1)$ in the symmetric group. The 1-skeleton of the polytope is the Cayley graph of the symmetric group with respect to this generating set. There is a similar polytope for each finite Coxeter group.<|endoftext|> TITLE: Global homological dimension of group rings QUESTION [14 upvotes]: In all that follows, let $k$ be a field and $G$ be a finite group. It is well-known that the order of $G$ is invertible in $k$ iff the group ring $k[G]$ is semisimple, which is equivalent, inter alia, to the fact that $\operatorname{Ext}^1_{k[G]}(V,W)$ vanish for all $V,W$ left $k[G]$-modules (= $k$-linear representations of $G$), or indeed, $\operatorname{Ext}^i_{k[G]}(V,W)$ for all $i\geq 1$, viꝫ. $k[G]$ has (left) global dimension zero. Today I learned that this is also equivalent to the (a priori weaker) condition that $k[G]$ be (left-)hereditary, which is equivalent, inter alia, to the fact that $\operatorname{Ext}^2_{k[G]}(V,W)$ vanish for all $V,W$ left $k[G]$-modules. (See, e.g., Dicks, “Hereditary Group Rings”, J. London Math. Soc. 20 (1979) 27–38, theorem 1.) This suggests the following question: what can be said about $G$ and $k$ if $\operatorname{Ext}^{d+1}_{k[G]}(V,W)$ vanish for all $V,W$ left $k[G]$-modules for a given $d\geq 2$? In other words, if we assume $k[G]$ has (left) global dimension $\leq d$? Does $k[G]$ having finite global dimension imply that the order of $G$ is invertible in $k$? REPLY [14 votes]: Here is another proof that the global dimension is infinite that is specific to groups and explicitly identifies a module of infinite projective dimension. Let $G$ be a finite group and suppose that the characteristic $p$ of $k$ divides the order of $G$. Then I claim that the trivial $kG$-module has infinite projective dimension. That is, the group $G$ has infinite mod $p$ cohomological dimension. First of all this follows when $G$ is a cyclic group of order $p$ from the very well known resolution of the trivial module (which can be obtained topologically using infinite lens spaces). If $t$ is the generator, you have a resolution where each module is $kG$ and you alternate between multiplying by $t-1$ and $1+t+\cdots+t^{p-1}$ (except for the augmentation $kG\to k$ at the beginning). When you hom into the trivial module $k$ you end up with a resolution with all the vector spaces $k$ and where all the maps are zero since $p$ is the characteristic of the field $k$ and so while $t-1$ always becomes zero after mapping into a trivial module, $1+t+\cdots+t^{-1}$ becomes multiplication by $p$, which is $0$, after mapping into a trivial module. This shows that $$H^n(C,k)=\mathrm{Ext}^n_{kG}(k,k)\cong k$$ for all $n\geq 0$. Next assume that $p\mid |G|$. Then $G$ has a cyclic subgroup $C$ of order $p$. Shapiro's lemma now implies that $$\mathrm{Ext}^n_{kG}(k,\mathrm{Coind}_C^G k)=H^n(G,\mathrm{Coind}_C^G k)\cong H^n(C,k)\neq 0$$ so again the trivial module has infinite projective dimension.<|endoftext|> TITLE: Induced subgraphs of the almost-disjointness graph QUESTION [6 upvotes]: Let $[\omega]^\omega$ denote the collection of infinite subsets on $\omega$, and let $$E=\big\{\{a, b\}:a,b\in [\omega]^\omega \text{ and } |a\cap b| \text{ is finite}\big\}.$$ Is every simple, undirected graph $G=(V,E)$ with $V\leq 2^{\aleph_0}$ isomorphic to an induced subgraph of $([\omega]^\omega, E)$? If not, what if $|V|\leq \aleph_0$? REPLY [4 votes]: I claim that every graph with $\leq \aleph_1$ vertices can be embedded in $[\omega]^\omega$ in the manner Dominic described. This means that we have a consistent answer to Dominic's question: the answer is yes assuming that the Continuum Hypothesis holds. Recall that $\mathcal P(\omega) / \mathrm{fin}$ denotes the Boolean algebra of all subsets of $\omega$ modulo the ideal of finite sets. I'm going to write my answer in terms of $\mathcal P(\omega) / \mathrm{fin}$, because I think that's a good way to think about the problem. (To translate: a subset of $\omega$ is infinite iff its equivalence class in $\mathcal P(\omega) / \mathrm{fin}$ is nonzero, two sets are almost disjoint iff their equivalence classes in $\mathcal P(\omega) / \mathrm{fin}$ are incompatible.) To prove the claim, we'll use something called the countable saturation of $\mathcal P(\omega) / \mathrm{fin}$. This means: Suppose $\{a_n :\, n \in \omega\}$ is a countable subset of $\mathcal P(\omega) / \mathrm{fin}$, and let $x$ be a variable. Suppose we have countably many statements (in the first-order language of Boolean algebras) involving $x$ and the $a_n$'s, and no finite subset of these statements is inconsistent. Then there is, in $\mathcal P(\omega) / \mathrm{fin}$, a solution to this infinite system of equations: i.e., there is a value we can assign to $x$ in order to make all the statements true simultaneously. (For example, if $a_n$ is (the equivalence class of) $\{2^nk :\, k \in \omega\}$ for each $n$, then we could have a countable sequence of statements asserting $0 \neq x < a_n$ for each $n$. No finite subset of these statements is inconsistent (indeed, the first $n$ statements are all satisfied by $x=a_{n+1}$). So countable saturation tells us that these statements are simultaneously satisfiable (e.g., by $\{2^k :\, k \in \omega\}$).) For a proof that $\mathcal P(\omega) / \mathrm{fin}$ has this property, I'll refer you to van Mill's article in the Handbook of Set Theoretic Topology (link), corollary 1.1.5. The proof is what you might imagine: a delicate diagonalization with lots of bookkeeping. Now suppose we have a graph $(V,E)$ with $|V| = \aleph_1$. Enumerate $V$ in type $\omega_1$, say $V = \{v_\alpha :\, \alpha < \omega_1 \}$. We'll build, by recursion, a sequence of $a_\alpha$'s such that the mapping $v_\alpha \mapsto a_\alpha$ is an embedding of $(V,E)$ into $\mathcal P(\omega) / \mathrm{fin}$. For the base case, let $a_0$ be any member of $\mathcal P(\omega) / \mathrm{fin}$ other than $0$ or $1$ (the equivalence classes of $\emptyset$ and $\omega$). At stage $\alpha > 0$, we've obtained a countable collection of members of $\mathcal P(\omega) / \mathrm{fin}$, namely $\{a_\beta :\, \beta < \alpha \}$. Let $x$ be a variable, and consider the following system of equations: $$x \neq 0,1 \quad \qquad \qquad \qquad \qquad \qquad$$ $$\qquad \quad x - \vee F \neq 0 \qquad \text{for any finite } F \subseteq \{a_\beta :\, \beta < \alpha\}$$ $$\qquad \quad x \vee (\vee F) \neq 1 \qquad \text{for any finite } F \subseteq \{a_\beta :\, \beta < \alpha\}$$ $$x - a_\beta \neq 0 \qquad \text{ for any } \beta < \alpha \qquad \ \quad$$ $$a_\beta \wedge x = 0 \qquad \text{whenever } \{ v_\beta,v_\alpha \} \in E$$ $$a_\beta \wedge x \neq 0 \qquad \text{whenever } \{ v_\beta,v_\alpha \} \notin E$$ This is a countable system of equations, and it's easy to check that any finite number of them are simultaneously satisfiable. By the countable saturation of $\mathcal P(\omega) / \mathrm{fin}$, there is some value we can assign to $x$ to satisfy all these equations simultaneously. This is our choice of $a_\alpha$. This recursion can continue through all countable ordinals. Once the recursion is done, it is easy to see that $a_\alpha \wedge a_\beta = 0$ if and only if $\{\alpha,\beta\} \in E$ for all $\alpha,\beta < \omega_1$. To translate this back into the language of Dominic's question: for each $\alpha < \omega_1$, let $A_\alpha$ be any representative of the equivalence class $a_\alpha$. Then $A_\alpha \cap A_\beta$ is finite if and only if $\{\alpha,\beta\} \in E$ for all $\alpha,\beta < \omega_1$. Finally, let me point out that a system of $\aleph_1$ equations like this is not generally satisfiable in $\mathcal P(\omega) / \mathrm{fin}$. That is, $\mathcal P(\omega) / \mathrm{fin}$ is not $\kappa$-saturated for any larger cardinals $\kappa$, not even consistently. The witness to this is Hausdorff gaps, which can be translated to a finitely satisfiable, unsatisfiable system of $\aleph_1$ equations in $\mathcal P(\omega) / \mathrm{fin}$. So to get an outright yes answer to Dominic's question, or even the consistency of a yes with the negation of CH, some new idea will be required.<|endoftext|> TITLE: A detail in the proof of Schur's lemma: the closures of the $\mathcal{Ker}$ and $\mathcal{Im}$ of the intertwiner QUESTION [10 upvotes]: $\renewcommand\Im{\operatorname{\mathcal{Im}}}\newcommand\Ker{\operatorname{\mathcal{Ker}}}$I was sure that this is a trivial question and placed it on Math Stackexchange https://math.stackexchange.com/questions/4136830/a-detail-in-the-proof-of-schurs-lemma-the-closures-of-the-cal-ker-and-cal Surprisingly, no one answered it there. So I am elevating it to MathOverflow. Consider two irreducibles of a topological group $G$, acting in respective Hilbert spaces $\mathbb V$ and $\mathbb V'$. Schur's lemma says: An intertwiner $M : \mathbb V\longrightarrow\mathbb V'$ of two irreducibles of a group is either zero or isomorphism. To prove it, we first show that both $\Im M$ and $\Ker M$ are invariant subspaces. Then two observations are made. Observation 1. For the invariance of $\Im M$ to agree with the irreducibility, $\Im M$ must coincide either with the space $\mathbb V'$ or with its zero subspace $\{\smash{\vec0}'\}$. Observation 2. For the invariance of $\Ker M$ to agree with the irreducibility, $\Ker M$ must coincide either with the space $\mathbb V$ or with its zero subspace $\{\vec0\}$. Summing up these observations, we conclude that $M$ is either zero or bijective and therefore invertible. Now, my question. Is it really true that $\Im M$ itself must coincide either with $\mathbb V'$ or $\{\smash{\vec0}'\}$? Or is it rather the closure of $\Im M$ that must satisfy this requirement? I am enquiring, because a representation is always defined in a Hilbert space or in a closed thereof. The same question about $\Ker M$. Should we prove that it is actually its closure that is invariant? If it is the closures of $\Im M$ and $\Ker M$ whose invariance needs to be proven, do we have to impose additional requirements on the representations and/or on the intertwiner $M$ (say, boundedness)? REPLY [6 votes]: Let $V$ and $W$ be Hilbert spaces with irreducible unitary $G$-actions and let $T:V \to W$ be a bounded intertwiner. Then the adjoint is an intertwiner as well and hence so are $T^\ast T$ and $TT^\ast$. Claim: These two are multiples of the identity. It follows that $T$ is zero or a multiple of an isometric isomorphism. To prove the claim, it is enough to show that a selfadjoint intertwiner $A$ of an irreducible represenation is a multiple of the identity. But the spectral projections of $A$ are intertwiners. By irreducibility, they are either 0 or 1. The spectrsl theorem then proves that $A$ is a multiple of the identity.<|endoftext|> TITLE: Does there exist an upper bound on the Fourier coefficients of the reciprocal theta function $\frac {1}{\theta}$? QUESTION [12 upvotes]: Define the theta function as $$ \theta(x) = \sum_{n=-\infty}^\infty e^{-\gamma(x+n)^2} $$ where $\gamma>0$. Clearly, $\theta$ is 1-periodic, non-zero and smooth. Therefore, the reciprocal map $x \mapsto \frac{1}{\theta(x)}$ is 1-periodic and smooth as well and can be expanded as a Fourier series: $$ \frac{1}{\theta(x)} = \sum_{n=-\infty}^\infty a_ne^{2\pi i nx}. $$ I'm interested in upper bounding the Fourier coefficients, i.e. I want to find a non-trivial map $f$ (with a fast decay) such that $$ |a_n| \leq f(n). $$ Are there any results in this direction? Thank you! REPLY [7 votes]: As @TerryTao stated correctly in the comments, a bound on the Fourier coefficients can be obtained by the saddle point method: The extension of $\frac 1 \theta$ to a meromorphic function on $\mathbb C$ has poles which are bounded away from the real axis. This implies that the Fourier coefficients of $\frac 1 \theta$ decay at least exponentially. In fact, the Fourier coefficients can be derived explicitly from which we can show that the decay is exactly exponentially: First note, that your map $\theta$ is essentially the Jacobi theta function of third kind. Using Poisson's summation formula we have $$ \theta(t) = \sqrt \frac \pi \gamma \sum_n \left ( e^{-\frac{\pi^2}{\gamma}} \right )^{n^2} e^{2\pi i n t} = \sqrt \frac \pi \gamma \vartheta_3(it, e^{-\frac{\pi^2}{\gamma}}). $$ Hence, for $q = e^{-\frac{\pi^2}{\gamma}}$ and $z=it$ we have $$ \frac 1 {\theta(t)} = \sqrt \frac \gamma \pi \frac 1 {\vartheta_3(z,q)}. $$ In the paper A.J.E.M. Janssen, Some Weyl-Heisenberg frame bound calculations, Indagationes Mathematicae, Volume 7, Issue 2, 1996, Pages 165-183, Janssen obtains on page 178 the Fourier expansion of $\frac 1 {\vartheta_3}$: $$ \frac 1 {\vartheta_3(z,q)} = \frac 1 C \sum_k (-1)^k a_k e^{2ikz} $$ $$ C = \sum_{n \in \mathbb Z} (-1)^n (2n+1)q^{(n+\frac 1 2)^2} $$ $$ a_k = 2 \sum_{m=0}^\infty (-1)^m q^{(m+\frac 1 2)(2|k|+m+\frac 1 2)} $$ From the value $|k|$ in the exponent in the definition of $a_k$ it follows directly that the $a_k$'s decay exponentially assuming that $|q|<1$. Applying some trivial bounds on $a_k$ gives you the desired map $f$ so that $|a_k| \leq f(k)$. To derive the above formula, Jenssen refers to an exercise in the book Whittaker, E., & Watson, G. (1927). A Course of Modern Analysis (4th ed., Cambridge Mathematical Library). He essentially solves this exercise which leads to the above expression. The above formula should agree with the one derived in the answer of @მამუკა ჯიბლაძე.<|endoftext|> TITLE: Weil pairing on abelian varieties and etale Chern classes QUESTION [13 upvotes]: Given a line bundle $L$ on an abelian variety $A/k$, there is an associated Weil pairing $e_L\colon\bigwedge^2V_pA\to\mathbb Q_p(1)$, where $p$ is a prime different from the residue characteristic of the base field $k$ and $V_pA$ is the $\mathbb Q_p$-linear Tate module of $A$. This is usually constructed by explicitly constructing a pairing between $A[p^n]$ and $A^\vee[p^n]$ using the interpretation of the latter as classes of divisors, and then pulling back along the polarisation induced by $L$. There is, however, another way to construct such a pairing, using the fact that $V_pA$ is dual to the etale cohomology $\mathrm H^1_{et}(A_{\bar k},\mathbb Q_p)$. Namely, the first etale Chern class $c^{et}_1(L)$ is an element of $\mathrm H^2_{et}(A_{\bar k},\mathbb Q_p)(1)=\mathrm{Hom}(\bigwedge^2V_pA,\mathbb Q(1))$, and we can just take the pairing corresponding to this element. What I want to know (and ideally would like a reference for) is whether these two pairings are the same. In other words, is the element of $\mathrm H^2_{et}(A_{\bar k},\mathbb Q_p)(1)$ corresponding to the Weil pairing equal to the first etale Chern class of $L$? REPLY [7 votes]: It seems that one of the pairings is the negative of the other (in char 0 this assertion is actually Lemma 2.6 of https://arxiv.org/pdf/1809.01440.pdf ).<|endoftext|> TITLE: Invariance of morse homology, doubt in proof in book "Morse Theory and Floer homology" QUESTION [5 upvotes]: I am reading the book "Morse theory and Floer Homology" by Michele Audin and Mihai Damian. Now I am reading the proof of the following theorem. Link to the statement of the theorem Basically we want to prove that the Morse homology is independent of the pseudo-gradient field and the morse function. Since the proof is a little bit long and I do not want to just summarize it, here is a link to the proof presented in the book. http://www.math.stonybrook.edu/~sunscorch/quals/Major/Morse_Invariance.pdf So, my confusion starts when the author proves that $(C_*(\tilde{F}|_{V \times A}),\partial _{\tilde{X}}) = (C_{*+1}(f_0), \partial_{X_0})$ (page 2 of pdf almost at the end of the page) and similarly for the other equality. After that he states that there are only two trajectories connecting critical points of $\tilde{F}$. So, I have three questions: The first type of trajectories are the ones staying always in $A$, and I argued that because of $(C_*(\tilde{F}|_{V \times A}),\partial _{\tilde{X}}) = (C_{*+1}(f_0), \partial_{X_0})$. So, like analyzing the trajectories in such section is just as analyzing the trajectories of $X_0$, similarly for $X_1$. Is this the correct idea? I do not get why there can only be trajectories from critical points of $f_0$ to those of $f_1$ (which are the second type of trajectories between critical points of $\tilde{F}$). My first thought about why we cannot have trajectories from critical points of $f_1$ to $f_0$ is because of how their indexes are related to the indexes of $\tilde{F}$ (almost at the beginning of page 2 of pdf) , but I am not sure why this would be true. Any suggestion? My final question is in the definition of $\partial_{\tilde{X}}$. I do not know why in the first row, second column of its matrix representation, it is the zero matrix. Any ideas? (My hypothesis is that since we can only have trajectories as those described in the second bullet point, the number of trajectories from critical points of $f_1$ to those of $f_0$ and by definition of $\partial_{\tilde{X}}$, then that is why it is the zero matrix). Again, I would really appreciate your help here. Maybe this is the easiest part of the proof and I am missing a basic fact, but I do not see it. Thanks in advance. REPLY [6 votes]: Oh, I happen to know the guy who wrote that PDF. As to your questions. Yes, that's the idea. In $V \times A$, $F = f_0$ so the critical points are in one-to-one correspondence with those of $f_0$. The shift in degree comes from this $g$ which has a maximum at $0$. Similarly in $V \times B$ but there is no shift in degree. Intuitively, the way $g$ and $\tilde{X}$ are chosen is so that the flow is going from a "high" position in the region of $V \times A$ to a "low" position in the region of $V \times B$. This is made precise by the fact that $g$ has very negative 1st derivative on $(0,1)$ and $\tilde{X}$ is a pseudo-gradient for $\tilde{F}$ so the trajectories cannot run the opposite way. The first row, second column entry has to be an operator $C_{k+1}(f_1) \to C_{k-1}(f_0)$ but it also has to use Morse trajectories since we're trying to describe $\partial_{\tilde{X}}$ which is defined in Morse terms. From (2), there are no trajectories running from the region associated to $f_1$ backwards to the region associated to $f_0$ so this operator must be zero.<|endoftext|> TITLE: Does perfect fraction field imply perfect residue field? QUESTION [7 upvotes]: Let $A$ be a local integral domain of characteristic $p$. Let $K$ be the fraction field and let $k$ be the residue field of $A$. If $K$ is perfect, is $k$ necessarily perfect? Thoughts: If $A$ is normal, then $K$ perfect implies that in fact $A$ itself is perfect, hence $k$ is perfect. If $A$ is essentially of finite type over a field, then $K$ perfect implies $\dim A = 0$. Here is a stronger question. For a field $F$ of characteristic $p$, the $p$-rank of $F$ is $p\operatorname{-rk}(F) := \log_{p} [F:F^{p}]$ (possibly infinite). So it would be enough to know: if $x_{1},x_{2}$ are points in an $\mathbb{F}_{p}$-scheme $S$ such that $x_{2} \in \overline{\{x_{1}\}}$ (i.e. "$x_{1}$ specializes to $x_{2}$"), then is $p\operatorname{-rk}(\kappa(x_{1})) \ge p\operatorname{-rk}(\kappa(x_{2}))$? Here is something related: For any integral domain $A$ with fraction field $K$, we can take an algebraic closure $\overline{K}$ and consider the integral closure $\overline{A}$ of $A$ in $\overline{K}$; then $\overline{A}$ is absolutely integrally closed (0DCL); in particular all residue fields of $\overline{A}$ are algebraically closed (0DCN). REPLY [12 votes]: The answer is no. Let $F$ be any imperfect subfield of a perfect field $F'$. Let $B$ be the ring of integral Puiseux series over $F'$ (integral meaning ones involving only nonnegative powers of $T$) and let $A$ be the subring consisting of those series whose coefficient of $T^0$ belongs to $F$. $A$ and $B$ have the same fraction field, which is perfect since $B$ is perfect. However the residue field of $A$ is isomorphic to $F$, and so isn't perfect.<|endoftext|> TITLE: Bounding the max-loaded bin using${m \choose k} \|A\|_k^k$ QUESTION [5 upvotes]: Throw $m$ balls into $n$ bins independently, each ball selecting a bin from the distribution $A \in \Delta_n$. This question is about lower-bounding the max-loaded bin. Background. In this MO answer I wrote about an upper bound based on collisions. Let $Z_k$ be all subsets of $k$ distinct balls. For $S \in Z_k$, let $1_S$ be the indicator that all balls in $S$ land in the same bin. Then $\mathbb{E} [1_S] = \sum_{i=1}^n A_i^k = \|A\|_k^k$. Let $C_k = \sum_{S \in Z_k} 1_S$, the number of $k$-way collisions. Then by Markov's inequality, $$ \Pr[ \text{max-loaded bin } \geq k] = \Pr[ C_k \geq 1] \leq \mathbb{E}[C_k] = {m \choose k} \|A\|_k^k . $$ For example, in the classic case of $m=n$ and uniform $A$, where $\|A\|_k^k = n^{1-k}$, we can use Stirling to get a bound closely approaching $\frac{n}{k^k}$, as expected. And this bound can be very tight: if we throw half as many balls (cut $m$ in half), the probability decreases by a factor of about $2^k$. Question. Is there a lower-bound on the max-loaded bin that matches this upper bound, in a sense? Or, what is the tightest non-asymptotic lower bound you know for this setting? Motivation. First, notice that if we pretended each collision $1_S$ were independent, we would obtain $$ \Pr[ \text{max-loaded bin} < k ] = \Pr[ C_k = 0] = \Pr[1_S = 0 ~ (\forall S \in Z_k)] \leq \left(1 - \|A\|_k^k\right)^{m \choose k} \leq \exp\left(- {m \choose k} \|A\|_k^k\right) . $$ That would be such a cool result, matching the upper bound so neatly. Unfortunately, the collisions are positively associated, not negatively: $\Pr[1_S = 1 \mid 1_{S'} = 1] \geq \Pr[1_S = 1]$. So I don't know of techniques to prove this. (Yet simulations suggest something not too far away could hold, at least for $A$ uniform...) What else I've tried. Well, if $X_i$ is the number of balls in bin $i$, then the $X_i$ are negatively associated, so I think we can get a bound by pretending the $X_i$ are independent Binomials, but it doesn't match. The standard approach would be to bound the variance of $C_k$ and use Chebyshev. I was able to get an only-somewhat-horrible expression for $\text{Var}(C_k)$, but I had trouble pushing it through to get a tight bound here. It might work. Finally, I'll mention that Raab and Steger is only asymptotic, whereas I'm hoping with this approach to get a concrete bound for any given $m,n,k$. Edit: esg once gave me this hint, but I was unable to prove it: one can show that $\mathbb{P}\big(C_k(m)\geq 1\big)\geq \mathbb{P}(\mathrm{Binom}\big(m,\lVert A\rVert_k\big)\geq k)$ where $\mathrm{Binom}\big(m,p)$ denotes a r.v. with has a Binomial distribution with parameters $m$ and $p$ Some more details: We know that ${m \choose k} = \left(\frac{\Theta(m)}{k}\right)^k$, so the upper bound above looks like $\left(\frac{\Theta(m) \|A\|_k}{k}\right)^k$. So what really matters is the ratio $\frac{m \|A\|_k}{k}$, and we get an exponential-in-$k$ bonus. My most optimistic hope is that the same expression provides a lower bound, where perhaps the constant factor of $m$ just decreases. From some simulations, I'm not sure this is true. We could instead hope to use $\left(\frac{\Theta(m) \|A\|_k}{k}\right)^{c}$, which it sounds like esg's approach can do at least for constant $c$. From simulations, I'm confident this dependence is true at least for the uniform distribution and $c=k/2$. Update: from esg's answer and a simple multiplicative Chernoff bound, I get that if $m \|A\|_k \geq \beta k$ and $\beta \geq 2$, then $\Pr[C_k \geq 1] \geq 1 - e^{-\beta k/8}$. Combining this with my fact above, rearranged, I get: If $m \leq \frac{\beta \cdot k}{\|A\|_k}$ for $\beta \leq \tfrac{1}{e}$, then $\Pr[\text{max load} \geq k] \leq e^{-k \ln(1/\beta)}$. If $m \geq \frac{\beta \cdot k}{\|A\|_k}$ for $\beta \geq 2$, then $\Pr[\text{max load} \geq k] \geq 1 - e^{-k (\beta/8)}$. REPLY [3 votes]: The following inequality holds: $$\mathbb{P}(C_k(m)\geq 1)\geq \mathbb{P}(\mathrm{Bin}(m, \lVert A\rVert_k)\geq k)$$ where here and in the sequel $\mathrm{Bin}(n,p)$ denotes a binomially distributed random variable with parameter $n$ and $p$. (I now change notation so I can use my old notes. In the sequel $r\geq 2$ is the "collision degree" (your $k$), $m$ is the number of bins (your $n$), $n$ is the time variable (number of balls, your $m$)) Situation: given are $m\geq 2$, a probability distribution $p_1,\ldots,p_m$ and an i.i.d. sequence $X_1, X_2,\ldots$ with $\mathbb{P}(X_1=i)=p_i$. Let for $n\geq 1, 1\leq i \leq m\;\;$ $B_i(n):=\sum_{j=1}^n 1_{\{X_j=i\}}$ the number of occurrences of $i$ at "time" n, and let $$T_r:=\inf\{n\geq 1\,|\,\exists\,i\in\{1,\ldots m\}\,:\,B_i(n)\geq r\}$$ the first time an element is observed $r$ times. We are interested in upper bounds for $\mathbb{P}(T_r>n)$ ( since $\{T_r\leq n\}=\{C_r(n)\geq 1\}$ in your notation). We use generating functions. Let $q_r(t):=\sum_{j=0}^{r-1} \frac{t^j}{j!}$ denote the $r$-th partial sum of the exponential series. The joint distribution of $(B_1(n),\ldots,B_m(n))$ is the multinomial distribution with parameters $n$ and $p_1,\ldots,p_m$. Since $$\{T_r>n\}=\{B_1(n)\leq r-1,\ldots, B_m(n)\leq r-1\}$$ we have $$\mathbb{P}(T_r>n)=n!\,[t^n]\prod_{i=1}^m q_r(p_it)\;.$$ Note also that $$\mathbb{P}(\mathrm{Bin}(n,p)\leq r-1)=n!\,[t^n] q_r(pt)\,e^{(1-p)t}$$ We first recall a well known way to rewrite binomial probabilities. Reminder: Let $00, p_1+p_2=1$ and $\lVert p\rVert_r:=(p_1^r+p_2^r)^{1/r}$. Then $$n![t^n] q_r(p_1t)q_r(p_2t)\leq n! [t^n] q_r(\lVert p\rVert_r, t)\,e^{(1-\lVert p\rVert_r)t}=\mathbb{P}(\mathrm{Bin}(n,\lVert p\rVert_r)\leq r-1)$$ Proof: Denote the coefficients on the left hand side resp. right hand side by $a_n$ resp. $b_n$. Clearly $a_n=b_n=1$ for $n\leq r-1$, and $a_n=02r-2$. Let $n=r-1+j, 1\leq j \leq r-1$, then on the left hand side \begin{align*} a_{r-1+j} &=\sum_{{k\leq r-1, i\leq r-1}\atop{ k+i=r-1+j}} \frac{(r-1+j)!}{k!j!} p_1^kp_2^i\\ &=\mathbb{P}(j\leq \mathrm{Bin}(r-1+j,p_1)\leq r-1)\\ &=1 -\mathbb{P}(\mathrm{Bin}(r-1+j,p_1)\leq j-1)-\mathbb{P}(\mathrm{Bin}(r-1+j,p_1)\geq r)\\ &=1 -\mathbb{P}(\mathrm{Bin}(r-1+j,p_1)\leq j-1)-\mathbb{P}(\mathrm{Bin}(r-1+j,p_2)\leq j-1)\\ &=1 -\sum_{k=0}^{j-1} {r+k-1 \choose k} (p_1^rp_2^k+p_2^rp_1^k)\end{align*} On the right hand side we have \begin{align*} b_{r-1+j} &= \mathbb{P}(\mathrm{Bin}(r-1+j,\lVert p\rVert_r)\leq r-1)\\ &= 1-\mathbb{P}(\mathrm{Bin}(r-1+j,\lVert p\rVert_r)\geq r)\\ &= 1-\mathbb{P}(\mathrm{Bin}(r-1+j,1-\lVert p\rVert_r)\leq j-1)\\ &= 1 -{\lVert p\rVert_r}^r \sum_{k=0}^{j-1} {r+k-1 \choose k} (1-\lVert p\rVert_r)^k\end{align*} where the reminder was used for the final equality. But $1-\lVert p\rVert_r\le \min\{p_1,p_2\}$ (since $\lVert p\rVert_r\ge \max\{p_1,p_2\}$) and thus for $k\ge 0$ $$(1-\lVert p\rVert_r)^k {\lVert p\rVert_r}^r\leq p_1^kp_2^r+p_2^kp_1^r$$ and the claim follows. End of proof Now to the general case: Theorem Let $\lVert p\rVert_r:=\left(p_1^r + \ldots + p_m^r\right)^{1/r}$ . Denote by $T_r:=T_r(p_1,\ldots,p_m)$ the time of the first occurrence of the first $r$-collision in $\{1,\ldots,m\}$. Then $$\mathbb{P}(T_r>n)\leq \mathbb{P}(\mathrm{Bin}(n,\lVert p\rVert_r)\leq r-1)$$ Proof: From the lemma above we get that for any $p_1,p_2>0$ and $k\geq 0$ $$[t^k] q_r(p_1 t)q_r(p_2 t)\leq [t^k] q_r(\lVert(p_1,p_2)\rVert_r t)\,e^{(p_1+p_2-\lVert(p_1,p_2)\rVert_r)t}$$ Hence \begin{align*} [t^n] q_r(p_1 t)q_r(p_2 t)q_3(p_3 t)&= \sum_{k=0}^n [t^{n-k}] q_r(p_3t)\, [t^k] q_r(p_1 t)q_r(p_2 t)\\ &\le \sum_{k=0}^n [t^{n-k}] q_r(p_3 t)\, [t^k] q_r(\lVert(p_1,p_2)\rVert_r t)\,e^{(p_1+p_2-\lVert(p_1,p_2)\rVert_r)t}\\ &=[t^n] q_r(p_3 t)q_r(\lVert(p_1,p_2)\rVert_r t)\,e^{(p_1+p_2-\lVert(p_1,p_2)\rVert_r)t}\\ &= \sum_{k=0}^n [t^{k}] q_r(p_3 t) q_r(\lVert(p_1,p_2)\rVert_r t)\,[t^{n-k}]e^{(p_1+p_2-\lVert(p_1,p_2)\rVert_r)t}\\ &\leq \sum_{k=0}^n [t^{k}] q_r(\lVert (p_1,p_2,p_3)\rVert_r t) e^{(p_3+\lVert(p_1,p_2)\rVert_r -\lVert (p_1,p_2,p_3)\rVert_r) t}\,[t^{n-k}]e^{(p_1+p_2-\lVert(p_1,p_2)\rVert_r)t}\\ &= [t^n] q_r(\lVert (p_1,p_2,p_3)\rVert_r t) e^{(p_1+p_2+p_3-\lVert (p_1,p_2,p_3)\rVert_r)t} \end{align*} and induction gives that $$[t^n] \prod_{i=1}^m q_r(p_i,t) \leq [t^n] q_r(\lVert p\rVert_r,t) e^{(p_1+\ldots+p_m-\lVert p\rVert_r)t}$$ Thus \begin{align*} \mathbb{P}(T_r>n)&= n! [t^n] \prod_{i=1}^m q_r(p_i,t) \leq n! [t^n] q_r(\lVert p\rVert_r,t) e^{(1-\lVert p\rVert_r)\,t} =\mathbb{P}(\mathrm{Bin}(n,\lVert p\rVert_r)\leq r-1) \end{align*} End of proof Remarks: (1) a completely different proof for the case $r=2$ was given in https://eprint.iacr.org/2005/318 (Theorem 3). (2) using $\mathbb{E}(X)=\sum_{n=0}^\infty \mathbb{P}(X>n)$ (for the expectation of a random variable with values in the nonnegative integers) and the inequality above gives $$\mathbb{E}(T_r)\leq \frac{r}{\lVert p\rVert_r}$$ (3) always $\mathbb{P}(\mathrm{Bin}(n,p)n)\leq (1-\lVert p\rVert_r^r)^{n \choose r}$$ is false. It can be violated if the $p_i$ are not uniformly small. Consider e.g. the case $p_1=\frac{1}{2}$, $p_1=\ldots=p_m=\frac{1}{2(m-1)}$ and $r=2, n=3$. For $m\longrightarrow \infty$ \begin{align*} \mathbb{P}(T_2>3)&\longrightarrow \mathbb{P}(\mathrm{Bin}(3,\frac{1}{2})\leq 1)=\frac{1}{2} \mbox{ and } \lVert p\rVert_2\longrightarrow \frac{1}{2}, \\(1- \lVert p\rVert_2^2)^3 &\longrightarrow \frac{27}{32} \end{align*} Thus the inequality is violated for all sufficiently large $m$.<|endoftext|> TITLE: Models with fixed cardinality of non-Lebesgue measurable sets QUESTION [7 upvotes]: In the usual $\mathsf{ZFC}$, we know that there are $2^\mathfrak{c}$ many subsets of $\mathbb{R}$ that are not (Lebesgue) measurable. On the other hand, the Solovay model also provides us a model of $\mathsf{ZF}$ which has $0$ non-measurable subsets of $\mathbb{R}$. I would like to know if the following in-between assertion has been established or is currently being researched on: For any cardinal $\kappa < 2^\mathfrak{c}$, is it consistent with $\mathsf{ZF}$ that there are exactly $\kappa$ non-measurable subset of $\mathbb{R}$? If this is false, can we classify the cardinals in which the assertion above holds? REPLY [6 votes]: Actually, $\mathsf{ZF}$ easily proves that if there exists at least one non-measurable subset of $\mathbb{R}$, then there must be $2^\mathfrak{c}$ non-measurable subsets of $\mathbb{R}$. Let $X \subseteq \mathbb{R}$ be non-measurable. Then $X = (X \cap [0,\infty)) \cup (X \cap (-\infty,0)) =: A \cup B$. If both $A$ and $B$ are measurable, then $X$ must also be measurable, at it is a finite union of measurable sets. Thus, we must have that either $A$ or $B$ is non-measurable. WLOG say $A$ is not measurable. Now there are $2^\mathfrak{c}$ measurable sets in $(-\infty,0)$, and for any of such set $C$, we have that $A \sqcup C$ is not measurable. This gives $2^\mathfrak{c}$ distinct non-measurable subsets of $\mathbb{R}$.<|endoftext|> TITLE: The verbs in combinatorics: Enumerating, counting, listing and all that QUESTION [24 upvotes]: Two closely related, but different tasks in combinatorics are determining the number of elements in some set $A$, and presenting all its elements one by one. Question: What are some works in combinatorics literature that explicitly consider the naming of these different tasks? As a background, it seems that the terminology is sometimes conflicting and confusing. In particular, enumerating can mean either task. In computer science and algorithms it often refers to task 2. In combinatorics it often refers to task 1, but not always. Pólya enumeration is definitely task 1, not task 2 (indeed in this MO question it is pointed out that Pólya enumeration is "not generally a good tool for actually listing"). For what it's worth, Merriam-Webster duly reports that enumerate has meanings 1. to ascertain the number of: COUNT; 2. to specify one after another: LIST. Task 1 is also called counting, which seems unambiguous. But I have seen "computing the number of elements without actually counting them"! Here counting seems to mean tallying, that is, keeping a counter and incrementing by one whenever a new object is seen. Task 2 admits many names, which also may indicate finer variations: listing the elements: Presenting a full listing, stored in some form (paper or computer file). generating the elements: A method that creates all the elements, one by one, but may not store them. Perhaps each element is examined, and then thrown away. visiting: similar to the previous, with a tone of computer science and data structures. constructing: similar, but with a more mathematical flavor. It suggests that creating even one object takes some effort, so it is not just "visiting". classifying: somewhat unclear, but often means something like generating the objects and counting how many of them have certain properties. But it might mean simply isomorph-free listing (in a sense, "classifying" the objects into isomorphism classes). Furthermore, task 2 is often emphasized with modifiers like "full", "explicit", "exhaustive", "actually", "one by one", "brute force" to set it apart from task 1. Enumerating may also mean a more abstract task where elements are equipped with indices and/or abstractly arranged in a potentially infinite list, but one never actually constructs the list (as in "enumerate all rational numbers"). To clarify my question: I am not asking for examples where the words are just used, as in "In this paper we enumerate all Schluppenburger contrivances of the second kind". I am interested in works that recognize the difference of these tasks and make a conscious effort in defining terminology, and perhaps explicitly comment on the usage. Here are some that I have found: Knuth (TAOCP 4B §7.2.1) considers many verbs: run through possibilities, look at permutations, enumerate, count, list, make a list, print, examine, generate, visit. He notes that enumerate may mean either task 1 or 2. He settles for generating and visiting for task 2, when the list is not explicitly stored. Cameron (Notes on Counting, p. 1–2) settles with counting for task 1 and generating for task 2. Later in the notes there are scattered instances of enumeration, which mostly seems to be synonymous with counting. Ruskey (Combinatorial Generation, 2003, p. iii) discusses the terminology for task 2. He mentions generate and enumerate but notes that both are overloaded with other meanings. For example, generate can mean generate uniformly at random, and enumerate can mean counting. Ruskey also considers listing but settles with generation. Kreher & Stinson (Combinatorial Algorithms, 1999, p. 1) defines: Generation, construct all the combinatorial structures of a particular type – – A generation algorithm will list all the objects. Enumeration, compute the number of different structures of a particular type – – each object can be counted as it is generated. REPLY [2 votes]: I would say, as a first distinction, enumerating and listing refer to ordinals (ex-numerare = producing an ordered bijection onto $[n]$, like a shepherd at sunset, emitting (ex-mittere) the numbers in order : $1,2,3,\dots$) while counting (cum-putare: putting together to reckon. I do not see etymologically the category of order in it) and measuring refer to cardinals. We may count a bunch of apples of equal size using a scale. In this sense a generating series $\sum_{k=0}^\infty a_k x^k$ enumerates; the inclusion-exclusion formula for $\big|\big(\cup_{i\in I} X_i\big)^c\big|$ counts. Indeed, the natural setting of it uses additive set functions, measures, probabilities, that are tools to weight sets.<|endoftext|> TITLE: A question about connected subsets of $[0,1]^2$ QUESTION [36 upvotes]: If $S⊂[0,1]^2$ intersects every connected subset of $[0,1]^2$ with a full projection on the $x$-axis, must $S$ have a connected component with a full projection on the $y$-axis? An equivalent form： If $S⊂[0,1]^2$ intersects every connected subset of $[0,1]^2$ with a full projection on the $x$-axis and $T⊂[0,1]^2$ intersects every connected subset of $[0,1]^2$ with a full projection on the $y$-axis, must $S\cap T\neq \emptyset$? The motivation of this question: The question came to me when I thought about the Brouwer fixed-point theorem: Let $f=(f_1,f_2)$ be a continuous function mappping $[0,1]^2$ to itself. Then $$S\triangleq\{(x,y)\in[0,1]^2:f_1(x,y)=x\}$$ intersects every connected subset of $[0,1]^2$ with a full projection on the $x$-axis and $$T\triangleq\{(x,y)\in[0,1]^2:f_2(x,y)=y\}$$ intersects every connected subset of $[0,1]^2$ with a full projection on the $y$-axis. My further question: If we assume that $S\subset [0,1]^2$ is a close set, what is the answer to my question, that is, if a close set $S⊂[0,1]^2$ intersects every connected subset of $[0,1]^2$ with a full projection on the $x$-axis, must $S$ have a connected component with a full projection on the $y$-axis? REPLY [4 votes]: This is an answer to the updated question. Proposition: If a closed $S\subset [0,1]\times [0,1]$ intersects every connected set with a full projection onto the $x$-axis, then it has a component with a full projection onto the $y$-axis. First, without loss of generality we may assume that $S$ does not intersect the left and the right sides of the square (otherwise, consider the same problem but for $[-1,2]\times [0,1]$. Let $\Pi=(0,1)\times (0,1)$, which is homeomorphic to the full plane. Pick a point on the left side take a disk $D$ around this point that does not intersect $S$. Take $x$ in $D\cap \Pi$. Do the same on the right side and get $E$ and $y$. Let $S'=S\cap \Pi$, which is closed in $\Pi$. Now $S'$ separates $x$ and $y$ within $\Pi$ (meaning any connected set in $\Pi$ that contains $x$ and $y$ has to intersect $S'$. Indeed, if a connected set $F\subset \Pi$ contains $x$ and $y$, then $F\cup D\cup E$ has a full projection onto the $x$-axis, and so has to intersect $S$, but since $D$ and $E$ do not, it follows that $F\cap S'=F\cap S\ne\varnothing$ (the first equality follows from $F\subset\Pi$). Since $S'$ is closed in $\Pi$, which is homeomorphic to the plane, by a Theorem V.14.3 in the book Newman - Elements of topology of planar sets of points, there is a component $C$ of $S'$ that separates $x$ and $y$ in $\Pi$. Clearly, $C$ has a full projection on the $y$-axis (and so it is $(0,1)$), since otherwise we could sneak in a horizontal segment between the left and right sides, which would contradict the separation. Since $S$ is closed, $\overline{C}$ is a connected compact subset of $S$. Hence, $C$ has a compact projection onto $y$ axis, and so this projection is $[0,1]$.<|endoftext|> TITLE: What is meant by a computational interpretation of univalence? QUESTION [12 upvotes]: In homotopy type theory the univalence axiom implies function extensionality. Suppose we have a recursive set we are not sure is empty (e.g. the set of even integers$\geq 4$ that are not a sum of two primes). Then via characteristic functions this becomes a claim about two functions $\mathbb{N}\to \mathbf{2}$ being equal. I would not expect a machine to settle this. Then what do people mean when they talk about a computational interpretation of univalence? REPLY [9 votes]: Roughly speaking, a type theory is computationally adequate if there is an algorithm that evaluates a term belonging to any type into a "normal form" of that type. The simplest form of this is when dealing with closed terms (not involving any variables or hypotheses) belonging to a "base" type such as $\mathbb{N}$ or $\mathbf{2}$, in which case the normal forms really are the obvious "canonical forms" such as numerals $s(s(s(\dots(s(0))))):\mathbb{N}$ or booleans ${\rm t}, {\rm f}:\mathbf{2}$. Thus, this aspect of computational adequacy (called canonicity) says that if you define a particular natural number without using any assumptions, the computer can (at least in principle) tell you exactly which natural number you've defined. Things get subtler when you talk about terms belonging to higher types such as a function-type $\mathbb{N}\to \mathbf{2}$. In general, the canonical forms of a type are those obtained from its introduction rule, which in the case of a function-type means a $\lambda$-abstraction. So canonicity implies that any function $\mathbb{N}\to \mathbf{2}$ can be evaluated to a $\lambda$-abstraction; but, unlike the situation for numerals in $\mathbb{N}$, it could still be the case that two syntactically-distinct $\lambda$-abstractions define extensionally equal functions. So computational adequacy doesn't give an algorithm for deciding whether two functions are equal, which as you noted would be impossible. As Reid pointed out in the comments, it's important to distinguish here between propositions, which are types, hence elements of a universe type $\mathcal{U}$, and booleans, which are elements of the type $\mathbf{2}$. Since $\mathbf{2}$ is a base type, any closed term of type $\mathbf{2}$ does evaluate, in a computationally adequate theory, to $\rm t$ or $\rm f$. But $\mathcal{U}$ is a higher type, and as with functions two propositions can be syntactically distinct and yet extensionally equal (have the same truth value). Thus, evaluating a proposition to a "canonical form" doesn't tell you whether that proposition is true or false -- unless you've proven that that proposition is decidable, in which case it can be represented by an element of $\mathbf{2}$. (Of course, in classical mathematics the Law of Excluded Middle asserts that every proposition is decidable -- and this is why LEM is a non-computational axiom.) Even subtler than this is what happens when you ask for computational adequality for open terms (those involving free variables and assumptions). This is a stronger property than canonicity, called normalization, and in this case the "normal forms" that terms are evaluated to may not be obtained from the introduction rule of their type; they could also be obtained from the elimination rule of the type of a free variable (a "neutral" term). For instance, in the context of a free variable $f:\mathbb{N}\to\mathbb{N}$, we have a term $f(0):\mathbb{N}$, which cannot be normalized any further, even though it is not a numeral. None of this is specific to univalence. The problem of giving a computational interpretation of univalence means to specify a type theory in which univalence is true (either as an axiom or as a theorem) and which is computationally adequate in one or more of these senses. With regard to the discusion in the comments, this implies that the univalence term $\rm ua$ cannot appear in a normalized closed term of base type, since any such term must be a canonical form such as a numeral; but it could still appear in a normalized closed term of higher type, or in a normalized open term. (This requires, of course, that there be some kind of "computation rules" that apply to $\rm ua$ enabling it to be reduced away in at least some cases, in contrast to how in Book HoTT computation "get stuck" when it encounters the bare axiom $\rm ua$.) This is not quite the same as giving a model of type theory in which everything computes, although practically and historically the two are often intertwined. Finally, just to note the current status of computational interpretations of univalence, canonicity for one form of cubical type theory was proven in 2016 by Simon Huber, while normalization for a different form of cubical type theory was proven in 2021 by Sterling and Angiuli.<|endoftext|> TITLE: Can $x^4+y^4+1$ be a perfect power? QUESTION [7 upvotes]: Recall that a perfect power has the form $x^m$ with $m,x\in\{2,3,\ldots\}$. Motivated by Fermat's result that the equation $x^4+y^4=z^2$ has no positive integer solution, here I ask the following question. Question 1. Can $x^4+y^4+1$ with $x,y\in\mathbb N=\{0,1,2,\ldots\}$ be a perfect power? Based on my computation, I conjecture that $x^4+y^4+1$ with $x,y\in\mathbb N$ can never be a perfect power. Question 2. Can we find $x,y\in\mathbb N$ such that $x^4+y^4+1=\prod_{i=1}^kp_i^{a_i}$ for some $a_1,\ldots,a_k\in\{2,3,\ldots\}$ and distinct primes $p_1,\ldots,p_k$? Via a computer I find no $x^4+y^4+1$ with $x,y\in\{0,1,\ldots,8000\}$ of the form $\prod_{i=1}^kp_i^{a_i}$ with $p_1,\ldots,p_k$ distinct primes and $a_1,\ldots,a_k\in\{2,3,\ldots\}$. Perhaps, Question 2 has a negative answer. Of course, a negative answer to Question 2 implies a negative answer to Question 1. Your comments are welcome! REPLY [23 votes]: To answer question 2: $$346^4+36788^4+1=1831575032204939793=3^3\cdot19^3\cdot179^2\cdot17569^2.$$<|endoftext|> TITLE: Is it always true that $\sum_{i=1}^{a-1}(-1)^i(a,i)\ge-1$? QUESTION [8 upvotes]: Following is an experimental math claim. We denote $(a,b)=\gcd(a,b)$. Let $$G(a)=\sum_{i=1}^{a-1}(-1)^i(a,i).$$ Note: $$ G(a) = \begin{cases} 0, & \text{if $a\equiv 1\pmod4$} \\ \text{odd}, & \text{if $a\equiv 2\pmod4$} \\ 0, & \text{if $a\equiv 3\pmod4$} \\ \text{even}, & \text{if $a\equiv 0\pmod4$.} \end{cases}$$ Can it be shown that for every $a\in\mathbb{Z}_{\ge2}$, $G(a)\ge-1$? Table $$\begin{array}{|c |c |} \hline a & G(a) \\ \hline 2 & -1 \\ \hline 3 & 0 \\ \hline 4 &0 \\ \hline 5 &0 \\ \hline 6 &-1 \\ \hline 7 &0 \\ \hline 8 &4 \\ \hline 9 &0 \\ \hline \vdots &\vdots \\ \hline \end{array}$$ source code PARI/GP for(a=1,10000,if(sum(i=1,a-1, (-1)^i*gcd(a, i))<-1,print([a]))) REPLY [5 votes]: $$G(a)+(-1)^aa=\sum_{i=1}^{a} (-1)^i (a,i)=\sum_{d|a} d \sum_{\substack{ l \\ (l, a/d)=1}} (-1)^{dl}=T(a).$$ When, $a$ is odd, $d, a/d$ is also odd. Hence, $$\sum_{\substack{l, \\ (l,a/d)=1 \\ d1$, because all $p_i$ are odd, hence each $p_i(r_i+1)-r_i$ is odd, and $a$ is even. More compactly, $G(a)=a[\frac{r_0}{2}\prod_{i=1}^{s} (r_i+1-\frac{r_i}{p_i})-1]$. Surely, $G(a) \geq -1$. It's $-1$, for $a=2p$ types, e.g. $a=6$.<|endoftext|> TITLE: Which unfoldings of the $d$-dimensional hypercube tile $(d{-}1)$-space? QUESTION [25 upvotes]: A six year old question, Which unfoldings of the hypercube tile $3$-space?, has just been answered by Moritz Firsching: All $261$ unfoldings tile space! So now we know: For $d=2$, the unfolding of the square tiles $\mathbb{R}$. For $d=3$, each of the $11$ unfoldings of the $\mathbb{R}^3$-cube tile $\mathbb{R}^2$. For $d=4$, each of the $261$ unfoldings of the $\mathbb{R}^4$-hypercube tile $\mathbb{R}^3$. The natural next question (also posed at MESE) is: Q. Is it true that, for every $d$, each of the unfoldings of the $d$-dimensional cube tiles $\mathbb{R}^{d-1}$? If not, up to which $d$ does this hold? Brute-force calculation may be limited by the combinatorics: Number of hypercube unfoldings, OEIS:A091159: $$1,11,261,9694,502110,33064966,2642657228 ,\ldots $$ Added. I should add that recently Satyan Devadoss and his students proved that no unfolding of a $d$-dimensional hypercube self-overlaps, i.e., each forms a net. DeSplinter, Kristin, Satyan L. Devadoss, Jordan Readyhough, and Bryce Wimberly. "Nets of higher-dimensional cubes." Canad. Conf. Comput. Geom. 2020, pp.114-120. CCCG proceedings link. REPLY [7 votes]: Linear unfoldings A category of unfoldings that can always be tiled are the linear ones. Let's say an unfolding is branching if there is a (d-1)-hypercube with at least 3 adjacent hypercubes. I then define linear unfoldings as those that don't branch. Examples: https://whuts.org/unfolding/1 is linear https://whuts.org/unfolding/8 is also linear https://whuts.org/unfolding/18 is branching Here is a table of how many unfoldings are linear $d$ Linear (A271215) Total (A091159) 2 1 1 3 4 11 4 24 261 5 184 9694 6 1911 502110 7 24252 33064966 Now we can use some results of the article in CCCG you linked. In particular lemma 2, which states: Let $T$ be a spanning tree of the $n$-Roberts graph with a coordinate system. If direction $x$ is used in the unfolding along some path of $T$, direction $-x$ will not be used in the unfolding along this path Because I have restricted to linear unfoldings, the tree $T$ is linear. Therefore there is a path trough all vertices in $T$. From the lemma it then follows: if we follow the unfolding from one end to another, we will only move in one direction of each dimension. Let's call these directions the positive direction of each dimension in (d-1)-space. Now lets associate each (hyper-)cube with a point in $\mathbb Z^{d-1}$. fix the start cube at $(0, 0, \ldots, 0)$, we can give the others coordinates as well. For the first example this will look like this: $$(0,0,0),(0,1,0),(1,1,0),(1,1,1),(2,1,1),(2,2,1),(3,2,1),(3,2,2)$$ Notice that between adjacent cubes exactly one coordinate increases by 1. Because of this the sum of the coordinates also increases by one each step. Now suppose we place a copy of the unfolding with it's start cube at $(0,1,-1)$: $$(0,1,-1),(0,2,-1),(1,2,-1),(1,2,0),(2,2,0),(2,3,0),(3,3,0),(3,3,1)$$ There cannot be any overlap with the other one. Suppose otherwise, there is a cube in common. Let's label the cubes of the unfolding from $0$ to $2d-1$, from start to finish. Now suppose the conflicting cube has label $i$ in the original unfolding, and $j$ in the copy. Then the sum of coordinates of that cube is $(0+0+0)+i=(0+1+-1)+j$, which gives $i=j$. That would imply all cubes of the unfolding conflict. But then we get $(0,0,0)=(0,1,-1)$ for the start cube, a contradiction. We get a space-filling tiling if we place copies at each point where the sum of coordinates is a multiple of $2d$. There is no overlap, by the argument as before. If two (hyper-)cubes of two different foldings conflict, they have the same coordinate sum modulo $2d$. This implies they have the same index in the unfolding, which implies the start cubes of both overlap leading to a contradiction. There is also no empty spaces. By definition each spot where the sum of coordinates is a multiple of $2d$ is filled by all start cubes. All spots with remainder $i$ modulo $2d$ then get filled with the cubes at index $i$. Expanding the method I have added this section to expand on my comments. It demonstrates how the method described can be expanded to proof more unfoldings tileable. As a running example I will use the class of crosses. A multi-dimensional cross is an unfolding of the $d$-cube, which has $2d$ dimension-$(d-1)$ cubes at the following coordinates: $$\begin{align}\{&(0,0,\ldots,0);& &(-1,0,\ldots,0);& &(0,-1,0,\ldots,0);& &\ldots;& &(0,0,\ldots,-1);&\\ &(2,0,\ldots,0);& &(\quad 1,0,\ldots,0);& &(0,\quad 1,0,\dots,0);& &\ldots;& &(0,0,\ldots,\quad 1)\}& \end{align}$$ If you tried to tile a cross without rotation/reflection, you will get stuck. This means we can't simply use the mod $2d$ method above. Therefore we need to expand the method to fit multiple orientations of an unfolding. For the cross it is sufficient to use two orientations, the original and the one with the long end the other way. I will show that these together tile the plane by an injective mapping of their cubes onto $0, \ldots, 4d-1$. This mapping will be a linear equation of the coordinates, modulo $4d$. For the cross I will use the following mapping: $f(x_1, x_2, \ldots, x_{d-1}) = 1\cdot x_1 + 3\cdot x_2 + 5 \cdot x_3 + \ldots + (2d-3)\cdot x_{d-1} \mod 4d$. So what will a cross look like after applying this mapping. Obviously the center cube, $(0,\ldots,0)$ is mapped to $0$. The extra cube $(2,0,\ldots,0)$ is mapped to $2$. All the other cubes will be mapped to $\pm (2i+1) \mod 4d$ for $i=0$ to $d-1$. The image of the mapping will look as follows (for d=5): ####.#.#.....#.#.#.# Now we do the same for the rotated cross. Note that inverting an even number of axes is equivalent to an 180° rotation. We say that the rotated cross is formed by inverting the cross along $x_1$ and $x_2$. The inversion along $x_2$ swaps the cubes at $(0,1,0,\ldots,0)$ and $(0,-1,0,\ldots,0)$; but that still gives the same shape. Similarly the inversion along $x_1$ swaps $(1,0,\ldots,0)$ and $(-1,0,\ldots,0)$, but it also moves $(2,0,\ldots,0)$ to $(-2,0,\ldots,0)$. If we apply the mapping to the rotated cross, the image will be mostly the same. There is only one difference, corresponding to the move of the extra cube. ##.#.#.#.....#.#.### If we then move the rotated cross in $(d-1)$-space, the image after mapping gets rotated. If we move the rotated cross in the right way (rotate 11), we can get the image to look as follows: ....#.#.#####.#.#.#. -- moved rotated cross ####.#.#.....#.#.#.# -- original cross which perfectly fills the gaps the original cross left behind. This implies we can fill space by placing a cross at each position where $f(x_1,\ldots, x_{d-1})=0$, and a rotated cross at each position where $f(x_1,\ldots, x_{d-1})=2d+1$. This method allows us to show that 248 / 261 hypercube unfoldings tile the plane with at most one extra unfolding rotated 180° from the original. I have not yet tried the 90° rotations as those are harder to implement. Instead I relaxed the "even number of axes reversed" to "any number of axes reversed", which then could show that all 261/261 unfoldings tile 3D space with at most one extra rotated/reflected piece. I have now run the method on all 5-cube unfoldings. They all (9694/9694) tile 4D space, using only a single extra rotated version, no reflections were needed. For every of those foldings it was tileable with the all-axis-flip rotation.<|endoftext|> TITLE: Functors between module categories that comes from restriction QUESTION [5 upvotes]: Suppose you have two $k$ algebras $A, B$ (say also finitely generated if this helps) and a functor $F: A-mod \to B-mod $ such that $| F(M) |= |M|$. Here $|U|$ denotes the underlying $k$ vector space. Can you find sufficient conditions so that $F$ is the restriction functor relative to $f:B \to A$? I suspect that a "nice" condition to start from is monoidality. Indeed, in the case in which the rings are the group rings of finite groups, by Tannaka reconstruction you could reconstruct the morphism. My motivation: I had to find an $\mathbb{R}$-algebra injective morphism from $\mathbb{C} \to \mathbb{R}[u]/(u^2+1) ^m$. It's really easy with Jordan form theory to give a complex structure on a module for the right hand ring, but constructing the actual morphism it's not easy at all, and some non trivial ad-hoc argument is needed to make " reconstruction ". REPLY [3 votes]: Let's assume that we are using right modules and that the isomorphism of underlying vector space functors is a natural isomorphism because we will just work up to isomorphism of functors. Note that $F$ is exact. To come from restriction $F$ should preserve all limits and colimits. Maybe this follows by functorially preserving underlying spaces but I'm not sure. If it doesn't you probably want to require that. It follows from Eilenberg-Watts that there is an $B$-$A$-bimodule $M$ such that $F\cong \hom_A(M,-)$. Moreover, since $A$ represents the underling vector space functor, we must have that $M\cong A$ as a right $A$-module. Since $End_A(A)\cong A$, it follows the left $B$-module structure comes from a homomorphism $F\colon B\to A$. So basically $F\cong \hom_A(A,-)$ via this bimodule structure. But then evaluating at $1$ gives an isomorphism of $F$ with the functor taking $N$ to $N$ with $B$-action $mb = mf(b)$. So $F$ is isomorphic to restriction along $f$.<|endoftext|> TITLE: Why is this function a modular function of level $5$? QUESTION [13 upvotes]: Suppose we have a function $\phi\colon \mathfrak H \longrightarrow \mathbb C$ such that $\phi^{24}$ is a modular function of level $5$. $\phi(\tau)=\sum_{n=-1}^{\infty}a_{n}q^{n/5}$, $a_{-1}\neq 0,q=e^{2\pi i\tau}$. Does it follow that $\phi$ is a modular function of level $5$? In particular, I am interested in the function $$\phi(\tau)=-\frac{1}{5^{1/2}}\frac{\eta(\tfrac{\tau}{5})\eta(\tfrac{\tau+1}{5})\eta(\tfrac{\tau-1}{5})}{\eta(5\tau)\eta(\tfrac{\tau+2}{5})\eta(\tfrac{\tau-2}{5})}.$$ Is there a a simple way to show that it is modular of level $5$? Siegel says that it is easy to show that it is so. We could compute the generators for $\Gamma(5)$ and then use the transformation formula for the Dedekind eta function, but this seems like a mess. REPLY [5 votes]: Here's a fairly straightforward way to show that $\phi$ is modular of level $5$ using Siegel functions. Claim: The function $f(\tau)$ is a modular function for $\Gamma(5)$ if and only if $f(5\tau)$ is a modular function for $\Gamma_{0}(25) \cap \Gamma_{1}(5)$. (This is straightforward to prove using properties of the slash operator.) Using this claim, it's not hard to see that $\eta(\tau/5)/\eta(5\tau)$ is a modular function of level $5$, using the standard result of Gordon, Hughes and Newman about when an eta quotient is modular of level $N$. (This result can be found in the Wikipedia article about the Dedekind $\eta$-function.) It remains to show that $\frac{\eta(\frac{\tau+1}{5})\eta(\frac{\tau-1}{5})}{\eta(\frac{\tau+2}{5})\eta(\frac{\tau-2}{5})}$ is a modular function of level $5$. The Siegel functions $g_{(a_{1}/N,a_{2}/N)}$ can be used to build modular functions of level $N$. (See Section 5 of the article here by Amanda Folsom that gives a product expansion of the Siegel functions and criteria of Kubert and Lang that indicate when a product of Siegel functions is modular.) The product formula implies that $$ g_{(0,a_{2}/N)}(\tau) = c \eta(\tau + a_{2}/N) \eta(\tau - a_{2}/N) $$ for some constant $c$. The modularity criteria imply that $$ h(\tau) = \frac{g_{(0,1/5)}(\tau)}{g_{(0,2/5)}(\tau)} = \frac{\eta(\tau + 1/5) \eta(\tau - 1/5)}{\eta(\tau + 2/5) \eta(\tau - 2/5)} $$ is a modular function for $\Gamma(5)$. Now $h(\tau) = h(\tau + 1)$ and so $h(\tau)$ is a modular function for $\left\langle \Gamma(5), \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\right\rangle = \Gamma_{1}(5)$. Since $\Gamma_{0}(25) \cap \Gamma_{1}(5) \subset \Gamma_{1}(5)$, the claim above implies that $h(\tau/5)$ is a modular function for $\Gamma(5)$ and this proves that $\phi$ is a modular function of level $5$. As a note related to your first question about $\phi^{24}$ being modular of level $5$ implying $\phi$ is modular of level $5$, Theorem 1 of Kubert and Lang's paper here is the following. Let $N$ be a prime power and $U_{N}$ be the group of modular units of level $N$ that are generated by the Siegel functions. If $g$ is a modular function and there is a positive integer $k$ so that $g^{k} \in U_{N}$, then $g \in U_{N}$.<|endoftext|> TITLE: CY fibration over $\mathbb P^1$ without any singular fibers QUESTION [7 upvotes]: Let's call a smooth projective vartiety $M$ as Calabi-Yau (CY) manifold if it has trivial canonical class and $h^i(M, \mathcal O_M ) = 0$ for $0 < i < \dim(M)$. In this definition, a CY 1-fold is an elliptic curve, a CY 2-fold is a projective $K3$ surface and etc. For each $n$, I looking for a smooth projective variety $X$ of dimension $n$ with a fibration $\pi: X \rightarrow \mathbb P^1$ such that $\pi$ has no singular fibers, any fiber of $\pi$ is a CY $(n-1)$-fold and $X$ is not a product of $\mathbb P^1$ and a CY $(n-1)$-fold. For $n=2$, it is known that such $X$ (an ellitic surface) does not exist. I put a question regarding the case of $n=3$ here but didn't get an answer. For some $n$, does such a variety $X$ exist? REPLY [6 votes]: To complement YangMills's answer: Viehweg and Zuo ("On the isotriviality of families of projective manifolds over curves") proved the following: Theorem. Let $X$ be a complex projective manifold of non-negative Kodaira dimension. Then a surjective morphism $X\to\mathbf{P}^1$ has at least 3 singular fibres. This predates Tosatti-Zhang and implies immediately what you want. The Theorem of the Fixed Part implies that the variation of Hodge structure induced by $X\to \mathbf{P}^1$ is constant. Therefore, by infinitesimal Torelli the fibration is formally locally trivial, and hence etale locally trivial. Again we can use Lemma 17 from Kollar-Larsen to conclude. The theorem is false in positive characteristic, see Schroer "Some Calabi-Yau threefolds with obstructed deformations over the Witt vectors" (the example, based on a construction of Moret-Bailly, is a K3 fibration over $\mathbf{P}^1$).<|endoftext|> TITLE: Irreducible components of an algebraic stack QUESTION [6 upvotes]: Let $\mathcal{X}$ be an algebraic stack of finite type over a (separably closed) field $ k$. Let's say that $\mathcal{X}$ has finite dimension $d \in \mathbb{Z}$. Is it still true that the number of irreducible components of dimension $d$ of $\mathcal{X}$ is the dimension of $H^{2d}_c(\mathcal{X},\bar{\mathbb{Q}}_{\ell})$ as in the case of schemes? (Here I'm referring to the lisse etale cohomology.) If this is not true in general, does it hold for a suitable class of stacks like the ones of the form $[X/G]$ where $X$ is a $k$ scheme? REPLY [4 votes]: I will say yes, although the level of generality is a bit scary and I hope I am not missing some stacky subtlety. I just took the standard argument for schemes, stared at it, and couldn't see anything that wouldn't work in the general case. Claim: if $X$ is an equidimensional finite type algebraic stack, and $d=\dim(X)$, then the rank of $H^{2d}_c(X)$ equals the number of irreducible components of $X$. [All coefficients are $\mathbf Q_\ell$ from now on.] Proof: Consider first the case $X$ smooth. In this case by Poincaré duality, $H^{2d}_c(X) \cong H^0(X)^\vee$. But in the smooth case we necessarily have connected components = irreducible components, so it's fine. In general choose $U \subset X$ smooth open with complement $Z$ of strictly smaller dimension. Then we have the long exact sequence $$ \ldots \to H^{k-1}_c(Z) \to H^{k}_c(U) \to H^{k}_c(X) \to H^k_c(Z) \to \ldots $$ and we win: $H^k_c(Z)$ vanishes for $k>2(d-1)$ by dimension reasons, so $H^{2d}_c(U)=H^{2d}_c(X)$.<|endoftext|> TITLE: 3D Edge matching puzzle generation QUESTION [6 upvotes]: I have this weird idea for a puzzle/toy (or torture device, depending on how you look at it) I've been trying to make for years now. I happen to be worse at this kind of math as I thought; and I'd be delighted to get some help. This puzzle is made of 64 cubes, with a peg or hole on each side (such as, you can connect them like Lego bricks). These pegs and holes have different shapes, call them, if you will, symbols (just bear in mind there's an "up" and "down" symbol pair on each connection). The goal of the puzzle is simple, build a solid $4\times 4\times 4$ cube, with no gaps in it. The problem is, how can I generate a set of these cubes with a low number of possible solutions (Ideally just 1), while using the fewest amount of symbols? P.S. Sorry if there was some weird grammar or tone here, I'm not a native speaker. REPLY [3 votes]: The size of the problem you're envisioning is so small that your best bet is probably to follow the suggestion in the comments. Just try lots of possibilities, using a (free) SAT solver such as clasp to count the number of solutions, until you find something you're happy with. If you want to study the work of other people who have worked on similar problems, you might want to look at Alex Selby's notes about how he and Oliver Riordan solved Christopher Monckton's Eternity puzzle. You'll see that Selby and Riordan thought pretty hard about how to estimate the difficulty of solving this type of puzzle. After losing money on Eternity, Monckton enlisted Selby and Riordan to help him design Eternity II, which in some ways is even closer in spirit to your problem, and which was designed to defeat the methods that Selby and Riordan had used to solve Eternity. A back-of-the-envelope calculation suggests that Eternity II is expected to have on the order of 1 solution. As of this writing, no solution to Eternity II has been published by anyone. The deadline for claiming the monetary prize has long since expired, but some people continue to work on it anyway. There is a discussion forum for the puzzle; the best partial result as of this writing is an arrangement found by Joshua Blackwood that contains only 10 mistakes.<|endoftext|> TITLE: Problems concerning subspaces of $M_{n}(\mathbb{Q}) $ QUESTION [11 upvotes]: Let $M_{n}(\mathbb{Q}) $ denote the $n$ times $n$ matrices over the rational number field. $N$ be a subspace of $M_{n}(\mathbb{Q}) $.Then if all the non-zero matrices in $N$ are invertible, what is the maximum the dimension of $N$ can be? We already know that if we take $M_{n}(\mathbb{R}) $ instead of $M_{n}(\mathbb{Q}) $ then the answer is $ \rho(n) $. where $ \rho(n) $ is Radon Hurwitz number i.e if $ n = (2a + 1 ) 2^{c+4d} $ where $ 0 \leq c \leq 3 $ then $ \rho(n) = 2^{c} + 8d $ . REPLY [21 votes]: Let's call this maximal dimension function $\rho_{\mathbb{Q}}:\mathbb{N}\to\mathbb{N}$, i.e., $\rho_{\mathbb{Q}}(n)$ is the largest possible dimension of a subspace $N\subset M_n(\mathbb{Q})$ such that all of the nonzero elements of $N$ are invertible. Then $\rho_{\mathbb{Q}}(n)\ge n$, as the following construction shows: Let $p(x)\in\mathbb{Q}[x]$ be a polynomial of degree $n$ that is irreducible over $\mathbb{Q}$, and let $A\in M_n(\mathbb{Q})$ be a matrix whose characteristic polynomial is $p$. (Such $A$ are easily constructed.) Let $N\subset M_n(\mathbb{Q})$ be the $\mathbb{Q}$-subspace spanned by the powers of $A$. Then $N$ is an $n$-dimensional over $\mathbb{Q}$ and $N$ is a field isomorphic to $\mathbb{Q}[x]/\bigl(p(x)\bigr)$, so every non-zero element of $N$ is invertible. Meanwhile, it's easy to see that $\rho_{\mathbb{Q}}(2)\le 2$, since any $3$-dimensional subspace of $M_2(\mathbb{Q})$ contains a nonzero element with vanishing determinant. Thus, $\rho_{\mathbb{Q}}(2) = 2$. Moreover, as Fedor points out in his comment below, this observation extends to all $n$ because, if $N\subset M_n(\mathbb{Q})$ had dimension greater than $n$ over $\mathbb{Q}$, then $N$ would have to intersect the codimension $n$ subspace of $M_n(\mathbb{Q})$ consisting of those matrices with first column equal to $0$. Thus, $\rho_\mathbb{Q}(n)=n$ for all $n$.<|endoftext|> TITLE: Is the Sato-Tate conjecture known for Bianchi modular forms? QUESTION [5 upvotes]: Originally formulated for elliptic curves, the Sato-Tate conjecture regarding the equidistribution of Frobenius trace values according to the Haar measure on a certain compact group (the Sato-Tate group) was generalised in 1994 by Serre to any motive over a number field - see 13.5? in [5]. As is now well known, the Sato-Tate conjecture has been proven for elliptic curves over totally real fields [3] as well as for all Hilbert modular forms [2]. More recently, in 2018, the authors in [1] prove modularity lifting theorems for Galois representations over CM fields which yield several attractive corollaries including the Sato-Tate conjecture for elliptic curves over CM fields. My main question is the following: Does it follow from the work of [1] that the Sato-Tate conjecture is known for some class of cuspidal automorphic forms for $\mbox{GL}_2$ over CM fields? I am chiefly concerned with the case of the CM field being an imaginary quadratic field, in which case the automorphic forms are often called Bianchi modular forms, hence the question in the title. A concise account of cusp forms for $\mbox{GL}_2$ associated to more general CM fields may be found in Section 2 of [4]. If I may hazard a guess at my question above, I would think that it is currently known only for automorphic $\mbox{GL}_2$ newforms over CM fields of parallel weight 2 with rational Fourier coefficients. But I am not an expert in this area, hence my question on this forum. I am also curious to know the following: What is the precise formulation of the Sato-Tate conjecture for general cuspidal automorphic forms for $\mbox{GL}_2$ over CM fields? I think [1] also establishes the Generalised Ramanujan Conjecture (GRC) in the parallel weight 2 setting. For general parallel weight $k$, does GRC imply that there is an inequality $$ \left|\frac{c(\frak{p})}{\mbox{Nm}({\frak{p}})^{\frac{k-1}{2}}}\right| < 2, $$ where the $c(\frak{p})$ are the Fourier coefficients of the (cuspidal new $\mbox{GL}_2$) automorphic form at the integral prime ideals of the CM field $K$? If so, then I can begin to see how one can at least start talking about equidistribution. References: [1]: P. B. Allen, F. Calegari, A. Caraiani, T. Gee, D. Helm, B. V. Le Hung, J. Newton, P. Scholze, R. Taylor, J.A. Thorne. Potential automorphy over CM fields. arXiv:1812.09999 (2018) [2]: T.Barnet-Lamb, T. Gee, D.Geraghty. The Sato-Tate cojecture for Hilbert Modular forms. JAMS (2011) [3]: T. Barnet-Lamb, D. Geraghty, M. Harris and R. Taylor . A family of Calabi-Yau varieties and potential automorphy II. P.R.I.M.S. 47 (2011) [4]: E. Ghate. Critical values of twisted tensor L-functions over CM-Fields. Proc. Symp. Pure Math. (1999) [5]: J.P. Serre. Propriétés conjecturales des groupes de Galois motiviques et des représentations $\ell$-adiques. Proc. Symp. Pure Math. (1994) REPLY [5 votes]: Does it follow from the work of [1] that the Sato-Tate conjecture is known for some class of cuspidal automorphic forms for GL2 over CM fields? Tautologically: "yes, those which correspond to modular elliptic curves". These are a (strict) subset of the forms which are cohomological of trivial weight (i.e. parallel weight 2 in your terminology) and have trivial character and Hecke eigenvalues in $\mathbf{Q}$. More relevantly: it looks to me as if their argument works for any newform of trivial weight, whatever the character and the coefficient field. These should still give rise to "very weakly compatible systems" as in Corollary 7.1.12 of op.cit., of which the Sato--Tate result is a special case. The restriction on the weight is more serious; one would certainly expect something of this sort to be true for arbitrary cohomological weights, but the ten-author paper does not cover this case, since they require that their Galois representations have all Hodge--Tate weights 0 or 1. It would take a genuine expert [not me!] to say how easy it would be to extend their results to cover this case. What is the precise formulation of the Sato-Tate conjecture for general cuspidal automorphic forms for GL2 over CM fields? For parallel weight and trivial character, the statement is exactly what you think it is. For more general (cohomological) weights, there are some fiddly but ultimately minor complications of book-keeping (which are already present over totally real fields too): you have to be careful about how to normalise $c(\mathfrak{p})$, with various conventions differing by integer or half-integer powers of $\mathrm{Nm}(\mathfrak{p})$. The "analytic" normalisation is such that the L-series $\sum c(\mathfrak{n}) \mathrm{Nm}(\mathfrak{n})^{-s}$ has its functional equation centred at $s = \tfrac{1}{2}$; then the Ramanujan conj. is that $|c(\mathfrak{p})| \le 2$ for all $\mathfrak{p}\nmid \mathfrak{N}$ where $\mathfrak{N}$ is the level. (Note that equality can occur, even in the parallel weight 2 case, although it occurs with "probability 0" in some sense.) From here, it should be clear how to state the Sato--Tate conj if $\pi$ has trivial character, since then $c(\mathfrak{p})$ is real, and we're just asserting that its distribution in $[-2, 2]$ is the expected one. To state the Sato–Tate conjecture for general characters $\chi$, one can argue as follows: $\chi$ will factor through the ray class group of level $\mathfrak{N}$; for each $x \in Cl(\mathfrak{N}) / \ker(\chi)$, choose some $\alpha_x$ such that $\alpha^2 = \chi(x)$. Then $c'(\mathfrak{p}) = c(\mathfrak{p}) / 2\alpha_{[\mathfrak{p}]}$ will be in the real interval $[-1, 1]$, and Sato--Tate predicts that the ratios $c'(\mathfrak{p})$ have the expected distribution.<|endoftext|> TITLE: What is the "Prikry–Silver collapse" when CH fails? QUESTION [14 upvotes]: We all know and love Cohen reals, and we can (and often do) define the Cohen forcing as partial functions $p\colon\omega\to 2$ with finite domain. The Prikry–Silver forcing is defined as partial functions $p\colon\omega\to 2$ with co-infinite domain. These two couldn't be any more different. For example, Cohen reals are aggressively non-minimal, whereas Prikry–Silver reals are minimal. We can look at a similar situation with other forcings that are given by finite conditions. For example $\operatorname{Col}(\omega,\omega_1)$ is a forcing notion whose conditions are finite partial functions $p\colon\omega\to\omega_1$. We can ask what would be the Prikry–Silver analogue of this forcing, then. That is, $\{p\colon\omega\to\omega_1\mid\operatorname{dom} p\text{ is co-infinite}\}$. Interestingly, assuming CH this is the same as the standard collapsing forcing. This follows from the fact that the cardinality of the partial order is $2^{\aleph_0}$, which under CH is just $\aleph_1$, and we know that any forcing of size $\aleph_1$ which collapses $\omega_1$ is equivalent to $\operatorname{Col}(\omega,\omega_1)$. Question. Is this so-called "Prikry–Silver collapse" provably equivalent to the standard collapsing forcing? REPLY [6 votes]: If $\mathrm{CH}$ fails then $\mathrm{Col}(\omega, \omega_1)$ does not add a generic for the "Prikry-Silver collapse" $\mathbb P$: Let $\mathbb U$ be $(\mathcal{P}(\omega)/I)^+$ where $I$ is the ideal of finite subsets of $\omega$. The map $$\pi:\mathbb P\rightarrow \mathbb U,\ p\mapsto [\omega\setminus\mathrm{dom}(p)]_I$$ is a projection so that $\mathbb P$ adds a $\mathbb U$-generic filter. It is hence enough to show that under $\neg\mathrm{CH}$, $\mathrm{Col}(\omega,\omega_1)$ does not add a $\mathbb U$-generic filter. Suppose it does. In this case, there is a projection $$\mu:\mathrm{Col}(\omega,\omega_1)\rightarrow \mathrm{RO}(\mathbb U)$$ but this implies that there is $p\in\mathbb U$ and a set $D\subseteq\mathrm{RO}(\mathbb U)$ of size $\omega_1$ that is dense below $p$. This contradicts the fact that $\mathbb U$ has an antichain of size continuum below every condition.<|endoftext|> TITLE: On Markoff-type diophantine equation QUESTION [6 upvotes]: Do there exist integers $x,y,z$ such that $$ x^2+y^2-z^2 = xyz -2 \quad ? $$ Why this is interesting? First, this equation arose in an answer to the previous Mathoverflow question What is the smallest unsolved diophantine equation? but was not asked explicitly as a separate question. The context is that, in a well-defined sense for the notion of "smallness", the equation above is the "smallest" open Diophantine equation. Second, this equation is one of the simplest non-trivial representative of the family of equations $ax^2+by^2+cz^2=dxyz+e$, which generalises a well-known Markoff equation $x^2+y^2+z^2=3xyz$. The well-known methods for the former (Vieta jumping) has been extended to the general case if $a,b,c$ are all natural numbers and are divisors of $d$ (see, for example, Fine, Benjamin, et al. "On the Generalized Hurwitz Equation and the Baragar–Umeda Equation." Results in Mathematics 69.1-2 (2016): 69-92). The question seems to be much more challenging when $a,b,c$ have different signs. The simplest case with different signs is $a=b=d=1$ and $c=-1$, which leads to the family of equations $x^2+y^2-z^2=xyz+e$. The equation above is the first non-trivial example from this family. REPLY [12 votes]: There is no solution. Fix a solution $(x,y,z)$ with $|x|+|y|+|z|$ minimal. We will show a contradiction. We can't have $xyz=0$ as we would then obtain one of the unsolvable equations $x^2+y^2= -2$, $x^2-z^2=-2$, $y^2-z^2=-2$. If $xyz>0$, then by swapping the signs of two of $x,y,z$ if necessary we can assume $x,y,z>0$, and switching $x$ and $y$ we can assume $x \geq y$. We have the Vieta jump $x \to yz-x$, so if this is minimal we have $x \leq yz/2$. Since $f(x)=x^2+y^2-z^2 - xyz + 2 $ is convex and vanishes at $x$, we must $f(y) \geq 0$ or $f(yz/2) \geq 0$. But $$f(y)= (2-z)y^2 -z^2 + 2$$ so $f(y) \leq 0$ imply $z<2$ and $z=1$ gives the impossible $x^2+y^2-xy=-1$ and $$f(yz/2) = y^2 -z^2 - y^2 z^2/4 +1= (1-z^2/4)y^2 - z^2 +1$$ which again is nonnegative only if $z<2$ which is impossible. If$xyz<0$, then by swapping the signs if necessary we can assume $x,y,z<0$. We have the Vieta jump $z \to -xy-z$, so if this is minimal we have $z \geq -xy/2$. We have $g(z)=z^2 + xyz - x^2-y^2-2$ is convex and vanishes at $z$, we must have $g(0) \geq 0$ or $g(-xy/2) \geq 0$. But $g(0) = -x^2 - y^2 - 2 <0$ and $g(-xy/2) = - x^2 y^2/4 - x^2 - y^2 -2 <0$. So neither case is possible.<|endoftext|> TITLE: Galois module theory: from global to local QUESTION [6 upvotes]: Let $L/\mathbb{Q}$ be a finite Galois extension with Galois group $G$. It is well known that the ring of integers $\mathcal{O}_L$ is free over its associated order $\mathfrak{A}_{L/\mathbb{Q}}=\{x\in \mathbb{Q}[G]\mid x\mathcal{O}_L\subseteq \mathcal{O}_L\}$ if $G$ is abelian (Leopoldt, 1959); $G$ is dihedral of order $2p$, where $p$ is a rational prime (Bergé, 1972); $G$ is the quaternion group of order $8$, and the extension is wild (Martinet, 1972). In some other papers, I have found written that also the local counterparts are true ($L/\mathbb{Q}_p$ finite Galois extension with the same Galois group as before), and it seems that the authors suggest that these results are naturally implied by the global ones. But while they seem "folklore", these implication are not immediate to me. A couple of observations (I wish to thank Fabio Ferri for the precious discussion). It seems to me that only the number fields case is considered in the papers. Probably the key is to repeat the proof as it is for the $p$-adic case; for example, this works in the Martinet's case. There are other ways to get these results, like for example using Lettl's work on absolutely abelian extensions for Leopoldt's local case, or realise local extensions as completion of global ones with the same Galois group (here $p\ne 2$, we refer to Henniart, 2001). But I am interested in knowing if there is an immediate and "direct way". Summarising, my question is: is it immediate that the global cases imply the local cases? (More generally, a question could be: if every Galois extension of number fields with a certain type of Galois group admits freeness of the ring of integers over the associated order, then the same also holds for every Galois extension of local fields with the same Galois group?) REPLY [3 votes]: There are a number of things one can say about this. First, some of the papers you mention do also cover the case of $p$-adic fields. For example, see Bergé's paper Sur l’arithmétique d’une extension diédrale Annales de l’institut Fourier, tome 22, no 2 (1972), p. 31-59. On page 32 you'll see that A can be a PID satisfying conditions 1), 2) & 3) listed there, so in particular one can take $A=\mathbb{Z}_{p}$. Second, I'm not sure what's "wrong" with the method of realising a Galois extension of $p$-adic fields as the completion of a Galois extension of number fields with the same Galois group. Given the result you are aiming to prove, this seems entirely natural to me. The result of Henniart you mention is described in Chapter IX, $\S$5 (page 574) of Cohomology of number fields, (available here https://www.mathi.uni-heidelberg.de/~schmidt/NSW2e/) and gives the result you want for $p$ odd. If this doesn't satisfy your definition of immediate and "direct", then I'm not sure what does. Can you clarify? Third, we know lots more in the $p$-adic case than we do for the number field case. For example, we know that for any (at most) tamely ramified finite Galois extension $L/K$ of $p$-adic fields, we have that $\mathcal{O}_L$ is free over the group ring $\mathcal{O}_K[\mathrm{Gal}(L/K)]$ (which is equal to the associated order in this case). There are lots more results, but I just give this as an example. But I think it's true that for (almost) any specific Galois group $G$ for which we know the freeness result for all $G$-extensions $K/\mathbb{Q}$, one can give a reference for a direct proof for the same result for $G$-extensions $L/\mathbb{Q}_p$ (assuming that these latter extensions even exist). Actually, perhaps one exception to this last claim is $Q_8$-extensions $L/\mathbb{Q}_2$. As you say, one can easily adapt the number field proof to the $2$-adic setting or use the method mentioned two paragraphs above (we can't use Henniart's result here, but as there are only finitely many $Q_8$-extensions $L/\mathbb{Q}_2$ one can, for example, use databases of extensions of $\mathbb{Q}_2$ and $\mathbb{Q}$.) I should say that Fabio Ferri (who is my PhD student), is aiming to post an preprint to the arXiv in the next couple of weeks, that in addition to its main results on the additive Galois module structure of $A_4$, $S_4$ and $A_5$-extensions of $\mathbb{Q}$, will also provide a nice summary / discussion of the relation between freeness in the number field and $p$-adic cases.<|endoftext|> TITLE: Subspaces of $ A_{n}(\mathbb {Q})$ in which all nonzero matrices are invertible QUESTION [8 upvotes]: Let $A_{n}(\mathbb{Q}) $ denote the $n$ times $n$ skew symmetric matrices over the rational number field. Let $N$ be a subspace of $A_{n}(\mathbb{Q}) $. If all the non-zero matrices in $N$ are invertible, what is the maximum the dimension of $N$ can be? You can assume $ n $ is even. Does there exist a subspace $M$ of $A_{n}(\mathbb{Q}) $ of dim $ n-1 $ with all the non-zero matrices in $M$ are invertible? You can assume $n$ is even. Note that if $A_{n}(\mathbb{R}) $ denotes the $n$-times-$n$ skew-symmetric matrices over the real number field, then, for $ n= 4 $ and $ n = 8 $, the answer of the second question is 'yes', but, for $ n= 6 $, there is no such subspace. REPLY [7 votes]: For a field $\mathbb{F}$, let $\mu_\mathbb{F}(n)$ denote the maximal dimension of a subspace $N\subset A_n(\mathbb{F})$ such that all the nonzero elements of $N$ are invertible. For simplicity, I will assume that the characteristic of $\mathbb{F}$ is not $2$. Then, because $\det(a) = (-1)^n\det(-a) = (-1)^n\det(a^\mathsf{T})= (-1)^n\det(a)$, we have $\mu_\mathbb{F}(2m{+}1) = 0$. Meanwhile, clearly, $\mu_\mathbb{F}(2m)\ge 1$, and, as the OP points out, $\mu_\mathbb{R}(4m)\ge 3$ and $\mu_\mathbb{R}(8m)\ge7$ due to the existence of normed division algebras $\mathbb{H}$, of dimension $4$ over $\mathbb{R}$, and $\mathbb{O}$, of dimension $8$ over $\mathbb{R}$. When $n=2m$, the polynomial function $\det:A_n(\mathbb{F})\to\mathbb{F}$ is the square of a polyomial $\mathrm{Pf}:A_n(\mathbb{F})\to\mathbb{F}$ homogeneous of degree $m=n/2$, unique up to a choice of sign. In fact, $\mathrm{Pf}$ is defined over the integers, $\mathrm{Pf}:A_n(\mathbb{Z})\to\mathbb{Z}$, as a polynomial with integer coefficients with the property that $\mathrm{Pf}(mam^\mathsf{T}) = \det(m)\,\mathrm{Pf}(a)$ for $a\in A_{2m}(\mathbb{Z})$ and $m\in M_{2m}(Z)$. Consequently, this property holds with $\mathbb{Z}$ replaced by $\mathbb{F}$ for any field $\mathbb{F}$. It follows that $\mu_\mathbb{R}(4m+2)=1$, since, in this case, $\mathrm{Pf}$ is a polynomial of odd degree, implying that, for any pair $a,b\in A_{4m+2}(\mathbb{R})$, the homogeneous polynomial $p(s,t)=\mathrm{Pf}(sa+tb)$ of odd degree $2m{+}1$ will vanish for some real ratio $[s:t]$. If $\mathbb{F}$ is an ordered field (more generally, if every nontrivial sum of squares in $\mathbb{F}$ is nonzero), then the standard Clifford algebra construction (using a definite quadratic form) shows that $\mu_\mathbb{F}(n)\ge\rho(n){-}1$, where $\rho(n)$ is the Radon-Hurwitz number. In particular, $\mu_\mathbb{Q}(n)\ge\rho(n){-}1$. Meanwhile, J. F. Adams has shown that $\mu_\mathbb{R}(n)=\rho(n){-}1$. Thus, $\mu_\mathbb{Q}(n)\ge\mu_\mathbb{R}(n)$, but, in general, equality does not hold. Claim: $\quad 2m{-}1\ge\mu_\mathbb{Q}(2m)\ge m$. In particular, $\mu_\mathbb{Q}(6)\ge 3 > \mu_\mathbb{R}(6) = 1$, thus verifying that $\mu_\mathbb{Q}(2m)$ can be strictly greater than $\mu_\mathbb{R}(2m)$. The claim follows from the fact that the characteristic polynomial of a generic element $a\in A_{2m}(\mathbb{Q})$ is irreducible over $\mathbb{Q}$. For, when the characteristic polynomial of $a$ is irreducible over $\mathbb{Q}$, then $I, a, a^2,\ldots, a^{2m-1}$ spans a field $\mathbb{Q}(a)\subset M_{2m}(\mathbb{Q})$, and hence every nonzero linear combination of these matrices is invertible. Meanwhile, $N(a) = \mathbb{Q}(a)\cap A_{2m}(\mathbb{Q})$ is a vector space with basis $a, a^3, \ldots a^{2m-1}$ and hence has dimension $m$. Thus, $\mu_\mathbb{Q}(2m)\ge m$. The upper bound follows from the fact that any subspace $N\subset A_{2m}(\mathbb{Q})$ of dimension greater than $2m{-}1$ must intersect nontrivially with the subspace of matrices with the first column equal to zero, since that subspace has codimension $2m{-}1$. Remark 1: It seems likely that the 'generic' $m$-dimensional subspace of $A_{2m}(Q)$ has all of its nonzero elements invertible, but, perhaps this depends on some carefully defined notion of 'generic'. Remark 2: Since $\mu_\mathbb{Q}(n)\ge \mu_\mathbb{R}(n)$, the lower bound in the Claim cannot always be strengthened to equality. For example, $\mu_\mathbb{Q}(4)\ge \mu_\mathbb{R}(4) = 3 > 2$. Thus, $\mu_\mathbb{Q}(4)=3$. Similarly, since $\mu_\mathbb{R}(8)=7$, we have $\mu_\mathbb{Q}(8)=7$. (This answers Question 1 for $n=4$ and $n=8$.) Note that the OP's Question 2 asks whether $\mu_\mathbb{Q}(2m)\ge 2m{-}1$, presumably provoked by the observation that $\mu_\mathbb{R}(2m) = 2m{-}1$, when $m=2$ and $m=4$. However, these low dimensions can be very misleading. For all other values of $m$, we have $\mu_\mathbb{R}(2m) < 2m{-}1$, and, in fact, for all but a finite set of values of $m$, we have $\mu_\mathbb{R}(2m) < m$, and in general, as $m$ grows, the lim inf of $\mu_\mathbb{R}(2m)/m$ equals $0$. On the other hand, $\mu_\mathbb{Q}(2m)/m\ge 1$ for all $m$. Remark 3: I'm including this last remark at the request of the OP, but, not being a number theorist, I do not have any realy confidence that this can be turned into a rigorous argument. I do not know whether $\mu_\mathbb{Q}(6)>3$, however, a very heuristic speculation leads me to suspect that this is true and that it might even be true that $\mu_\mathbb{Q}(6)=5$. The Grassmannian $G_4(15)$ of $4$-dimensional subspaces of $A_\mathbb{Q}(6)$ is a rational variety of dimension $4\cdot (15-4) = 44$. Meanwhile, the group $\mathrm{SL}(6,\mathbb{Q})$ has dimension $35$ and it acts on $A_\mathbb{Q}(6)$ via $m\cdot a = mam^\mathsf{T}$ preserving $\mathrm{Pf}:A_\mathbb{Q}(6)\to\mathbb{Q}$. The induced action of $\mathrm{SL}(6,\mathbb{Q})$ on $G_4(15)$ has generic orbits of dimension $35$, so the 'moduli space' $\mathscr{M}$ of orbits has formal dimension $44-35 = 9$. Meanwhile, the restriction of $\mathrm{Pf}$ to a subspace $N\subset A_\mathbb{Q}(6)$ is a rational cubic form on $N$, generically nondegenerate. The moduli of cubic forms of rank $4$ under $\mathrm{GL}(4,\mathbb{Q})$ equivalence has formal dimension $20 - 16 = 4 < 9$, and it is known that there are rational cubic forms of rank 4 that do not represent $0$ rationally. It seems that the map assigning to a generic $4$-plane $N\subset A_6(\mathbb{Q})$ the rational cubic form $\mathrm{Pf}:N\to\mathbb{Q}$ is 'dominant'. For this reason, it seems likely to me that a 'generic' $4$-plane $N\subset A_6(\mathbb{Q})$ will have the property that $\mathrm{Pf}:N\to\mathbb{Q}$ will not represent $0$ rationally (and hence the nonzero elements of $N$ would all be invertible). However, it's not that easy to determine whether a given rational cubic forms of rank $4$ represents $0$ rationally, so just choosing a $4$-plane $N$ 'at random' and testing whether its Pfaffian represents $0$ seems to be a very labor intensive way to try to find an example. All of the above is very speculative, but one could go on to make a similar argument for $5$-planes in $A_6(\mathbb{Q})$. There, it seems even harder to test for when a given rational cubic form of rank $5$ represents $0$ rationally, though.<|endoftext|> TITLE: Question about the theory behind an important footnote to entry 109 in Gauss's diary QUESTION [16 upvotes]: Entry 109 in Gauss's diary (and the related material in Gauss's Nachlass) is the main subject of David Cox's famous article "The Arithmetic-Geometric mean of Gauss". This entry deals with the agm of two complex numbers $a,b$, taken as a multi-valued function. It reads as follows: Between two given numbers there are always infinitely many means both arithmetic-geometric and harmonic-geometric, the observation of whose mutual connection has been a source of happiness for us. The harmonic-geometric mean of two numbers $a,b$ (which we denote as $H(a,b)$) is $M(a^{-1},b^{-1})^{-1}$. According to Cox, this entry is a continuation of a previous study of the agm, which resulted in the following relation between the infinitely many values of the agm: $$ (1)\qquad \frac{1}{(\mu)} = \frac{1}{M(a,b)}+\frac{4ik}{M(a,\sqrt{a^2-b^2})}. $$ Relation (1) is not the most general formula for the different values of the agm - this is written as: $$ (2) \qquad\frac{1}{(\mu)} = \frac{d}{M(a,b)}+\frac{ic}{M(a,\sqrt{a^2-b^2})}, $$ where d and c are relatively prime numbers and $d \equiv 1\pmod{{4}}, c\equiv 0\pmod{{4}}$. The main purpose of part 3 of Cox's article is to check if Gauss only knew (1) or that he even knew (2) (Gauss never stated (2)). Looking into diary entry 109 in volume 10-1, I saw an additional piece of evidence that is not considered at Cox's article; this is a footnote that Gauss writes next to this entry, and reads in Latin as: Terminus constans expressionis $$\frac{Ad\phi}{\sqrt{f+2g\cos\phi + h\cos^2\phi}}$$ est Medium Geometrico harmonicum inter $$\frac{A}{\sqrt{\frac{\sqrt{(f+h)^2-4g^2} + f-h}{2}}}$$ et $$\frac{A}{((f+h)^2-4g^2)^{\frac{1}{4}}}$$. Now, after a lot of effort, I deciphered what was Gauss's intention in his cryptic Latin expression "Terminus constans expressionis" - the Latin phrase "Terminus constans expressionis" seems to mean the "constant term" $A$ in the Fourier expansion of the expression $\frac{1}{\sqrt{f+2g\cos\phi+h\cos^2\phi}}$, (the trigonometric series $A+A'\cos\phi+A''\cos2\phi+...$) which is actually equal to $\frac{1}{\pi}\int_0^{\pi}\frac{d\phi}{\sqrt{f+2g\cos\phi+h\cos^2\phi}}$. This interpretation of Gauss's Latin is consistent with Gauss's notation in other places in his writings - for example, in p.52 of Cox's article, Cox mentions Gauss's result that $\frac{2}{\pi}\int_0^{\pi/2}\frac{d\phi}{\sqrt{1+\mu \cos^2\phi}} = M(\sqrt{1+\mu},1)^{-1}$ - a result that Gauss describes by using the same Latin expression (the meaning of this result is that the constant term in the Fourier expansion of $\frac{1}{\sqrt{1+\mu \cos^2\phi}}$ is $M(\sqrt{1+\mu},1)^{-1}$). Note that the Gaussian result that is mentioned in Cox's article is a very special case of Gauss's much more general result stated in his Latin footnote; put $f = 1, g =0 , h = \mu$, and you will get this special case. Classification of the integral in Gauss's footnote By employing the substitution $t = \tan(\frac{\phi}{2})$ (the "universal trigonometric substitution") Gauss's integral can be brought into the form: $$ \int_0^{\pi}\frac{d\phi}{\sqrt{f+2g\cos\phi + h\cos^2\phi}}= \int_0^{\infty}\frac{2dt}{(1+t^2)\sqrt{f+2g\frac{1-t^2}{1+t^2} + h(\frac{1-t^2}{1+t^2})^2}} = \int_0^{\infty}\frac{2dt}{\sqrt{(f+2g+h)+(2f-2h)t^2+(f-2g+h)t^4}}. $$ The denominator of the integrand is therefore a square root of a fourth degree polynomial - and this is the definition of a specific case of elliptic integral. In addition, substitution $x = t^2, 2dt = \frac{dx}{\sqrt{x}}$, brings the denominator into the form of a square root of a cubic: $$\int_0^{\infty}\frac{dx}{\sqrt{(f+2g+h)x+(2f-2h)x^2+(f-2g+h)x^3}}$$ and every cubic of the form $a_0x^3+a_1x^2+a_2x$ can be transformed into Weierstrass normal form $4s^3-g_2s-g_3$ under the following suitable change of variables: $$s = (\frac{a_0}{4})^{\frac{1}{3}}x+\frac{a_1}{3(\frac{a_0}{4})^{\frac{2}{3}}}$$ $$g_2 = \frac{\frac{4a_1^2}{3a_0}-a_2}{(\frac{a_0}{4})^{\frac{1}{3}}}$$ $$g_3 = \frac{36a_0a_1a_2-32a_1^3}{27a_0^2}$$ Therefore, Gauss's result is possibly about broadening the concept of the AGM algorithm to a wider class of integrals. Special cases of Gauss's hgm formula, numerical verification Note also that in the special cases where $h=1, f = g^2$, than we have to calculate the harmonic-geometric mean of two equal numbers equal to $\frac{1}{\sqrt{g^2-1}}$, a prediction that was verified by me by calculating specific integrals on Wolfram Alpha; for example, if $h=1, g=2, f = 4$, than the result of the integral is $\frac{1}{\sqrt{3}}$. This shouldn't come as a surprise, because in this case the indefinite integral is elementary (one can see it directly via the "universal trigonometric substitution"), and it's also evident from the fact that the hgm algorithm was not really needed here. The significance of Gauss's formula appears clearly in cases when there is a need to run the hgm algorithm. For example, if we take $h = 1, g =5, f = 24$, than: $$ M(\sqrt{\frac{\sqrt{(f+h)^2-4g^2}+f-h}{2}},((f+h)^2-4g^2)^{1/4})^{-1} = M(\sqrt{\frac{\sqrt{525}+23}{2}},(525)^{1/4})^{-1} = 0.208811, $$ and this numerical result is in exact accordance with a calculation of the integral $\frac{1}{\pi}\int_0^{\pi}\frac{d\phi}{\sqrt{24+10cos\phi+cos^2\phi}}$ on Wolfram Alpha. Questions regarding Gauss's hgm formula Therefore my questions are: I'm trying to put Gauss's result in context, but I didn't find articles connecting this specific elliptic integral to the AGM algorithm. Therefore, I'd like to know if this result is known to the mathematical community. Maybe some of the mathematicians here who are familiar with the theory of elliptic integrals can see the connection. Schlesinger remarks that the expression "has been a source of happiness for us" was a kind of code-expression that Gauss used to refer to new results in number theory, and speculates that the appearence of this expression in diary entry 109 means that Gauss understood the connection between this note and the theory of binary quadratic forms. Also, the notation of Gauss in his footnote (he uses $f,g,h$) reminds one of his notation for quadratic forms. So, is there any relation of this footnote to binary quadratic forms? REPLY [4 votes]: I asked David Cox about how to prove Gauss's AGM formula, and he showed me that despite that at first glance the integral in Gauss's footnote looks more general than the usual version, a few algebraic operations can transform the elliptic integral in Gauss's footnote into the standard form. His proof consists of two steps: Transforming the standard form of the elliptic integral into an alternative form by using the universal trigonometric substitution. Transforming the integral in Gauss's footnote, again by the universal trigonometric substitution, into the alternative form of standard integral, ad than settling a few algebraic details. Step one The well-known formula $$\frac{2}{\pi}\int_0^{\pi/2}\frac{d\phi}{\sqrt{1+\mu cos^2\phi}} = M(\sqrt{1+\mu},1)^{-1}$$ can be written $$\frac{1}{\pi}\int_0^{\pi}\frac{d\phi}{\sqrt{1+\mu cos^2\phi}} = M(\sqrt{1+\mu},1)^{-1}$$ The substitution $t = \mathbb{tan(\phi/2)}$ turns it into: $$ \frac{1}{\pi}\int_0^{\infty}\frac{dt}{\sqrt{(1+\mu) + 2(1-\mu)t^2 +(1+\mu)t^4}} = M(\sqrt{1+\mu},1)^{-1} $$ Under the radical, factor out $1+\mu$ to get $$ \frac{1}{\sqrt{1+\mu}}\frac{1}{\pi}\int_0^{\infty}\frac{dt}{\sqrt{1+2\lambda t^2 + t^4}} = M(\sqrt{1+\mu},1)^{-1}, \lambda = \frac{1-\mu}{1+\mu} $$ Hence: (1) $$\frac{1}{\pi}\int_0^{\infty}\frac{dt}{\sqrt{1+2\lambda t^2 + t^4}} = \frac {\sqrt{1+\mu}}{M(\sqrt{1+\mu},1)}, \mu = \frac{1-\lambda}{1+\lambda}$$ Step two The goal is to express (2) $$\frac{1}{\pi}\int_0^{\pi}\frac{d\phi}{\sqrt{f+2gcos\phi+hcos^2\phi}}$$ in terms of the AGM. This is done by transforming Gauss's integral into the form (1). First use $t = \mathbb{tan(\phi/2)}$ to transform the integral into $$\frac{1}{\pi}\int_0^{\infty}\frac{dt}{\sqrt{(f+h+2g)+2(f-h)t^2+(f+h-2g)t^4}}$$ If we set $a=f+h+2g,b = f-h, c = f+h-2g$, then this integral can be written $$\frac{1}{\pi}\int_0^{\infty}\frac{dt}{\sqrt{a+2bt^2+ct^4}}$$ Now define $\delta = (a/c)^{1/4}$ so that $\delta^4 = a/c$. The substitution $t = \delta \tau $ turns the above integral into $$\frac{1}{\pi}\int_0^{\infty}\frac{\delta d\tau}{\sqrt{a+2b\delta^2\tau^2+c\delta^4\tau^4}} = \frac{1}{\pi}\int_0^{\infty}\frac{d\tau}{\sqrt{(a/\delta^2)+2b\tau^2+(c\delta^2)\tau^4}}$$ The definition of $\delta$ implies $a/\delta^2 = c\delta^2$. Factoring out $c\delta^2$ under the radical transforms this integral into (3) $$\frac{1}{\sqrt{c}\delta}\frac{1}{\pi}\int_0^{\infty}\frac{d\tau}{\sqrt{1+2\lambda\tau^2+\tau^4}}, \lambda = \frac{b}{c\delta^2} = \frac{f-h}{\sqrt{(f+h)^2-4g^2}}$$ since $b=f-h$ and $c\delta^2 = c\sqrt{a/c} = \sqrt{ac} = \sqrt{(f+h+2g)(f+h-2g)} = \sqrt{(f+h)^2-4g^2}$ Note that $f-h$ and $\sqrt{(f+h)^2-4g^2}$ appear in Gauss's formula, as does $\sqrt{c}\delta = ((f+h)^2-4g^2)^{1/4}$. The integral (3) can now be computed using (1). Concluding remarks Gauss's integral can be turned into the standard one without too much difficulty. Therefore, the main importance of Gauss's footnote is that it demonstrates the great scope of the AGM algorithm; even elliptic integrals that are not given in the standard form can be transformed into a form which can then be computed using the AGM. Gauss's footnote has, however, no connection to the deeper issues concerning multiple values of the arithmetic-geometric and harmonic-geometric means, which diary entry 109 is about. Schlesinger probably mentioned Gauss's formula in the subtext to entry 109 just to give an example of Gauss's use of the term "harmonic-geometric mean".<|endoftext|> TITLE: Is "There exists an unbounded non-measurable set but no bounded non-measurable set" consistent with $\mathsf{ZF}$? QUESTION [12 upvotes]: This is a follow-up to this question. We say that a set $A \subseteq \mathbb{R}$ is bounded if there exists a finite interval $(a,b)$ such that $A \subseteq (a,b)$. Working in $\mathsf{ZFC}$, the existence of a (Lebesgue) non-measurable set (of $\mathbb{R}$) easily implies the existence of a bounded non-measurable set. A proof is as follows - Let $X \subseteq \mathbb{R}$ be non-measurable. Then $X = \bigsqcup_{n \in \mathbb{Z}} X \cap (n,n + 1]$. If all $X \cap (n,n+1]$ are measurable, then $X$ must also be measurable, at it is a countable union of measurable sets ($\mathsf{AC}$ is used here). Thus, there exists an $n \in \mathbb{Z}$ in which $X \cap (n,n+1] \subseteq (n,n+2)$ is not measurable. However, it appears to not be so clear if we only work in $\mathsf{ZF}$, since we cannot guarantee that the countable union above is measurable. I also can't seem to see an easy way around choice here. Thus, if we write: $\mathsf{NM}$ as "there exists a non-measurable set". $\mathsf{BNM}$ as "there exists a bounded non-measurable set". $\mathsf{M}_\omega$ as "countable union of measurable sets is measurable". My questions are (assuming $\mathrm{Con}(\mathsf{ZFC})$): Is $\mathsf{ZF} + \neg\mathsf{M}_\omega$ consistent? Is $\mathsf{ZF} + \mathsf{NM} + \neg\mathsf{BNM}$ consistent? REPLY [7 votes]: The answer to your first question is yes, and the answer to your second question is no, under any of the multiple definitions of "measurable" in choiceless contexts. We will prove a theorem relating various measure-theoretic consequences of countable choice. (ZF) The following are equivalent. Note that (1)-(6) are about the algebra of subsets of $[0,1]$ which satisfy $\lambda^*(X)+\lambda^*([0,1] \setminus X)=1$ while (7) refers to the algebra of subsets of $\mathbb{R}$ which satisfy $\lambda^*([-n, n] \cap X) + \lambda^*([-n, n] \setminus X) = 2n$ for all $n.$ Lebesgue measure is $\sigma$-additive. A countable union of measurable sets is measurable. A countable union of null sets is measurable. A countable union of null sets is null. Every null set is contained in a null $G_{\delta}$ set. For every measurable set $X,$ there is an $F_{\sigma}$ set $A$ and a $G_{\delta}$ set $B$ such that $A \subset X \subset B$ and $B \setminus A$ is null. Any of the above but for measurable subsets of $\mathbb{R}.$ Proof: (1) $\rightarrow$ (2) Clear. (2) $\rightarrow$ (3) Clear. (3) $\rightarrow$ (4) Suppose towards contradiction $X_i$ are null sets with $\lambda(\bigcup_{i<\omega} X_i)>0.$ Let $Y_i = \{x+\frac{m}{2^i}: m < 2^i, \exists j < i (x \in X_j)\}$ (note that addition is mod 1) and $Z_i = Y_i \setminus Y_{i-1}.$ In particular, the $Z_i$ are disjoint null sets whose union has positive measure and is closed under translation by dyadic rationals. For $A \subset \omega,$ let $H_A=\bigcup_{n \in A} Z_n.$ By (3), each $H_A$ is measurable. We will show that for every $A \subset \omega,$ either $H_A$ or $H_{\omega \setminus A}$ has measure 1. Suppose $H_A$ has positive measure (otherwise $H_{\omega \setminus A}$ has positive measure). Fix $\epsilon>0.$ By Lebesgue density theorem, there is some interval $I$ of length $\frac{1}{2^n}$ such that $\lambda(H_A \cap I) > \frac{1-\epsilon}{2^n}.$ Clearly $H_{A \setminus (n+1)}$ also satisfies this inequality, and is furthermore closed under translation by $\frac{1}{2^n}.$ Thus, $\lambda(H_A) = \lambda(H_{A \setminus (n+1)}) > 1-\epsilon,$ so $\lambda(H_A)=1.$ Since $H_{\omega}$ has measure 1, we see that $[0,1]$ is a countable union of null sets. By (3), every subset of $[0,1]$ is measurable. However, $\{A \subset \omega: H_A \text{ is measure 1} \}$ is a non-principal ultrafilter, so there is a nonmeasurable subset of $[0,1],$ contradiction. (4) $\rightarrow$ (1) Let $X_i$ be measurable sets. Let $U_i$ enumerate the basic open sets. Define $S_n \subset \omega$ recursively by having $i \in S_n$ iff there is $j$ such that $\lambda(U_i \cap X_j \setminus \bigcup_{k \frac{n}{n+1} \lambda(U_i).$ Let $V_n = \bigcup_{i \in S_n} U_i.$ Then $V:=\bigcap_{n < \omega} V_n$ is a $G_{\delta}$ set such that $\lambda(V)=\sup_{n<\omega} \lambda(\bigcup_{i < n} X_i)$ and $V \triangle \bigcup_{i<\omega} X_i$ is null. (4) $\rightarrow$ (5) Let $X$ be null. By (4) we can assume $X$ is closed under translation by rational numbers. Let $U$ be an open cover of $X$ of measure less than $\frac{1}{2}.$ We can canonically cover $X \cap [0, \frac{1}{n}]$ with an open set of measure $\frac{1}{2n}$ by considering the least $m$ such that $\lambda(U \cap [\frac{m}{n}, \frac{m+1}{n}])<\frac{1}{2n}.$ We can thus recursively construct open covers of $X$ of measure less than $\frac{1}{2^n}.$ (5) $\rightarrow$ (6) Similarly to in (4) $\rightarrow$ (1), there is a $G_{\delta}$ set $B_1$ with null symmetric difference from $X.$ Let $B_2$ be a null $G_{\delta}$ set containing $X \setminus B_1.$ Then $B:=B_1 \cup B_2$ is a $G_{\delta}$ set with $X \subset B$ and $B \setminus X$ null. We can similarly construct such a $B'$ for $[0, 1] \setminus X$ and set $A = [0, 1] \setminus B'.$ (6) $\rightarrow$ (4) Let $X_i$ be null sets. Let $X=\bigcup_{i<\omega} X_i.$ Consider $Y = \{2^{-i-1}(1+x): x \in X_i\}.$ It is easy to see $Y$ is null, and thus contained in some null $G_{\delta}$ set $Y'.$ Let $U_n$ be a sequence of open sets covering $Y,$ each satisfying $\lambda(U_n)<\frac{1}{n}.$ Fix $\epsilon>0$ and $i<\omega.$ Let $n$ be least such that $\frac{1}{2^n}<\frac{\epsilon}{4^{i+1}}.$ Then $X_i$ is covered by a translation of $U_n$ scaled up by $2^{i+1},$ which has measure less than $\frac{\epsilon}{2^{i+1}}.$ Applying this construction to all $i,$ we get a cover of $X$ of measure less than $\epsilon.$ Thus, $X$ is null. (7) Finally, it is routine to verify that assertions (1)-(6) collectively prove their generalizations to $\mathbb{R}.$ E.g., one could verify $\frac{1}{x}$ on $(0, \infty)$ sends null sets to null sets and measurable sets to measurable sets using the fact that it's Lipschitz on each $[2^{-n}, 2^n]$ and that $\frac{1}{x}$ sends $G_{\delta}$ null sets to $G_{\delta}$ null sets. $\square$ Thus, $\text{M}_{\omega}$ fails in any model of ZF + "$\mathbb{R}$ is a countable union of countable sets" since this theory negates (1), providing an affirmative answer to question 1. As for question 2, if $\neg \text{BNM}$ holds, then we immediately have (2), so every subset of $\mathbb{R}$ is measurable. In particular, all interpretations of "all sets are measurable" are equivalent.<|endoftext|> TITLE: In the Oldenburger-Kolakoski sequence, is #1s = #2s infinitely many times? QUESTION [5 upvotes]: The Oldenburger-Kolakoski sequence, $OK$, is the unique sequence of $1$s and $2$s that starts with $1$ and is its own runlength sequence: $$OK = (1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1,2,1,1,\ldots).$$ For $n \geq 1$, let $a(n)$ be the number of indices $h \leq n$ such that $OK(h) =1$, and let $b(n)$ be the number of $h \leq n$ such that $OK(h) = 2.$ QUESTION. Is $a(n) = b(n)$ for infinitely many $n$? For example, the first ten runs are $1,22,11,2,1,22,1,22,11,2$, and the lengths of these runs are $1,2,2,1,1,2,1,2,2,1$. In the OEIS, the sequence is A000002. There are several related easily stated unsolved problems, such as the conjecture that the limiting density of $1$'s is $1/2$. REPLY [2 votes]: The short answer is: nobody knows. To me this problem looks even harder than the problem of limit densities or other open problems like whether the sequence is recurrent and mirror/reversal invariant. The general opinion among the ones who studied the problem is that it's very unlikely that $d(1)$ doesn't exists (this is in fact another open question), and provided that it exists, your claim is actually much stronger than $d(1)=1/2$. Some heuristics: let $T(w)$ be the operator associating to every word in $\{1,2\}^*$ the word $v$ starting with 1 and such that $w$ lists the lengths of the runs of $v$ (example: $T(1221)=122112$). If you believe that the parity of the lengths $\ell(T^k(w))$ "behaves like" a Bernoulli variable with probability $1/2$, then "it follows" that the sequence is recurrent and that $d(1)$ equals 1/2 (if it exists). But even assuming this strong heuristic viewpoint, your claim would still be far from obvious.<|endoftext|> TITLE: Looking for a $q$-analogue of a binomial identity QUESTION [7 upvotes]: The following identity is well-known and there are a few proofs to it (see Bijective proof problems, by R P Stanley, for this and similar formulae): $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}=4^n \qquad \iff \qquad \sum_{j+i=n}\binom{2j}j\binom{2i}i\frac1{4^n}=1.\tag1$$ QUESTION. Is there a $q$-analogue to this "innocent-looking" identity in equation (1)? REPLY [4 votes]: We start with $q$-binomial theorem $$ (x+y)(x+qy)\cdots(x+q^{n-1}y)=\sum q^{k\choose 2}{n\choose k}_qy^k x^{n-k}.\quad\quad\quad(\heartsuit) $$ Put $n=a+b$ in $(\heartsuit)$ and consider separately the first $a$ multiples in LHS and the last $b$ multiples. We get $$ \left(\sum q^{j\choose 2}{a\choose j}_qy^j x^{a-j}\right)\cdot \left(\sum q^{{i\choose 2}+ai}{b\choose i}_qy^i x^{b-j}\right)= \sum q^{k\choose 2}{a+b\choose k}_qy^k x^{a+b-k}. $$ Taking coefficients of $x^ky^{n-k}$ we get $$ \sum_{j+i=k} q^{{j\choose 2}+{i\choose 2}+ai}{a\choose j}_q {b\choose i}_q=q^{k\choose 2}{a+b\choose k}_q.\quad\quad\quad\quad\quad\quad\quad\quad(\clubsuit) $$ For fixed $k$ (and $q$) both LHS and RHS of $(\clubsuit)$ are polynomials in $q^a$ and $q^b$. Thus we may substitute $q^a=q^b=q^{-1/2}$ (I do this because in the ordinary version the identity is equivalent to the $a=b=-1/2$ version of the Vandermonde--Chu convolution $\sum_{a+b=k} {a\choose i}{b\choose j}={a+b\choose k}$). We have $${-1\choose k}_q=\frac{(q^{-1}-1)(q^{-2}-1)\cdots(q^{-k}-1)}{(q^k-1)\cdots(q-1)}=q^{-{k+1\choose 2}}(-1)^k.$$ Next, denote $q=\tau^2$, then $${-1/2\choose j}_q=\frac{(q^{-1/2}-1)(q^{-3/2}-1)\cdots(q^{-(2j-1)/2}-1)}{(q^j-1)\cdots(q-1)}\\=\tau^{-j^2}(-1)^j\frac{(\tau-1)(\tau^3-1)\cdots (\tau^{2j-1}-1)}{(\tau^2-1)(\tau^4-1)\cdots (\tau^{2j}-1)}\\=\tau^{-j^2}(-1)^j{2j\choose j}_{\tau}\frac1{(1+\tau)^2(1+\tau^2)^2\cdots (1+\tau^j)^2}.$$ So $(\clubsuit)$ reads as $$ \sum_{j+i=k} \tau^{j} {2j\choose j}_{\tau}{2i\choose i}_\tau\frac1{(1+\tau)^2(1+\tau^2)^2\cdots (1+\tau^j)^2\cdot (1+\tau)^2(1+\tau^2)^2\cdots (1+\tau^i)^2}=1. $$ This is an analogue of the identity you ask about in the sense that for $\tau=1$ we get it. Note by OP: If we write the $j$-factorial analogue as $(q)_j=\prod_{k=1}^j\frac{1-q^k}{1-q}$, then the last identity takes the form $$ \sum_{j+i=n} \tau^{j} {2j\choose j}_{\tau}{2i\choose i}_\tau \frac{(\tau)_j^2(\tau)_i^2}{(\tau^2)_j^2(\tau^2)_i^2}=1. $$<|endoftext|> TITLE: If the Grothendieck ring of a semiring on a free commutative monoid is unital, is the original semiring unital? QUESTION [8 upvotes]: Suppose $S$ is an associative semiring whose underling commutative monoid is free (in particular, cancellative) and that its Grothendieck ring $G(S)$ is a unital ring. Can we conclude that $S$ must be unital, and if not, is there a nice counter-example? Alternatively, are there additional suppositions we can put on $S$ which would allow us to conclude the unitality of $S$ from that of $G(S)$? The context is that I am struggling to disprove that a particular ring I have constructed is unital, and I have proven it is of the form $G(S)$ for $S$ which is not unital. I am hoping this helps me somehow. Cross-posted on math.stackexchange. REPLY [8 votes]: The answer is no. Let $S$ be a finite meet semilattice without maximum. For concreteness, take $S$ to be the proper subsets of $\{1,2\}$ under intersection. Let $\mathbb NS$ be the semigroup semiring of $S$. Then the underlying additive monoid is free on $S$. The Grothendieck ring is the semigroup ring $\mathbb ZS$ and this ring has an identity, but the identity cannot be expressed as a non-negative linear combination of elements of $S$. One writes it using inclusion-exclusion (or Mobius inversion). For my particular $S$, the proper subsets of $\{1,2\}$, the identity element of $\mathbb ZS$ is $e=\{1\}+\{2\}-\emptyset$, which does not belong to $\mathbb NS$. Recalling that my semigroup multiplication is intersection, you can just check that $e$ multiplied by an element of $S$ returns that element. For a more general meet semilattice $S$, the identity is $\sum_{s\in S}\sum_{t\leq s}t\mu(t,s)$ where $\mu$ is the Mobius function of $S$. I think for general semirings there is no hope of a reasonable condition. For example if $S$ is a finite left regular band (that is a semigroup satisfying the identities $xyx=xy$ and $x^2=x$), I showed with Margolis and Saliola that $\mathbb ZS$ has an identity if and only if certain posets associated to $S$ have connected Hasse diagrams. But if $S$ is not a monoid, the identity will not belong to $\mathbb NS$. This is already a hard result. The meet semilattice case is just a commutative left regular band. I also showed that for any finite inverse semigroup $S$, the semigroup ring $\mathbb ZS$ has an identity, but $\mathbb NS$ will not have an identity unless $S$ has an identity. This also generalizes the meet semilattice case. Update: Here are some more observations. First, if $S$ is a semigroup, then the semigroup ring $\mathbb NS$ is unital if and only if $S$ is a monoid. That is because you have the surjective augmentation homomorphism $\mathbb NS\to\mathbb N$ summing up the coefficients and since the only nonzero idempotent of $\mathbb N$ is $1$, we must have that the support of an identity element has one element of $S$ with coefficient $1$ and so $S$ is a monoid. Second, let $S$ be a meet semilattice (possibly infinite) viewed as a semigroup under meet. Then $\mathbb NS$ has an identity iff $S$ has a maximum. But $\mathbb ZS$ is isomorphic to the locally constant functions with compact support from $\widehat{S}\to \mathbb Z$ under pointwise operations. Here $\widehat{S}$ is the space of all non-zero semigroup homomorphisms $\phi\colon S\to \{0,1\}$ with the topology of pointwise convergence, where $\{0,1\}$ is a semigroup under usual multiplication. Thus $\mathbb ZS$ has an identity if and only if $\widehat{S}$ is compact. This is obviously a topological issue.<|endoftext|> TITLE: Refinement of mean value conjecture for complex polynomials? QUESTION [20 upvotes]: I was playing around with Smale's Mean Value Conjecture and found a curious formulation of it which would be stronger (and which may simply be false). It seems to hold for `generic' random polynomials but of course this may mean very little. I'd be interested in whether there is a chance it might be true or whether there are counterexamples (it isn't really clear to me how to try to construct one). Smale's Mean Value Conjecture. Smale proved in 1981 that if $f:\mathbb{C} \rightarrow \mathbb{C}$ is a polynomial with $f(0) = 0$ and $|f'(0)| = 1$, then there exists a critical point of the polynomial ($z \in \mathbb{C}$ such that $f'(z) = 0$) satisfying $$ |f(z)| \leq 4 |z|.$$ One question is whether 4 can be replaced by 1 (which would be best possible if one is interested in universal constants that do not depend on the degree). The best bound depending on the degree is conjectured to be $(d-1)/d$. I was looking at some examples and noticed a curious pattern which seems to hold at least for polynomials with randomly generated roots with very high probability (meaning I did not find a counterexample but maybe random polynomials are the wrong place to look). A Stronger Conjecture? Let $g:\mathbb{C} \rightarrow \mathbb{C}$ be a polynomial with $|g(0)|=1$. Consider the subset $$ A = \left\{z \in \mathbb{C}: |g(z)| < 1 \right\} \subset \mathbb{C}$$ and let $B$ denote the connected component of $A$ whose closure contains the point 0. Is it true that the number of critical points of $z g(z)$ in $B$ is exactly the same as the number of roots of $z g(z)$ in $B$? Or, a weaker formulation, that $B$ contains at least one critical point? Note that $\log |g(z)|$ is analytic and constant on the boundary, therefore $B$ is simply connected and $B$ contains at least one root of $g(z)$. If this statement were true, it would therefore imply Smale's conjecture. It would also be a bit finer since Smale's conjecture is a priori a statement about the set $A$ containing a critical point (and $B \subset A$). The subsequent picture gives an example: the red dots are the roots of $z g(z)$, the black dots represent critical points of $z g(z)$. The set $A$ is shown in blue, the set $B$ is the big blue set in the middle. There is one root of $z g(z)$ on the boundary of $B$ (that is the origin). The set then contains 4 roots and 4 critical points. This seems to be true at least for polynomials with randomly chosen roots. A more complicated example is shown below. What is nice about the formulation is that it seems somewhat more tractable: the function $g$ maps the set $B$ to the unit disk and the boundary to the boundary. So the question is really whether $g(z)$ and $g(z) + z g'(z)$ have the same number of roots. I like this formulation because it is much closer to the usual Rouche-type questions and seems a little bit easier to approach (if it were true). REPLY [10 votes]: It is indeed true that among random polynomials only a very tiny fraction of them are counterexamples. Even after a few days my program found only very few examples among thousands of tested cases. In all findings it turned out that the critical point which is supposed to lie outside of $B$ actually is very close to the border of $B$. The following is a degree $4$ example which we check purely algebraically without worrying about numerical issues. For $S=\{10, 53-21i, 31-67i, -\frac{41483090}{1864897}-\frac{121204944}{1864897}i\}$ set $g(z)=\prod_{u\in S}(1-z/u)$. Then in the same coloring as in the question the picture looks as follows: We prove that all four roots are inside $B$ by picking a point $\beta=15-50i$ and checking that the (green) open straight lines connecting $\beta$ with the zeros $\alpha$ of $g(z)$ do not intersect the curve $|g(z)|=1$, and neither does the (magenta) line connecting $\beta$ and $0$. We do this using Sturm sequences as follows: Pick points $\alpha$ and $\beta$ with exact coordinates in $\mathbb Q(i)$. Let $F(t)$ be the real polynomial $F(t)=g(t\alpha+(1-t)\beta)\bar g(t\bar\alpha+(1-t)\bar\beta)-1$. So the straight open line from $\alpha$ to $\beta$ lies completely in a connected component of the set $A$ of those $z$ with $|g(z)|<1$ if $F(t)$ hat no real root $01$, so $\gamma$ lies outside $B$. Here is the SageMath code which algebraically verifies the example: k. = QuadraticField(-1) ki. = k[] def conj(f): """ complex conjugate of polynomial f """ return ki(str(f).replace('^', '**').replace('I', '-I')) def line_in_B(g, a, b): """ check if the line [a, b] intersects curve |g(z)=1| """ fab = g(z=t*a+(1-t)*b) f = fab*conj(fab)-1 f_pari = pari(PolynomialRing(QQ, 'T')(f.polynomial(t))) return f_pari.polsturm([0,1]) == 0 roots = [10, -21*I + 53, -67*I + 31, -121204944/1864897*I - 41483090/1864897] g = prod(r-z for r in roots)/prod(roots) b = 15-50*I # Check if the lines connecting b with the roots a of g(z) are in B: for a in roots: print(line_in_B(g, a, b)) # Check that a critical point lies outside B crit = -54*I - 8 print((z*g).derivative(z)(z=crit) == 0) # crit is critical point critval = g(z=crit) print(critval*conj(critval) > 1) There are also real counterexamples $g(z)\in\mathbb R[z]$. The images for the degree $6$ and degree $7$ polynomials $\frac{1}{244925}(z^2 - 8z + 97)(z^2 + 2z + 101)(z^2 + 8z + 25)$ and $\frac{1}{12873250}(z^2 + 28z + 221)(z^2 + 16z + 233)(z^2 + 4z + 125)(z + 2)$ are: Furthermore, there are counterexamples where the set $A$ is disconnected, like $g(z)=\prod_{u\in S}(1-z/u)$ for $S=\{12 i - 20, -21 i + 1, -20 i - 12, -15 i + 20, -10 i - 18, -3 i + 4, 9 i + 7\}$ or $g(z)=\prod_{u\in S}(1-z/u)$ for $S=\{-7 i - 10, -8 i + 8, -7 i + 5, -5 i - 1, -4 i - 8, -4 i + 5, 6 i + 8, 7 i + 1, 7 i + 6\}$<|endoftext|> TITLE: List of problems for graduate topics? QUESTION [11 upvotes]: When I study a new topic, I never feel satisfied until I have spent some time solving a long list of problems. I am looking for either a problem book or a list of problems on graduate math topics. While there is an abundance of problem books on undergraduate math topics (such as various websites on quals or books like Berkeley Problems in Mathematics), there seems to be fewer books at the graduate level with a lot of problems. There are books like Evan's PDE book or do Carmo's Riemannian Geometry book which has a good number of problems, but again, I feel like they are in the minority. The closest thing to what I am looking for is the Cambridge Tripos III. To clarify, the following is what I am looking for: A book with ≥ 10 problems for a particular topic. By "graduate topic," I mean anything that requires standard undergraduate curriculum (single/multivariate calculus, basic/Fourier analysis, ODEs, linear/basic algebra, point set topology, basic manifold theory, curves and surfaces, say) as a prerequisite. I am particularly interested in problem books for "advanced topics" whose prerequisites are standard graduate topics (algebraic/differential topology/geometry, measure theory, real/complex analysis, commutative algebra, representation theory, say). REPLY [7 votes]: László Lovász, Combinatorial Problems and Exercises: Second Edition. Over 600 pages, divided into Problems, Hints, and Solutions.<|endoftext|> TITLE: What is the largest subcategory $C$ of a module category over an Artin algebra, such that $C$ is Krull-Schmidt (and abelian)? Does $C$ exist? QUESTION [9 upvotes]: Let $A$ be an Artin algebra, $\text{Mod}\,A$ the category of $A$-modules and $\text{mod}\,A$ the category of finitely generated $A$-modules. It is well-known that $\text{mod}\,A$ is a Krull-Schmidt category. Can we find a larger full subcategory $\mathcal{C}$ of $\text{Mod}\,A$, such that it remains a Krull-Schmidt category? Is there a largest $\mathcal{C}$ and if yes, how does $\mathcal{C}$ look like? What happens if we additionally demand $\mathcal{C}$ to be an abelian category? REPLY [6 votes]: To answer the first question, there is a larger Krull-Schmidt category than $\text{mod}\,A$ unless $A$ has finite representation type. Every indecomposable pure-injective module has local endomorphism ring, and an Artin algebra of infinite representation type has an infinitely generated indecomposable pure-injective module. So you could take $\mathcal{C}$ to be the full subcategory of finite direct sums of pure-injective modules. [The claims above about pure-injective modules can be found, for example, as Theorem 4.3.43 and Theorem 5.3.40 of 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.] I doubt this is often maximal. As Uriya First pointed out in comments, the maximal Krull-Schmidt subcategory of $\text{Mod}\,A$ will consist of all finite direct sums of modules with local endomorphism ring. I also doubt that it will often (or ever, maybe) be abelian, for the following reason. If $\mathcal{C}$ is a subcategory of $\text{Mod}\,A$ containing $\text{mod}\,A$ that is abelian, then since $\mathcal{C}$ contains $A$ and $DA$ it follows that exact sequences in $\mathcal{C}$ are exact in $\text{Mod}\,A$, so $\mathcal{C}$ is closed under kernels and cokernels in $\text{Mod}\,A$. Often this can be used to show that if $\mathcal{C}$ contains an infinitely generated module then it contains an infinite direct sum of simple modules, and so cannot be Krull-Schmidt. I don't know if this is always the case, but here's one illustrative case: Suppose $A$ is local, and $\mathcal{C}$ is an abelian subcategory of $\text{Mod}\,A$ that contains an infinitely generated module $X$. Then if $\operatorname{rad}^iX$ is the last nonzero power of the radical, there is a map $X\to X$ whose cokernel is $X/\operatorname{rad}^iX$. By induction, $\mathcal{C}$ contains $X/\operatorname{rad}X$, which is an infinite direct sum of simple modules.<|endoftext|> TITLE: Is the tensor product of distributions a continuous bilinear map with respect to the weak topology? QUESTION [6 upvotes]: Let $X$ and $Y$ be smooth manifolds. The map $\mathcal{D}'(X)\times\mathcal{D}'(Y)\to\mathcal{D}'(X\times Y)$ given by $(S,T)\mapsto S\boxtimes T$ is continuous with respect to the strong topology. Is it continuous with respect to the weak-* topology? REPLY [7 votes]: I will use the convention $\mathbb{N}=\{1,2,\ldots\}$ and denote by $s(\mathbb{N})$ the space of (real) sequences $(\mu_i)_{i\ge 1}$ of rapid decay, i.e., such that for all integer $k\ge 0$, $$ \|\mu\|_k:=\sup_{i}i^k|\mu_i|\ <\infty\ . $$ We give it the locally convex topology defined by the norms $\|\cdot\|_k$, indexed by integers $k\ge 0$. The dual can be identified with the space $s'(\mathbb{N})$ of sequences $(a_i)_{i\ge 1}$ which grow at most polynomially in $i$. We will write $$ \langle \mu,a\rangle:=\sum_i \mu_i a_i $$ for the duality pairing. We can similarly define the two index generalizations $s(\mathbb{N}^2)$ and $s'(\mathbb{N}^2)$. The analogue of the tensor product of distributions is the map $s'(\mathbb{N})\times s'(\mathbb{N})\rightarrow s'(\mathbb{N}^2)$ given by $(a,b)\mapsto a\otimes b$ where $$ (a\otimes b)_{i,j}:=a_i b_j\ . $$ Saying that this bilinear map is continuous, where all spaces carry the weak-$\ast$ topology amounts to saying that $\forall \rho\in s(\mathbb{N}^2)$, $\exists p, q$, $\exists \mu^{(1)},\ldots,\mu^{(p)},\nu^{(1)},\ldots,\nu^{(q)}\in s(\mathbb{N})$, such that $\forall a,b\in s'(\mathbb{N})$, $$ \left|\sum_{i,j}\rho_{i,j}a_i b_j\right|\le\left(\sum_{m=1}^{p}|\langle\mu^{(m)},a\rangle|\right) \times \left(\sum_{n=1}^{q}|\langle\nu^{(n)},b\rangle|\right)\ . $$ Lemma: Let $\mu^{(1)},\ldots,\mu^{(p)}$ and $\mu$ be elements of $s(\mathbb{N})$. Suppose that for all $a\in s'(\mathbb{N})$, if $\langle\mu^{(1)},a\rangle=\cdots=\langle\mu^{(p)},a\rangle=0$ then $\langle\mu,a\rangle=0$. This hypothesis implies that $\mu$ is in the linear span of $\mu^{(1)},\ldots,\mu^{(p)}$. For the proof, package the $\mu$'s into a $(p+1)\times\infty$ matrix and do row operations to reduce to row echelon form. The above inequality and the lemma immediately show that the rows of the matrix $\rho_{i,j}$ must span a finite-dimensional space. So, pick say $\rho_{i,j}=e^{-i}\delta_{i,j}$ and this proves that the tensor product is not continuous for the weak-$\ast$ topology. Now, as I said in my comment, the correct topology to use is the strong topology. If all spaces carry it then saying that $(a,b)\mapsto a\otimes b$ is continuous amounts to saying $\forall \rho\in s(\mathbb{N}^2)$, $\exists p, q$, $\exists \mu^{(1)},\ldots,\mu^{(p)},\nu^{(1)},\ldots,\nu^{(q)}\in s(\mathbb{N})$, such that $\forall a,b\in s'(\mathbb{N})$, $$ \sum_{i,j}\left|\rho_{i,j}a_i b_j\right|\le\left(\sum_{m=1}^{p}\sum_i|\mu^{(m)}_i a_i|\right) \times \left(\sum_{n=1}^{q}\sum_j|\nu^{(n)}_j b_j|\right)\ . $$ See the difference? One has traded absolute values of sums for sums of absolute values. The statement about the strong topology is true, with $p=q=1$, because given $\rho\in s(\mathbb{N}^2)$, $\exists \mu,\nu\in s(\mathbb{N})$, such that for all $i,j$, $$ |\rho_{i,j}|\le |\mu_i|\times|\nu_j|\ . $$ Finally, using isomorphisms with sequence spaces the above proofs also shows continuity of the tensor product operation $\mathscr{S}'(\mathbb{R})\times\mathscr{S}'(\mathbb{R})\rightarrow \mathscr{S}'(\mathbb{R}^2)$.<|endoftext|> TITLE: Software for recognizing algebraic or D-finite formal power series QUESTION [11 upvotes]: I have a formal power series in one variable that I think might be algebraic (or perhaps just D-finite). Is there software that could help me explore this? By way of comparison, there’s a very simple way to see if a formal power series appears to be rational: for small values of $n$, compute the determinant of the $(n+1)$-by-$(n+1)$ Hankel matrix whose entries are the first $2n+1$ coefficients of the formal power series. If the determinant is 0, then nontrivial elements of the nullspace correspond to possible $n$th order recurrence relations. (I’m including the combinatorics tag since this sort of pattern-finding is sometimes an important early step in a combinatorial research project.) REPLY [9 votes]: Fricas is good at that. It can be accessed via sage, once installed. sage: L=[catalan_number(i) for i in range(20)] sage: fricas.guessHolo(L) [ n 2 , [[x ]f(x): (4 x - x)f (x) + (2 x - 1)f(x) + 1 = 0, 2 3 4 f(x) = 1 + x + 2 x + 5 x + O(x )] ] sage: and also sage: fricas.guessAlg(L) n 2 2 3 4 [[[x ]f(x): x f(x) - f(x) + 1 = 0, f(x) = 1 + x + 2 x + 5 x + O(x )]]<|endoftext|> TITLE: Endomorphisms of Artinian modules QUESTION [5 upvotes]: The following claim is from a paper [On the moduli spaces of bundles on K3 surfaces, I, p. 358] of Mukai. Consider an artinian module $\mathrm{M}$ over a local ring, and let $\mathrm{M}_0$ be the submodule of all $x\in\mathrm{M}$ annhilated by the maximal ideal of the local ring. Then every endomorphism of $\mathrm{M}$ preserves $\mathrm{M}_0$, and the natural map $$\mathrm{Hom}(\mathrm{M},\mathrm{M})\rightarrow\mathrm{Hom}(\mathrm{M}/\mathrm{M}_0,\mathrm{M}/\mathrm{M}_0)$$ is surjective. I can't quite see why this map is surjective - this is not justified in the paper, so I may be missing something obvious here (a comment may be enough). REPLY [10 votes]: This is false. Take $A=k[x,y]/(x^2,y^2,xy)$ and $M=\omega_A$. Then $\operatorname{Hom}(\omega,\omega)$ has length 3 and $\operatorname{Hom}(M/M_0,M/M_0)$ has length 4, so the natural map cannot be surjective.<|endoftext|> TITLE: Spectrum $E$ with $H^\bullet(E,\mathbb{Z}/2)=\mathcal{A}//\mathcal{A}(n)$ QUESTION [10 upvotes]: Let $\mathcal{A}$ be the Steenrod Algebra and $\mathcal{A}(n)$ be the subalgebra generated by $Sq^1, Sq^{2}, Sq^{2^2},\ldots, Sq^{2^n}$. It is known that $H^*(H\mathbb{Z},\mathbb{Z}/2)=\mathcal{A}//\mathcal{A}(0)$, $H^*(ko,\mathbb{Z}/2)=\mathcal{A}//\mathcal{A}(1)$, $H^*(tmf,\mathbb{Z}/2)=\mathcal{A}//\mathcal{A}(2)$. Now, I have read somewhere that it is known that there is no spectrum $E$ such that $H^*(E,\mathbb{Z}/2)=\mathcal{A}//\mathcal{A}(n)$ with $n\ge 3$. Where can I find the proof? REPLY [11 votes]: This is the same answer as Tyler's but formulated in a different way. In the Adams spectral sequence for the sphere, there is a differential $d_2(h_{n + 1}) = h_0 h_n^2$ for $n \geq 3$. If such a spectrum $E$ existed, then the inclusion of the bottom cell $S \hookrightarrow E$ induces the obvious map $A/\!/A(n) \twoheadrightarrow k$. By inspecting the cobar complexes for $A$ and $A(n)$, we see that the image of $h_{n + 1}$ is trivial in $\operatorname{Ext}_A(A/\!/A(n), k) = \operatorname{Ext}_{A(n)}(k, k)$ but that of $h_0 h_n^2$ is not. This is a contradiction.<|endoftext|> TITLE: Residual finiteness for modules over group rings QUESTION [9 upvotes]: Let $G$ be a finitely generated residually finite group and let $M$ be a finitely generated $\mathbb{Z}[G]$-module. Question: Must $M$ be residually finite in the sense that for all nonzero $x \in M$, there exists some submodule $N$ of $M$ such that $x \notin N$ and $M/N$ is finite? If this is not true in general, is it true if $G$ is also assumed to be nilpotent? REPLY [4 votes]: Here is another example with different groups. Formanek showed that the group ring over any field of a free product of non-trivial groups (and not both order 2) is primitive, has a faithful simple module. That means that $\mathbb F_pG$ is primitive whenever $G=A\ast B$. And of course $G$ is residually finite if both $A$ and $B$ are. Obviously a faithful simple $\mathbb F_pG$-module is cyclic as a $\mathbb ZG$-module and since $G$ is infinite, it cannot be finite. So this gives lots of examples including the free group on two generators.<|endoftext|> TITLE: Category with binary biproducts but no zero morphism QUESTION [5 upvotes]: Is there a category with binary biproducts but no zero morphism? I'm wondering if the definition of biproducts as objects that are simultaneously products and coproducts that obey some identities on the projections/injections is 'different' than the definition involving a zero object. Whenever a zero object exists the definitions are trivially equivalent, and an empty biproduct is a zero object, so we would need a category with binary biproducts but no nullary ones. Any assistance is appreciated. REPLY [2 votes]: Karvonen has an article on arXiv showing that if a category has all binary biproducts (using the alternative definition Tom mentioned) then it has zero morphisms (Corollary 3.3).<|endoftext|> TITLE: Is there a monoidal analogue of equalizers? QUESTION [9 upvotes]: There are three different kinds of finite limits in categories: terminal objects, binary products, and equalizers. In a category $C$, these define functors $1_{C},\times,\mathrm{Eq}\colon\mathrm{Fun}(I,C)\to C$ where $I=\emptyset,\{\bullet\ \ \bullet\}$, and $\{\bullet\rightrightarrows\bullet\}$ respectively. Monoidal categories generalise the first two in that we now have a functor $1_{C}$ from $\mathcal{C}^{\emptyset}=*$ to $\mathcal{C}$ and a functor $\otimes_C$ from $C^{\{\bullet\ \ \bullet\}}=C$ to $C$, i.e.\ functors \begin{align*} 1_C &\colon * \to \mathcal{C}\\ \otimes_C &\colon \mathcal{C}\times\mathcal{C} \to \mathcal{C} \end{align*} together with associativity and unitality natural isomorphisms satisfying compatibility conditions. What about equalizers? Has the notion of a category $C$ equipped with a functor $\rm{Eq}\colon\rm{Fun}(\{\bullet\rightrightarrows\bullet\},C)\to C$, a unit functor, and unitality/associativity natural isomorphisms satisfying coherence conditions been studied before? Moreover, are there any examples of such structures "found in nature"? REPLY [3 votes]: Here is a way to write down what a symmetric monoidal category is starting just from the idea of finite (co)products. We have $$SymMonCat = Fun^\times_{(2,1)}(Span^{fin}_{(2,1)}, Cat_{(2,1)})$$ That is, a symmetric monoidal category is equivalent to the data of a finite-product-preserving $(2,1)$-functor from the $(2,1)$-category $Span^{fin}_{(2,1)}$ of finite sets and spans between them, to the $(2,1)$-category $Cat_{(2,1)}$ of categories. I can think of at least two ways to understand the significance of $Span^{fin}_{(2,1)}$ from the point of view of finite (co)products. There is a functor $Span^{fin}_{(2,1)} \to Cat^{\amalg}_{(2,1)}$ (where $Cat^{\amalg}_{(2,1)}$ is the $(2,1)$-category of categories with finite coproducts and finite-coproduct preserving functors). The functor sends $n \mapsto Set^{fin}/n$. This functor is fully faithful. Moreover, $Set^{fin}/n$ is itself the free coproduct completion of the discrete category $n$. So from this perspective, we have boiled it all down to finite (co)products, but the picture is a bit odd and I'm not sure how to complete the thought from this perspective. Alternatively, Note that the $(2,2)$-category $Span^{fin}_{(2,2)}$ of finite sets, spans between them, and maps of spans has the following universal property. For any 2-category $\mathcal K$ with finite products, we have that the 2-category $Fun_{(2,2)}^\times(Span^{fin}_{(2,2)}, \mathcal K)$ of all finite-product preserving 2-functors, is equivalent to $FinProd(\mathcal K)$, the 2-category of objects of $\mathcal K$ which internally have finite products. (2) suggests the following generalization. Let $J$ be a set of small categories, and let $2Cat(J)$ be the category of 2-categories with $J$-limits. I believe that the functor $Comp_J: 2Cat(J) \to 2Cat$ carrying $\mathcal K$ to the 2-category of objects in $\mathcal K$ which internally have $J$-limits, is corepresentable by some $\mathcal J \in 2Cat(J)$. Define a $J$-symmetric monoidal category to be an object of $Fun^J_{(2,1)}(\mathcal J_{(2,1)}, Cat)$, where $\mathcal J_{(2,1)}$ is the maximal sub $(2,1)$-category of the $(2,2)$-category $\mathcal J$, and $Fun^J_{(2,1)}$ means we take $(2,1)$-functors which preserve $J$-limits in the $(2,1)$-categorical sense. Then in the case where $J$ consists of the finite discrete categories, a $J$-symmetric monoidal category is a usual symmetric monoidal category. When $J$ consists of just the empty category, a $J$-symmetric monoidal category is an $E_0$-category. You could try out other classes of $J$, such as equalizers, but I'm not sure what you get!<|endoftext|> TITLE: Representing a symmetric polynomial as a conical sum of squares QUESTION [6 upvotes]: This question in inspired by the recent solution to another question. The following inequality for monomial symmetric polynomials in 4 positive variables $x_1,x_2,x_3,x_4$: $$m_{(4, 3, 2, 1)} + m_{(4, 4, 2)} \geq 3m_{(4, 2, 2, 2)} + m_{(4, 3, 3)} + 2 m_{(4, 4, 1, 1)}$$ seems to be hard to prove directly, but easily follows from a rather unobvious identity: \begin{split} &m_{(4, 3, 2, 1)} + m_{(4, 4, 2)} - (3m_{(4, 2, 2, 2)} + m_{(4, 3, 3)} + 2 m_{(4, 4, 1, 1)})\\ =&\frac12\left( f(x_1,x_2,x_3,x_4)^2 + f(x_1,x_3,x_2,x_4)^2 + f(x_2,x_3,x_1,x_4)^2 \right), \end{split} with $$f(x,y,z,t) := (x-y) (z-t) (xy(z+t) - (x+y)zt).$$ Relatedly, my questions are Q1: How to find representation of a given symmetric polynomial in a fixed number of variables as a conical sum of squares of polynomials if one exists? Q2: Is it any simpler if we restrict Q1 to the case when all the coefficients in the sum are the same, and the polynomials being squared represent the same polynomial with just somehow permuted arguments? (like in the above example) Q3: Is there a direct "obvious" proof of the above inequality. REPLY [3 votes]: For four variables $a$, $b$ $c$ and $d$ we'll use the following natation. $$\sum_{sym}a=6(a+b+c+d),$$ $$\sum_{sym}ab=4(ab+ac+bc+ad+bd+cd),$$ $$\sum_{sym}a^2b=2\sum_{cyc}a^2(b+c+d).$$ Number of addends is equal to $4!=24$. Now, let $x_1=\frac{1}{a},$ $x_2=\frac{1}{b},$ $x_3=\frac{1}{c}$ and $x_4=\frac{1}{d}.$ Thus, we need to prove that: $$\sum_{sym}\left(\frac{2}{a^4b^3c^2d}+\frac{1}{a^4b^4c^2}-\frac{1}{a^4b^2c^2d^2}-\frac{1}{a^4b^3c^3}-\frac{1}{a^4b^4cd}\right)\geq0$$ or $$\sum_{sym}(2a^3b^2c+a^4b^2-a^2b^2c^2-a^4bc-a^3b^3)\geq0$$ which is $$\sum_{sym}(a-b)^2(a-c)^2(b-c)^2\geq0$$ because for three variables $a$, $b$ and $c$ we have: $$\prod_{cyc}(a-b)^2=\sum_{cyc}(a^4b^2+a^4c^2-2a^3b^3-2a^4bc+2a^3b^2c+2a^3c^2b-2a^2b^2c^2)=$$ $$=\sum_{sym}(a^4b^2-a^4bc-a^3b^3+2a^3b^2c-a^2b^2c^2).$$ The last identity we can get by the following way: $$\sum_{cyc}(a^4b^2+a^4c^2-2a^4bc-2a^3b^3+2a^3b^2c+2a^3c^2b-2a^2b^2c^2)=$$ $$=\sum_{cyc}(c^4a^2-2c^4ab+c^4b^2-c^3(a^3+b^3-a^2b-ab^2)+abc(a^2c+b^2c-2abc))=$$ $$=\sum_{cyc}(a-b)^2(c^4-c^3(a+b)+c^2ab)=\sum_{cyc}(a-b)^2c^2(c-a)(c-b)=$$ $$=\sum_{cyc}(a-b)(b-c)(c-a)c^2(b-a)=\prod_{cyc}(a-b)\sum_{cyc}(a^2c-a^2b)=\prod_{cyc}(a-b)^2.$$ Your linked inequality: For positive variables $x_1$, $x_2$, $x_3$ and $x_4$ we have $$24m_{(3, 3, 3, 1)} + 3 m_{(4, 3, 2, 1)} + 9 m_{(4, 4, 2)} \geq 12 m_{(3, 3, 2, 2)} + 3 m_{(4, 2, 2, 2)} + 9 m_{(4, 3, 3)} + 14 m_{(4, 4, 1, 1)}$$ We can prove by the following way. Again, let $x_1=\frac{1}{a},$ $x_2=\frac{1}{b},$ $x_3=\frac{1}{c}$ and $x_4=\frac{1}{d}.$ Thus, in the previous notation we need to prove that: $$\sum_{sym}(9a^4b^2-9a^4bc+8a^4bcd+6a^3b^2c-6a^2b^2cd-7a^3b^3-a^2b^2c^2)\geq0.$$ Now, let $$a+b+c+d=4u,$$ $$ab+ac+bc+ad+bd+cd=6v^2,$$$$abc+abd+acd+bcd=4w^3$$ and $$abcd=t^4.$$ Thus, we need to prove that: $$3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6\geq0.$$ Now, $a$, $b$, $c$ and $d$ are roots of the polynomial $$f(p)=(p-a)(p-b)(p-c)(p-d)$$ or $$p^4-4up^3+6v^2p^2-4w^3p+t^4,$$ which by the Rolle's theorem says that the polynomial $$f'(p)=4p^3-12up^2+12v^2p-4w^3$$ has three roots. Let $x$, $y$ and $z$ be these roots. Thus, $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$, which says that it's enough to prove our inequality for three variables $x$, $y$ and $z$. But it's really incredible here that: $$(x-y)^2(x-z)^2(y-z)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ and we are done!<|endoftext|> TITLE: Distribution over Penrose Tilings? QUESTION [5 upvotes]: The set of possible kit-and-dart Penrose tilings is uncountably infinite. It would be very helpful to have some natural probability distribution $\mu$ over this set; such a distribution would allow us to make statements like "Viewed as an infinite planar graph, the expected degree of a Penrose tiling is...." and be a little more precise (see, e.g., papers like this). To address some symmetry issues, let us restrict to the set of Penrose tilings with a tile corner at the origin and an edge between tiles along the positive x-axis ("an edge laying due east"). We will refer to this (also uncountably infinite) set as $\mathcal{P}$. We would really like $\mu$ to cover all of $\mathcal{P}$ in some sense. So, does there exist a probability distribution $\mu$ over $\mathcal{P}$ that satisfies the following? Consider $F$, an arbitrarily large, finite subset of tiles in the plane selected from any Penrose tiling in $\mathcal{P}$. Let $P_F\subseteq \mathcal{P}$ be the set of Penrose tilings that contains that fragment in the same location. Then $\mu(P_F)>0$. If there were an efficient algorithm for sampling from $\mu$, that would be even more helpful, but perhaps I'm hoping for too much. Edit: Per Steven Stadnicki's and Mateusz Kwaśnicki's comment's, I've clarified that the fragment $F$ occurs in the same location. REPLY [5 votes]: The tiling space of the penrose tilings is uniquely ergodic with respect to the translation action. As is any reasonable notion of the 'canonical transversal' (also sometimes called the discrete hull) which is essentially the one you describe as the set of tilings with a vertex at the origin (the action on this set is a little more difficult to describe though and really needs groupoids). This is all in Aperiodic Order, Vol. 1 by Baake and Grimm. By Birkhoff's ergodic theorem then, the unique measure is the one which assigns to the set of tilings with a patch $P$ at the origin, the frequency of that patch $P$ in a Penrose tiling (which is independent of the choice of tiling). These frequencies can all be calculated as entries of the corresponding right Perron-Frobenius eigenvector of the associated substitution matrix (or the matrix of the collared substitution with collaring radius suitably large to contain the patch). It's easier to do this with the Robinson triangles because they give an exact Stone inflation, but the same method does work for the kite-and-dart substitution. Another method for calculating the frequencies is to put everything in terms of the cut-and-project method, in which case the relative volume of the 'acceptance domain' of a particular patch $P$ (compared to the volume of the entire window) is the frequency of the patch $P$.<|endoftext|> TITLE: What's the bijection between reals and infinite sequences of integers? QUESTION [12 upvotes]: In Descriptive Set Theory we often see the notion of encoding a real as a sequence of integers or natural numbers -- i.e. there obviously is a bijection according to ZF axioms. But how does it look like concretely? Anybody has seen a simple construction? My own approach is by chain-fractions: Let $q\in\mathbb{R}$ be the given real and now define the sequence $(z_i,q_i)$ by $$z_{i+1}=\begin{cases}[q_i]&\text{if } \{q_i\}\leq\frac{1}{2}\\ [q_{i}]+1&\text{else} \end{cases}$$ $$q_{i+1}=(q_i-z_{i+1})^{-1}$$ where $[q]$ is the next lower integer and $\{q\}=q-[q]$. (Hence $(q_i-z_{i+1})\in(-\frac12,\frac12]$ thereby absolute value of $q_{i+1}$, its reciprocal, is bigger than 2.) Now my bijection is mapping $q$ to the sequence: $$m_i=\begin{cases}z_i-2&z_i>0, i>1\\ z_i&i=1\\ z_i+2&z_i<0, i>1\end{cases}$$ with $i$ starting at 1 and above $q_0$ becomes the initial $q$. And the inverse of my bijection just calculates the chain-fraction:$q_{i-1}\in(z_i-\frac12,z_i+\frac12]$ with $q_{i-1}=z_i+q_i^{-1}$ step-wise narrowing down the real by a sequence of intervals each containing the next. is there a paper or book covering my example? any other simple constructions? REPLY [43 votes]: [Note: this answer uses the convention where $\mathbb{N} := \{ 0, 1, 2, \dots \}$ contains zero.] There's an elegant explicit order-preserving bijection between the Baire space $\mathbb{N}^{\mathbb{N}}$ (under lexicographical order) and $\mathbb{R}_{\geq 0}$ (under the usual order) described here. In particular, we define the image of: $$ (a_0, a_1, a_2, a_3, \dots) $$ to be the generalised continued fraction: $$ a_0 + \cfrac{1}{1 + \cfrac{1}{a_1 + \cfrac{1}{1 + \cfrac{1}{a_2 + \ddots}}}} $$ This order-preserving bijection shows that $\mathbb{R}_{\geq 0}$ and $\mathbb{N}^{\mathbb{N}}$ are not only isomorphic as sets (i.e. equinumerous), but also isomorphic as totally-ordered sets. Topologically, this bijection from $\mathbb{N}^{\mathbb{N}}$ to $\mathbb{R}_{\geq 0}$ is continuous, meaning that every open subset of the nonnegative reals corresponds to an open subset of Baire space. The converse is not quite true (if it were, the two spaces would be homeomorphic, which they're not); continuity of the inverse map fails exactly at the positive rationals.<|endoftext|> TITLE: Littlewood-Paley theory and dense property of Sobolev spaces QUESTION [6 upvotes]: I'm learning the Littlewood-Paley theory by myself and I encounter the following claim: Pick a smooth function $\chi$ such that: $$\chi(\xi) = \begin{cases} 1 &|\xi| \leq \frac{1}{2}\\ 0 &|\xi| \geq 1 \end{cases}$$ Based on the function $\xi$ picked above, let's define an operator $S_{k} \ (\forall \ k \in \mathbb{Z})$ acting on an arbitrary function $f$ as follows: $$S_{k}f(x) = \left(\chi\Big(\tfrac{\xi}{2^{k}}\Big)\hat{f}(\xi)\right)^{\vee}(x)$$ where $\hat{}$ and $\vee{}$ above denotes Fourier transform and inverse Fourier transform, respectively. Moreover, based on $\xi$, we can also define the Littlewood-Paley projection operators $P_{l} \ (\forall \ l \in \mathbb{Z})$. Now for any $s \in \mathbb{R}$ and any $k \in \mathbb{Z}^{+} \cup \{0\}$, we have the following inequality (equivalence of norms) for any function $f$ in the Sobolev space $H^s$: $$c^{-1}\|f-S_{k}f\|_{H^{s}}^2 \leq \sum_{l \geq k}2^{2ls}\|P_{l}f\|_{L^2}^2 \leq c\|f-S_{k}f\|_{H^{s}}^2$$ Above $c=c(s) \geq 1$ is some constant dependent on $s$. From the inequality above (equivalence of norms), we can further deduce that $\cap_{t \in \mathbb{R}}H^{t}$ is dense in $H^{s}$. I roughly get the idea of deducing the dense property from the inequality listed above. However, I'm bit stuck with proving the inequality... it feels that $S_{k}f$ is some other form of "projection", so I guess probably "almost orthogonality" will be useful here? Any help/hint will be appreciated! REPLY [2 votes]: By Plancherel $$ \| f - S_k f\|_{H^s}^2 \approx \int (1 + |\xi|^2)^{s} (1 - \chi(2^{-k}\xi))^2 |\hat{f}|^2 ~d\xi $$ By almost orthogonality you can write $$ (1 - \chi(2^{-k}\xi))^2 \approx \sum_{\ell = k}^{\infty} (\underbrace{\chi(2^{-\ell-1}\xi) - \chi(2^{-\ell}\xi)}_{\phi_\ell(\xi)})^2 $$ as the telescoping sum. On the support of $\phi_\ell(\xi)$, you have that $$ |\xi| \approx 2^{\ell} $$ So $$ \| f - S_k f\|_{H^s}^2 \approx \int \sum_{\ell = k}^\infty (1 + 2^{2s\ell}) \phi_\ell^2 |\hat{f}|^2 ~d\xi $$ As each $\ell \geq k \geq 0$ you have that $1 + 2^{s\ell} \approx 2^{s\ell}$ and so you finally get $$ \| f- S_k f\|_{H^s}^2 \approx \sum_{\ell = k}^\infty 2^{2sl} \| P_\ell f\|_{L^2}^2 $$<|endoftext|> TITLE: Properties (T) and (FA) QUESTION [7 upvotes]: I have been thinking a lot recently about Property (T) and Property (FA) for discrete groups. I understand that the prior implies the latter, but not the other way around, and I have also seen one or two ad-hoc examples that illustrate this failure. I was just wondering if anything else is known. Is there a sense for how "rare" (FA)-but-no-(T) is? For instance, is there a theorem that provides a sufficient condition for an (FA) group to have (T)? Otherwise, is there a place I can see a large collection of (FA)-but-no-(T) groups? Thanks for your time. REPLY [7 votes]: Here is a point of view which justifies why Property $(FA)$ is a very particular case of Property $(T)$. First, Chatterji-Drutu-Haglund proved that: Theorem: A discrete group has $(T)$ iff all its isometric actions on metric median spaces have bounded orbits. A metric space $(X,d)$ is median if, for every triple $x,y,z \in X$, there exists one and only one point $m \in X$ satisfying $$\left\{ \begin{array}{l} d(x,y)=d(x,m)+d(m,y) \\ d(x,z)=d(x,m)+d(m,z) \\ d(y,z)=d(y,m)+d(m,z) \end{array} \right.$$ If $(X,d)$ is geodesic, it amounts to saying that $m$ is the unique point that belongs to the intersection between three geodesics connecing $x,y,z$. Therefore, you can "discretise" Property $(T)$ by introducing: Definition: A group has Property $(FW)$ if all its actions by automorphisms on median graphs have bounded orbits. As proved independently by Chepoï and Roller, median graphs coincide with one-skeleta of CAT(0) cube complexes, so you can replace "median graphs" with "CAT(0) cube complexes" in the previous definition. As a consequence, you can introduce a hierarchy of properties: Definition: Given an $n \geq 1$, a group has Property $(FW_n)$ if all its actions by automorphisms on $n$-dimensional CAT(0) cube complexes have bounded orbits. If you want, you can also introduce $(FW_\infty)$ for finite-dimensional CAT(0) cube complexes or $(FW_\omega)$ for CAT(0) cube complexes without infinite cubes. In each case, there is something interesting to say. But the key point is that one-dimensional CAT(0) cube complexes coincide with simplicial trees, so $(FW_1)$ actually coincides with $(FA)$. Conclusion: Property $(FA)$ is the one-dimensional discrete version of Property $(T)$. $$\begin{array}{ccc} (FA) & & (T) \\ \Updownarrow & & \Updownarrow \\ (FW_1) & \Leftarrow \cdots \Leftarrow (FW_n) \Leftarrow \cdots \Leftarrow (FW_\infty) \Leftarrow (FW_\omega) \Leftarrow(FW) \Leftarrow & (FMed) \end{array}$$<|endoftext|> TITLE: A remarkable identity involving $\chi^2$ random variables QUESTION [11 upvotes]: In the process of computing inclusion constants for the complex matrix cube (which is a free spectrahedron), the following identity was proven: for all $n \geq 1$, $$\mathbb E \Big| \sum_{i=1}^{2n} x_i^2 - \sum_{j=1}^{2n} y_j^2 \Big| = \mathbb E \Big| \sum_{i=1}^{2n} x_i^2 - \sum_{j=1}^{2n-2} y_j^2 \Big| = 4^{1-n}n \binom{2n}{n},$$ where $x_i$,$y_j$ are i.i.d. standard real Gaussian random variables. The proof we have at this moment is by using the explicit form of the density of the difference of two $\chi^2$ random variables, see here and also Klar, Bernhard, A note on gamma difference distributions, ZBL07183251. Since the result is so simple, there should be a more direct and more insightful proof of it. Question 1: give an easy, conceptual proof of the identity above. Consider now the function $$k \mapsto \mathbb E \Big| \sum_{i=1}^{2k} x_i^2 - \sum_{j=1}^{2(n-k)} y_j^2 \Big|$$ from $\{0,1,\ldots, n\} \to \mathbb R_+$; as above, the $x_i$ and $y_j$ are standard i.i.d. Gaussians. Question 2: Show that the function above is unimodular, and that its minimum is attained at $k = \lfloor n/2 \rfloor$. There is a pretty involved proof of the second fact above in Helton, J. William; Klep, Igor; McCullough, Scott; Schweighofer, Markus, Dilations, linear matrix inequalities, the matrix cube problem and beta distributions, Memoirs of the American Mathematical Society 1232. Providence, RI: American Mathematical Society (AMS) (ISBN 978-1-4704-3455-7/pbk; 978-1-4704-4947-6/ebook). vi, 106 p. (2019). ZBL1447.47009. Question 3: Does the claim in ``Question 2'' hold for arbitrary probability distributions? Any help or insight about these questions would be appreciated! REPLY [10 votes]: I think I found an elementary proof of Question 2/3 for arbitrary probability distributions. In fact, it is not required that the components in the sums are squares, but general i.i.d. non-negative random variables work. Further, the requirement that both sums have an even number of terms ($2k$ and $2(n-k)$ in the question) is not required. Let $X_1, ..., X_N$ be i.i.d. non-negative, integrable random variables. Let $T_k := \mathbb{E}[|\sum_{i=1}^k X_i - \sum_{i=k+1}^N X_i|]$. Lemma 1: For $k \geq \lfloor N/2\rfloor$ it holds $T_{k+1} \geq T_k$. By symmetry (for $k \leq \lfloor N/2\rfloor$) this Lemma yields the question. To prove Lemma 1, we require the two following supplementary statements Lemma 2: Let $C$ be a symmetric, integrable random variable, and $a, b\in\mathbb{R}$, $|b| \geq |a|$. Then $\mathbb{E}[|C+b|] \geq \mathbb{E}[|C+a|]$. Hereby, $C$ being symmetric means that $C$ and $-C$ have the same distribution. Lemma 3: Let $C$ be a symmetric, integrable random variable and $A$, $B$ non-negative and integrable random variables and assume that $A, B, C$ are independent. Then $\mathbb{E}[|C+A-B|] \leq \mathbb{E}[|C+A+B|]$. Proof of Lemma 2: Without loss of generality, by symmetry of $C$, let $0\leq a\leq b$. Then one calculates \begin{align*} \mathbb{E}[|C+a|] &= \mathbb{E}[\mathbb{1}_{a \geq |C|} (C+a) + \mathbb{1}_{a < |C|} (\mathbb{1}_{C > 0} + \mathbb{1}_{C < 0}) |C+a|]\\ &= \mathbb{P}(|C| \leq a) a + \mathbb{E}[\mathbb{1}_{a < |C|} (\mathbb{1}_{C > 0} (C+a) + \mathbb{1}_{C < 0} (-C-a)]\\ &=\mathbb{P}(|C| \leq a) a + \mathbb{E}[\mathbb{1}_{|C| > a} |C|] \\ &=\mathbb{E}[\max\{a, |C|\}] \end{align*} and this term is obviously increasing in $a$. (in general, it simply holds $\mathbb{E}[|C+a|] = \mathbb{E}[\max\{|C|, |a|\}]$, which can also be proved via the identity $|C+a| + |-C+a| = 2\max\{|C|, |a|\}$ as pointed out by Fedor.) Proof of Lemma 3: Let $C \sim \mu, A \sim \nu, B \sim \theta$. Then $$ \mathbb{E}[|C+A-B| - |C+A+B|] = \int \int \Big(\int |c+a-b| - |c+a+b| \mu(dc)\Big)\nu(da)\theta(db) \leq 0 $$ since the term inbetween the large brackets is non-positive since $|a+b| \geq |a-b|$ (since $a, b \geq 0$) and by Lemma 2. Proof of Lemma 1: Let $C := \sum_{i=1}^{N-k-1} X_i - \sum_{i=k+1}^N X_i$ and note that $C$ is symmetric, $A := \sum_{i=N-k}^k X_i$ (where the sum is understood to be 0 if $k < N-k$), and $B := X_{k+1}$. Note that, for $k \geq \lfloor \frac{N}{2}\rfloor$, it holds $T_k = \mathbb{E}[|C+A-B|]$ and $T_{k+1} = \mathbb{E}[|C+A+B|]$. The claim now follows from Lemma 3.<|endoftext|> TITLE: Presentations of mapping class groups in dimension $3$ QUESTION [10 upvotes]: For any closed oriented surface $M$, its mapping class group $MCG(M)$ can be generated by Dehn twists along certain curves on $M$. A presentation for the group $MCG(M)$ was found in [1] and then simplified in [2]. How about in dimension $3$? The first question is that, unlike in dimension $2$ in which the classification of spaces is easy (genus and number of boundaries), in dimension $3$ the classification is complicated. One either uses Thurston's geometrization Lickorish and Kirby's presentation using links Neither of them is easy, so I'd expect it much harder to get presentation for the mapping class group. Question: Nevertheless, is there any presentation known? In the first view-point, the closest answer I have seen is this, in which Allen Hatcher claims that the MCG of three-manifolds are essentially known [3]. There, the natural map from MCG to $Out(\pi_1(M))$ is considered. While the kernel of this map is understood in the nonprime case [3. section 2], it is not necessarily onto. Even if it's onto, we still do not have a presentation for MCG. In the second view-point, I have not heard of any result. Reference [1] A presentation for the mapping class group of a closed orientable surface-[Hatcher and Thurston] [2] A simple presentation for the mapping class group of an orientable surface-[Bronislaw Wajnryb] [3] Stabilization for mapping class groups of 3-manifolds-[Allen Hatcher and Nathalie wahl] REPLY [2 votes]: For certain 3-manifolds (irreducible), if you are willing to take finite index subgroups, the description is relatively easy, and has to do with dehn-twists too. Let me add the following paper by McCullough which was useful to me when considering this question: https://projecteuclid.org/journals/journal-of-differential-geometry/volume-33/issue-1/Virtually-geometrically-finite-mapping-class-groups-of-3-manifolds/10.4310/jdg/1214446029.full I think this goes in line with the comments by HJRW which are more detailed, but I thought this might be useful.<|endoftext|> TITLE: Mechanical systems with their configuration space being a Lie group QUESTION [8 upvotes]: Cross-posted from Physics.SE In Marsden, Ratiu - Introduction To Mechanics And Symmetry there is a certain focus on reducing cotangent bundles of Lie groups. More precisely, if $G$ is a Lie group, then $T^*G$ is naturally a symplectic manifold. Then, Lie-Poisson reduction provides a Poisson structure on the Lie coalgebra $\mathfrak{g}^*$ and a reduced Hamiltonian $\mathfrak{g}^* \to \mathbb{R}$. In other words, the dynamics can be analyzed on the reduced space $\mathfrak{g}^*$. symplectic reduction allows us to reduce the phase space further and identify the reduced phase space with a coadjoint orbit $\mathcal{O} \subset \mathfrak{g}^*$, where $\mathcal{O}$ is equipped with the KKS symplectic structure. However, apart from the configuration space $G = SO(3)$ which is the configuration space for the rigid body, all the examples provided by Marsden and Ratiu are infinite-dimensional. I failed to come up with any other non-trivial finite-dimensional example, i.e. $\mathbb{R}^n \times (SO(3))^k$ doesn't count. Are there any other naturally occurring mechanical systems, whose configuration space is a Lie group $G$ and that the Hamiltonian $H\colon\,T^*G \to \mathbb{R}$ is $G$-invariant with respect to the natural action of $G$ on $T^*G$? REPLY [2 votes]: If one dimensional $n$ body systems fit your definition of a "naturally occurring mechanical system", then the paper by M.A. Olshanetsky and A.M. Perelomov, "Completely Integrable Hamiltonian Systems Connected with Semisimple Lie Algebras", Inventiones mathematicae (1976) Volume: 37, page 93-108 provides examples apart from just $SO(3)$.<|endoftext|> TITLE: Characterizations of metric trees QUESTION [8 upvotes]: Let $X$ be a geodesic space. Then the following conditions are equivalent: For any $x,y\in X, x \neq y$, there is a unique arc (homeomorphic to the interval $[0,1]$) with endpoints $x$ and $y$. No subset of $X$ is homeomorphic to $S^1$. $X$ is simply connected and its topological dimension (small inductive) equals $1$. Every geodesic triangle is isometric to a tripod. $X$ is $0$-hyperbolic in the sense of Gromov. Intersection of any two closed balls is a closed ball or an empty set. For every Lipschitz maps $\gamma:S^1\to X$ and $\pi:X\to\mathbb{R}^2$, $$ \int_{S^1}(\pi\circ\gamma)^*(x\, dy)=0. $$ A metric space that satisfies any of the above equivalent conditions is known as the metric tree or an $\mathbb{R}$-tree. The problem is that it is very difficult to find a single place where one could find proofs of such equivalences. Thus my question is: Question. Is there a single paper that where one would find seld-contained proofs of such characterizations of metric trees? I do not necessarily mean exactly the same characterizations. Just many equivalent characterizations. I wanted to ask my student to write a survey paper that would cover in particular all such proofs. I think it would be a useful reference for people working in analysis on metric spaces, and for him it would be a good way to learn this material, but if there is already a good reference for such characterizations, then perhaps it would be a pointless task. This is why I am asking that question. REPLY [2 votes]: Some equivalent characterizations of metric trees can be extracted from Theorem 8 of the paper "Trees, tight extensions of metric spaces, and the cohomological dimension of certain groups: A note on combinatorial properties of metric spaces" by Dress (1984) DOI link. The focus of the theorem is more on subsets of metric trees and injective hulls/tight spans, but the proof is self-contained.<|endoftext|> TITLE: Fubini's theorem for Hausdorff measures QUESTION [5 upvotes]: $B\subset \mathbb{R}^2$ is a Borel set. Define the slices $B_x:= \{y \in \mathbb{R}: (x,y) \in B \}$. If $\lambda$ denotes the Lebesgue measure on $\mathbb{R}$, presentations of Fubini's theorem often include that fact that the function $\lambda(B_x)$ is measurable. Question: If $H^s$ denotes the $s$th Hausdorff measure, how do I see that the function $H^s(B_x)$ is measurable? $(\star)$ I came across this while looking at Marstrand's slice theorem in a book. The authors suggest to use a Monotone class argument wherein one would have to show the following If $B=U\times V$ then, $H^s(B_x)= \mathbb{1}_U(x)\cdot H^s(V)$ which is measurable. If $B$ is a finite union of disjoint rectangles then again $H^s(B_x)$ is measurable. If $B_n$ is an increasing family of sets each of which satisfies $(\star)$ then $H^s\left((\cup B_n)_x\right) = H^s(\cup (B_n)_x) = \lim H^s((B_n)_x)$ which is again measurable. If $B_n$ is a decreasing family of sets each of which satisfies $(\star)$, one would like to show the same for $H^s((\cap B_n)_x)$. However, this is equal to $H^s(\cap (B_n)_x)$. This in general won't be $\lim H^s((B_n)_x)$ since we don't know if any of the terms has finite $H^s$ measure. I don't see how to prove the last point in the absence of $\sigma$-finiteness. Am I missing something easy? REPLY [6 votes]: If $s>1$, then clearly $H^s(B_x)=0$ so there is nothing to do. If $s=1$, $H^1$ is just the Lebesgue measure so measurability follows. If $0 TITLE: Books/References for Inequalities that take advantage of orders QUESTION [8 upvotes]: Are there any good references/papers/books that specifically address inequalities that take advantage of orders or monotonicity? I have already browsed through the classical Cauchy-Schwarz Master Class by Steele as well as Inequalities by Hardy, Littlewood, and Polya, and while they were amazing books, I wanted to dive deeper. Thank you very much!! REPLY [5 votes]: There is also Barry Arnold's Majorization and the Lorenz Order: A Brief Introduction published in the Springer Lectures in Statistics series.<|endoftext|> TITLE: Isomorphic endomorphism algebras implies isogenous (for abelian varieties over finite fields)? QUESTION [8 upvotes]: $\newcommand{\F}{\mathbb{F}} \newcommand{\End}{\mathrm{End}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}}$ I would like to know if the following is true: Proposition A : Let $A_1, A_2$ be two abelian varieties over a finite field $k$. If $\End_{\overline k}(A_1) \otimes_{\Z} \Q$ and $\End_{\overline k}(A_2) \otimes_{\Z} \Q$ are isomorphic (as $\Q$-algebras), then $A_1, A_2$ are isogenous over $\overline k$. The converse holds: let $\phi : A_1 \to A_2$ be an isogeny of degree $m$. There is an isogeny $\psi : A_2 \to A_1$ such that $psi \circ \phi = [m]$ (see Poonen "Lectures on rational points", Proposition 4.1.19 ; this is not the dual isogeny!). Define a map $\End_{\overline k}(A_1) \otimes_{\Z} \Q \to \End_{\overline k}(A_2) \otimes_{\Z} \Q$ via $f \longmapsto \dfrac{1}{m} \phi \circ f \circ \psi $. It is an algebra isomorphism. The result holds over $\mathbb C$, see here or Prop. 1.2.17. It holds for elliptic curves over a finite field: the supersingular case can be proved "by hand", using Tate isogeny theorem, while the ordinary case follows from Deuring's work on CM (lifting, etc.). Note that the curves do not need to be isogenous over the base field $k$. I skimmed through the paper Endomorphisms of Abelian Varieties over Finite Fields of Tate, but I did not find such a statement. Maybe one could use Serre–Tate theory instead of Deuring, but it seems to be only available for ordinary abelian varieties. I am not aware of the details here anyway. If the claim does not hold, is there a suitable hypothesis to make it true? REPLY [9 votes]: Here is a counterexample to Proposition A provided by 8-dimensional abelian varieties $A_1$ and $A_2$ over a finite field $F_{p^2}$ where $p$ is any prime that is congruent to $1$ modulo $17$ (e.g., $p=103$). The corresponding endomorphism algebra is the $17$th cyclotomic field $E=Q(\zeta_{17})$. The congruence condition means that $p$ splits completely in $E$. It is known that $E/Q$ is a cyclic extension, the class number of $E$ is $1$ (see ``Introduction to cyclotomic fields" by Larry Washington) and any proper subfield of $E$ is totally real (because $17$ is a Fermat prime). Let $O_E$ be the ring of integers in $E$ and $\iota: E \to E$ be the complex conjugation, which is the only element of order 2 in the cyclic Galois group $G:=Gal(E/Q)$ of order 16=2^4. (In particular, every nontrivial subgroup of $G$ contains $\iota$.) Let S be the 16-element set of maximal ideals $\mathfrak{P}$ of integers $O_E$ with residual characteristic $p$. The group $G$ acts freely transitively on the set $S$ of maximal ideals in $O_E$ and $\Pi_{\mathfrak{P}\in S}\mathfrak{P}=p \cdot O_E$. Let H be the set of ideals $\mathfrak{B}$ of $O_E$ such that $\mathfrak{B}\cdot \iota(\mathfrak{B})=p \cdot O_E$. The set $H$ has $2^8=16^2$ elements. I claim that the natural action of $16$-element $G$ on $H$ is free. Indeed, if it's not free then there is $\mathfrak{B}\in H$ such that $\iota(\mathfrak{B})=\mathfrak{B}$ and therefore $$p \cdot O_E=\mathfrak{B}\cdot \iota(\mathfrak{B})=\mathfrak{B}^2,$$ which implies that $p$ is ramified in $E$, which is not the case. So, the action is free and therefore $H$ consists of $16$ orbits of $G$. Now let's construct Weil's $p^2$-numbers $\pi_1$ and $\pi_2$, using $\mathfrak{B}_1, \mathfrak{B}_2 \in H$ that belong to different orbits of $G$. Let $z_1, z_2 \in O_E$ be generators of ideals $\mathfrak{B}_1$ and $\mathfrak{B}_2$ respectively. Then both $v_1=z_1 \iota(z_1)$ and $v_2=z_2 \iota(z_2)$ are ``real" (i.e., $\iota$-invariant) generators of $p\cdot O_E$, i.e., there are exist units $u_1,u_2 \in O_E^*$ such that $$v_1=p u_1, \ v_2=p u_2.$$ Clearly, both $u_1$ and $u_2$ are real totally positive. Now let us put $$\pi_1=z_1^2/u_1 \in O_E, \ \pi_2=z_2^2/u_2\in O_E.$$ Clearly, $$\pi_1\cdot \iota(\pi_1)=p^2=\pi_2\cdot \iota(\pi_2)$$ (recall that $\iota(u_1)=u_1$ and $\iota(u_2)=u_2$). Taking into account that $$\pi_1 O_E=\mathfrak{B}_1^2, \pi_2 O_E=\mathfrak{B}_2^2,$$ we conclude that $\pi_1$ and $\pi_2$ are not Galois-conjugate (and the same is true for powers $\pi_1^m$ and $\pi_2^m$ for any positive integer $m$). Clearly, $$\pi_1 \ne \iota(\pi_1), \ \pi_2 \ne \iota(\pi_2)$$ and therefore $$Q(\pi_1)=E= Q(\pi_2).$$ Now if $A_1$ (resp. $A_2$) is a simple abelian variety over $F_{p^2}$ attached (by Honda-Tate) to $\pi_i$ then the center of the division algebra $End(A_i)\otimes Q$ is isomorphic to $E$. Since $\pi_1$ and $\pi_2$ (and even their powers) are not Galois-conjugate then $A_1$ and $A_2$ are not isogenous over $F_{p^2}$ (and even over its algebraic closure). On the other hand, both $A_1$ and $A_2$ are obviously ordinary. Since they are simple, their endomorphism algebras are commutative, i.e., coincide with their centers and therefore both $End(A_1)\otimes Q$ and $End(A_2)\otimes Q$ are isomorphic to $E$, Hence, $$\dim(A_1)=[E:Q]/2=\dim(A_2),$$ i.e., both $A_1$ and $A_2$ are $8$-dimensional and $End(A_1)\otimes Q$ is isomorphic to $End(A_2)\otimes Q$. Let me stress that both $A_1$ and $A_2$ remain simple over an algebraic closure of $F_{p^2}$ and their endomorphism algebras remain isomorphic to $E$.<|endoftext|> TITLE: Reference request: the fixed category of an adjunction QUESTION [9 upvotes]: Let $F: A \to B$ and $G: B \to A$ be adjoint functors, with $F \dashv G$. There is a full subcategory $A'$ of $A$ consisting of those objects $a$ for which the unit map $a \to GF(a)$ is an isomorphism, and there is a dually-defined full subcategory $B'$ of $B$. It is an elementary exercise to show that $F$ and $G$ restrict to an equivalence $A' \simeq B'$. Either of the equivalent categories $A'$ and $B'$ is called the invariant part or fixed category of the adjunction. There are other names too; the terminology hasn't settled down. Q. Where did this general construction first appear in print? Adjoint functors were introduced by Kan in 1958. I don't see this construction in his paper. But I guess someone must have mentioned or used it quite soon thereafter. I want to know who I should cite. Let me make clear that I'm not asking about particular instances of this construction. It's the general construction I'm after. I'm also interested in early references in which the general construction plays a significant part, even if they're not the original source. REPLY [2 votes]: The best reference I know of so far is this: J. Lambek and B. A. Rattray. Localization and duality in additive categories. Houston Journal of Mathematics 1 (1975), 87-100. The result I mentioned appears as part of Theorem 1.1. 1975 seems late for this to be the first place where this result is stated explicitly, but I'm not aware of anything earlier.<|endoftext|> TITLE: Is there a Chebotarev‘s theorem for non-Galois extension over Q？ QUESTION [6 upvotes]: For a Galois extension $K/\mathbb{Q}$, the Chebotarev Density Theorem predicts the density of primes with a certain splitting type. I'm wondering if there is a similar result for non-Galois extension? Any references are welcome! REPLY [10 votes]: Actually the usual Chebotarev density theorem in the Galois case can also be applied to the non-Galois case. For example, consider a non-Galois cubic extension $K=\mathbb{Q}[x]/(f)$. I claim that the following splitting types of unramified primes $p$ occur with the following densities $\delta$: $$\delta(1,1,1) = 1/6, \quad \delta(1,2) = 1/2 , \quad \delta(3) = 1/3.$$ To see this, let $L$ be the Galois closure of $K$, which is an $S_3$-extension of $\mathbb{Q}$. The Galois group $S_3$ acts on the roots of $f$ and the splitting type is determined by the action of the Frobenius element $\mathrm{Frob}_p$. We obtain splitting type $(1,1,1)$ (completely split) if and only if $\mathrm{Frob}_p$ fixes all the roots, which means exactly that $\mathrm{Frob}_p$ is trivial. So Chebotarev gives the expected $1/6$. We obtain splitting type $(1,2)$ if and only if $\mathrm{Frob}_p$ is a transposition in $S_3$, of which there are $3$, so we get density $3/6 = 2/3$. Finally we obtain splitting type $(3)$ if and only if $\mathrm{Frob}_p$ is a $3$-cycle in $S_3$, of which there are $2$, giving density $2/6 = 1/3$, as claimed. In general, a splitting type corresponds to some conjugacy invariant subset of the Galois group of the splitting field, as can be seen by considering the action on the roots of the defining polynomial. The density of this is then calculated using the usual Chebotarev for Galois extensions.<|endoftext|> TITLE: Power of $2$ dividing a specialized Mittag-Leffler polynomial QUESTION [5 upvotes]: While studying the so-called Mittag-Leffler Polynomials, denoted $M_n(x)$, I was looking into the sequence $\frac1{n!}M_n(n)$ which takes the following form $$a_n:=\sum_{k=1}^n\binom{n-1}{k-1}\binom{n}k2^k.$$ QUESTION 1. Let $\nu_2(m)$ denote the $2$-adic valuation of $m\in\mathbb{N}$. Is this true? $$\nu_2(a_n)=\begin{cases} \,\,\,\,\,\,1 \qquad \,\,\,\,\,\text{if $n$ is odd} \\ 3\nu_2(n) \qquad \text{if $n$ is even}. \end{cases} $$ QUESTION 2. Is it true that $a_n$ is never divisible by $5$, for $n\geq1$? Postscript. I've recently solved QUESTION 2, so only QUESTION 1 remains. REPLY [6 votes]: I will confine myself to Question 1 since you mentioned that you know how to do Question 2. Also the case when $n$ is odd is easy, and let us restrict to $n$ being divisible exactly by $2^r$ with $r\ge 1$, and we need to show that the exact power of $2$ dividing $a_n$ is $3r$. Thus in what follows we may discard any terms that are divisible by a power of $2$ larger than $3r$. For example, we can restrict attention to the terms with $k\le 3r$ in the sum. The case $r=1$ can be easily checked by hand, and we assume below that $r\ge 2$, and $k \le 3r$. Consider the $k$-th term in the sum defining $a_n$: $$ 2^{k} \binom{n}{k} \binom{n-1}{k-1} = 2^k \frac{n}{k} \prod_{j=1}^{k-1} \Big(1 -\frac{n}{j}\Big)^2. $$ Let us now expand the product above, discarding any terms that are divisible by $2^{3r+1}$. Note that $n/j$ is a $2$-adic integer, since the power of $2$ dividing $j$ is at most $\lfloor \log_2(k-1)\rfloor \le \lfloor \log_2 (3r-1)\rfloor \le r$. We first observe that when expanding the product out terms that ``have $n^4$" in them (so using $3$ factors of $n/j$ in the product) can be omitted, and therefore terms that have higher powers of $n$ may also be omitted. Indeed a term using $3$ factors of $n/j$ in the product is divisible by a power of $2$ $$ \ge k + r -v_2(k) + 3r - 3\lfloor \log_2(k-1)\rfloor. $$ A small calculation shows that $k-v_2(k)-3\lfloor \log_2(k-1) \rfloor \ge -1$ for all $k\ge 1$, and therefore the above is $\ge 3r+r-1 \ge 3r+1$, as desired. Now consider terms with $n^3$ in them (so using $2$ terms of the form $n/j$ from the product). We claim that there is exactly one such term that is relevant, and this happens only in the case $k=4$, and the two terms from the product are (both) $n/2$. Indeed the power of $2$ in term using two factors $n/j$ is $$ \ge k + r-v_2(k) + 2 r -2 \lfloor \log_2 (k-1)\rfloor, $$ and we can check that $k-v_2(k) -2 \lfloor \log_2(k-1)\rfloor \ge 1$ for $k\ge 1$, except in the case $k=4$. In the case $k=4$ we can quickly check that the only relevant term with two factors $n/j$ must be $(n/2)(n/2)$. Putting all these observations together, we find that $a_n$ equals (after omitting any terms divisible by $2^{3r+1}$) \begin{align*} &\sum_{k\le 3r} 2^k \frac{n}{k} \Big( 1 - 2 \sum_{j=1}^{k-1} \frac{n}{j}\Big) + 2^4 \frac{n}{4} \frac n2 \frac n2 \\ &= n \sum_{k\le 3r} \frac{2^k}{k} -n^2 \sum_{k\le 3r} \frac{2^k}{k} \sum_{j=1}^{k-1}\Big( \frac 1j + \frac{1}{k-j}\Big) +n^3\\ &= n \sum_{k\le 3r} \frac{2^k}{k} -n^2 \sum_{k\le 3r} \sum_{j=1}^{k-1} \frac{2^k}{j(k-j)} + n^3. \end{align*} Now here it is convenient to extend the sums over $k$ to infinity, noting that the extra terms are divisible by $2^{3r+1}$. (Thus one checks that for $k>3r$ one has $r+k -v_2(k) \ge 3r+1$, and that $2r + k -2 \lfloor \log_2 (k-1)\rfloor \ge 3r+1$.) Now comes the magical bit. In the $2$-adics one has $$ \sum_{k=1}^{\infty} \frac{2^k}{k} =0, $$ and therefore also $$ \sum_{k=1}^{\infty} 2^k \sum_{j=1}^{k-1} \frac{1}{j(k-j)} = \Big( \sum_{k=1}^{\infty} \frac{2^k}{k} \Big)^2 =0. \tag{*} $$ We are then left with $a_n$ being $n^3$ up to multiples of $2^{3r+1}$, which is what we want. For the evaluation of $\sum_{k=1}^{\infty} 2^k/k$, note that if $|x|_2 <1$ then the series $$ \sum_{k=1}^{\infty} \frac{x^k}{k} $$ converges in the $2$-adics, and this clearly "looks like" $-\log(1-x)$. In our case we'd get $-\log (1-2) = \log (-1) =0$ (upon "using" $\log (-1) + \log (-1) = \log 1 = 0$). This last bit can all be made precise; a lovely write up of what is involved can be found in Keith Conrad's notes (see Example 8.10 there).<|endoftext|> TITLE: Polynomial with many integer but no other rational solutions? QUESTION [29 upvotes]: Is it true that if a (multivariate) polynomial has infinitely many integer solutions, then it also has a non-integer rational solution? My motivation comes from this approach to Hilbert's tenth problem for the rationals. There it was shown that the (non-polynomial) $2^{x^2}-y$ only has integer solutions over $\mathbb Q$. I'm also interested to know what happens if we only want to force some of the variables to be integers, like just $x$ in $2^x-y$, but $y$ can take other rational values too. REPLY [4 votes]: Expanding on some of the comments, I think that Vojta's conjecture will imply a strong result related to your question. First I'm going to reformulate the question in terms of Zariski density, which I think is more natural, since it can be applied inductively to reduce to the case of lower dimensional varieties. Set-Up: Let $K$ be a number field, let $R$ be a ring of $S$-integers in $K$, let $X/K$ be a (smooth) projective variety, let $H\subset X$ be an ample divisor defined over $K$, and let $U=X\smallsetminus H$ be the affine subvariety of $X$ that is the complement of $H$. Choose an affine embedding $\phi:U\hookrightarrow \mathbb A^n$ defined over $K$. This allows us to talk about the $R$-integral points of $U$ relative to $\phi$, which we denote by $$ U_\phi(R) = \phi^{-1}\bigl( \phi(U)(R) \bigr). $$ Question: Does $$ \text{$U_\phi(R)$ is Zariski dense in $U$} \quad\Longrightarrow\quad \text{$X(K)\smallsetminus U_\phi(R)$ is Zariski dense in $X$?} $$ Observation: Assume that Vojta's conjecture is true. If $X$ has Kodaira dimension $\kappa(X)\ge0$, then the question has an affirmative answer, for the silly reason that the conjecture implies in this case that $U_\phi(R)$ is never Zariski dense in $U$! More generally, again assuming Vojta's conjecture, the question has an affirmative answer for the same silly reason if there is an integer $m\ge1$ such that $\mathcal K_X\otimes\mathcal O_X(nH)$ is ample. So if there are any examples for which the question has a negative answer, they must (conjecturally) be found using a variety of Kodaira dimension $-\infty$ and an effective ample divisor $H$ that isn't "too" ample.<|endoftext|> TITLE: If I have zeros at the vertices of an icosahedron, where should the poles go? QUESTION [14 upvotes]: I've been tinkering with Newton's method applied to polynomials. E.g., Newton's method for $z^5 - 1 = 0$ gives: There aren't a lot of symmetric patterns of finite sets of points in the plane, so I decided to rerun the code on the Riemann sphere using a version of Newton's method with spherical symmetry. I thought this would allow me to use point patterns corresponding to all the Platonic solids. However, only the tetrahedral case works: This image has tetrahedral symmetry using the identification of $\mathbb{C}_\infty$ with $S^2$. However, in order to get a rational function with tetrahedral symmetry, we need to decide where to put the poles. A polynomial no longer works: all of its poles are at $\infty$, which is not symmetric. Happily, for a tetrahedron there is a convenient place to put the poles: in the middle of each face. The face centers are at the antipodes $-1/\bar{z}$ of the vertices, so our rational function is $$f(z) = \prod_{k=1}^4 \frac{z-z_k}{z+1/\bar{z_k}}$$ where $z_{1\ldots4}$ are the stereographically projected vertices of the tetrahedron. This rational function has tetrahedral symmetry, so a spherically symmetric version of Newton's method applied to it produces a tetrahedrally symmetric image. Unfortunately, all of the other Platonic solids have vertices at the antipodes of their vertices (they are symmetric under $p \mapsto -p$). Let's consider the icosahedron specifically: Question: If we put a simple zero at each of the 12 vertices of the icosahedron, what's the most symmetric place to put 12 poles (not necessarily simple)? One answer is to inscribe a tetrahedron inside the icosahedron with tetrahedron vertices at face centers, and put order 3 poles at each tetrahedron vertex. This produces an image with tetrahedral symmetry: Is that the best one can do? Edit: Here's a version of the icosahedral symmetry one using @pregunton's linked rational function $F_{3,5}(z)$, but taking the fifth root to turn the zeros back into simple zeros and the order 3 poles into order $3/5$ singularities since that produces a slightly nicer picture: Higher resolution versions of these images: quintic, tetrahedron, icosahedron, modified icosahedron. REPLY [10 votes]: You can color the icosahedron red and blue, such that three red faces and two non-adjacent blue faces meet at each vertex. Then you can put poles in the middle of the twelve red faces, which treats all of the vertices symmetrically. Geoffrey's edit: The resulting render is (Higher resolution)<|endoftext|> TITLE: Self homeomorphism of $\mathbb CP^1$ holomorphic a.e QUESTION [8 upvotes]: Suppose $\varphi:\mathbb CP^1\to \mathbb CP^1$ is a homeomorphism holomorphic on a connected open subset $U\subset \mathbb CP^1$ with $\mathbb CP^1\setminus U$ of zero measure. Is it true that $\varphi$ is holomorphic on the whole $\mathbb CP^1$ (so it is a projective transformation)? If no, what kind of assumptions of $U$ would suffice? (for example $\mathbb CP^1\setminus U$ has Hausdorff dimension $\le 1$?) REPLY [3 votes]: Following the links of Wojowu, the answer to this question is negative for the case of self-homeomorphisms of $\mathbb C^1$, here it is: Functions holomorphic on a region minus a Cantor set So by extending the self-homeo to $\mathbb CP^1$ the answer is negative for $\mathbb CP^1$ as well. To have a positve answer, one has to require indeed that the Hausdorff dimension of $\mathbb CP^1\setminus U$ is less than $1$. (the case of $\dim=1$ seems to be still open)<|endoftext|> TITLE: Background for Varifold theory QUESTION [5 upvotes]: I noticed this question posted on MO, hence I estimated that this may be an acceptable question even in MO (and not for MSE). I studied the notion of current and in a nutshell I understood "varifolds are weaker objects than currents." My question is what kind of prerequisites one needs to have in order to study varifold theory and varifold geometry. For example, as far as currents are concerned, one needs to have grasped the notions of geometric measure theory and a bit of multilinear algebra and the introduction to distribution theory in functional analysis. I would like to ask the principal prerequisites one needs to know to fully grasp varifold theory. Geometric measure theory, of course, but how much Riemannian geometry? How much partial differential equation theory for the theory of regularity as well? Thank you very much in advance! REPLY [8 votes]: The general prerequisites are almost the same as for currents, mainly a strong understanding of measure theory and a bit of geometrical intuition. There is an aspect of multilinear algebra and some functional analysis involved as well, but a lot of that can be studied at the same time. The need for Riemannian geometry depends mostly on what you want to study. For varifolds on Riemannian manifolds, the need is kind of obvious, but for varifolds on $\mathbb{R}^n$, you can do completely without. The notion of mean curvature is involved in many problems involving varifolds, but not much more, as they simply lack the regularity needed for some of the more fancy ideas from differential geometry. Anything else, such as knowledge of PDEs and Calculus of variations is nice to have with regards to context, but the same is true in reverse and you have to start somewhere. Fundamentally, varifolds in $\mathbb{R}^n$ are just Radon measures on $\mathbb{R}^n \times Gr(n,m)$, where $Gr(n,m)$ is the space of $m$-dimensional linear subspaces of $\mathbb{R}^n$. If you understand all the words in that sentence and know a bit about rectifiability, then you have the prerequisites to study varifolds. P.S. I wouldn't say that they are either weaker or stronger than currents though. In the technical sense, there are currents that cannot be reasonably expressed as varifolds and varifolds that cannot be expressed as currents (even ignoring the issue of orientation). Similarly, in the practical sense, each has its advantages and its downsides and which one is better is entirely problem-dependent.<|endoftext|> TITLE: Eigenvalue pattern QUESTION [13 upvotes]: We consider a matrix $$M_{\mu} = \begin{pmatrix} 1 & \mu & 1 & 0 \\ -\mu & 1 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 &-1 & 0 & 0 \end{pmatrix}$$ One easily checks that $\operatorname{det}(M_{\mu})=1$. I however noticed something peculiar: Consider a sequence of real numbers $\mu_i$ then the four eigenvalues $\lambda_1,..,\lambda_4$ of $$ A=\prod_{i=1}^n M_{\mu_i}$$ have the property that they can be chosen to satisfy $\lambda_1 = \overline{\lambda_2}$ and $\lambda_3 = 1/\lambda_1$ and consequently by the determinant $\lambda_4 = 1/\lambda_2.$ I tried to explicitly study the product and see if I can see some structure explaining all this, but so far only with limited luck. Question: Show that the eigenvalues of $A$ for every $n$ and $\mu_i \in \mathbb R$ satisfy $$\lambda_1 = \overline{\lambda_2 } = 1/\lambda_3 = 1/\overline{\lambda_4}.$$ Update: What we know so far is that $\lambda_1 = \frac{1}{\lambda_3}$ and $\lambda_2 = \frac{1}{\lambda_4}$ and if any of the eigenvalues is non-real then we have it. However, we cannot exclude $\lambda_1>\lambda_2$ both real. REPLY [20 votes]: The explanation is pretty simple with a suitable change of basis. Letting $$B = \begin{pmatrix} 1 & 0 & 1 & 0 \\ i & 0 & -i & 0 \\ 0 & 1 & 0 & 1 \\ 0 & i & 0 & -i \end{pmatrix}$$ we have $$B^{-1}M_{\mu}B = \begin{pmatrix} 1+i\mu & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 1-i\mu & 1 \\ 0 & 0 & -1 & 0 \end{pmatrix}$$ Letting $N_\mu = \begin{pmatrix} 1+i\mu & 1 \\ -1 & 0 \\ \end{pmatrix}$, we thus have $$B^{-1}AB = \begin{pmatrix} \prod N_{\mu_i} & 0 \\ 0 & \overline{\prod N_{\mu_i}} \end{pmatrix}$$ where the bar denotes entry-wise complex conjugation. Thus the eigenvalues of $A$ are those of $\prod N_{\mu_i}$ plus those of $\overline{\prod N_{\mu_i}}$, which are their complex conjugates. Moreover, since $N_\mu$ has determinant $1$, so does $\prod N_{\mu_i}$, so its two eigenvalues are inverses of each other.<|endoftext|> TITLE: Riemann rearrangement theorem with restricted choices QUESTION [5 upvotes]: Note: For convenience, all sequences will be indexed by the positive integers $\mathbb Z_+$. Definitions and some motivation: The Riemann rearrangement theorem says that if we have a sequence that is conditionally but not absolutely convergent, we can rearrange it to converge to any desired value. Looking at the proof a little, we can get the following statement: Let $a_n$ be a sequence of real numbers with $|a_n| \to 0$. If $\sum a_n$ diverges, then there exists a $\{-1, 1\}$-valued sequence $\varepsilon_n$ such that $\sum \varepsilon_n a_n$ converges. What if we only allow a "small" set of sequences $\varepsilon_n$? Is it still possible to get the sum to converge after multiplication by $\varepsilon_n$? Question set up: Let $E = \{-1, 1\}$. Define the map $H: (0, 1) \to E^{\mathbb Z_+}$ as follows - expand $x$ in binary form, where we always take the expansion that ends in an infinite number of $1$'s whenever a choice is to be made. Then the $i$'th coordinate of $H(x)$ is $1$ if the $i$'th binary decimal digit of $x$ is $1$, and $-1$ otherwise. Question: For every subset $S$ of $(0, 1)$ with Lebesgue measure $0$, does there always exist a sequence $a_n$ of real numbers with $|a_n| \to 0$, and $\sum a_n$ divergent such that for any sequence $\varepsilon_n$ in $H(S)$, $\sum \varepsilon_n a_n$ fails to converge? REPLY [7 votes]: No. For example, we can define $S$ by insisting that $\epsilon_{2n+1}=\epsilon_{2n}$. If we have no restrictions, then a good strategy to make $\sum \epsilon_n a_n$ convergent would be to choose the $\epsilon_n$ recursively in such a way that the partial sums stay as close to $0$ as possible. A slightly modified version of this strategy still works with our restriction; we now just let $b_n=a_{2n}+a_{2n+1}$ take the role of the $a_n$'s.<|endoftext|> TITLE: $\mathrm{Li}(x)$ vs $x/\log x$ QUESTION [8 upvotes]: I need an explicit lower bound for $\mathrm{Li}(x)$ in terms of $x$ and $\log x$. Say, Wikipedia gives $$ \mathrm{li}(x) >\frac x{\log x}+\frac x{(\log x)^2} $$ for $x>e^{11}$, see the logarithmic integral entry, and so $$ \mathrm{Li}(x)= \mathrm{li}(x)-\log 2 >\frac x{\log x}+\frac x{(\log x)^2}-\log 2 $$ but it does not quote any sources. I need a similar bound but with exact quotation. Of course, I can easily do it myself (and I will if I find nothing to quote) but I prefer having a reference. REPLY [11 votes]: Using ${\rm Li}(x) := \int_2^x dt/\log t$, as usual, here is an elementary argument that ${\rm Li}(x) > x/\log x$ for $x \geq 7$, so no need to appeal to a lower bound valid only starting at $e^5 \approx 148.4$ as in the Rosser-Schoenfeld paper from the comments. For $x > 1$, let $f(x) = {\rm Li}(x) - x/\log x$. Then $$ f'(x) = \frac{1}{\log x} - \left(\frac{\log x - 1}{(\log x)^2}\right) = \frac{1}{(\log x)^2} > 0, $$ so $f$ is increasing on $(1,\infty)$. Since $f(7) \approx .1145$, for $x \geq 7$ we have $f(x) \geq f(7) > 0$, so ${\rm Li}(x) > x/\log x$. Since $f(6) \approx -.1716$, $f$ vanishes somewhere between 6 and 7. That happens slightly below 6.58: from PARI, I find $f(6.58) \approx .000076$. Similarly, for $x > 1$ let $g(x) = {\rm Li}(x) - x/\log x - x/(\log x)^2$. Then $$ g'(x) = \frac{2}{(\log x)^3} $$ so $g$ is increasing on $(1,\infty)$. From PARI, $g(20) \approx -.044$ and $g(21) \approx .028$, so ${\rm Li}(x) > x/(\log x) + x/(\log x)^2$ for $x \geq 21$. Since $g(20.65) \approx .0030005$, $g$ vanishes slightly below 20.65. More generally, from repeated integration by parts $$ {\rm Li}(x) = \sum_{n=1}^N \frac{(n-1)!x}{(\log x)^n} + \int_2^x \frac{N!\, dt}{(\log t)^{N+1}} + c_N $$ for $N \geq 0$ and $x > 1$, where $c_N$ is a constant. Therefore the difference $$ g_N(x) := {\rm Li}(x) - \sum_{n=1}^N \frac{(n-1)!x}{(\log x)^n} $$ has $g_N'(x) = N!/(\log x)^{N+1} > 0$, so $g_N$ is increasing on $(1,\infty)$. Find an $x_0$ where $g_N(x_0) > 0$ and then ${\rm Li}(x) > \sum_{n=1}^N (n-1)!x/(\log x)^n$ for $x \geq x_0$.<|endoftext|> TITLE: Is there an elementary proof that distal maps are invertible? QUESTION [39 upvotes]: Let $T: X \to X$ be a continuous map on a compact metric space $X$. We say $T$ is distal if $\inf_n d(T^n x, T^n y) = 0$ implies $x = y$. Then it is true that $T$ is bijective. Question: Is there an elementary proof of this fact? (Injectivity clearly follows, surjectivity is the issue.) The two proofs I know go through the enveloping semigroup, and the Stone-Čech compactification $\beta \mathbb N$ respectively. REPLY [7 votes]: The answer is yesser than I thought. I mentioned this issue at http://eventos.cmm.uchile.cl/edynamicsxiii/, since the proximality lemma from my previous answer was discussed there. Someone pointed out that Hindman's original proof of his famous theorem is at least somewhat elementary in some technical sense, and implies a significant part of the proximality theorem, so the answer must be "yes" at least in some technical sense. After a bit of searching I found the relevant paper [1]: in the sense of reverse mathematics, your theorem is provable in $\mathrm{ACA}_0^+$, a certain fragment of second-order arithmetic. I quote the relevant theorem in the form stated in this paper. They call this the Auslander-Ellis theorem. Theorem. Let $X$ be a compact metric space and let $T : X \to X$ be continuous. Regard $(X,T)$ as dynamical system. Given $x \in X$, there exists $y \in X$ such that $y$ is uniformly recurrent and proximal to $x$. They show that this theorem is provable in $\mathrm{ACA}_0^+$. Let me recall how to conclude your result (this deduction seems very elementary, so I guess it needs much less than $\mathrm{ACA}_0^+$). Corollary. Every distal system is invertible. Proof. Injectivity is clear. Take $x \in X$, and apply the previous theorem, to get that $x$ is proximal to some uniformly recurrent $y$. Then $x = y$, so every point in $X$ is uniformly recurrent. Clearly this implies surjectivity, since if $U$ is a neighborhood of any $x \in X$, we have $T^n(x) \in U$ for some positive $n$, and then $T^{n-1}(x)$ maps to $U$ in $T$, so $TX$ is dense in $X$ and by compactness is equal to $X$. Square. I am not an expert, but as far as I understand, second-order arithmetic in itself is weaker than Zermelo-Frankel set theory (without choice), and $\mathrm{ACA}_0^+$ is about medium strength as far as the most commonly studied fragments of second-order arithmetic go. Roughly, the logic $\mathrm{ACA}_0$ says that a set of natural numbers exists if you can define it by an arithmetical formula, and $\mathrm{ACA}_0^+$ adds the $\omega$th Turing jump of each set already definable. Nevertheless, to quote [1], "It is well known that all existing proofs of HT are nonconstructive. One of the goals of this paper is to delimit the degree of nonconstructivity which is inherent in Hindman's Theorem." To summarize the logical connections obtained: Write AET for the theorem about proximality, HT for Hindman's theorem, DT for OP's theorem about distal systems. We have $$\mathrm{ACA_0^+} \implies \mathrm{AET} \implies \mathrm{HT} \implies \mathrm{ACA_0} \wedge \mathrm{DT}$$ Intuitively (and again I am not an expert), if you believe results of a certain flavor of infinite computation are well-defined, you can prove AET and therefore the theorem about distal systems. But in theory, the problem about distal systems could be much easier (I have a hard time imagining a proof not going through proximal pairs, but I've been wrong before). In the reverse math framework, one could ask if it is in $\mathrm{ACA}_0$, or $\mathrm{WKL}_0$, or $\mathrm{RCA}_0$, in decreasing order of strength. Reference [1]: Blass, Andreas R.; Hirst, Jeffry L.; Simpson, Stephen G., Logical analysis of some theorems of combinatorics and topological dynamics, Logic and combinatorics, Proc. AMS-IMS-SIAM Conf., Arcata/Calif. 1985, Contemp. Math. 65, 125-156 (1987). ZBL0652.03040.<|endoftext|> TITLE: On the Fourier coefficients at cusps of a modular form QUESTION [7 upvotes]: Given a holomorphic modular form $f(z)$ of integral weight for $\Gamma_{1}(N)$ with integer Fourier coefficients (at $i\infty$), is it true that the Fourier coefficients at other cusps of $f(z)$ are all algebraic integers? Can anyone help me with this problem? Highly appreciated! REPLY [5 votes]: In general no. Take any newform $f$ of weight $k$ on $\Gamma_0(N)$. Then $f$ is an eigenvector of the Atkin-Lehner involution $W_N$, with eigenvalue $\pm 1$. Then by definition of $W_N$, we have \begin{equation*} f |_k \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} (\tau) = i^{-k} N^{-k/2} (W_N f)\bigl(\frac{\tau}{N}\bigr) = \pm i^k N^{-k/2} f\bigl(\frac{\tau}{N}\bigr), \end{equation*} whose first Fourier coefficient is not an algebraic integer. More conceptually, the $q$-expansion of modular form is obtained by evaluating algebraic modular forms (in the sense of Katz, see $p$-adic Interpolation of Real Analytic Eisenstein Series) at the Tate curve. The $N$-torsion subgroup of Tate curve is isomorphic to $\mu_N \times (\mathbb{Z}/N\mathbb{Z})$ as a group scheme over $\mathbb{Z}((q^{1/N}))$. To get an action of $\mathrm{GL}_2(\mathbb{Z}/N\mathbb{Z})$, one needs to identify $\mu_N$ with $\mathbb{Z}/N\mathbb{Z}$, and this can only be done over $\mathbb{Z}\bigl[\zeta_N, \frac{1}{N}\bigr]$, because the discrete Fourier transform has denominators.<|endoftext|> TITLE: Triangular numbers of the form $x^4+y^4$ QUESTION [7 upvotes]: Recall that triangular numbers are those $T(n)=n(n+1)/2$ with $n\in\mathbb N=\{0,1,2,\ldots\}$. Fermat ever proved that the equation $x^4+y^4=z^2$ has no positive integer solution. So I think it's natural to investigate triangular numbers of the form $x^4+y^4$ with $x,y\in\mathbb N$. Clearly, $$T(0)=0^4+0^4\ \ \mbox{and}\ \ T(1)=1^4+0^4.$$ Via a computer, I find that $$15^4+28^4=665281=T(1153).$$ Question. Is $T(1153)=665281$ the only triangular number greater than one and of the form $x^4+y^4$ with $x,y\in\mathbb N$? My computation indicates that $\{x^4+y^4:\ x,y=0,\ldots,5100\}$ contains no other triangular numbers greater than one. I guess that the above question has a positive answer. Can one prove this? Your comments are welcome! REPLY [20 votes]: An exhaustive search up to $10^7$ finds that the only solutions of $x^4+y^4=T(n)$ in integers with $0 \leq x \leq y \leq 10^7$ are: $$ 0^4 + 0^4 = T(0), \quad 0^4 + 1^4 = T(1), \quad 15^4 + 28^4 = T(1153), $$ noted by OP Zhi-Wei Sun; $$ 3300^4 + 7712^4 = T(85508608), $$ already found by Tomita; and the new example $$ 6848279^4 + 6896460^4 = T(94462407145153). $$ Such searches can be done efficiently using the gp function hyperellratpoints, recently ported into gp from Stahlke and Stoll's C program ratpoints; the search up to $10^7$ took a few hours. Probably there are infinitely many solutions, but quite sparse, because one expects the number of solutions with $0 \leq x,y \leq N$ to be asymptotically proportional to $$ \sum_{x=0}^N \sum_{y=0}^N \left(8(x^4+y^4)+1\right)^{-1/2} $$ which grows as a multiple of $\log N$. Proving anything like that seems intractable by present-day methods. This is typical of Diophantine equations leading to "log-K3 surfaces" such as $x^4 + y^4 = n(n+1)/2$.<|endoftext|> TITLE: Almost Arzela Ascoli QUESTION [8 upvotes]: Definitions: We say a sequence of continuous functions $f_n: [0, 1] \to \mathbb R$ is equicontinuous on average if for every $x \in [0, 1]$ and $\varepsilon > 0$ there exists some $\delta > 0$ such that $\limsup_{N \to \infty} \frac{1}{N} \sum_{n = 0} ^{N-1} |f_n (x) - f_n (y)| < \varepsilon$ whenever $|x - y| < \delta$. Suppose $f_n$ are continuous functions that are equicontinuous on average and uniformly bounded - that is, $\sup_n \sup_x |f_n(x)| < \infty$. Question: Does there exist a subsequence $f_{n_k}$ that converges in measure to some continuous $f$? REPLY [6 votes]: $\newcommand{\ep}{\epsilon}\newcommand{\de}{\delta}$The answer is no. To show this, let us use a slightly modified suggestion in a comment by Mateusz Kwaśnicki, as follows. Let \begin{equation} f_n:=g_{a_n,1/n}, \end{equation} where \begin{equation} g_{a,h}(x):=1(aa+h) \end{equation} and $(a_n)$ is any sequence uniformly distributed on $[1/4,3/4]$, so that \begin{equation} \frac1N\sum_1^N 1(aa)$ for some $a\in[1/4,3/4]$ and almost all $x\in[0,1]$. So, the limit cannot be continuous.<|endoftext|> TITLE: Integration of a function over 7-sphere QUESTION [13 upvotes]: Suppose we have $x_1^2 + y_1^2 + x_2^2 + y_2^2 + x_3^2 + y_3^2 + x_4^2 + y_4^2 = 1$ and we define $z_j = x_j + iy_j$, where $j = 1,\,2,\,3,\,4$. The problem is finding or approximating the following integral (The actual problem is more complex than this !) $$I(k) = \int\limits_{\mathbb{S}^7}\lvert(z_1z_4 - z_2z_3)^k\rvert,$$ where $\mathbb{S}^7$ is the $7$-sphere, and the integral is over its surface measure. The ultimate aim is to find or get an estimate of the ratios of the type $\displaystyle \frac{I(k)}{I(k+1)}$. I have never done an integration over such a surface, the hypersphere and don't know where to begin. Any hint or any pointers would really help me to get started. (I have been suggested to make use of the fact that the points $x_i$'s and $y_i$'s lie on a sphere; hence we can use it to reduce one variable and make use of Fubini's Theorem with proper limits for integration to proceed. But I can't see how to make use of this.) What have I done till now? As the problem say, it would be acceptable if we can approximate the ratios or the integral, so $$\lvert z_1z_4 - z_2z_3\rvert \, \leq \, \lvert z_1z_4\rvert + \lvert z_2z_3\rvert \, = \, \lvert z_1\rvert\lvert z_4\rvert + \lvert z_2\lvert\rvert z_3\rvert.$$ This converts everything into real variables, and then I have used Mathematica to calculate this. But the problem is Mathematica gives the solution for small values of $k$, but as I have to find it for the general case, it is not of much use. Also, I have tried using the spherical coordinates to help with the integral, but they make it more complicated. Any pointers on this front are also welcome. Here are the values of $I(k)$ for $k=1,\,\cdots,\,10$. $z_1 = x_1 + ix_2, \ \ z_2 = x_3+ix_4, \ \ z_3 = x_5+ix_6, \ \ z_4 = x_7+ix_8$ then $$\mid z_1z_4 - z_2z_3| \ \ = \ \ \sqrt{\left(x_4 x_5+x_3 x_6-x_2 x_7-x_1 x_8\right){}^2+\left(x_3 x_5-x_4 x_6-x_1 x_7+x_2 x_8\right)^2} $$ To integrate over $\mathbb{S}^7$, I have used two methods, one is using NIntegrate and the second using integrateSphere function from HFT Software package. (Note : HFT package doesn't work for odd values of $k$, but for even values of $k$, the results match perfectly with that of INtegrate) $$\begin{align} I(1) \ \ &\approx \ \ 9.54903\\ I(2) \ \ &= \ \ \frac{\pi^4}{3\times 10} \ \ \approx \ \ 3.24697\\\\ I(3) \ \ &\approx \ \ 1.19691\\ I(4) \ \ &= \ \ \frac{\pi^4}{3\times70} \ \ \approx \ \ 0.463853\\\\ I(5) \ \ &\approx \ \ 0.188085\\ I(6) \ \ &= \ \ \frac{\pi^4}{3\times420} \ \ \approx \ \ 0.0773088\\\\ I(7) \ \ &\approx \ \ 0.0329181\\ I(8) \ \ &= \ \ \frac{\pi^4}{3\times2310} \ \ \approx \ \ 0.0140561\\\\ I(9) \ \ &\approx \ \ 0.00620447\\ I(10) \ \ &= \ \ \frac{\pi^4}{3\times12012} \ \ \approx \ \ 0.0027031\\\\ \end{align}$$ REPLY [6 votes]: $\newcommand{\R}{\mathbb{R}}\newcommand{\C}{\mathbb{C}}$Here's an approach I like that takes advantage of the fact that the integrands are homogeneous. I'll just show how to compute $$ \int_{S^7} |z_1z_4|. $$ First, denote $\hat{z} \in \C^4$ and $|\hat z|^2 = |\hat z_1|^2 + |\hat z_2|^2 + |\hat z_3|^2 + |\hat z_4|^2$, then, using polar coordinates \begin{align*} \int_{\R^8} e^{-|\hat z|^2}|\hat z_1\hat z_4|\,d\hat{z} &= \int_{z \in S^7}\int_{r=0}^{r=\infty} e^{-r^2}r^2|z_1z_4|r^{7}\,dr\,dz\\ &= \left(\int_{r=0}^{r=\infty} r^9e^{-r^2}\,dr\right)\int_{S^7} |z_1z_4|\,dz, \end{align*} where $dz$ is the surface area measure on $S^7$. The second factor on the right is what we want to calculate. The value of the first factor on the right is easy to evaluate in terms of the gamma function. So it remains only to evaluate the left integral. But this is straightforward. For convenience, I'll write $z \in \C^4$ instead of $\hat{z}$. Then \begin{align*} \int e^{-|z|^2}|z_1z_4|\,dz &= \int_{\C} e^{-|z_1|^2}|z_1|\,dz_1\int_{\C} e^{-|z_4|^2}|z_4|\,dz_4 \int_{\C} e^{-|z_2|^2}\,dz_2\int_\C e^{-|z_3|^2}\,dz_3\\ &= \left(\int_{\C} e^{-|z|^2}|z|\,dz\right)^2\left(\int_{\C} e^{-|z|^2}\,dz\right)^2. \end{align*} The two integrals on the right are also easily evaluated.<|endoftext|> TITLE: Anomalous phenomena QUESTION [6 upvotes]: What are examples of strikingly anomalous phenomena in mathematics, where just one or a rather small number of cases stand out because they don't fit a general pattern? This is most interesting when the situation considered is very simple and basic and where the exceptional cases are not merely the lowest-numbered ones: for example, the outer automorphism of the symmetric group $S_{6}$ (which exists for no other $S_{n}$), and the existence of non-standard differentiable structures on $\mathbb{R}^{4}$ (but no other $\mathbb{R}^{n}$). REPLY [11 votes]: The integral of $x^r$ is another power of $x$, for any value of $r$ except $r=-1$, when it's a natural logarithm. That still amazes me. REPLY [5 votes]: Besides the extra automorphism of $S_6$, both $S_6$ and $S_7$ have exceptional triple covers. REPLY [5 votes]: The special orthogonal group $\operatorname{SO}_8$ is the only $\operatorname{SO}_n$ that has an outer automorphism of order $3$.<|endoftext|> TITLE: Is a Hopf algebra a group object of some category? QUESTION [8 upvotes]: The page of ncatlab on group object states that: A group object in $\mathrm{CRing}^{\mathrm{op}}$ is a commutative Hopf algebra. Question: Is a (noncommutative) Hopf algebra a group object of some category? [let assume finite dimensional, if necessary.] The page of Wikipedia on group object states that: Hopf algebras can be seen as a generalization of group objects to monoidal categories. It is not clear to me how this sentence answers the above question. REPLY [7 votes]: Cocommutative Hopf algebras are group objects in the cartesian category of cocommutative coalgebras. There is no such description in the non-cocommutative case. Also since the antipode does not need to be invertible, which is certainly true for group objects.<|endoftext|> TITLE: Imaginary eigenvalues QUESTION [11 upvotes]: Consider the matrix $$A(\mu) = \begin{pmatrix} 0 & 1& 0 & 0 \\ -1 & -i\mu & 0 & i \\ 0 & 0 & 0 & 1 \\ 0 &i & -1 & i\mu \end{pmatrix}.$$ This matrix is for $\mu \in \mathbb R$ skew hermitian, i.e. all the eigenvalues are imaginary. Let $(\mu_i)_i$ be a sequence of real numbers. We consider the product $$M=\prod_{i=1}^n A(\mu_i).$$ I claim the following two facts are true (observed numerically): 1.) If $n$ is odd, then all eigenvalues are imaginary (this is non-trivial for $n\ge 3$ since the matrix $M$ is in general not skew hermitian anymore) 2.) Show that the eigenvalues satisfy for $n \in 2\mathbb N_0+1$ that $\lambda$ is an eigenvalue of $M$ if and only if $-\lambda$ is. If you show this for one eigenvalue it will hold for all eigenvalues of $M$. REPLY [15 votes]: Define the unitary and Hermitian matrices $$U=\left( \begin{array}{cccc} 0 & 0 & -i & 0 \\ 0 & 0 & 0 & -i \\ i & 0 & 0 & 0 \\ 0 & i & 0 & 0 \\ \end{array} \right),\;\; V=\left( \begin{array}{cccc} 0 & 0 & -i & 0 \\ 0 & 0 & 0 & i \\ i & 0 & 0 & 0 \\ 0 & -i & 0 & 0 \\ \end{array} \right), \;\;U^2=I=V^2,$$ and note that, for $\mu\in\mathbb{R}$, $$UA(\mu)U=\bar{A}(\mu),\;\;VA(\mu)V=-A(\mu).$$ $\bullet$ Hence if $\lambda$ is an eigenvalue of $\prod_{i=1}^n A(\mu_i)$, then $$0=\overline{\det\bigl(\lambda I-\prod_i A(\mu_i)\bigr)}=\det\bigl(\bar{\lambda} I-\prod_i\bar{A}(\mu_i)\bigr)=\det\bigl(\bar{\lambda} I-\prod_i UA(\mu_i)U\bigr)=\det\bigl(\bar{\lambda} I-\prod_i A(\mu_i)\bigr)=0.$$ So the eigenvalues come in complex conjugate pairs: if $\lambda$ is an eigenvalue of $\prod_i A(\mu_i)$, then also $\bar{\lambda}$ is an eigenvalue. (This holds irrespective of whether $n$ is even or odd.) $\bullet$ Similarly, if $\lambda$ is an eigenvalue of $\prod_{i=1}^n A(\mu_i)$ and $n$ is an odd integer, then $$0=\det\bigl(\lambda I-\prod_{i=1}^n VA(\mu_i)V\bigr)=\det\bigl(\lambda I-(-1)^n\prod_{i=1}^n A(\mu_i)\bigr)=\det\bigl(\lambda I+\prod_{i=1}^n A(\mu_i)\bigr)=0,$$ so the eigenvalues come in inverse pairs for odd $n$: if $\lambda$ is an eigenvalue then also $-\lambda$ is an eigenvalue. This proves property 2. $\bullet$ Since $\det A(\mu)=1$ for any $\mu$, the product of the four eigenvalues of $\prod_{i=1}^n A(\mu_i)$ equals unity. This gives for odd $n$ the following three possibilities (with real $c$ and $\phi$): A. $\lambda_1=ic$, $\lambda_2=-ic$, $\lambda_3=i/c$, $\lambda_4=-i/c$ (this is property 1), B. $\lambda_1=c$, $\lambda_2=-c$, $\lambda_3=1/c$, $\lambda_4=-1/c$, C. $\lambda_1=e^{i\phi}$, $\lambda_2=-e^{i\phi}$, $\lambda_3=e^{-i\phi}$, $\lambda_4=-e^{-i\phi}$. The eigenvalues are either all four on the imaginary axis, or on the real axis, or on the unit circle.<|endoftext|> TITLE: A smooth function such that the second derivative of its absolute value is a distribution of positive order QUESTION [6 upvotes]: Let $f\in C^\infty(\mathbb R;\mathbb R)$ and let us define $g(x)=\vert f(x)\vert$. It is easy to verify that $g$ is locally Lipschitz-continuous function, but I would like to find an example of a smooth function $f$ where $g''$ is a distribution with positive order, that is a distribution which not a Radon measure. REPLY [4 votes]: No, this is not possible. Clearly, the singular part of $g''$ results from points where $f(x)=0$ and $f'(x)\neq 0$. A given interval $[a,b]$ can contain at most countably many such points. Denote the set of these points by $X$. We need to show that $\sum_{x\in X} |f'(x)|$ is finite. Take any finite subset of $X$, ordered such that $a\le x_1<... TITLE: Are rigid-analytic spaces obsolete, since adic spaces exist? QUESTION [37 upvotes]: Recently in a seminar the following question was raised and, despite my familiarity with theory, I couldn't come up with a good answer: Are there any good reasons to use Tate's theory of rigid-analytic spaces, given that Huber's theory of adic spaces seems to be superior in all regards? The only possible advantage I can think of is that of simplicity — only having classical points to worry about may be conceptually simpler. This would be similar to treatments of classical algebraic geometry using maximal spectra (as done e.g. by Milne), although having to work with the G-topology seems to offset any pedagogical benefit to me. For some time I also thought the development of rigid cohomology is where rigid spaces would be advantageous, but as discussed in Lazda and Pál, "Rigid Cohomology over Laurent Series Fields", rigid cohomology can be, and for their purposes has to be, developed with adic generic fibers instead. For contrast, let me mention that I am aware of some ways in which say Berkovich spaces have advantages despite also being subsumed by adic spaces, coming from their "Euclidean" nature, for instance the theory of integration on them, the theory of skeleta or some relations to tropical geometry. Are there any contexts like that where classical rigid varieties shine? REPLY [27 votes]: There are two questions here: which version of the theory is easiest to get off the ground axiomatically, and which version is more convenient to work with in applications? For the first question, it's very much a matter of taste. Adic spaces are genuinely topological spaces, whereas Tate's G-topological spaces aren't; but there are many more, and weirder, points in them whose geometric significance takes some getting used to. At risk of treading on some toes, I'm going to point out that Huber's adic spaces were available in the literature for at least 20 years before they started to become really popular. If there were a decisive advantage to setting up the theory in terms of adic spectra rather than G-topologies, then number theorists would have abandoned Tate's theory en masse in 1995 or so; and they didn't. The benefits offered by Huber's foundations weren't persuasive enough to to outweigh the "first-mover advantage" given by 30 years' worth of literature written in Tate's language. Adic spaces caught on because Scholze showed they could be used to do radically new things that were impossible in Tate's rigid geometry -- not because they allowed you to re-prove or re-visualize existing theorems in nicer ways. As for the second question, I think the correct answer is "both". There are some (mostly younger) mathematicians who, whenever rigid spaces are mentioned, smile indulgently at the folly of their elders and assume that anything written in this language is obsolete or misguided, a bit like teenagers laughing at their parents' CD collection. This is a misconception, since rigid spaces over a nonarchimedean field K can be identified with a subcategory of adic spaces over K, and a rigid space and its corresponding adic space have the same sheaf theory (equivalent as topoi). Hence, when you want to apply p-adic analytic geometry to actually do something, it very often doesn't matter whether you write $Max(A)$ or $Spa(A, A^+)$ -- their underlying sets are hugely different, but that's usually not relevant if you're writing a research paper as opposed to a textbook. Indeed, in recent literature it seems to be quite common to simply redefine "rigid space over K" to mean an adic space which is locally of finite type over K. So the large corpus of work written in Tate's language remains useful, and a working number theorist nowadays who isn't familiar with the older language is at risk of needlessly reinventing the wheel. [I'm sure Wojowu already knows everything I've written in this paragraph, but I'm putting it in for the benefit of other readers of this question.]<|endoftext|> TITLE: Rigorous construction of fermionic field theory? QUESTION [7 upvotes]: In section X.7 of Reed & Simon's book there is a nice rigorous construction of the free scalar field theory which applies to the Klein-Gordon field. Question: Are there references which discuss, in analogous fashion, the construction of (rigorous) fermionic field theories? At the moment, I'm more interested in the Dirac field theory because it includes the spin and two kinds of creation and annihilation operators. REPLY [12 votes]: There is the construction of the C${}^*\!$-algebra of canonical anticommutation relations (CAR's), which is actually somewhat easier than the construction of free bosonic fields: given any complex pre-Hilbert space $\mathfrak{h}$, which may be thought of as our "one-particle" space, define the unital *-algebra $\text{CAR}_0(\mathfrak{h})$ given by the generators $a(f),a^*(f)=a(f)^*$, $f\in\mathfrak{h}$ (respectively, the annihilation and creation operators for the particle with wave function $f$), and the following relations: The map $f\mapsto a(f)$ is anti-linear (hence the map $f\mapsto a^*(f)$ is linear); $\{a(f),a(g)\}=0$, where $\{A,B\}=AB+BA$ is the anti-commutator of two elements $A,B$ in an associative algebra (hence $\{a^*(f),a^*(g)\}=\{a(f),a(g)\}^*=0$); $\{a(f),a^*(g)\}=\langle f,g\rangle\mathbf{1}$, where we assume that the scalar product $\langle\cdot,\cdot\rangle$ of $\mathfrak{h}$ is anti-linear in the first variable and linear in the second variable. Notice that the CAR's imply that $$(a^*(f)a(f))^2=a^*(f)\{a(f),a^*(f)\}a(f)=\|f\|^2a^*(f)a(f)\ ,\quad f\in\mathfrak{h}$$ hence if there is a C${}^*\!$-norm $\|\cdot\|$ on $\text{CAR}_0(\mathfrak{h})$, we must have $$\|(a^*(f)a(f))^2\|=\|a^*(f)a(f)\|^2=\|f\|^2\|a^*(f)a(f)\|\ ,$$ therefore $\|a(f)\|=\|a^*(f)\|=\|f\|$ for all $f\in\mathfrak{h}$. In other words, unlike for bosonic fields, fermionic creation and annihilation operators are necessarily bounded, thanks to the Pauli exclusion principle encoded in the CAR's. To show that such a C${}^*\!$-norm actually exists, notice that there is a nontrivial *-representation of $\text{CAR}_0(\mathfrak{h})$ in the fermionic (i.e. anti-symmetric) Fock space $\mathfrak{F}_-(\mathfrak{h})$ generated by $\mathfrak{h}$: $$\mathfrak{F}_-(\mathfrak{h})=\bigoplus^\infty_{n=0}\wedge^n\mathfrak{h}\ ,$$ where $\wedge^0\mathfrak{h}=\mathbb{C}$, $\wedge^1\mathfrak{h}=\mathfrak{h}$, and $\wedge^n\mathfrak{h}$ for $n>1$ is the vector space generated by the $n$-fold wedge (i.e. anti-symmetrized tensor) product of elements of $\mathfrak{h}$: $$f_1\wedge\cdots\wedge f_n=\frac{1}{n!}\sum_{\sigma\in\mathbb{S}_n}\epsilon_\sigma f_{\sigma(1)}\otimes\cdots\otimes f_{\sigma(n)}\ ,$$ where $\mathbb{S}_n$ is the group of permutations of $n$ elements and $\epsilon_\sigma$ ( = $(-1)^{inv(\sigma)}$, $inv(\sigma)=$number of inversions of $\sigma\in\mathbb{S}_n$) is the sign of the permutation $\sigma$, endowed with the scalar product $$\langle f_1\wedge\cdots\wedge f_n,g_1\wedge\cdots\wedge g_n\rangle=\det[\langle f_i,g_j\rangle]\ .$$ The direct sum is assumed to be orthogonal. The action of $a(f),a^*(f)$ on $\mathfrak{F}_-(\mathfrak{h})$ is given by $$a(f)\lambda=0\ ,\,a^*(f)\lambda=\lambda f\ ,\quad\lambda\in\mathbb{C}\ ,$$ $$a(f)f_1\wedge\cdots\wedge f_n=\frac{1}{\sqrt{n}}\langle f,f_1\rangle f_2\wedge\cdots\wedge\cdots\wedge f_n\ ,$$ $$a^*(f)f_1\wedge\cdots\wedge f_n=\sqrt{n+1}f\wedge f_1\wedge\cdots\wedge f_n\ ,\quad f_1,\ldots,f_n\in\mathfrak{h}\ .$$ It is easy to verify that this defines a *-representation of $\text{CAR}_0(\mathfrak{h})$ by bounded linear operators on $\mathfrak{F}_-(\mathfrak{h})$ (boundedness is guaranteed by the CAR's as shown above). This ensures that the completion of $\text{CAR}_0(\mathfrak{h})$ w.r.t. the maximal C${}^*\!$-norm $\|\cdot\|$ is a C${}^*\!$-algebra $\mathfrak{A}$ acting on the Fock Hilbert space $\overline{\mathfrak{F}_-(\mathfrak{h})}$, called the CAR algebra associated to $\mathfrak{h}$. As written, one is only able to define Majorana (i.e. Hermitian) fermionic field operators $$\psi_R(f)=\frac{1}{\sqrt{2}}(a^*(f)+a(f))\ ,$$ from which $a(f)$ and $a^*(f)$ may be recovered through the formula $$a(f)=\frac{1}{\sqrt{2}}(\psi_R(f)+i\psi_R(if))\ ,\,a^*(f)=\frac{1}{\sqrt{2}}(\psi_R(f)-i\psi_R(if))\ ,\quad f\in\mathfrak{h}\ .$$ We remark that the map $f\mapsto\psi_R(f)$ is only real linear. To obtain the actual Dirac fields, $\mathfrak{h}$ needs to have the form $\mathfrak{h}=\mathfrak{k}\oplus\mathfrak{k}$, where this direct sum is orthogonal. These direct summands correspond to the particle and antiparticle sectors. Defining $$b(f)=a((f,0))\ ,\,c(f)=a((0,f))\ ,\quad f\in\mathfrak{k}\ ,$$ one can define the Dirac field operators $$\psi(f)=\frac{1}{\sqrt{2}}(b(f)+c^*(f))\ ,\psi^*(f)=\psi(f)^*=\frac{1}{\sqrt{2}}(c(f)+b^*(f))\ ,\quad f\in\mathfrak{k}\ .$$ As before, we point that $f\mapsto\psi(f)$ and $f\mapsto\psi^*(f)$ are only real linear. The above yields the CAR's in Dirac form $$\{b(f),b(g)\}=\{b(f),c(g)\}=\{c(f),c(g)\}=\{b(f),c^*(g)\}=\{c(f),b^*(g)\}=0\ ,$$ $$\{b(f),b^*(g)\}=\{c(f),c^*(g)\}=\langle f,g\rangle\mathbf{1}$$ for all $f,g\in\mathfrak{k}$, hence $$\{\psi(f),\psi(g)\}=0\ ,\,\{\psi(f),\psi^*(g)\}=\langle f,g\rangle\mathbf{1}\ .$$ We remark that, as before with the Majorana field operators, $b(f)$, $c(f)$ (and therefore also $a((f_1,f_2))=b(f_1)+c(f_2)$) can be recovered from $\psi(f)$ and $\psi^*(f)$ for each $f(,f_1,f_2)\in\mathfrak{k}$ in a way similar to the recovery of the bosonic annihilation operators from the free bosonic field operators - to wit, $$b(f)=\frac{1}{\sqrt{2}}(\psi(f)+i\psi(if))\ ,\,c(f)=\frac{1}{\sqrt{2}}(\psi^*(f)+i\psi^*(if))\ ,$$ hence $$b^*(f)=\frac{1}{\sqrt{2}}(\psi^*(f)-i\psi^*(if))\ ,\,c^*(f)=\frac{1}{\sqrt{2}}(\psi(f)-i\psi(if))\ .$$ All that is left is the construction of the one-particle pre-Hilbert space $\mathfrak{k}$, which can be identified with the space of initial data for positive-energy solutions of the Dirac equation. We point that a similar procedure to that used to obtain Dirac fields can be employed for the free complex (bosonic) scalar field, which also enjoys a similar concept of particle and antiparticle sectors. Apart from the last section on Dirac fields, all of the above may be found e.g. in Section 5.2 of the book by Ola Bratteli and Derek W. Robinson, Operator Algebras and Quantum Statistical Mechanics 2. Equilibrium States, Models in Quantum Statistical Mechanics (2nd. ed., Springer-Verlag, 1997).<|endoftext|> TITLE: Reference request: Different definitions of Big O notation QUESTION [5 upvotes]: This question might sound strange, but I would like to settle this problem once and for all. For as long as I can remember, I was introduced to the Big O notation by this definition: Def. 1: Let $f, g$ be some real (or complex) functions defined on a set $A$. Then the notation “$f(x) = O(g(x))\text{ for } x \in A$” means that there exists a constant $C_{f,g} > 0$, depending only on $f$ and $g$, such that $|f(x)| \leq C_{f,g} |g(x)|$ for every $x \in A$. Therefore, for me it makes perfect sense (and it is often useful) to write statements like: $$ \frac1{1 - x} = 1 + x + O(x^2) \quad\text{ for } x \in [0, 1/2],\label{1}\tag{1} $$ or $$ \lfloor x \rfloor = x + O(1) \quad\text{ for } x \in \mathbb{R}. \label{2} \tag{2} $$ However, when talking to colleagues, it often turns out that they are used to a different definition of the Big O notation, that is: Def. 2: Let $f, g$ be some real (or complex) functions defined on some topological space containing the point $x_0$. Then the notation “$f(x) = O(g(x))\text{ as } x \to x_0$” means that there exist a constant $C_{f,g} > 0$ and a neighborhood $U_{f,g}$ of $x_0$, depending only on $f$ and $g$, such that $|f(x)| \leq C_{f,g} |g(x)|$ for every $x \in U_{f,g}$. Thus, if anything, they would rewrite \eqref{1} as: $$\frac1{1 - x} = 1 + x + O(x^2) \quad\text{ for } x \to 0,$$ (which is a weaker statement) and they would say that \eqref{2} is wrong notation / meaningless. Although I'm sure that other people than me use Def. 1 daily, I could not find a single book defining the Big O notation as in Def. 1 - but only books defining the Big O notation similarly to Def. 2. What is a reference to a book defining the Big O notation as in Def. 1 ? Thanks REPLY [2 votes]: Def. 1 can be found on pag. 8 of D. Koukoulopoulos, The Distribution of Prime Numbers (Luckily, pag. 8 can be read on the Google Books preview: https://www.google.it/books/edition/The_Distribution_of_Prime_Numbers/me7CDwAAQBAJ?hl=it&gbpv=0). Although Def. 1 and Def. 2 are equivalent on $\mathbb{N} \cup \{\infty\}$, I agree that Def. 1 is superior to Def. 2 in may ways, which include: Def. 1 is easier (no topology required) and more general. Def. 1 works in summations, while Def. 2 don't. For example, knowing that $1/(1-x) = 1 + x + O(x^2)$ for $x \in [0, 1/2]$, we can find that $$\sum_{n = 1}^N \frac1{1 - x / n} = \sum_{n = 1}^N \left(1 + \frac{x}{n} + O\left(\frac{x^2}{n^2}\right)\right) = N + \left(\sum_{n=1}^N \frac1{n}\right) x + O\left(\sum_{n=1}^N \frac1{n^2} x^2\right)$$ $$ = N + \left(\sum_{n=1}^N \frac1{n}\right) x + O\left(x^2\right) \quad\text{ for } x \in [0, 1/2].$$ No way to do that using Def. 2. Def. 1 can express boundness: $f(x) = O(1)$ for $x \in A$ is indeed equivalent to $f$ is bounded on $A$. On the other hand, $f(x) = O(1)$ for $x \to x_0$ only means that $f$ is bounded around $x_0$, and may have singularities somewhere else.<|endoftext|> TITLE: What is the top cohomology group of a non-compact, non-orientable manifold? QUESTION [12 upvotes]: Let $M$ be a connected, non-compact, non-orientable topological manifold of dimension $n$. Question: Is the top singular cohomology group $H^n(M,\mathbb Z)$ zero? This naïve question does not seem to be answered in the standard algebraic topology treatises, like those by Bredon, Dold, Hatcher, Massey, Spanier, tom Dieck, Switzer,... Some remarks. a) Since $H_n(M,\mathbb Z)=0$ (Bredon, 7.12 corollary) we deduce by the universal coefficient theorem: $$ H^n(M,\mathbb Z) =\operatorname {Ext}(H_{n-1}(M,\mathbb Z), \mathbb Z)\oplus \operatorname {Hom} (H_n(M,\mathbb Z),\mathbb Z)=\operatorname {Ext}(H_{n-1}(M,\mathbb Z),\mathbb Z )$$ But since $H_{n-1}(M,\mathbb Z)$ need not be finitely generated I see no reason why $\operatorname {Ext}(H_{n-1}(M,\mathbb Z),\mathbb Z)$ should be zero. b) Of course the weaker statement $H^n(M,\mathbb R) =0$ is true by the universal coefficient theorem, or by De Rham theory if $M$ admits of a differentiable structure. c) This question was asked on this site more than 8 years ago but the accepted answer is unsubstanciated since it misquotes Bredon. Indeed, Bredon states in (7.14, page 347) that $H^n(M,\mathbb Z)\neq0$ for $M$ compact, orientable or not, but says nothing in the non-compact case, contrary to what the answerer claims. REPLY [8 votes]: I think it's worth pointing out that at least for smooth or PL $n$-manifolds $M$ that are connected but not compact, something much stronger holds than the $n$th homology vanishing -- the manifold is actually homotopy equivalent to an $(n-1)$-dimensional simplicial complex! This is a theorem of Whitehead from J. H. C. Whitehead, The immersion of an open 3-manifold in euclidean 3-space, Proc. London Math. Soc. (3) 11 (1961), 81–90. I gave a modern treatment of it in my note here. In that note, I say that the manifold is smooth, but really all the proof uses is PL (I should fix this sometime). I don't know if this holds if the manifold is not smooth or PL, but I suspect it does.<|endoftext|> TITLE: Was there a time in mathematics when a counterexample was wrong? QUESTION [8 upvotes]: I am doing an essay on the knowledge of Mathematics and how we know what we know to be true. I was just wondering if there was an example in mathematics of some theorem that was disproven by a counterexample, but then the counterexample was wrong and the theorem stayed true after all? Thanks for your time. :) REPLY [23 votes]: A very famous and important example of a counterexample that was found to be defective occurred in set theory. As Georg Cantor developed the theory of infinite sets, he proved that some infinite sets are larger than others, by showing that there was not a one-to-one function between them. Most famously, he showed that the real numbers $a\leq x\leq b$ are more numerous (have greater "cardinality") than the natural numbers $\{1,2,\ldots\}$. However, he also proved a more general statement, Cantor's Theorem—that if $X$ is a set (any set, finite or infinite), and ${\cal P}(X)$ is the set of all subsets of $X$ [we call ${\cal P}(X)$ the "power set" of $X$], then ${\cal P}(X)$ has a greater cardinality than $X$. Crudely put, this says that any set has more subsets than it has elements. That is certainly true for finite sets; a set with $n$ elements has $2^{n}$ subsets, and $2^{n}>n$ for all integers $n\geq0$. However, Cantor showed it was true for arbitrary sets, even infinite ones. There were many objections to Cantor's theory of infinite sets, and one important objection was to the theorem that ${\cal P}(X)$ has a greater cardinality than $X$. This seems to be impossible, because it cannot be true of the set of all sets (usually denoted by ${\bf U}$, for the "universe" all sets). For, if ${\bf U}$ contains all sets, then it necessarily contains all elements of its own power set ${\cal P}({\bf U})$, meaning ${\cal P}({\bf U})$ cannot be more numerous than ${\bf U}$. The resolution of this apparent paradox is that the supposed universal set ${\bf U}$ cannot exist, at least not as a set. (The universe ${\bf U}$ does exist as a more general kind of object, a proper class.) This was not evident in Cantor's approach, since he did not provide a axiomatic basis for his set theory. (His approach is sometimes therefore known as "naive set theory.") However, with the development of a rigorous foundation for set theory, based on axioms (such the Zermelo-Fraenkel axioms), it became clear that the object ${\bf U}$ is simply not definable as a set in the theory. While this might seem to be a problem, since it means that a rather natural-seeming object cannot exist, it is actually extremely advantageous, since it means that the counterexample to Cantor's Theorem does not actually exist, leaving the theory (so far as we can tell) consistent.<|endoftext|> TITLE: Closure of the locus of polynomials vanishing to a given order at two points QUESTION [6 upvotes]: I have an elementary question concerning zeros of polynomials, which must be well-known. Fix a base field $ k$ (can assume to be characteristic zero if it makes a difference). Consider the affine space $ P_n \times \mathbb A^1_k $, where $ P_n $ denotes the space of polynomials of degree $ n $ over $ k$ (so $ P_n $ is an affine space of dimension $ n+1 $ given by the coefficients). Given $ p \in P_n $ and $ z \in k $, let $ ord_z(p) $ denote the order of vanishing of the polynomial $ p $ at $z $. Fix $ m \in \mathbb N $ and consider the closed subset of $ X_m \subset P_n \times \mathbb A^\times $ given as $$ X_m := \{ (p,s) : ord_s(p) + ord_0(p) \ge m \}$$ In other words, we study those polynomials whose vanishing orders at $ 0 $ and $ s $ add up to at least $ m$ (where $ s $ is non-zero). Then I take $ \overline{X_m} \subset P_n \times \mathbb A $. Question: Is it true that $ \overline{X_m} \cap P_n \times \{0\} $ consists of those polynomials which vanish to order at least $ m $ at $ 0$? This is just a set-theoretic question. Scheme-theoretically, this statement seems to be false. REPLY [2 votes]: For a commutative ring $R$, write $P_{n,R}$ for the affine $(n+1)$-space of polynomials of degree $\leq n$ over $R$. In other words, its $S$-points for an $R$-algebra $S$ are given by $S[x]_{\leq n}$. Note that $P_{n,R} = P_{n,\mathbf Z} \times \operatorname{Spec} R$. Lemma. Let $R$ be a domain, and $g \in R[x]$ a monic polynomial of degree $d$. Then the map \begin{align*} \phi \colon P_{n,R} &\to P_{n+d,R} \\ f &\mapsto fg \end{align*} is a closed immersion. For any $R$-algebra $S$, the $S$-points of the image of $\phi$ are the polynomials in $S[x]$ that are divisible by $g$. Proof. Both $P_{n,R}$ and $P_{n+d,R}$ are affine spaces over $R$, and $\phi$ is a linear map with fibrewise trivial kernel. Such a map is a closed immersion (exercise), and the second statement is obvious. $\square$ Applying this to $R = k[s]$ and $g = x^i(x-s)^j$ for $i+j = m$, we see that the locus $$Z_{i,j} := \left\{(f,s) \in P_n \times \operatorname{Spec} R : x^i(x-s)^j \mid f\right\}$$ is closed, hence the same goes for $Z_m = \bigcup_{i+j=m} Z_{i,j}$. The restiction of $Z_m$ to $P_n \times \operatorname{Spec} R[1/s]$ is $X_m$, which shows $\bar X_m \subseteq Z_m$. But if $Z = \{f \in P_n : x^m \mid f\}$, then $Z_m \cap \big(P_n \times \{0\}\big)$ is just $Z \times \{0\}$. Since $Z \times \operatorname{Spec} R[1/s] \subseteq X_m$, we get $Z \times \{0\} \subseteq \bar X_m$, hence $Z_m \subseteq \bar X_m$. We conclude that $Z_m = \bar X_m$, which proves the required statement. $\square$<|endoftext|> TITLE: What is Young measure? QUESTION [5 upvotes]: I read about Young measures from the book, Weak convergence methods for nonlinear partial differential equations by L.C. Evans. He introduces the concept by the following theorem: Theorem. Assume that the sequence $\left\{f_{k}\right\}_{k=1}^{\infty}$ is bounded in $L^{\infty}\left(U ; \mathbb{R}^{m}\right) .$ Then there exists a subsequence $\left\{f_{k_{j}}\right\}_{j=1}^{\infty} \subset\left\{f_{k}\right\}_{k=1}^{\infty}$ and for a.e. $x\in U$ a Borel probability measure $\nu_{x}$ on $\mathbb{R}^{m}$ such that for each $F \in C\left(\mathbb{R}^{m}\right)$ we have $$ F\left(f_{k_{j}}\right) \stackrel{*}{\longrightarrow} \bar{F} \quad \text { in } \quad L^{\infty}(U), $$ where $$ \bar{F}(x) \equiv \int_{\mathbb{R}^{m}} F(y) d \nu_{x}(y) \quad(\text { a.e. } x \in U) $$ Definition. We call $\left\{\nu_{x}\right\}_{x \in U}$ the family of Young measures associated with the subsequence $\left\{f_{k_{j}}\right\}_{j=1}^{\infty}$. My question is: what does $\nu_x(E)$ signify for any Borel set $E\subset \mathbb{R}^m$? Does it measure, for a fixed $x$, how often $f_k(x)$ takes value in $E$ as $k$ changes? REPLY [4 votes]: Related to Nate River's answer, I personally prefer to think of Young measures as single measures $\nu$ on $U \times \mathbb{R}^m$, which have the condition that their projection on the first component (the pushforward along $(x,y) \mapsto x$) is the Lebesgue measure on $U$. For any reasonable function $f:U\to \mathbb{R}^m$, you can define such a measure "living on" the graph of $f$ in $U \times \mathbb{R}^m$.¹ If you have a bounded sequence of functions $f_k$, using the usual compactness theorems for measures, there is a weakly converging subsequence of the corresponding measures (call them $\nu^{k_j}$) and because of the $L^\infty$ bounds, there is no mass escaping towards infinity, so the projection condition is conserved. The limit $\nu$ then is the Young measure limit (in the above sense) for that sequence. In particular, now for $G \in C(U\times \mathbb{R}^m)$, we have $$\lim_{k_j \to \infty} \int_U G(x,f_{k_j}(x)) dx = \lim_{k_j \to \infty} \int_{U\times \mathbb{R}^m} G(x,y) d \nu^{k_j} = \int_{U\times \mathbb{R}^m} G(x,y) d \nu $$ just by definition and weak convergence. If you take $G(x,y) = F(y) \phi(x)$ for fixed $F$ and all $\phi$, you recover precisely the theorem in the question. Now the "classic" Young measure $(\nu_x)_{x\in U}$ then consists of "vertical slices" of this measure, i.e. a disintegration. Since $\{x\} \times \mathbb{R}^m$ always has measure $0$ wrt. to any Young measure defined as above, you can also immediately see, that this is not well defined for any single $x$, but only if you consider enough of them. If you think graphically about the measures $\nu^{k_j}$ and how they converge you can also see what the measure $\nu$ at $(x_0,y_0)$ (and thus the classic Young measure $\nu_{x_0}$ at $y_0$) represents, namely the limit of how often $f_{k_j}(x)$ for $x$ close to $x_0$ takes values $y$ close to $y_0$.² There are many other interpretations of Young measures, but most of them are problem specific. If you know what $f_k$ represents, then this gives you a context of what the resulting Young measure limit is supposed to mean. (e.g. a classical example: If $f_k$ represents regularised solutions to a problem, where you force the oscillating stuff to stay on a scale $\frac{1}k$, then you can argue that $\nu_x$ represents the local microstructure, which is too small to resolve on the scale of functions) There are also a lot of people that offer probabilistic interpretations of Young measures. Personally I am not a fan of those, because there is no randomness involved here. The only reason that $\nu_x$ is a probability measure is because that is the name we give to measures normalized to unit mass. ¹Roughly it is the $n$-dimensional Hausdorff measure restricted to the graph $\{(x,f(x)), x \in U\}$, locally scaled by a factor (something like $\sqrt{1+|\nabla f|^2}^{-1}$) to make the projection work. Alternatively, consider it the pushforward of the Lebesgue-measure on $U$ by $x\mapsto (x,f(x))$ ²In particular the $x$ close to $x_0$ part is something that is often missed. Even if $f_{k_j}(x_0)$, converges to $y_0$, the Young measure $\nu_{x_0}$ can be completely different from $\delta_{y_0}$, because of what happens around it for different $x$. (Though only in a sense, as it is not strictly well defined in the first place).<|endoftext|> TITLE: Cotangent bundles of surfaces as varieties QUESTION [7 upvotes]: As far as I understand, it is easy to see (and find in the literature) that the affine variety $$z_1^2+z_2^2+z_3^2=1$$ with the restriction of the standard $\omega_{std}$ of $\mathbb{C}^3$ is symplectomorphic to $T^*S^2$ with the standard complex structure. Now, as mentioned here (https://www.math.stonybrook.edu/~markmclean/talks/cotangentaffine.pdf), also $T^*T^2$ can also be written explicitly as an affine variety. So my question is: Is there a general way to write $T^*\Sigma_g$ as the zero set of some polynomial in $\mathbb{C}^3$? REPLY [8 votes]: This is not possible. There is something called the growth rate of symplectic cohomology which is subexponential for affine varieties and exponential for cotangent bundles of higher genus surfaces (amongst other things). This was proved by McLean: https://arxiv.org/abs/1011.2542<|endoftext|> TITLE: Looking for source: "How not to be a graduate student" QUESTION [25 upvotes]: I remember having read, about 15 years ago, a transcript of a lecture given by Richard K. Guy, titled "How not to be a graduate student". He gave lots of advice, mostly humorous, concealing sharp and deep observations. Unfortunately, I haven't been able to locate this text on the Internet; I have the feeling I had read it in a newsgroup list, and these things don't seem to exist anymore. Does anyone perchance have a copy of this? REPLY [25 votes]: I was able to locate it, \TeX ed it, and uploaded it to https://www.math.uni-sb.de/ag/bartholdi/pub/Guy-NotGrad.pdf<|endoftext|> TITLE: Are there always more conjugacy classes in the kernel of a morphism to $Z_2$ than not? QUESTION [28 upvotes]: Let $G$ be a finite group and let $\phi:G\to Z_2$ be a homomorphism to the group with two elements. Is it always the case that there are more conjugacy classes in the kernel of $\phi$ than conjugacy classes not in the kernel of $\phi$? I've tried a little bit of messing around algebraically and written down some exact sequences of $G$-modules to try to apply the methods of group cohomology, but I haven't gotten anything to work. My inspiration here is the special case when $G$ is the symmetric group $S_n$ and $\phi$ is the sign homomorphism. In this case conjugacy classes of $G$ correspond to partitions, and the problem becomes about counting partitions of $n$ with an even number of even parts versus an odd number of even parts. I was able to prove (via generating functions and also bijectively) that the number of partitions of $n$ with an even number of even parts minus the number of partitions of $n$ with an odd number of even parts is equal to the number of partitions of $n$ with all parts odd and distinct. I could not find a reference for this fact after some googling, so I would be interested to know if this is a well-known partition identity. I'm also interested in possible extensions of this problem where $Z_2$ is replaced by another group $H$ (possibly required to be abelian). (I just posted this on stackexchange and then learned that mathoverflow might be more appropriate but I'm not sure.) REPLY [10 votes]: This question was the motivation for two papers of mine with John R. Britnell. In Commuting conjugacy classes, an application of Hall's Marriage Theorem to group theory, J. Group. Th, 12 (2009) 795–802, we showed that there is a bijection between the non-split classes in the kernel and those not in the kernel, such that if $C \subseteq \ker \phi$ is paired with $D \subseteq G \setminus \ker \phi$ then $C$ and $D$ have representatives that commute. Here a class $C$ is non-split if its centralizer is not contained in $\ker \phi$; clearly this is a necessary condition for the claimed bijection. This immediately implies that there are at least as many conjugacy classes in $\ker \phi$ as conjugacy classes in $G \setminus \ker \phi$. The proof is elementary, by counting the number of conjugacy classes that can be paired with each $C$, and showing that this number satisfies the conditions for Hall's Marriage Theorem. In the case of the symmetric group this implies the identity counting partitions into an odd/even number of even parts mentioned in the question. (But while it says a bijection exists, it does not give one explicitly.) In Combinatorial proof of a theorem on partitions into an even or odd number of parts, J. Comb. Th. Ser A. 21 (1976) 100–103, Gupta gave an explicit bijective proof of this identity that happens to have the commuting property: see Section 3.1 of my joint paper. In On the distribution of conjugacy classes between the cosets of a group in a cyclic extension, Bull. Lond. Math. Soc. 40 (2008) 897–906, we generalised the counting part of this result to arbitrary cyclic quotients. REPLY [5 votes]: I will restrict to the case that $G$ is finite. There are many interesting results of this type already in the literature, some well-known, some not so well-known (some repeatedly re-discovered, so I will do my best with attributions). Of great relevance are Clifford's theorem, Frobenius reciprocity, and Brauer's permutation lemma, among other things. Brauer's permutation lemma plays a role in changing this to a question about complex irreducible characters. Let $N$ be a normal subgroup of a finite group $G$. Then $G$ acts by conjugation on $N$, and there is a natural action of $G$ on (complex) irreducible characters of $N$ (in both cases, $N$ is acting trivially, and the action is really one of $G/N$). Brauer's permutation lemma tells us that for any $x \in G$, the number of conjugacy classes of $N$ fixed by $x$ is the same as the number of irreducible characters fixed by $x$. The usual orbit counting formula then tells us that $G$ has the same number of orbits on conjugacy classes of $N$ as it does on irreducible characters of $N$. Hence the number of conjugacy classes of $G$ contained in $N$ is the number of $G$ orbits on irreducible characters of $N$. Before I continue, I will fix some notation. Let $G$ be a finite group, $H$ be a (not necessarily normal) subgroup of $G$ and $N$ be a normal subgroup of $G$. We let $k(G)$ denote the number of conjugacy classes of $G$, and we let $k_{G}(N)$ denote the number of $G$-orbits on conjugacy classes of $N$, which is the same as the number of conjugacy classes of $G$ contained in $N$. Notice that it is not generally true that $k(N) = k_{G}(N).$ Frobenius reciprocity is relevant because (even in the case that $H$ is not normal in $G$ ), we may note that for each irreducible character $\mu$ of $H$, ${\rm Ind}_{H}^{G}(\mu)$ has at most $[G:H]$ irreducible constituents (counting multiplicity), while each irreducible character $\chi$ of $G$ occurs as an irreducible constituent of some such induced character. It follows that $k(G) \leq [G:H]k(H)$. Similarly, we find that if $\chi$ is an irreducible character of $G$, then ${\rm Res}^{G}_{H}(\chi)$ has at most $[G:H]$ irreducible constituents, so we obtain $k(H) \leq [G:H]k(G).$ Hence we have $\frac{k(H)}{[G:H]} \leq k(G) \leq [G:H]k(H)$ whenever $H$ is a subgroup of the finite group $G$. These inequalities were proved by P.X. Gallagher around 1962. We will see that they are also relevant to the application of Clifford's theorem, as Gallagher himself knew. Clifford's theorem is relevant for the following reason. If $\chi$ is an irreducible character of $G$, and $N \lhd G$, then the irreducible constituents of ${\rm Res}^{G}_{N}(\chi)$ lies in a single $N$-orbit (and all occur with equal multiplicity). Furthermore, if $\mu$ is an irreducible character of $N$, then there is a bijection between irreducible characters $\chi$ of $G$ such that $\mu$ occurs with non-zero multiplicity in ${\rm Res}^{G}_{N}(\chi)$, and irreducible characters of the $G$-stabilizer of $\mu$, usually denoted $I_{G}(\mu)$, which restrict to $N$ as a multiple of $\mu$. Using projective representations (in Schur's sense), it can be proved that the number of irreducible characters of $I_{G}(\mu)$ which restrict to a multiple of $\mu$ is at most $k(I_{G}(\mu)/N).$ The main idea is to reduce to the case that $N$ is central and $\mu$ is a linear character of $N$ by another part of Clifford's theorem(s). Applying Gallagher's inequalities with $I_{G}(\mu)/N$ in the role of $H$, and $G/N$ in the role of $G$, , we obtain that $k(I_{G}(\mu)/N) \leq [G:I_{G}(\mu)]k(G/N).$ Notice that $[G:I_{G}(\mu)]$ is the length of the $G$-orbit of $\mu$. Letting $\mu$ run through orbit representatives for the action of $G$ on irreducible characters of $N$, we obtain another inequality of P.X. Gallagher (also proved independently by H. Nagao around 1962): $k(G) \leq k(N)k(G/N).$ A variant of the above argument which is sometimes useful, and is relevant to the present question was proved by L.G. K'ovacs and myself in 1993: if $N$ is a normal subgroup of $G$, then we have $k(G) \leq m k_{G}(N)$, there $m$ is the maximum number of conjugacy classes of any subgroup of $G/N$. Notice that if $G/N$ is Abelian, then $m = [G:N] = k(G/N).$ In particular, when $G/N$ is Abelian, we obtain a sharpening of the Gallagher-Nagao bound, for then we have $k(G) \leq k(G/N)k_{G}(N).$ Recalling that $k_{G}(N)$ is the number of $G$-conjugacy classes contained in $N$, we see that whenever $N$ is a normal subgroup of $G$ with $G/N$ Abelian, then we have $k(G) \leq k(G/N)$ $\times$ (the number of conjugacy classes of $G$ contained in $N$). This answers the question in the case $[G:N] =2$ Notice that when $N$ is a normal subgroup of $G$ with $G/N$ non-Abelian, it may not be straightforward to obtain the precise value of $m$, the maximum number of conjugacy classes of any subgroup of $G/N$ (though of course we always have $m < [G:N]$ given that $G/N$ is non-Abelian. Later edit: in fact, it is an easy exercise to check that a non-Abelian group $X$ has at most $|X|-3$ conjugacy classes, and if it has $|X|-3$ conjugacy classes, then $|X| \in \{6,8 \}).$<|endoftext|> TITLE: An integral identity involving cotangents and Bessel functions QUESTION [10 upvotes]: Numerical experiments suggest that the following integral identity holds for Bessel functions of the first kind, $$J_2(t) = 12 \int_0^{1/2}\mathrm{d}x\,\cot \pi x \int_0^x \mathrm{d}y\, \cot \pi y \, J_0(ty)\big[J_0(tx)\,J_0(t(1-x-y))-J_0(t(1-x))\,J_0(t(x-y))\big],$$ but so far I have been unable to prove this. Does anyone have a suggestion how to go about this? Note that by symmetry one could equivalently drop the second term in the brackets and extend the range of $x$ to $(0,1)$, but then one should take the principal value of the integration. REPLY [10 votes]: With hindsight the identity is not that magical: the Bessel functions play only a secondary role, in the sense that there is a more general identity for arbitrary differentiable functions $f : [0,1] \to \mathbb{R}$, namely \begin{align} &12 \int_0^{1/2}\mathrm{d}x\,\cot \pi x \int_0^x \mathrm{d}y\, \cot \pi y \, f(y)\big[f(x)\,f(1-x-y)-f(1-x)\,f(x-y)\big] \\ &= -f(1)\,f(0)^2+ 2 \int_0^1\mathrm{d}x\int_0^{1-x} \mathrm{d}y\, f(x)f(y)f(1-x-y).\tag{1} \end{align} Substituting $f(x) = J_0(t\,x)$ and using the Laplace transform $\mathcal{L}\{J_0(t\, x)\}(\xi) = (t^2 + \xi^2)^{-1/2}$, we find that the double convolution integral satisfies $$\int_0^1\mathrm{d}x\int_0^{1-x} \mathrm{d}y\, J_0(t\,x)J_0(t\,y)J_0(t(1-x-y)) = -\frac{1}{t}J_0'(t).$$ Together with $J_2(t) = -J_0(t)-2 J_0'(t)/t$ and $J_0(0)=1$, this reproduces the claimed identity $$J_2(t) = 12 \int_0^{1/2}\mathrm{d}x\,\cot \pi x \int_0^x \mathrm{d}y\, \cot \pi y \, J_0(ty)\big[J_0(tx)\,J_0(t(1-x-y))-J_0(t(1-x))\,J_0(t(x-y))\big].$$ Let's see why (1) holds. For this it is convenient to introduce the integral $$I(\epsilon):=2\int_\epsilon^{1-2\epsilon}\mathrm{d}x\int_\epsilon^{1-x-\epsilon} \mathrm{d}y\, f(x)f(y)f(1-x-y)$$ with $0<\epsilon <1/4$. The meat of the argument is Hermite's cotangent identity, $$\cot(\pi x)\, \cot(\pi y) + \cot(\pi y)\,\cot(\pi(1-x-y)) + \cot(\pi(1-x-y))\cot(\pi x)= 1,$$ which after insertion in the integrand and using the symmetry under permutation of $x$, $y$ and $1-x-y$, implies $$I(\epsilon) = 6\int_\epsilon^{1-2\epsilon}\mathrm{d}x\int_\epsilon^{1-x-\epsilon} \mathrm{d}y\,\cot(\pi x)\cot(\pi y)\, f(x)f(y)f(1-x-y).$$ Since the integrand is symmetric in $x$ and $y$ we can further reduce the domain to $$I(\epsilon) = 12\int_\epsilon^{1-2\epsilon}\mathrm{d}x\int_\epsilon^{\min(1-x-\epsilon,x)}\!\!\!\! \mathrm{d}y\,\cot(\pi x)\cot(\pi y)\, f(x)f(y)f(1-x-y).$$ Folding the region $1/2< x < 1$ onto $0 < x < 1/2$ via the substitution $x\to 1-x$ gives \begin{align} I(\epsilon) =& 12\int_{2\epsilon}^{1/2}\mathrm{d}x\int_\epsilon^{x-\epsilon}\! \mathrm{d}y\,\cot(\pi x)\cot(\pi y)\, f(y)\big[f(x)f(1-x-y)-f(1-x)f(x-y)\big]\\ &+ 12 \int_\epsilon^{1/2}\mathrm{d}x\int_{\max(x-\epsilon,\epsilon)}^{\min(1-x-\epsilon,x)}\! \mathrm{d}y\,\cot(\pi x)\cot(\pi y)\, f(x)f(y)f(1-x-y). \end{align} Both $I(\epsilon)$ and the first term on the right-hand side have limits given by the corresponding terms in (1) as $\epsilon \to 0$. It remains to examine the last integral on the right-hand side. By the change of variables $x = \epsilon\, \hat{x}$ and $y = \epsilon\,\hat{y}$ we see that it has a limit as $\epsilon\to 0$ given by \begin{align} &\frac{12 f(1)f(0)^2}{\pi^2}\int_1^\infty \mathrm{d}x\int_{\max(x-1,1)}^x \mathrm{d}y \frac{1}{x\,y} \\ &=\frac{12 f(1)f(0)^2}{\pi^2}\int_1^\infty \mathrm{d}y\int_{y}^{y+1} \mathrm{d}x \frac{1}{x\,y} \\ &= \frac{12 f(1)f(0)^2}{\pi^2} \int_{1}^\infty \frac{1}{y^2}\log\left(\frac{y+1}{y}\right) = f(1)f(0)^2. \end{align} This finishes the proof of (1).<|endoftext|> TITLE: Changes forced by the pandemic QUESTION [51 upvotes]: The Covid-19 pandemic has changed our work-lives in ways few of us could have anticipated. These exceptional circumstances have forced each one of us and each one of our institutions to adapt, sometimes in creative ways. I would like to compile a list of those changes and adaptations at all levels of the mathematical ecosystem (all the way from the lives of math undergrads to the working of national funding bodies). For each entry, please discuss the advantages and disadvantages of the new setup, compared to the previous way of doing things. Be as specific as possible. Where relevant, please discuss issues of accessibility of events/resources to people who would otherwise have less access to them, and issues of climate change (less traveling means fewer emissions). REPLY [6 votes]: Interactive virtual poster sessions In this video, Joel Rosenfeld conpares his experience at two different online conferences. ► He describes the fist conference as disappointing because he wasn't able to establish any of the connections that he had hoped to be able to establish through the workshop. In his words: Speakers come, give their talk, and then "poof", they're gone. ► He then proceeds to discuss another online conference that he attended, with special attention to a certain poster session that was organised in a particularly creative way. He calls it "that spatial chat thing" (I don't know the official name). It's really hard to describe the setup using words, so I won't describe it here${}^\dagger$ and thus force you to you watch Joel Rosenfeld's video. He says: I initially thought that this virtual setting was absolutely ludicrous. But later, he says: It turned out to be a lot closer to a regular poster session than I had anticipated. Being able to isolate yourself from the rest of the room and talk to somebody made a huge difference as far as having one-on-one conversations. And finally: Honestly, it was surprisingly effective. [...] At first, I thought this was a circus. [...] It was better than most of the other conferences I've been to. [...] This added a whole degree of interactivity and engagement that I didn't really think was going to be possible. ${}^\dagger$If someone could edit my answer to include the official name of "that spatial chat thing", it would be great.<|endoftext|> TITLE: Is the unit sphere of a Banach space dense in the unit sphere of its second dual with respect to the weak-$\ast$ topology QUESTION [8 upvotes]: To be a bit more precise and fix notations, let $X$ be a Banach space (over $\mathbb{R}$ or $\mathbb{C}$), $X^{\ast\ast}$ its second dual (as a Banach space). Here and in the following we identify $X$ as a (norm) closed subspace of $X^{\ast\ast}$ via the canonical embedding $J : X \hookrightarrow X^{\ast\ast}$. Now let $S_X$ (resp. $S_{X^{\ast\ast}}$) denote the unit sphere (the set of vectors with norm $1$) of $X$ (resp. $X^{\ast\ast}$). Question 1 Is it true that $S_X$ is dense in $S_{X^{\ast\ast}}$ with respect to the $\sigma(X^{\ast\ast}, X^\ast)$ topology? Question 2 Do the above have an affirmative answer in the special case where $X = A$ is a $C^\ast$-algebra, hence $X^{\ast\ast}=A^{\ast\ast}$ is its enveloping von Neumann algebra? Question 3 This is a little more general than Question 2. Let $M$ be a von Neumann algebra acting on some Hilbert space $H$, $A$ a $\ast$-subalgebra of $M$ (not necessarily norm closed) such that $A$ is non-degenerate as an algebra of operators on $H$ and such that the double commutant of $A$ is $M$. Is it true that $S_A := \{a \in A \mid \|a\| = 1\}$ is dense in $S_M := \{x \in M \mid \|x\| = 1\}$ with respect to the weak operator topology? If we replace unit spheres by closed unit balls, all of the above questions have an affirmative answer (Question 1 is Goldstine's theorem, and Question 2 and 3 part of Kaplansky's density theorem). I was wondering whether the above finer statements still hold. Do we have (counter-)examples? REPLY [9 votes]: I think a simple rescaling argument works. I do Question 1, Q3 being similar. Given $F\in S_{X^{**}}$, by Goldstine there is a net $(x_i)$ in $B_X$ converging weak$^*$ to $F$. Given $\epsilon>0$ there is $f\in S_{X^*}$ with $|F(f)|>1-\epsilon$ and so $$ 1-\epsilon < |F(f)| = \lim_i |f(x_i)| \leq \|f\|\liminf_i\|x_i\| \leq \liminf\|x_i\| \leq 1, $$ from which it follows that $\liminf \|x_i\|=1$. So for $\epsilon>0$ there is $i_0$ with $\|x_i\| > 1-\epsilon$ for $i\geq i_0$, but as $\|x_i\|\leq 1$, we conclude that actually $\lim_i \|x_i\| = 1$. Thus, wlog $x_i\not=0$ for all $i$, and so we can define $y_i = \|x_i\|^{-1} x_i \in S_X$ for each $i$. Then $\lim_i \|x_i - y_i\| = \lim_i \|x_i\| |1-\|x_i\|^{-1}| = 0$, and so also $(y_i)\rightarrow F$ weak$^*$, as required.<|endoftext|> TITLE: sci.math.research archive? QUESTION [20 upvotes]: Does there exist an archive somewhere of posts to the USENET newsgroup sci.math.research? The best approximation I'm aware of is Google Groups. However, despite the Google brand name, the search capability of Google Groups is abysmal, and it is hard to tell how much content is really archived there. I have an old bookmark to a Math Forum URL that currently seems to be dead. An old website called the The Mathematical Atlas indicates that the Math Forum moved to forum.swarthmore.edu but that site currently seems not to be working either. The Usenet Archives website, as of this writing, seems to have archived a grand total of two (2) posts from sci.math.research. This question was prompted in part by another MO question, Looking for source: “How not to be a graduate student”, which hints that the desired transcript may have been posted to a USENET newsgroup such as sci.math.research or sci.math. In my opinion there was some valuable content posted to sci.math.research and it would be a shame if it were lost forever. REPLY [13 votes]: This is probably better suited as a comment to Gro-Tsen's answer, but I do not have an account on this site. A relatively complete archive (including messages spanning 1991 through 2013) appears to be available on the Archive.org site at: https://archive.org/details/usenet-sci and in particular at the link: https://archive.org/download/usenet-sci/sci.math.research.mbox.zip REPLY [10 votes]: There's a bit of content on Archive.org: the link says sci.math, but among the files in the collection there is a sci.math.research.20140626.mbox.gz which contains ~11k messages apparently posted between 2003 and 2014. Sadly, this does not seem to contain the specific content you were looking for. You could try contacting Archive.org: even though they don't have the data, they might be in position to persuade Google to give them a full dump of a low-volume high-quality newsgroup like sci.math.research since it's part of their service mission to archive this sort of data. Another option is to ask Daniel Grayson, who was moderator for sci.math.research (he wasn't the only one, but there seems to be a connection between the newsgroup and UIUC so maybe the chances are higher): it isn't unlikely that the moderation process kept a complete archive at some level.<|endoftext|> TITLE: Why stable $\infty$-categories? QUESTION [24 upvotes]: I begin by saying that while I understand what a triangulated / derived category is pretty well, I know nothing about Higher Algebra stuff and not even $\infty$-categories. I've heard some people say that stable $\infty$-categories are a "more natural" point of view to derived categories. Can anyone explain in simple terms (given my background) to me why this is so? I study algebraic and arithmetic geometry but all possible motivations are welcome. REPLY [39 votes]: I already answered some version of this question in this answer, but let me try to expand a bit on the concrete advantages in mathematical practice. For understanding the following you need to take on faith that ∞-categories exist and have roughly the same properties as ordinary categories (which in fact are just a special example of ∞-categories). The derived $\infty$-category of a scheme One of the biggest advantages of stable ∞-categories compared to triangulated categories is that they work well in families. This already should be appealing to an algebraic geometer. But it turns out that this makes constructing stable ∞-categories easier than triangulated categories (contrary to what you might be expecting). For example let me sketch a construction of the symmetric monoidal ∞-category of a scheme (you can easily adapt it to algebraic spaces/stacks etc..) First you need to do the affine case: the derived category of a ring. There are many ways of doing it. The most concrete is probably taking the 1-category of chain complexes and inverting quasi-isomorphisms (in the ∞-categorical sense) $$\mathscr{D}(R):=\operatorname{Ch}(R)[q.iso^{-1}]$$ Then you need to prove that it is functorial, symmetric monoidal etc. I personally favour another (equivalent) definition, which uses more technology but it makes all properties follow formally. I am taking as the definition of the derived ∞-category the stabilization of the animation of the category of finitely generated projective $R$-modules $$\mathscr{D}(R):=\operatorname{Sp}(\mathcal{P}_\Sigma(\operatorname{Proj}_R))\,.$$ Then $\mathscr{D}(R)$ has all the properties we want because $\operatorname{Proj}_R$ does. Moreover you can still show easily that $\mathscr{D}(R)$ is localization of the category of chain complexes at quasi-isomorphisms by using a clever trick, so you lost nothing in concreteness. Our next step is to show that $\mathscr{D}(-)$ as a functor on affine schemes satisfies Zariski descent. Concretely it boils down to showing that the following square of stable ∞-categories is cartesian $$\require{AMScd} \begin{CD} \mathscr{D}(R) @>>> \mathscr{D}(R[1/f])\\ @VVV @VVV\\ \mathscr{D}(R[1/g]) @>>> \mathscr{D}(R[1/fg])\end{CD}$$ and you can prove this is true again because it is true for $\operatorname{Proj}_R$ (where it is elementary). NOTE THAT THIS IS FALSE FOR THE CORRESPONDING TRIANGULATED CATEGORIES This is one of the amazing advantages of stable ∞-categories compared to triangulated categories: you can glue their objects. Using this now it is clear how to define the derived $\infty$-category of a scheme $X$ $$\mathscr{D}(X)=\lim_{\operatorname{Spec}R\subseteq X} \mathscr{D}(R)$$ where the limit is taken over the poset of open affine subsets of $X$. That is, we're saying that an object of $\mathscr{D}(X)$ is just the collection of an object of $\mathscr{D}(R)$ for each open affine subscheme plus suitable gluing data. Now by formal properties of limits this is automatically a symmetric monoidal $\infty$-category: we can take tensor products of elements in $\mathscr{D}(X)$ by taking them in any affine open and gluing back the resulting objects. If all you have are triangulated categories, constructing the symmetric monoidal structure on $h\mathscr{D}(X)$ is highly non-trivial. Stability is a property One great advantage of stable $\infty$-categories is that stability is a property, not a structure. To construct a triangulated category it's not enough to construct the category: you also have to come up with a shift functor, and a family of exact triangles etc. In stable ∞-categories you don't have to worry about that: you just construct a certain ∞-category and then check that, say the (canonically defined) functor $\Omega$ is an equivalence. This has advantages because for example it's clear what should be a stable symmetric monoidal ∞-category: it's just a symmetric monoidal ∞-category which is stable and such that the tensor product is exact in each variable. Try instead to come up with the notion of symmetric monoidal triangulated category. I strongly suspect you would not come up with all the required axioms (there is a compatibility between the shift and the tensor product which has some tricky signs). Homotopy (co)limits Triangulated categories rarely have colimits. They have direct sums and usually little more. Stable ∞-categories instead have all finite limits and colimits (and most of those you'll encounter in practice have all small limits and colimits). This gives you a huge flexibility in working in them. For example you can say that you can reconstruct the sheaves (say on a space $X$) from a sheaf on a neighborhood of a closed subset of $Z$ and a sheaf on the complement, plus some gluing data. This is classical for sheaves of abelian groups, but it gets tricky (although not impossible) to state for derived categories of sheaves. For derived ∞-categories, however, the same statement as for sheaves of abelian groups work, with the same proof. Algebraic K-theory A powerful motivation for me is that you cannot define the higher algebraic K-theory of a triangulated category. There are stable ∞-categories that have the same underlying triangulated category but not the same higher algebraic K-theory. As a person that finds algebraic K-theory very interesting, this would be enough for me to not want to work with triangulated categories altogether Sheaves You can actually have sheaves with values in an ∞-category. It makes perfect sense, say, to say that algebraic K-theory gives you a Zariski (or Nisnevich) sheaf of spectra on the category of schemes. This ended up allowing wonderful computations that would have been hard to do without stable ∞-categories. Perhaps not impossible, but most of the older work in the same vein used secretly something like dg-categories which has many of the same features (although not the same versatility), and in any case not triangulated categories.<|endoftext|> TITLE: Trisecting $3$-fold sumsets: is the middle part always thick? QUESTION [6 upvotes]: Here is a truly minimalistic and seemingly basic question which should have a simple solution (I hope it does). Let $A$ be a finite set of integers with the smallest element $0$ and the largest element $l$. The sumset $C:=3A$ resides in the interval $[0,3l]$, and we let $C_1:=C\cap[0,l]$, $C_2:=C\cap[l,2l]$, and $C_3:=C\cap[2l,3l]$. Is it true that $|C_2|\ge\frac12\,(|C_1|+|C_3|)$, for any choice of the set $A$? Computations seem to suggest that the answer is in the affirmative. REPLY [6 votes]: No. Take $A = \{0,1,\ldots,9,10,20,30,\ldots,90,100,200,300,\ldots,900,1000\}$. Then $|C_1|=1001$, $|C_2|=272$ and $|C_3|=29$. A smaller counterexample in the same spirit is $\{0,1,2,3,4,5,10,15,20,25,50,75,100\}$, with sizes $101, 53, 13$.<|endoftext|> TITLE: How complicated can the path component of a compact metric space be? QUESTION [6 upvotes]: Let $X$ be a compact metric space and $P$ be a path component of $X$. Since we are not assuming $X$ is locally path connected, $P$ must need not be open nor closed. Certainly, $P$ must be separable but how bad can $P$ be? Is every separable path-connected metric space, homeomorphic to the path component of some compact metric space? I'm curious about this question because subspaces like $w(Y)=\{x\in X\mid \text{$X$ is not semilocally simply connected at $x$}\}$ are important homotopy-invariants in the study of Peano continua. If $Y$ is compact and locally path-connected, $X=w(Y)$ may be any compact metric space. Iteration of $w(-)$ becomes relevant. By this, I mean take a path component $P$ of $X$ and then take $w(P)$. This is why I want to know how bad $P$ can be. However, I don't know much about metric compactifications that would be relevant. REPLY [4 votes]: Not every path connected separable metric space is homeomorphic to a path component of a compact metric space. The following cardinality arguments can be used: Fact 1. There is up to homeomorphism more than $|\mathbb R|$-many path-connected separable metric spaces: Just consider all the subspaces of the form $X_A=(\mathbb R\times (0,1))\cup (A\times\{0\})$ for $A\subseteq \mathbb R$. Each of them is path-connected and every space $X_A$ is homeomorphic to at most $|\mathbb R|$-many spaces $X_B$. Fact 2. There is up to homeomorphism only $|\mathbb R|$-many analytic subsets of compact metric spaces. Fact 3. Every path-connected component of a compact metric space is analytic. Hence there is at least one path connected space of the form $X_A$ which is not homeomorphic to a path component of a compact metric space.<|endoftext|> TITLE: Peer review 2.0 QUESTION [92 upvotes]: I have an idea for a website that could improve some well-known difficulties around peer review system and "hidden knowledge" in mathematics. It seems like a low hanging fruit that many people must've thought about before. My question is two-fold: Has someone already tried this? If not, who in the mathematical community might be interested in creating and maintaining such a project or is working on similar projects? Idea A website dedicated to anonymous discussions of mathematical papers by experts. Motivation 1: Hidden knowledge Wilhelm Klingenberg's "Lectures on closed geodesics" can be found in every university's math library. One of the main theorems in the book is the following remarkable result, a culmination of decades of work on Morse theory of the loop space by many mathematicians: Every compact Riemannian manifold contains infinitely many prime closed geodesics. Unfortunately, there is a mistake in the proof. 44 years after the book's publication the statement is still a widely open problem. The reason I know this is because when I was in grad school I mentioned the book to my adviser and my adviser told me about it. If I tried to look for this information online I wouldn't find it (in fact, I still haven't seen it written down anywhere). This is one of many examples of "hidden knowledge", information that gets passed from adviser to student, but is inaccessible to an outsider. In principle, a new Ramanujan can open arxiv.org and get access to the cutting edge mathematical research. In reality, the hidden knowledge keeps many mathematical fields impenetrable to anyone who is not personally acquainted with one of a handful of experts. Of course, there is much more to hidden knowledge than "this paper form 40 years ago actually contains a gap". But I feel that the experts' "oral tradition" on papers in the field is at the core of it. Making it common knowledge will be of great benefit to students, mathematicians from smaller universities, those outside of North America and Europe, people from adjacent fields, to the experts themselves and to the mathematical progress. Motivation 2: Improving peer review Consider the following situations: You are refereeing a paper and get stuck on some minor issue. It will take the author 5 minutes to explain, but a few hours for you to figure it out on your own. But it doesn't quite feel worth initiating formal communication with the author through the editor over this and you don't want to break the veil of anonymity by contacting the author directly. You are being asked to referee a paper, but don't have time to referee the whole paper. On the other hand, there is a part of it that is really interesting to you. Telling the editor "yes, but I will only referee Lemma 5.3" seems awkward. You are refereeing a paper that is at the intersection of your field and a different field. You would like to discuss it with an expert in the other field to make sure you are not missing anything, but don't know a colleague in that area or feel hesitant revealing that you are a referee for this paper. These are some of many situations where ability to anonymously discuss a paper with the author and other experts in a forum-like space would be helpful in the refereeing process. But also outside of peer review mathematicians constantly find small errors, fillable gaps and ways to make an old paper more understandable that they would be happy to share with the others. At the same time, they often don't have time to produce carefully polished notes that they would feel comfortable posting on arxiv, or if they post notes on their website they may not be easy to find for anyone else reading the paper. It would be helpful to have one place where such information is collected. How will it work? The hope is to continue the glorious tradition of collaborative anonymous mathematics. One implementation can work like this: Users of the website can create a page dedicated to a paper and post questions and comments about the paper on that page. To register on the website one needs to fill in a form asking for an email and two links to math arxiv papers that have the same email in them (this way registration does not require verification by moderators) and choose their fields of expertise. When a user makes a comment or question only their field/fields are displayed. REPLY [18 votes]: Imho, mathematics has already a good culture and a relatively rich informational ecosystem to detect, document, and disseminate these issues (many options are mentioned above); what might be lacking is a kind of (semi)automated system of interlinked communication that helps spreading the information further. Judging from the several examples we had to deal with at zbMATH Open (and the discussions involved), I would certainly agree that it seems less feasible to have a single seemingly authoritative source, but we need to employ and coordinate our existing procedures. A natural problem would be, e.g., the scope - there is a number of problematic (mostly, APC) journals around where one has doubts about the proper peer review - would you like to have a website that addresses all of them (including simple Fermat/Goldbach proofs etc.)? Likely, serious mathematicians lack the resources to do that. But also with restricted scope, every single issue is a tedious work. To give a recent example, even in a rather clear-cut case Ramm, Alexander G., Solution of the Navier-Stokes problem, Appl. Math. Lett. 87, 160-164 (2019). ZBL1410.35100. it was impossible to convince the author of the reviewer's arguments (although we employed another expert as arbiter who went through the arguments and clearly ruled in favour of the reviewer). Every such issue is different and needs to be handled carefully. On the technical side, e.g., if we detect such an issue with a widely accepted older article via zbMATH Open (documented usually as an additional "Editorial remark" in the original) which is lacking an erratum/corrigendum for this, I currently post this also manually as an answer at the standard question at MathOverflow Widely accepted mathematical results that were later shown to be wrong? which has the additional advantage that an automated backlink in the respective zbMATH entry is generated (e.g., see the entry Saito, Masa-Hiko, On the infinitesimal Torelli problem of elliptic surfaces, J. Math. Kyoto Univ. 23, 441-460 (1983). ZBL0532.14019.) Of course, this doesn't mean everyone is aware of this information, and there are likely various ways to improve/automatize things further. One could think about creating a specific MathOverflow entry/tagging for each issue/publication that needs discussion/updates which could trigger an activity on our and other sites. But the main effort will likely always be mathematically, not technically, and require mathematicians who are willing to explore things deeply. The platforms are there, we should just make sure that the documentation is appropriate, accessible, and ideally increasingly interoperable.<|endoftext|> TITLE: Composite knots and their braid words QUESTION [5 upvotes]: Given a composite knot K = K_1 # K_2, I wonder how the braid word looks like. Is it possible to see from the word that the knot is composite? I am not aware of a statement such as "the closure of a braid of a given form is a composite knot" and wouldn't hope for it. But then there is Vogel's algorithm, perhaps something similar more advanced can give an organised way to keep the sum visible somehow when bringing the knot K into a braid presentation? REPLY [5 votes]: Birman and Menasco show that the `composite braid representation' described by Marco Golla in his comment is in some sense canonical. That is, you can get from an arbitrary braid word to the one that Marco describes by a series of moves. See Birman-Menasco, Studying links via closed braids. IV. Composite links and split links, Invent. Math. 102 (1990), no. 1, 115–139. There is also a later erratum in Invent. Math. 160 (2005), no. 2, 447–452 that corrects the proof; the statement remains the same. The erratum refers to a lot of subsequent work that might be useful.<|endoftext|> TITLE: The number of chains of chordal graphs QUESTION [9 upvotes]: Consider a saturated chain $G_0 \subset G_1 \subset \cdots \subset G_m$ of graphs on $n$ labelled vertices, where $G_i$ has $i$ edges, and $m = {{n}\choose {2}}$. Altogether there are $m!$ such chains of graphs (they are also known as "graph processes") and they form an interesting topic of study. Recall that $G$ is chordal if all induced cycles in $G$ are 3-gons. The question Let $f(n)$ be the number of such chains where all graphs $G_i$ are chordal. What is the value of $f(n)$? a) for small values of $n$? b) what is its asymptotic behaviour? c) Is there an exact formula? Of course $f(n) \le {{n} \choose {2}}!$, and it is easy to see that $f(n) \ge \prod_{k=1}^{n-1} k!$. Variations: We can also ask about a) saturated chains of perfect graphs, b) saturated chains of chordal graphs whose complement is also chordal. Motivation For a saturated chain of graphs $G_0 \subset G_1 \subset \cdots \subset G_m$ we can look at the vector $(v_0,v_1,...,v_{n-1})$ where $v_k$ is the number of indices $j$ where $G_{j+1}$ has $k$ additional triangles compared to $G_j$. Such vectors introduced by Beus in 1970 (in the context of sorting algorithms) are interesting parameters of saturated chains of graphs and the case of chordal graphs is precisely the case where the vector is $(n-1,n-2,\dots,1)$. Remark: $f(3)=6$, and $f(4)=576$. I am curious to know the values of $f(5)$ and $f(6)$. REPLY [12 votes]: This is not a general answer but the big numbers don't fit properly into a comment. f(1) = 1 f(2) = 1 f(3) = 6 f(4) = 576 f(5) = 1416960 f(6) = 120678543360 f(7) = 455010170456862720 f(8) = 95371866538619173904056320 f(9) = 1383866987105877308750365304858542080 f(10) = 1716187027583005555045945024371317843956845772800 f(11) = 221917018834976627508152930913765491170568412125060985539788800 f(12) = 3598055237740601485367382153175891099609454479883844294426214728495086488780800 f(13) = 8665460290021468438320782226358244848272843476236433280013965605190231652374443764439998581964800 The last two took about 10 minutes and 5 hours. For a graph $G$ let $g(G)$ be 0 if $G$ is not chordal and equal to the number of chordal chains back to the empty graph otherwise. Set $g({\rm empty})=1$ and for one member $G$ of each isomorphism class of nonempty chordal graph in non-decreasing order of the number of edges, do $$ g(G) := \sum_{e\in E(G)} g(G-e). $$ The answer is $g(K_n)$. No adjustment for automorphism group size is needed. The only technical requirement is recognising $G-e$ as isomorphic to a previous graph. The program can handle any class of graphs closed under isomorphism provided they all fit into memory at once. On request, here is the corresponding table for split graphs. s(1) = 1 s(2) = 1 s(3) = 6 s(4) = 480 s(5) = 719040 s(6) = 28111985280 s(7) = 39667596799259520 s(8) = 2716101119587792215121920 s(9) = 11750142295253741381979240922398720 s(10) = 4059370170952132363824590307446791630779187200 s(11) = 138004666315436722628999805261994204164032807656029840998400 s(12) = 557103455087735168484078548670473120844063643381325957791547628642631680000 s(13) = 316753104615638650562235298836069531430557783996203420700809420563227053308369342951115980800 s(14) = 29665849491651526562732309913886504922801500810240504259322068041948739753073885726711112718885233432115281920000<|endoftext|> TITLE: Verifying the Lefschetz Conditions for crystalline cohomology QUESTION [7 upvotes]: For context, I am rather new to the whole business of abstract Weil cohomology theories and motives in general, so if I am not making sense somewhere, do let me know! In many of the literature that I am consulting while trying to learn about Weil cohomology theories and motives, it is often said that these cohomology theories must, in particular, satisfy the Weak and Hard Leftschetz Conditions. However, The Stacks Project apparently does not impose this axiom upon Weil cohomology theories (see their Tag 0FHA). Am I misinterpreting The Stacks Project, or can the Lefschetz Conditions be deduced from the other Weil cohomology axioms ? The other thing that I have come across is that the theory of crystalline cohomology is a Weil cohomology theory, but for some reason I can not find any source which confirms that the Lefschetz Conditions are satisfied here. Can anyone point me to such a source, or alternatively, does crystalline cohomology even satisfy the Lefschetz Conditions at all ? (I should note that I'm not yet too familiar with crystalline cohomology.) Thank you! REPLY [11 votes]: The Hard Lefschetz theorem can certainly not be deduced formally from the axioms of a Weil cohomology theory given in the Stacks Project. The reason it is called "hard" Lefschetz is that it is a really hard theorem, and deeper than the other properties. The situation is rather that the axiomatization of what it means to be a Weil cohomology theory is perhaps not completely standardized; it is sometimes sloppily used as a catch-all term for "a cohomology theory with cycle maps, duality, traces, and all of the usual things that go with such things". I believe that it is more standard to not include the hard and weak Lefschetz theorems as axioms for a Weil cohomology theory. The Hard Lefschetz theorem holds in crystalline cohomology. This was first proven by Katz--Messing ("Some consequences of the Riemann hypothesis for varieties over finite fields"), who deduced it from the weak Lefschetz theorem in crystalline cohomology, the Hard Lefschetz in étale cohomology, and Deligne's proof of the Weil conjectures.<|endoftext|> TITLE: Trisecting $3$-fold sumsets, II: is the middle part ever thin? QUESTION [6 upvotes]: This is a refined version of the question I asked yesterday. Let $A$ be a finite set of integers with the smallest element $0$ and the largest element $l$. The sumset $C:=3A$ resides in the interval $[0,3l]$, and I write $C_1:=C\cap[0,l)$, $C_2:=C\cap[l,2l]$, and $C_3:=C\cap(2l,3l]$. Is it true that $|C_2|\ge\min\{|C_1|,|C_3|\}$, for any choice of the set $A$? A brute force computation shows that there are no counterexamples with $l\le 22$. REPLY [6 votes]: The title and body are asking the question in opposite senses. For the title, the answer is "yes" (it can be thin), and for the body, the answer if "no" (it is not true that it is never thin). An example is $$ A = \{0,1,2,3,8,11,26,38,56,69,85,89,179,189,197,221,226,243,254,257,264,266,269,270\} $$ where the sumset parts $C_1,C_2,C_3$ have sizes $270, 268, 269$. Perhaps a word about the construction. The underlying idea was to take two shorter sets $P,Q$ that are as thin (small) as possible, while their sumsets $3P, 3Q$ would cover some interval $[0,b]$ fully. Then make $A = P \cup (l-Q)$ with some conveniently chosen $l$. Hopefully then $3A$ covers the left and right thirds almost fully, while the middle third might have enough holes to make a counterexample. Note the "might" – this is all heuristic, and it is not done until you just try it out. I took $$P=\{0,1,2,3,8,11,26,38,56,69,85,89\}$$ $$Q=\{0,1,4,6,13,16,27,44,49,73,81,91\}$$ from Challis and Robinson (2010), Some Extremal Postage Stamp Bases, Journal of Integer Sequences 13, Article 10.2.3; Table "h=3" on page 9. Here $P$ and $Q$ are minimal-cardinality additive bases (of order 3) for the interval $[0,186]$, so that part of the construction is done. Then a brute force search over some possible values of $l$ gave a hit at $l=270$. Indeed, $C_1$ has no "holes" (it covers the integer interval $[0,269]$ fully). $C_2$ has three holes in interval $[270,540]$, and $C_3$ has one hole in interval $[541,810]$.<|endoftext|> TITLE: Are there any papers about this observation of the distribution of the zeros of the zeta function? QUESTION [13 upvotes]: Choose some $x > 1$. Then $$ \lim_{T\to\infty} \sum_{\Im(\rho) TITLE: Transcendence of values of Fredholm series at algebraic arguments QUESTION [6 upvotes]: Let $d$ be an integer greater than $1$ and let $f(z)=\sum_{n\ge0}z^{d^n}$ be the Fredholm series. It is well-known that the Roth-Ridout theorem implies that $f(r)$ is a transcendental number for all non-zero rational $r$. By Mahler’s method, it is even true for any nonzero algebraic $r$. Can one extend the arguments of the proof based on the Roth-Ridout theorem to recover Mahler's result? Thank you in advance. REPLY [3 votes]: It seems that the answer is no. From this paper, the Roth-Ridout theorem is not sufficient to establish transcendence for algebraic $r$. Indeed, it's not enough to completely cover the rational case. To obtain the same result as the Mahler's method, one should invoke the more general ($p$-adic) Subspace Theorem, as shown in the Corvaja-Zannier paper.<|endoftext|> TITLE: A basis of holomorphic differentials on Fermat curves QUESTION [5 upvotes]: I am currently reading the paper "Holomorphic Differentials of Generalized Fermat Curves" by Rubén Hidalgo. The case I am interested in is that of a classical Fermat curve $F_k$, which in his terminology is a generalized Fermat curve of type $(k, 2)$. This is the curve in $\mathbb{P}^2(\mathbb{C})$ defined by the homogeneous equation $x_1^k + x_2^k + x_3^k = 0$. With regard to this special case, Hidalgo defines meromorphic maps on $F_k$ via $y_2 = \dfrac{x_2}{x_1}$, $y_3 = \dfrac{x_3}{x_1}$ (note that Hidalgo writes $z:=y_2$). In section 3.2, he constructs a set of holomorphic differential forms given by the formula $\theta_{r;\alpha} = \dfrac{y_2^r dy_2}{y_3^{\alpha}}$, where $0 \le \alpha \le k-1$, and $0 \le r \le \alpha - 2$, for a total of $g(F_k) = \dfrac{(k-1)(k-2)}{2}$ distinct forms. Hidalgo also defines symmetries of the curve by \begin{align} a_1([u:v:w] = [\omega_k u:v:w]) \\ a_2([u:v:w] = [u:\omega_k v:w]) \\ a_3([u:v:w] = [u:v:\omega_k w]) \end{align} where $\omega_k$ is a primitive $k$-th root of unity. He remarks that $\theta_{r;\alpha}$ pulls back under these symmetries according to \begin{equation} a_j^*(\theta_{r;\alpha}) = \begin{cases} \omega_k^{r + 1 - \alpha} \theta_{r;\alpha} & j = 1 \\ \omega_k^{-r-1} \theta_{r;\alpha} & j = 2 \\ \omega_k^{\alpha} \theta_{r;\alpha} & j = 3 \end{cases} \end{equation} He denotes the divisor with a value of 1 at each fixed point of $a_j$ by $\textrm{Fix}_{div}(a_j)$, so that the divisor of $\theta_{r;\alpha}$ is given by \begin{equation} (\theta_{r;\alpha}) = (\alpha - 2 - r) \textrm{Fix}_{div}(a_1) + r \textrm{Fix}_{div}(a_2) + (k-1-\alpha) \textrm{Fix}_{div}(a_3) \end{equation} His theorem 3.1 states that these $\theta_{r;\alpha}$ form a basis for the space of holomorphic differentials on $F_k$. He claims that linear independence follows immediately from the above pullback and divisor formulas for $\theta_{r;\alpha}$, and does not elaborate the details. I have as of yet been unable to see why this is true. I have been trying to see if there's a simple linear algebraic approach to the group generated by the pullbacks, to show that a single nontrivial linear dependence equation will transform into enough additional constraints under the pullbacks to be unsatisfiable and produce a contradiction. From what I have found so far it looks like I will get at most $k$ independent constraints using this method, so this alone does not seem promising. I do not see how the divisors come into it. I understand that there is general classical theory behind constructing holomorphic differentials on projective curves, and have found similar-looking results in references such as Plane Algebraic Curves by Brieskorn and Knorrer, section 9.3, theorem 1. However, my knowledge of algebraic geometry is limited to Riemann surfaces. I think Hidalgo is indicating that the proof in this special case may be simpler. Can anyone please point me in the right direction? REPLY [5 votes]: The automorphisms $a_1,a_2,a_3$ are part of a $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$ action on $F_k$, which in turn induces an action on the space $H^0(\Omega^1_{F_k})$ of holomorphic differentials on $F_k$. Let me explain how the representation-theory of this group gives the desired independence. The $\theta_{r,\alpha}$ define a linear map $\displaystyle\bigoplus_{r,\alpha}\mathbb{C}\rightarrow H^0(\Omega^1_{F_k}),$ and their linear independence is equivalent to this map being injective. The pullback rules give each $\mathbb{C}$-summand above a $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$-representation structure so that the map is equivariant. Call this representation (for a fixed $(r,\alpha)$) $\mathbb{C}_{r,\alpha}.$ Importantly, for different $(r,\alpha)$, the $C_{r,\alpha}$ are different representations. Now assume that $\oplus_{r,\alpha}\mathbb{C}_{r,\alpha}\rightarrow H^0(\Omega^1_{F_k})$ is not injective. The kernel, which must be a $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$-subrepresentation of $\displaystyle\bigoplus_{r,\alpha}\mathbb{C}_{r,\alpha},$ must in fact take the form $$\displaystyle\bigoplus_{(r,\alpha)\in S}\mathbb{C}_{r,\alpha}$$ for some set $S$. But if $S$ is nontrivial, then that implies that some $\theta_{r,\alpha}$ is trivial, which is clearly not true. If you wish to remove the representation theory from this argument, here is an elementary translation. (But I still recommend learning the representation theory, because it's broadly useful for this kind of study.) Assume the $\theta_{r,\alpha}$ are not linearly independent. Then take a relation $$\displaystyle\sum_{(r,\alpha)\in S} c_{r,\alpha}\theta_{r,\alpha}=0$$ with $|S|$ minimal. Pulling back by $a_2$, we get another relation $$\displaystyle\sum_{(r,\alpha)\in S} \omega_k^{-r-1}c_{r,\alpha}\theta_{r,\alpha}=0.$$ Now, unless these two relations are multiplies of each other, we can take some linear combination of them to get a relation with less terms, a contradiction. So they must be multiples, i.e., all the $(r,\alpha)\in S$ must have the same value of $r$. Repeating this argument but for the pullback by $a_3$ we see that they must also all have the same value of $\alpha$, i.e., $S$ must be a one element set. But that would imply that $\theta_{r,\alpha}$ is trivial, a contradiction.<|endoftext|> TITLE: on a strange character sum QUESTION [7 upvotes]: Recently while studying cubic residues modulo a prime $p$ with $p\equiv1\pmod 3$, I met the following character sum: $$\sum_{0\le x\le p-1}\left(\frac{x}{p}\right)\left(\frac{x+1}{p}\right)\left(\frac{x+\omega}{p}\right)\left(\frac{x+\omega^2}{p}\right)\chi_p(x^3),$$ where $\omega\in\mathbb{Z}$ with $\omega$ mod $p$ a primitive cubic root of unity in $\mathbb{Z}/p\mathbb{Z}$, $\chi_p$ is a generator of the group of all multiplicative characters over $\mathbb{Z}/p\mathbb{Z}$, and $(\frac{\cdot}{p})$ is the Legendre symbol. Are there any references concerning the character sums of this type? Your comments are welcome. REPLY [13 votes]: Your sum is expressible as a linear combination of Jacobi sums. More precisely, let us notice that $$ \left(\frac{x^3}{p}\right)=\left(\frac{x}{p}\right) $$ for all $x$ and also that $(x+1)(x+\omega)(x+\omega^2)=x^3+1$. So, the sum $S$ in question is equal to $$ S=\sum_{0\leq x\leq p-1}\left(\frac{x^3}{p}\right)\left(\frac{x^3+1}{p}\right)\chi_p(x^3). $$ Now, let $\psi$ be one of two non-principal characters modulo $p$ with $\psi^3=\chi_0$. It is easy to see that $$ \#\{x:x^3\equiv y \pmod p\}=1+\psi(y)+\psi^2(y) $$ (i.e. it is $1$ when $y=0$, $3$ when $y$ is a nonzero cubic residue and $0$ if it is a cubic non-residue) Arranging the sum according to the value of $x^3$, we obtain $$ S=\sum_{0\leq y\leq p-1}\left(\frac{y}{p}\right)\left(\frac{y+1}{p}\right)\chi_p(y)(1+\psi(y)+\psi^2(y)). $$ To transform this into Jacobi sums, let us make a substitution $y=-z$ and notice that $\psi(-1)=1$, $\chi_p(-1)=-1$ and $\left(\frac{-1}{p}\right)=(-1)^{(p-1)/2}:=\varepsilon_p$. We get $$ S=-\varepsilon_p\sum_z \left(\frac{z}{p}\right)\left(\frac{1-z}{p}\right)\chi_p(z)(1+\psi(z)+\psi^2(z)). $$ Next, if $\eta_1,\eta_2$ are characters modulo $p$, then the Jacobi sum $J(\eta_1,\eta_2)$ is given by $$ J(\eta_1,\eta_2)=\sum_z \eta_1(z)\eta_2(1-z). $$ For instance, we get $$ S=-\varepsilon_p(J(\chi_1,L_p)+J(\chi_2,L_p)+J(\chi_3,L_p)), $$ where $L_p$ is the Legendre symbol and $\chi_i=\psi^iL_p \chi_p$. It is also known that if $\eta_1\eta_2$ is non-principal, then $$ J(\eta_1,\eta_2)=\frac{g(\eta_1)g(\eta_2)}{g(\eta_1\eta_2)}, $$ (here $g$ is the Gauss sum $g(\eta)=\sum_n \eta(n)e^{2\pi i n/p}$) so in this case we obtain $$ S=-\varepsilon_p g(L_p)\sum_i \frac{g(\chi_i)}{g(\psi^i\chi_p)} $$ and as a consequence of the identity $|g(\chi)|=\sqrt p$ for non-principal $\chi$, we prove Weil-like estimate $$ |S|\leq 3\sqrt p. $$<|endoftext|> TITLE: What's the cohomology ring structure of a blow-up? QUESTION [10 upvotes]: Let $X$ be a compact Kähler manifold, with $j_Z: Z\hookrightarrow X$ a submanifold of complex codimension $r$, $\tau: \widetilde{X} \to X$ the blow-up of $X$ along $Z$, with exceptional divisor $j: E \hookrightarrow \widetilde{X}$. The restriction $\tau_E: E\to Z$ is then a $\mathbb{CP}^{r-1}$-bundle. Let $h=c_1(\mathcal{O}_E(1)) \in H^2(E,\mathbb{Z})$ be the first Chern class of the tautological bundle $\mathcal{O}_E(1)$ of the $\mathbb{CP}^{r-1}$-bundle $\tau_E$. One may compute the cohomology groups of $\widetilde{X}$ (see for example [1, Theorem 7.31]) : $$H^k(X,\mathbb{Z}) \oplus \Big(\oplus_{i=0}^{r-2} H^{k-2i-2}(Z,\mathbb{Z} ) \Big) \simeq H^k(\widetilde{X}, \mathbb{Z} ), $$ where the isomorphism is given by the sum of $\tau^*: H^k(X,\mathbb{Z}) \to H^k(\widetilde{X}, \mathbb{Z} )$ and $$j_* \circ (\cup h^i) \circ \tau_E^*: H^{k-2i-2}(Z,\mathbb{Z} ) \to H^k(\widetilde{X}, \mathbb{Z} ). $$ I want to compute the cohomology ring sturcture of $H^*(\widetilde{X}, \mathbb{Z})$. For example, for any $\alpha \in H^{2k}(X, \mathbb{Z})$, how can I compute $[E]^{n-k} \cup \tau^*\alpha$? I'd like some references about such computations. Thanks a lot for your answers! [1] C. Voisin, Hodge Theory and Complex Algebraic Geometry I. Cambridge University Press, 2003 REPLY [10 votes]: This is straightforward. Using $j^*[E]=-h$ and the projection formula $j_*(x\cdot j^*y)=j_*x\cdot y$, we get $[E]^p=j_*1\cdot [E]^{p-1}= (-1)^{p-1}j_*(h^{p-1})$. Then $$[E]^{p}\cdot \tau ^*\alpha = (-1)^{p-1}j_*(h^{p-1}\cdot j^*\tau ^*\alpha )= (-1)^{p-1}j_*(h^{p-1}\cdot p^*\alpha_{|Z} )\,,$$ where $p:E\rightarrow Z$ is the projection. You can compute all products in this way.<|endoftext|> TITLE: Why is it easier to prove $e$ is transcendental than $\pi$? QUESTION [13 upvotes]: Why is it easier to prove $e$ is transcendental than $\pi$? I noticed that the proofs of $\pi$'s transcendence are much longer and have more details to check than those of $e.$ My guess is that it's simply because the infinite series for $e$ is much simpler and easier to work with than any series for $\pi$, but I wanted to see if there was more to it, so I asked my professor a few weeks ago as he was going over some bonus proofs of transcendence (including $\pi$ and $e$) thrown in at the end of a Galois theory textbook. He said he couldn't really see a philosophical justification for why one should be easier than the other. Has this question been addressed in the literature, and if so, what do the transcendental number theory experts have to say about it? REPLY [11 votes]: This is not a complete answer, but I would say that the "standard" way to prove the transcendence of $\pi$ is as a corollary of the more general fact that $e^\alpha$ is transcendental for all nonzero algebraic $\alpha$. For general $\alpha$, one has to come up with a general method for dealing with those pesky algebraic numbers in the exponent. But for $\alpha=1$, clever ad hoc arguments are possible. For example, in the book Making Transcendence Transparent by Burger and Tubbs (which I highly recommend as a source for further details), they show how to write down explicitly a polynomial $\mathcal{P}_n(z)$ such that $\mathcal{P}_n(1), \mathcal{P}_n(2), \ldots, \mathcal{P}_n(d)$ provide exceptionally good rational approximations to $e, e^2, \ldots, e^d$ respectively. This proof does exploit special properties of the series $\sum_n z^n\!/n!$ so this perhaps vindicates your intuition that $e$ is easier because we have a nicer series for $e$ than for $\pi$. On the other hand, this argument is a bit circular, because isn't $${\pi \over 4} = 1 - {1\over 3} + {1\over 5} - {1\over 7} + \cdots$$ a "nice" formula for $\pi$? Well, maybe, but it's not "nice" in a way that lets us prove transcendence! Hmmm… So I think that the answer is that we don't know of a way to prove the transcendence of $\pi$ that is significantly simpler than a proof of a more general result, whereas we do know some ad hoc tricks that work for $e$. In principle this could change in the future if, for example, someone finds an amazingly simple ad hoc proof for the transcendence of $\pi$, or (less likely) a dramatic simplification of the proof of the transcendence of $e^\alpha$ for all nonzero algebraic $\alpha$.<|endoftext|> TITLE: Are there three non-commutative polynomials in three variables with finite dimensional quotient? QUESTION [11 upvotes]: $\newcommand\la{\langle}\newcommand\ra{\rangle}$Let $K$ be a field and $K\la x,y,z\ra$ the non-commutative polynomial ring in 3 variables. Question 1: Are there three (fewer is probably not possible?!) polynomials $f,g,h \in K\la x,y,z\ra$, which are sums of monomials of degree at least two, such that the algebra $K\la x,y,z\ra/(f,g,h)$ is finite dimensional? Is there a systematic way to construct such polynomials? For two variables we can take for example $f=xy+yx$ and $g=x^3+y^3$, but I do not know (or forgot) such examples for more than 2 variables. (If possible we should also have (this is equivalent to $(f,g,h)$ being an admissible ideal) that there exists an $n \geq 2$ such that $J^n \subseteq (f,g,h)$ where $J=\la x,y,z\ra$ is the ideal generated by $x,y,z$.) Question 2: Let $f_i$ for $i=1,\dotsc,n-1$ be polynomials spanning an admissible ideal in the non-commutative polynomial ring in $n-1$ variables $x_i$. Can we find $g$ such that $\la f_i,g\ra$ is an admissible ideal in $K\la x_i,y\ra$? REPLY [7 votes]: Q1. Yes, such 3 monomials exist. Moreover, the generic homogeneous quadratic polynomials $f,g,h$ are such that the dimension of the quotient algebra $K⟨x,y,z⟩/(f,g,h)$ is 28. This follows, in particular, from Th.1.3 of [Natalia Iyudu and Stanislav Shkarin. The Golod-Shafarevich inequality for Hilbert series of quadratic algebras and the Anick conjecture]. Q2. In contrast, such an ideal does not exist if $n\ge 5$. If follows from the Vinberg version of the Golod--Shafarevich theorem that if an ideal $I$ is generated by $r$ elements (which are linear combinations of at least quadratic monomials) in a free associative algebra $F$ with $g$ generators, then the quotient algebra is infinite-dimensional provided that the polynomial $1-gz+rz^2$ has a positive root (cf, for example, Prop. 2.7 in [M.Ershov, Golod-Shafarevich groups: a survey]). In your case we have $r=g=n-1$, so that the quotient algebra is infinite if $n\ge 5$.<|endoftext|> TITLE: What is so geometric about symplectic geometry? QUESTION [37 upvotes]: Symplectic geometry is often motivated by the Hamilton's equation which in turn are a reformulation of Newton's third law. But the subject itself is of independent mathematical interest. What I don't understand is why symplectic geometry is, in fact, geometry. Compared with Riemannian geometry where angles, lengths and geodesics can be defined and which can be used to provide models for classical geometry, the notion of a smooth manifold equipped with some two-form doesn't really seem geometric to me at all. In fact I would argue that the aspect of providing a model for classical geometries is the defining feature of Riemannian geometry making it "geometry". The only "geometric interpretion" of symplectic geometry I always see is "symplectic geometry is the geometry of phase spaces" with no mention what the geometric content of phase spaces actually is. Notions like a two-form or a Liouville form seem to me much more algebraic than geometric in nature. Is there a more conceptual way of understanding symplectic structures by motivating them through geometric properties? Or perhaps even some axioms of some geometric space for which symplectic geometry provides a concrete framework to construct a model for? A remark: One could answer this by defining geometry as the study of some space with an associated symmetry group (in the Riemannian case isometries and in the symplectic case symplectomorphisms), but this only moves the question to why one would care about symplectomorphisms from a geometric viewpoint. The case of isometries again is rather clear…. REPLY [8 votes]: I think the question is about "geometric intuition" and "geometric understanding", ill defined as those may be. Perhaps that can be answered with some comparisons. I'll be a bit lazy and just take examples from or around my work. 1. A classic and undeniably geometric result by Cauchy says that the surface area of a convex body $K$ in Euclidean 3-space (to fix dimensions) is $4$ times the average of the areas of its orthogonal projections onto planes. Equivalently, it is equal to $\frac{1}{\pi}$ the volume of the set of oriented straight lines that cross it. Is that more geometric than: the open unit codisc bundle for the Riemannian metric on $\partial K$ is symplectomorphic to the set of oriented straight lines that cross the interior of $K$? The latter statement is more basic and not only says more but can be easily generalized to normed spaces and other Finsler manifolds giving you integral geometric formulas in all those spaces. 2. A lovely result by Banchoff states that if a surface $S \subset \mathbb{R}^3$ is diffeomorphic to a sphere and no plane or (round) sphere separates it into more than two connected pieces, then $S$ is itself a round sphere. That has to be as geometric as one can hope for, but I don't think it is much more geometric than Let $L_S$ be the set of oriented lines perpendicular to $S$ seen as an immersed submanifold in the space of oriented lines in $\mathbb{R^3}$ and let $\hat{x}$ denote the sphere of all oriented lines passing through a point $x \in \mathbb{R}^3$. If every sphere $\hat{x}$ that intersects $L_S$ transversally intersects it in exactly two points, then $L_S$ is itself the set of all oriented lines passing through some point in 3-space. The second (symplectic) version can be immediately generalized not only to more general Lagrangian submanifolds than normal congruences to smooth surfaces, but also to normed spaces, Hadamard manifolds, metrics arising as solutions of Hilbert's fourth problem, etc. 3. A nice problem by Juan-Jorge Schäffer in the theory of Banach spaces asked whether the girth of a Banach space (the infimum of the lengths of centrally symmetric curves on its unit sphere) equaled the girth of its dual. It turned out the (positive) solution was easy, IF you had the idea to think the problem "symplectically". To see the "geometry" in symplectic geometry, you also have to look at its impressive array of applications to other geometries that may look more geometric to you at first, and start building "dictionaries" in your mind. The many books and papers by V. Arnold are a great place to start.<|endoftext|> TITLE: Lifting $\mathfrak{sl}_2$-triples QUESTION [5 upvotes]: Let $k$ be an algebraically closed field, $G$ a (smooth, connected) reductive algebraic group over $k$, $H$ a (smooth, connected) reductive group of semisimple rank 1, and $T$ a maximal torus in $H$. I am specifically interested in the case where the characteristic of $k$ is a bad prime for $G$. Suppose that we are given a group embedding $F_T : T \to G$ and a $T$-equivariant Lie-algebra embedding $f_H : \mathfrak h \to \mathfrak g$, such that the restrictions to $\mathfrak t$ of $f_H$ and the derivative of $F_T$ agree. Can we extend $F_T$ to an embedding $H \to G$ whose derivative is $f_H$? (I know that most results in this area are stated under the assumption of good characteristic, or even of characteristic 0, but I don't actually know any of the counterexamples or where to look for them, so I can't tell if the hypothesis of the existence of $T$ is strong enough to allow me to overcome bad characteristic.) REPLY [4 votes]: Typically not, even for good characteristic. This comes up indirectly in my paper with Adam Thomas The Jacobson–Morozov theorem and complete reducibility of Lie subalgebras, JLMS arxiv version. The nilpotent orbits of $G$ on $\mathfrak g$ in characteristic $0$ have representatives $e_{\mathbb Z}$ which are sums of root elements and which remain in distinct orbits when base-changed to a field of good characteristic. Furthermore, each can be completed to an $\mathfrak{sl}_2$-triple $(e_{\mathbb Z},h_{\mathbb Z},f_{\mathbb Z})\in\mathfrak{g}_{\mathbb Z}^3$—this is a result of Pommerening, though Thomas and I extend it slightly. Denote by $(e,h,f)$ the elements $(e\otimes 1,h\otimes 1,f\otimes 1)$ of $(\mathfrak{g}_{\mathbb Z}\otimes k)^3$. We have $h$ is toral, so $\mathfrak t=\langle h\rangle$ is a torus of $\mathfrak{g}$. Since $h$ can be expressed as an integral combination of coroots, $\mathfrak t$ can be lifted to a torus $T$ such that $T$ normalises $\langle e\rangle$ and $\langle f\rangle$. Let $\mathfrak{h}=\langle e,h,f\rangle\cong \mathfrak{sl}_2$. We have that $T$ normalises $\mathfrak{h}$, so that the embedding $\mathfrak{h}\to\mathfrak{g}$ is $T$ equivariant. Suppose we could lift this to $F_H:H\to G$ for some reductive $H$. (So $H\cong \operatorname{SL}_2$ or $\operatorname{PGL}_2$.) Take the case that $e$ is a regular nilpotent element of $\mathfrak g$. Then a non-trivial unipotent element of $H$ must be regular too. But for $p$ less than the Coxeter number of $G$, a regular unipotent element has order at least $p^2$, so cannot be embedded into an $SL_2$ or $PGL_2$ (whose unipotents have order $p$). EDITED: I initially expected this problem goes away if you insist $p$ is bigger than the Coxeter number of $G$, because I said there is a 1-1 correspondence between conjugacy classes of $\mathfrak{sl}_2$-subalgebras of $\mathfrak g$ and subgroups of $G$ of type $A_1$. This is obviously wrong however, since in positive characteristic one gets many more irreducible representations for $SL_2$ than for $\mathfrak{sl}_2$, giving many more $SL_2$ or $PGL_2$ subgroups of $GL_n$ than $\mathfrak{sl}_2$ subalgebras, c.f. Steinberg's tensor product theorem. By way of correction, if $p$ is bigger than the Coxeter number, all unipotent elements have order $p$. Then one can cite Theorem 4.2 of Lawther, R.; Testerman, D. M., (A_1) subgroups of exceptional algebraic groups, Mem. Am. Math. Soc. 674, 131 p. (1999). ZBL0936.20039. in which an $A_1$ subgroup is guaranteed such that the Lie algebra of its root group contains nilpotent elements of the same Bala–Carter type; a fortiori one that corresponds under a Springer isomorphism.<|endoftext|> TITLE: Smoothness of coordinates in the rectification theorem for ODE QUESTION [6 upvotes]: The rectification theorem says that near a regular point every vector field $v$ is standard, that is, there exists a local coordinate system such that $v=\frac{\partial }{\partial x_1}$. In the textbooks (e.g. of Arnold and Hartmann) the vector field is assumed to be $C^r$ with $r\ge 1$ and the local coordinate system such that $v=\frac{\partial }{\partial x_1}$ is also of smoothness $C^r$. Why the smoothness of the coordinate system is not $C^{r+1}$? What (counter)example shows this? REPLY [10 votes]: In dimension $1$, it's true that a flowboxing change of coordinates for a $C^r$ vector field is $C^{r+1}$, but this is no longer true in dimensions greater than $1$. Basically, the reason is this: If $V$ is a $C^r$ vector field on $\mathbb{R}^2$ and $V(0,0)\not=0$, then there exist local $C^r$ coordinates $(x^1,x^2)$ centered on $(0,0)$ such that $V = \partial/\partial x^1$. Any other set $(y^1,y^2)$ of such $C^r$ 'flowbox' coordinates is locally of the form $$ (y^1,y^2) = \bigl(x^1 + f(x^2), \ g(x^2)\bigr) $$ for some $C^r$ functions $f$ and $g$ of one variable, that satisfy $f(0)=g(0)=0$. Generally, if $(x^1, x^2)$ is only $C^r$, there will not exist such functions $f$ and $g$ that will make $y^1$ and $y^2$ be $C^{r+1}$. For a specific example, let $h:\mathbb{R}\to\mathbb{R}$ be a $C^r$ function that is not $C^{r+1}$ and satisfies $h(0)\not=0$ and consider the vector field $V$ defined on a neighborhood of $(0,0)$ in the $uv$-plane by $$ V := \frac{1}{h(v)}\,\frac{\partial}{\partial u} + 0\,\frac{\partial}{\partial v}. $$ The $C^r$ local flowbox coordinates for $V$ on a neighborhood of $(0,0)$ are of the form $$ (x^1,x^2) = \bigl(\ u\,h(v)+f(v),\ g(v)\ \bigr) $$ where $f$ and $g$ are $C^r$ functions. Clearly, it is not possible to choose $f$ and $g$ so that $x^1$ will be of class $C^{r+1}$.<|endoftext|> TITLE: Third Galois cohomology group QUESTION [10 upvotes]: It is well known that when $K$ is a local or global field the Galois cohomology group $H^{3}(K,K_{\text{sep}}^{\times})=0$ where $K_{\text{sep}}$ denotes the separable closure of $K$. Could someone give an example of a field $K$ where $H^{3}(K,K_{\text{sep}}^{\times}) \neq 0$ and why it is non-zero in this case? REPLY [16 votes]: The group $H^3(K,\bar{K}^\times)$ naturally arises when trying to calculate the Brauer group of a variety. Explicitly, the Hochschild-Serre sequence yields the exact sequence $$0 \to \mathrm{Br}_1(X)/\mathrm{Br}(K) \to H^1(K,\mathrm{Pic}(X_{\bar{K}})) \to H^3(K,\bar{K}^\times)$$ for a projective variety $X$ over a perfect field $K$, where $\mathrm{Br}_1(X) = \ker(\mathrm{Br}(X) \to \mathrm{Br}(X_{\bar{K}}))$ is the algebraic part of the Brauer group of $X$. Over number fields the vanishing you mention allows one to calculate the Brauer group using $H^1(K,\mathrm{Pic}(X_{\bar{K}}))$, which is often easier to understand. But there are examples where this doesn't happen where $K$ is a function field. One of the only examples I know of is due to Umenatsu- On the Brauer group of diagonal cubic surfaces. This concerns the variety $$X: \quad x^3 + by^3 + cz^3 + dt^3 = 0$$ over the function field $K=k(b,c,d)$ where $k$ is any field containing a third root of unity. Then in the above cited paper it is shown that $\mathrm{Br}(X)/\mathrm{Br}(K) = 0$ but $H^1(K,\mathrm{Pic}(X_{\bar{K}})) \cong \mathbb{Z}/3\mathbb{Z}$. Thus $H^3(K,\bar{K}^\times)$ is non-trivial, and gives a geometric interpretation of this through the failure of elements of $H^1(K,\mathrm{Pic}(X_{\bar{K}})$ to come from Brauer group elements.<|endoftext|> TITLE: A tensor category need not be isomorphic to a strict tensor category QUESTION [5 upvotes]: This question was originally posted on MSE, but got no answer even after putting a bounty on it, so I'll try here. I'm reading the book "Tensor categories" by Etingof, Gelaki, Nikshych, and Ostrik. In remark 2.8.6 (posted below), it is claimed that the category $\mathcal{C}_G^\omega$ (defined in example 2.3.8, also below) is not isomorphic to a strict tensor category whenever $\omega$ is not cohomologically trivial. Can someone explain this a bit more? If someone knows another example of a tensor category that is not isomorphic to a strict tensor category, I'm also interested in that. REPLY [6 votes]: First consider the category $\mathcal{C}_G$ with its bifunctor $\otimes$ and unit. How many ways are there to enhance this to a monoidal category structure? The missing data are precisely the associators. The associators will be a choice of map $$a_{\delta_g, \delta_h, \delta_m}: (\delta_g \otimes \delta_h) \otimes \delta_m \to \delta_g \otimes (\delta_h \otimes \delta_m)$$ for each $g,h,m \in G$. However in $\mathcal{C}_G$, the hom spaces are identified with $A$. Thus the associator amounts to an assignment of elements $\omega(g,h,m) \in A$ for each triple $g,h,m \in G$. Now the associator is required to satisfy some equations, such as the pentagon equation. Written in terms of the $\omega$, the pentagon equation is equivalent to saying that $\omega$ satisfies the cocycle condition, which we might write as $\delta \omega= 0$. (It should actually be a normalized cocycle because of the unit triangle equations, but lets ignore that). When are two of these structures isomorphic? For that we need a tensor isomorphism. Assume for simplicity that it is the identity on objects and morphisms and preserves the $\otimes$ bifunctor strictly. This is not really a restriction as we could use the isomorphism to make this identification. There is still another piece of data, which is a collection of isomorphisms: $$f_{\delta_g, \delta_h}: F(\delta_g) \otimes F(\delta_h) \to F(\delta_g \otimes \delta_h)$$ for each pair $g,h \in G$. As before we can identify this with a collection of elements $c(f,g) \in A$ for each pair $g,h \in G$. These are required to satisfy a hexagon identity which involves the two associators. When written in terms of these elements, this becomes: $$\omega_1 - \omega_2 = \delta c$$ So that the two cocycles $\omega_1$ and $\omega_2$ differ by a cochain. In summary if two of these monoidal structures are related by an isomorphism of tensor categories, the two cocycles must be cohomologous. Note: to be strict, means that the corresponding cocycle is the trivial cocycle. Putting this together we see that to be isomorphic to a strict tensor category requires that the underlying cocycle be cohomologically trivial.<|endoftext|> TITLE: Strongly compact categories (reference request) QUESTION [8 upvotes]: The notion of a "compact category" was introduced by Isbell$\color{red}{^{1,2}}$. A locally small category $\mathcal{C}$ is called compact when every functor $\mathcal{C} \to \mathcal{D}$ into any category $\mathcal{D}$ which preserves all (possibly large!) colimits is a left adjoint. Equivalently, every presheaf $\mathcal{C}^{\mathrm{op}} \to \mathbf{Set}$ preserving all (possibly large!) limits is representable. Personally, I find large limits and colimits a bit awkward (but maybe you can convince me why they are important?), and in my research I need a property which is perhaps stronger: Let's call a category $\mathcal{C}$ strongly compact when every functor $\mathcal{C} \to \mathcal{D}$ which preserves small colimits is a left adjoint (and hence preserves all colimits). This is stronger because we deal with a a priori larger class of functors. Equivalently, every presheaf on $\mathcal{C}$ preserving small limits is representable. I have done quite a bit of literature research, but did not find this property elsewhere. But it is very natural, in particular in the context of the special adjoint functor theorem, which says that every wellcopowered cocomplete category with a generator is strongly compact. So I suspect that this notion appears somewhere? If not, what name do you suggest? An alternative would be "realized-sketchable" since I can show that a category is strongly compact if and only if it is the category of models of a (possibly large) realized limit sketch for which every model is small. *Edit. In my paper I am using the name strong compact. $\color{red}{^1}$ J. R. Isbell, Small subcategories and completeness, Mathematical systems theory 2.1 (1968): 27-50 $\color{red}{^2}$ R. Börger, W. Tholen, M. B. Wischnewsky, H. Wolff, Compact and hypercomplete categories, Journal of Pure and Applied Algebra 21.2 (1981): 129-144 REPLY [2 votes]: In Adjoints to functors from categories of algebras Rattray defines that a category $\mathcal{A}$ has LAP (left adjoint property) when every continuous functor on $\mathcal{A}$ has a left adjoint, or equivalently every continuous functor $\mathcal{A} \to \mathbf{Set}$ is representable. This is exactly the dual notion to what I called "strongly compact". So Rattray would probably call this property RAP (right adjoint property). He shows that every monadic category over a RAP category is again a RAP category. (He proves the dual version: every comonadic category over a LAP category is a LAP category.) It is a bit confusing that it is claimed in the mentioned paper "Compact and hypercomplete categories" by R. Börger, W. Tholen, M. B. Wischnewsky, H. Wolff that Rattray actually proves that every monadic category over a compact category is again compact: compactness involves hypercontinuous functors. But probably Rattray's proof can be used in both settings.<|endoftext|> TITLE: Why does $\iota_4^2 \in H^8(K(\mathbb Z/2,4);\mathbb Z/2)$ not come from $H^8(K(\mathbb Z/2,4);\mathbb Z)$? QUESTION [7 upvotes]: In Hatcher's Chapter 5 (https://pi.math.cornell.edu/~hatcher/AT/ATch5.pdf) on page 574 (page 57 in the pdf), he states that $\iota_4^2 \in H^8(K(\mathbb Z/2,4);\mathbb Z/2)$ is not in the image of $H^8(K(\mathbb Z/2,4);\mathbb Z)\to H^8(K(\mathbb Z/2,4);\mathbb Z/2)$. The argument for the lower classes ($\iota_4, Sq^2\iota_4, Sq^2Sq^1\iota_4$) which don't come from $H^*(K(\mathbb Z/2,4);\mathbb Z)$ is that the Bockstein homomorphism $Sq^1\colon H^*(K(\mathbb Z/2,4);\mathbb Z/2) \to H^{*+1}(K(\mathbb Z/2,4);\mathbb Z/2)$ applied to those classes is nonzero, so it can't come from $H^*(K(\mathbb Z/2,4);\mathbb Z)$. This argument doesn't work for $\iota_4^2 = Sq^4 \iota_4$ because $Sq^1Sq^4 \iota_4 = Sq^5 \iota_4 = 0$. How can one argue that $\iota_4^2 \in H^8(K(\mathbb Z/2,4);\mathbb Z/2)$ doesn't come from $H^8(K(\mathbb Z/2,4);\mathbb Z)$? Is it even true? REPLY [10 votes]: It is true, and follows from results of Browder on the mod 2 Bockstein spectral sequence for $K(\mathbb{Z}/2,4)$. (We can replace $4$ by any even integer $k$ and conclude that $\iota_k^2$ ia not the reduction of an integral class.) An argument is spelled out in Section 3 of Grant, Mark; Szűcs, András, On realizing homology classes by maps of restricted complexity, Bull. Lond. Math. Soc. 45, No. 2, 329-340 (2013). ZBL1270.57068 available in preprint form at https://arxiv.org/abs/1111.0249. It is based on Theorem 5.5 of Browder, W., Torsion in (H)-spaces, Ann. Math. (2) 74, 24-51 (1961). ZBL0112.14501.<|endoftext|> TITLE: Books in advanced differential topology QUESTION [16 upvotes]: I am looking for books or other sources in differential topology that include topics like: vector bundles, fibration, cobordism, and other related topics. In general, if anyone has recommendation of books of advanced differential topology I would like to hear (I've already read Bott&TU, Warner). Thanks! REPLY [2 votes]: Let me mention a recent book: Differential Cohomology: Categories, Characteristic Classes, and Connections, edited by Araminta Amabel, Arun Debray and Peter J. Haine, available on arXiv. It is assembled from talks in a graduate student seminar. Although most of content is classical, the treatment is relatively modern, using $\infty$-categories, $\infty$-topoi and methods in motivic cohomology à la Voevodsky.<|endoftext|> TITLE: Practical Benefits of HTT/univalent foundations for assisted proofs QUESTION [7 upvotes]: I'm trying to understand what the claimed practical benefits of HTT/univalent foundations are for doing computer assisted proofs and while I've seen a lot of claims of benefits they all seem to be spelled out in a way that requires learning HTT/univalent to evaluate. I'd like to know if it really has much of practical value to add to computer assisted proofs before I spend the time digging into a bunch of category theory to learn it. So could someone please spell out these benefits in a way that doesn't require learning a bunch of category theory to understand and, ideally, which brings it down to what kind of benefits in end-user interface/automation we should expect these new techniques to enable? Just for context/clarification let me explain that I read similar claimed benefits for the type theory (Curry–Howard) based approach and after a bunch of time learning enough to understand what was going on they weren't that impressive and there were drawbacks as well. Are the benefits of HTT/univalent just more stuff like using Curry–Howard to construct proofs: kinda elegant and neat but actually not something someone interested just in efficently formalizing classical proofs about impredicative structures which all live inside ZFC would care about? Or does it really have benefits which promise to increase the ability to enter math in a natural fashion by mathematicians doing normal math? For instance, are the claims suggesting that this lets one reuse the same proof in multiple contexts (I assume this is meant to solve the choice of representation issues when doing math in Coq) just a theoretical niceity or is it plausible this allows the computer to do all the work under the hood? I mean if, as it seems for many benefits claimed for the standard type theory approach, these benefits require deep knowledge of the system to achieve in each instance or add more complexity than they eliminate via deduplication they aren't so helpful. For concreteness, one of the places I ran into discussions of the benefits of HTT/univalent is Escardo - Computing an integer using a Grothendieck topos (you have to scroll down a bit) but this is just one example but I often see somewhat vague statements suggesting this is the right or best way to formalize math with an explanation that involves colimits, sheafs, Grothendieck universes or other stuff I don't understand. Also, this answer to a question on HTT was nice but I don't know if that actually means the computer can (or even might in the future) efficently translate work entered in one context to another without carefully setting things up that way before entering the proof or human hinting. REPLY [12 votes]: You didn't specify exactly what "claimed benefits" for non-univalent type theory in general you're referring to, and I happen to believe that even non-univalent type theory does have substantial benefits for the formalization of mathematics. So let me start by recapping those, not to defend type theory in general, but because some of the benefits of univalent type theory expand on them. One of these benefits is explained eloquently by Andrej Bauer in his answer to the question linked by LSpice in the comments: namely, since mathematics as it is done by mathematicians is essentially typed, and type theory has much more robust error-detection, whether or not the foundational kernel of your proof assistant is type theory, the vernacular that you actually use will almost inevitably be a form of type theory; and in that case why bother putting a layer of set theory under it? Another benefit of a type-theoretic proof assistant is that type theory has many models, so that a proof formalized in (constructive) type theory actually proves much more than an analogous proof formalized in ZFC. This is where things like categories, sheaves, and so on generally enter the picture, but the basic point can be appreciated without any of those words. For example, instead of interpreting "type" as "set", one can additionally interpret "type" as "topological space", and the same theorems and proofs will be valid. (Technically, he textbook definition of "topological space" isn't quite nice enough for this to be precisely true, but there are well-known modifications of it that are.) Thus, when you prove something in type theory, you haven't just proven the theorem about sets that you would have proven in the analogous ZFC proof, but you've also proven a theorem about topological spaces at the same time. For instance, your theorem about groups is automatically also a theorem about topological groups. My paper The logic of space is an introduction to type theory (including univalence) from this perspective. So what about univalent type theory? Like non-univalent type theory, it also has many models. Roughly speaking, every model of univalent type theory is an "extension" of some model of non-univalent type theory, and most interesting models of non-univalent type theory can be extended to some model of univalent type theory. But the two classes of models are not equivalent, because a given non-univalent model could be extended to more than one univalent model, and often we may want to access univalent features that depend on the choice of model. To be sure, the difference lies entirely in the world of homotopy theory / higher category theory, so this is not presently a feature that's of much relevance to mathematicians who don't yet care about such things. However, with the increasing importance of such "homotopification" in mathematics, it seems like a good thing to accomodate in the design of a purportedly generic proof assistant, even if it will be invisible to many of its users. More importantly for the average mathematician, univalence also makes type theory an even closer match for ordinary mathematical practice, and hence a better vernacular for a proof assistant; and also fixes some of the traditional problems of type theory relative to set theory. Firstly, mathematics is naturally univalent: group theorists do not distinguish between isomorphic groups, and so on. Thus, one naturally wants a vernacular for a proof assistant that can also seamlessly replace one structure with an isomorphic one. Plenty of work has gone into finding ways to build such invariance on top of existing set-theoretic or type-theoretic foundations, which more or less amounts to building a univalent context on top of a non-univalent one; so we may ask, in parallel with Andrej's point, once we have a univalent vernacular, why bother putting a non-univalent layer under it? Secondly, while you didn't say what drawbacks of non-univalent type theory you have in mind, traditionally many of those drawbacks have involved the unnatural treatment of quotients. Non-univalent type theory doesn't really have a good way to deal with quotients, leading to the introduction of things like "setoids" that look very unnatural to an ordinary mathematician. But univalent type theory solves this problem, with a very natural presentation of quotients that has all the properties an ordinary mathematician would expect. Similarly, impredicative notions such as powersets are often klunky to deal with in non-univalent type theory, but work much better in univalent type theory. I should say, though, as I said in my answer to the question linked by Wowoju in the comments, that these benefits are not currently huge. The usefulness of univalence for isomorphism-invariance, in particular, is limited in the absence of proof assistants that can actually use univalence to transfer a definition across an isomorphism automatically and compute the result, rather than forcing the user to laboriously unwind the transport to figure out the result. Cubical Agda, and other more recent proof assistants based on cubical type theories, have the promise of behaving in this way, but we still have a lot to learn about how best to design such proof assistants and what can be done with them. Moreover, we are still working out how to reconcile these computational features with the desired many models of type theory. So while I believe the potential benefits are substantial, many of them are still in the future.<|endoftext|> TITLE: Why is the Simple Zeros Conjecture said to be stronger than the Riemann Hypothesis? QUESTION [8 upvotes]: Let the "Simple Zeros Conjecture (SZC)" be the statement that all zeros of the Riemann zeta function are simple. I have often heard of the statement that the SZC is stronger than the Riemann Hypothesis (RH). However, I have never seen or heard of any justification of this claim, and a quick internet search doesn't seem to reveal any result like SZC $\implies$ RH. Therefore, can someone explain why the SZC is said to be stronger than the RH ? REPLY [26 votes]: As Peter Humphries points out, the precise claim is that "RH + Simple Zeroes" is stronger than "RH". Of course, this is formally trivial. So what's really meant is that "RH + Simple Zeroes" is a natural strengthening of RH The reason for this is a generalization of the following simple fact: Let $f$ be a monic polynomial in one variable of degree $n$ with real coefficients. Then $f$ has all roots real if and only if the coefficients of $f$ lie in a certain closed subset of $\mathbb R^n$, and $f$ has all roots real and simple if and only if the coefficients of $f$ lie in the interior of that closed subset. So real roots + simple is just slightly stronger than real roots alone in a very natural way. REPLY [3 votes]: SZC is thought to be stronger than RH not because any proof exists that SZC implies RH but because all existing hypotheses implying SZC are stronger than RH. The most important of these involve the Mertens function $M(x)=\sum_{n\leq x}\mu(n)$ and include the generalised Mertens Hypothesis or GMH ($M(x) = O(x^{\frac{1}{2}}$)) and the slightly less drastic hypothesis that $\int_{1}^{X}(\frac{M(x)}{x})^2dx = O(\log(X)$ which GMH obviously implies. As far as is known however, neither of these follows from RH.<|endoftext|> TITLE: Hamming distance between $a+b$ and $a \oplus b \oplus ((a \land b) \ll 1)$ QUESTION [5 upvotes]: Motivation. In their paper about the cryptographic scheme NORX, the authors use a fast approximation of + by bitwise operations (taking fewer CPU cycles than proper addition) using the formula $$a+b "=" a \oplus b \oplus ((a \land b) \ll 1)$$ where $\oplus$ is bitwise XOR and $\land$ is bitwise AND, and $\ll$ is left-shift by 1 position. (The purpose of $((a \land b) \ll 1)$ is to simulate the "carry-bit" operation.) I was wondering "how well" this approximation worked in terms of the Hamming distance between $a+b$ and $a \oplus b \oplus ((a \land b) \ll 1).$ Let's make this precise. Formal version. Let $\{0,1\}^\mathbb{N}$ denote the collection of functions $f:\mathbb{N}\to \{0,1\}$ and let $$\{0,1\}^* = \{x \in \{0,1\}^\mathbb{N}: \exists N\in\mathbb{N}(\forall k\in\mathbb{N}(k\geq N\implies x(k)=0))\}.$$ Every member of $\{0,1\}^*$ can be viewed as the binary expansion of a natural number; this is a unique correspondence. This correspondence gives rise to the addition $+:\{0,1\}^* \times \{0,1\}^* \to \{0,1\}^*$. Denote by $\ll 1$ the left-shift by one position, i.e. $\ll 1 : x \in \{0,1\}^* \to x'\in \{0,1\}^*$ where $x'(0) = 0$ and $x'(n+1) = x(n)$ for all $n\in \mathbb{N}$. We usually write $x \ll 1$ instead of $\ll 1(x)$. For $x\in\{0,1\}^*$ we set $\text{len}(x) = \max\{k\in\mathbb{N}:x(k) = 1\}$. For $a,b\in\mathbb{N}$ we denote by $\text{diff}(a,b)$ the Hamming distance between $a+b$ and $a \oplus b \oplus ((a \land b) \ll 1)$. Question. Do we have $\max\{\text{diff}(a,b): a,b\in \{0,1\}^*\} < \infty$? If not, what is $\lim \sup_{n\to\infty} D_n$ where $$D_n=\frac{1}{n} \max\{\text{diff}(a,b): a,b\in\{0,1\}^*, \max\{\text{len}(a),\text{len}(b)\} = n\}$$ for $n\geq 1$? REPLY [5 votes]: The lim sup is exactly $1$. Almost certainly the exact value of $D_n$ comes from $a = (1,1,1,\ldots,1)$ and $b = (1,0,0,\ldots,0)$, and even if not, it’s off by at most $O(1/n)$ which is inconsequential. In such case we have about the maximum number of carries which are undetected by the $a \land b$. So $a+b = 2^n$ but $(a \land b) \ll 1 = 2$, so the approximate sum is $(0,0,1,1,1,\ldots,1)$, making the Hamming distance equal to $n-1$ (there are $n+1$ bits in $a+b$ and only the lowest two agree). So $D_n \ge 1 - \frac1n$. Update: thanks to user44191’s observation, this bound is in fact precise, so $D_n = 1 - \frac1n$.<|endoftext|> TITLE: What is the best known upper bound for $\frac{1}{\zeta'(\rho)}$ assuming the SZC but not the RH? QUESTION [5 upvotes]: Let $\zeta$ denote the Riemann zeta function and let $\rho$ denote one of its complex zeros. What is the best known upper bound for $\frac{1}{\zeta'(\rho)}$ assuming that all zeros are simple (SZC), but not assuming Riemann Hypothesis ? Assuming both the RH and the SZC, one can mimick the proof of Theorem 15.6 of Montgomery-Vaughan's Multiplicative Number Theory and show that $$ \frac{1}{\zeta'(\rho)} \ll X,\label{1}\tag{1} $$ where $X$ is any real number $\geq |\rho| $ (actually the Montgomery-Vaughan argument seems to yield an upper bound of the form $o(X)$). However, it looks like the bound \eqref{1} could hold assuming the SZC alone, as it appears that Montgomery-Vaughan only invoked the RH on bounding the $S(T)=\arg \zeta(\sigma + iT)$. On the RH, it is a classical fact that $$ S(T) \ll \frac{\log T}{\log \log T} $$ whilst $S(T) \ll \log T$ unconditionally. The unconditional bound seems sufficient for the purposes of showing that \eqref{1} comes from the Montgomery-Vaughan argument. REPLY [15 votes]: We have an exact formula \begin{align*} \frac{1}{\zeta'(\rho)} &= \lim_{s \to \rho} \frac{s-\rho}{\zeta(s)} \\ &= \lim_{s \to \rho} \frac{(s-\rho) (s-1) \Gamma(1+s/2) \pi^{-s/2}}{\xi(s)} \\ &= (\rho-1) \Gamma(1+\rho/2) \pi^{-\rho/2} \lim_{s \to \rho} \frac{s-\rho}{\frac{1}{2} e^{Bs} \prod_{\rho'} (1-\frac{s}{\rho'}) e^{s/\rho'}} \\ &= -\frac{ 2 e^{1-B\rho} \rho(\rho-1) \Gamma(1+\rho/2) \pi^{-\rho/2}}{\prod_{\rho' \neq \rho} (1-\frac{\rho}{\rho'}) e^{\rho/\rho'}} \end{align*} where $B = -0.0230957\dots$ is the constant in Theorem 10.12 of Montgomery-Vaughan. All of the factors in the above formula are well understood except for the terms $1-\frac{\rho}{\rho'}$ for nearby zeroes $\rho' = \rho+O(1)$, which are proportional in magnitude to the distances from $\rho$ to the nearby zeroes $\rho'$. So the problem of upper bounding $1/\zeta'(\rho)$ is more or less equivalent to that of lower bounding the distance $|\rho-\rho'|$ to the nearest zero $\rho'$ (or more precisely the product of the distances to those zeroes $\rho'$ within $O(1)$ of $\rho$). An assumption of simple zeroes merely says that this distance is positive, but a more quantitative version of this hypothesis would be needed to get any quantitative upper bound.<|endoftext|> TITLE: Reference for proof of an integral from the "Tables of Integral Transforms" involving a Gaussian and a Laguerre polynomial QUESTION [7 upvotes]: I am looking for a proof of one of the integrals presented in Harry Bateman's Tables of Integral Transforms. The specific integral in question is presented on page 42 in chapter 8.9 as equation (3): $$ \int_0^{\infty} \left[x^{\nu+\frac{1}{2}} \mathrm{e}^{-\frac{1}{2} x^2} L^{\nu}_n\!\left(x^2\right) \right]J_{\nu}\!\left(xy\right) \sqrt{xy}\,\mathrm{d}x = \left(-1\right)^n \mathrm{e}^{-\frac{1}{2} y^2} y^{\nu+\frac{1}{2}} L^{\nu}_n\!\left(y^2\right) $$ where $J_{\nu}$ are the Bessel functions of the first kind and $L^{\nu}_n$ are the associated Laguerre polynomials. This integral is basically the Hankel transform of the function $f\!\left(x\right) = x^{\nu} \mathrm{e}^{-\frac{1}{2} x^2} L^{\nu}_n\!\left(x^2\right) $ with an additional factor $\sqrt{y}$ (depending on the definition). A similar integral also appears in this question, but the integral in question here is a bit different. The full reference of the table entry is: Harry Bateman, Tables of Integral Transforms, Volume 2, p. 42 (ch. 8.9, eq. (3)) REPLY [6 votes]: The question asks for a reference; the body asks for proof. Here's a proof that's really more of a verification. From the proposed formula make a generating formula: $$\sum_{n=0}^\infty z^n \int_0^\infty x^{\nu+1}\exp{(-x^2/2)}L_n^{\nu}(x^2) J_{\nu}(xy)\,dx \overset{?}{=}$$ $$ y^{\nu}\exp{(-y^2/2)} \sum_{n=0}^\infty (-z)^n L_n^{\nu}(x^2).$$ Use the generating formula for Laguerre polynomials (Gradshteyn and Rhyzik 8.975.1), $$\sum_{n=0}^\infty z^n L_n^{\nu}(x) = (1-z)^{-\nu-1}\exp[-x \ z/(1-z) ].$$ Use G&R 6.631.4 $$\int_0^\infty x^{\nu+1}\exp{(-ax^2)}J_{\nu}(xy)dx = \left( \frac{y}{2a} \right)^{\nu} \frac{1}{2a} \exp\left[-y^2/(2a)\right].$$ The prior two formulas are well known. Algebra completes the verification. For a proof, work backwards.<|endoftext|> TITLE: Which homotopy classes $S^3 \to S^2$ lift to embeddings $S^3 \to S^2 \times D^3$? QUESTION [23 upvotes]: The question is, for a smooth embedding $$f : S^3 \to S^2 \times D^3$$ one can compose the map $f$ with projection $\pi : S^2 \times D^3 \to S^2$, giving the map $\pi \circ f : S^3 \to S^2$. Which homotopy classes of maps are realizable? The analogous question where you vary the dimension of the disc, i.e. $$ f : S^3 \to S^2 \times D^k $$ has immediate answers when $k \neq 3$. When $k < 3$ the answer is only the constant map, and when $k > 3$ the answer is all homotopy classes, since $S^3$ lives in the boundary of $D^k$. The $k < 3$ argument is a basic cut-and-paste topology argument, to argue that the projection must be null-homotopic. I suspect the answer to this question exists in the literature, as you can view this as the problem of if one can "link" a linearly-embedded $S^2$ with a non-linearly embedded $S^3$ in $S^5$. But I have looked through the old Haefliger-Zeeman literature on mixed-dimensional links, without much luck. A short summary of the theory is here, written-up by Skopenkov: http://www.map.mpim-bonn.mpg.de/High_codimension_links A closely-related MO question is on the $k=2$ case, but where you let $f$ be an immersion. This was answered positively in the comments. Hopf fibration inside the retraction of R^4 minus line -> S^2? edit: I suppose I could add I have a suspicion the answer is at least the subgroup of index two in $\pi_3 S^2 \simeq \mathbb Z$. At present I do not see how to obstruct a generator of $\pi_3 S^2$ being realizable. REPLY [2 votes]: Earlier I thought I had an argument that twice the Hopf map was realizable for an embedding $S^3 \to S^2 \times D^3$, but there was a mistake in my argument -- the map I suggested failed to be a smooth embedding. Here is a more elementary observation on the problem. Proposition: Given a smooth embedding $f : S^3 \to S^2 \times D^3$, then the projection map $\pi : S^2 \times D^3 \to S^2$ can not restrict to a locally-trivial fiber bundle on $f(S^3)$. This proposition does not answer the question, but it provides some restrictions on answers. The idea is to consider the simplest case. Assume $\pi$ restricts to a locally-trivial fiber bundle on $f(S^3)$, and that $f(S^3)$ intersects the fibers of $\pi$ in unknotted loops. You could cite Hatcher's work here and conclude they must be linearly embedded loops. This implies the Hopf fibration is classified by a map $S^2 \to V_{3,2} / SO_2$. We know the Hopf fibration is classified by the generator of $\pi_2 (V_{4,2} / SO_2)$, and $\pi_2 (V_{3,2}/SO_2)$ is the index two subgroup. I believe this argument extends to the case the fibers are non-linearly embedded -- in that case you can argue the Euler class must be zero. That said, there is probably a simpler way to rule out non-linearly embedded fibers. Comment: There is a map $f : S^3 \to S^2 \times D^3$ that is a 2-to-1 immersion, with the projection $S^3 \to S^2$ equal to the Hopf fibration. The idea is that $S^3$ modulo antipodal points has a canonical identification with the unit tangent bundle of $S^2$, which is a subset of $S^2 \times D^3$. I suppose using Neil's notation this would be the map $(\eta(u), \tau(u))$ where $\tau(u) = uj\overline{u}$.<|endoftext|> TITLE: Dense subcategory of measurable spaces QUESTION [5 upvotes]: Recall the notion of a dense subcategory $\mathcal{D}$ of a category $\mathcal{C}$. It means that the restricted Yoneda functor $\mathcal{C} \to \mathrm{Hom}(\mathcal{D}^{op},\mathbf{Set})$, $A \mapsto \mathrm{Hom}(-,A)|_{\mathcal{D}}$ is fully faithful. Roughly, it means that $\mathcal{D}$ "detects morphisms" in $\mathcal{C}$. One can show that $\mathbf{Meas}$, the category of measurable spaces$^1$, has no small dense subcategory. Trivially, $\mathbf{Meas}$ is a dense subcategory of $\mathbf{Meas}$, but that is not very interesting. Question. What is an example of a "quite small" proper dense full subcategory of $\mathbf{Meas}$? By "quite small" I mean that we are not just removing a bunch of measurable spaces, but rather that the objects of the dense subcategory are parametrized by a very simple structure. Imagine, very informally, there was a measure on $\mathbf{Meas}$, then I want the dense subcategory to be of measure $0$. We can assume that the one-point measurable space belongs to the subcategory. If $\mathcal{K}$ denotes the rest, we have the following characterization of density: If $X,Y$ are measurable spaces, then a map $f : X \to Y$ is measurable iff for every measurable map $a : A \to X$ for $A \in \mathcal{K}$ the composition $f \circ a : A \to Y$ is measurable. (This is what I meant above with "detecting morphisms"). The question asks for such a class of measurable spaces. At first you might think that this is completely impossible. I had the same suspicion for $\mathbf{Top}$, but it turns out that for $\mathbf{Top}$ it is possible: take the one-point-space and the topological spaces of the form $P \cup \{\infty\}$ for directed sets $P$, where the sets $P_{\geq p} \cup \{\infty\}$ form a local base at $\infty$. This subcategory is dense: This is just a fancy way of saying that a map is continuous iff it preserves net convergence. Maybe there is some similar theory of "net convergence" for measurable spaces? I found the related discussion What properties are preserved under a measurable mapping?, but I am not sure if Eric Wofsey's answer settles my question, because convergent filters cannot be seen as maps. $^1$ Since Dmitri Pavlov's notion of a measurable space has become quite prominent on mathoverflow, let me mention that I use the "classical" definition here. It's just a set with a $\sigma$-algebra. However, if there was a very good answer for Pavlov's measurable spaces, I would be happy to hear about that too. REPLY [2 votes]: A rather satisfying answer to this question can be given if one is willing to equip measurable spaces with a σ-ideal of negligible sets (i.e., sets of measure 0, except that we need not choose any specific measures). This is an extremely natural choice to make, since the resulting category is contravariantly equivalent to the category of commutative von Neumann algebras. Assuming we are working in the category CSLEMS of compact strictly localizable enhanced measurable spaces described there, one can give a complete classification (essentially due to von Neumann and Maharam) of objects in CSLEMS up to an isomorphism. Specifically, any objects of CSLEMS is canonically isomorphic to the disjoint union of its atomic and nonatomic parts, and the nonatomic part is canonically isomorphic to the disjoint union of nonempty measurable spaces $F_κ$, where $κ$ is an infinite cardinal and $F_κ$ is noncanonically isomorphic to $I⨯2^κ$, where $I$ is an infinite set and $2^κ$ is interpreted as the product of $κ$ copies of measurable spaces $2=\{0,1\}$. (And for $κ=0$ we recover the atomic part mentioned above, so it is also covered by this construction if we allow $I$ to be finite in this case.) Furthermore, the classification works in the relative case, i.e., for morphisms in CSLEMS. Indeed, mapping to a disjoint union of measurable spaces amounts to partitioning the domain and mapping each part separately. Thus, it suffices to describe maps of the form $F_κ→F_λ$ for some infinite cardinals $κ$ and $λ$ (we can also allow $κ=0$ or $λ=0$). Such morphisms exist if and only if $κ≥λ$. Furthermore, after performing a further (canonical) partition of the domain and codomain, we can make the resulting parts noncanonically isomorphic to the projections $I⨯J⨯2^κ→I⨯2^λ$, given by the product of the projection $I⨯J→I$ and the projection $2^κ→2^λ$. This relative Maharam theorem allows us to easily identify dense subcategories of CSLEMS: these are precisely those subcategories that have objects with nonempty components $F_κ$ for arbitrary large cardinals $κ$. In particular, no such subcategory can be small. For example, we could take the spaces $2^κ$ for all infinite cardinals $κ$.<|endoftext|> TITLE: Spin connection in the tetradic Palatini-formalism of general relativity QUESTION [5 upvotes]: $\DeclareMathOperator\SO{SO}$I am trying to understand the tetradic Palatini-formalism of general relativity from a mathematical point of view. I am graduate student and quite new to mathematical gauge theory and principal bundles in general yet, therefore I need some help. To be precise, I have the following question: First of all, let us fix some notation and terminology. Let $(\mathcal{M},g)$ be a $4$-dimensional Lorentzian manifold of signature $(+,-,-,-)$ and let $\mathcal{F}_{\mathrm{Ort}}(T\mathcal{M})$ denote the bundle of orthonormal (co-)frames on $(\mathcal{M},g)$. It can be shown that this bundle is in fact a principal $\SO(1,3)$-bundle, where $\SO(1,3)$ denotes the Lorentz group, as usual. Now let us consider the associated fibre bundle \begin{align*}E:=\mathcal{F}_{\mathrm{Ort}}(T\mathcal{M})\times_{\rho}\mathbb{M}^{4},\end{align*} where $\rho:\SO(1,3)\to\mathrm{Aut}(\mathbb{M}^{4})$ denotes the fundermental representation of the Lorentz group and where $\mathbb{M}^{4}:=\mathbb{R}^{4}$ is the Minkowski space in four dimension with signature $(+,-,-,-)$. Now as far as I know, the spin connection is usually defined to be a connection 1-form in the orthonormal frame bundle, i.e. $\omega\in \Omega^{1}(\mathcal{F}_{\mathrm{Ort}}(T\mathcal{M}),\mathfrak{so}(1,3))$. Choosing a local gauge, i.e. a local section $s:U\subset\mathcal{M}\to\mathcal{F}_{\mathrm{Ort}}(T\mathcal{M})$, we can define a local gauge field $A\in\Omega^{1}(U,\mathfrak{so}(1,3))$ on $\mathcal{M}$ via $A:=s^{\ast}\omega$. So far so good. Now many texts mention (e.g. arXiv:gr-qc/9410018) that we can also view the connection $1$-form as a $\bigwedge^{2}E$-valued 2 form on $\mathcal{M}$, i.e. an element of $\Omega^{1}(U,\bigwedge^{2}E)$ where $U\subset\mathcal{M}$ is an open subset. This is needed, because in the end the first-order tetradic Palatini action for general relativity is defined via $$S[e,\omega]:=\int_{\mathcal{M}}\operatorname{tr}(e\wedge e\wedge F)$$ where $F$ denotes the corresponding curvature $2$-form and where $\operatorname{tr}$ has to be understood as some kind of "internal volume form", which is a section of $\bigwedge^{4}E$ and maps a $\bigwedge^{4}E$-valued form into an ordinary (real-valued) form. The field $e$ above is a bundle isomorphism of the form $e:T\mathcal{M}\to E$, called the "cotetrad", which can also be viewed as an element of $\Omega^{1}(\mathcal{M},E)$. Can anyone explain me how we can view the connection $1$-form as a $\bigwedge^{2}E$-form as explained above? Of course, if something I have explained above is totally wrong, I am happy about every error pointed out. REPLY [3 votes]: For a finite dimensional inner product space $(V,\eta)$, $\bigwedge^2 V \cong_\eta \mathfrak{so}(\eta) \subset \operatorname{End}(V) \cong V\otimes V^* \cong_\eta V\otimes V$. The antisymmetry condition appears when expanding the identity $\eta(e^{tA}v,e^{t A}u) = \eta(v,u)$ to first order in $t$, to get $\eta(Av,u)+\eta(v,Au)=0$.<|endoftext|> TITLE: Generation of permutation groups by fixed elements subgroups QUESTION [5 upvotes]: Suppose $(H,X)$ is a permutation group (with $H$ a group acting faithfully on the set $X$). Under what circumstances is $H$ generated by its subgroups $H_x$, where $H_x$ is the subgroup of $H$ fixing $x \in X$, and $x$ varies over $X$ ? Of course, there are plenty of examples where this cannot work, such as sharply transitive permutation groups (and certainly many many more in the non-transitive case). Side questions: If $H$ is transitive, but not sharply transitive, is the answer: "always" (even in the infinite case) ? How important are finiteness conditions ? REPLY [6 votes]: We looked at this question during our research retreat and obtained the following characterisation: If $H$ is transitive on $X$ then it will be generated by its point stabilisers if and only if it does not have a proper system of imprimitivity upon which it acts regularly. By proper I mean one with at least two parts and by regular I mean that any element of $H$ that fixes one block in the partition fixes all of the blocks in the partition. The proof is as follows: Let $N$ be the normal subgroup of $H$ that is generated by all the point stabilisers. If $N\neq H$ then since $H_x\leqslant N< H$ for any $x\in X$ it follows that $N$ is intransitive and so its set of orbits forms a proper system of imprimitivity. If $B$ is one orbit of $N$ then $H_B$ acts transitively on $B$. Since $N$ also acts transitively on $B$ we have that $H_B=NH_x$ for some $x\in B$. Thus $H_B=N$ and so $H_B$ fixes each $N$-orbit, that is, $H$ acts regularly on the set of $N$-orbits. Conversely, suppose that $\mathcal{B}$ is a proper system of imprimitivity upon which $H$ acts regularly. Let $K\neq H$ be the kernel of this action. Let $x\in X$ and $B$ be the block of $\mathcal{B}$ containing $x$. Then $H_x\leqslant H_B=K$. Since $K$ is normal and all point stabilisers are conjugate to $H_x$ it follows that $K$ contains all point stabilisers and so the subgroup generated by all point stabilisers is contained in $K$ and so is not equal to $H$.<|endoftext|> TITLE: Suspension of a topological space QUESTION [17 upvotes]: Let $X$ be a topological space such that its suspension is a topological manifold. Can we prove that $X$ itself is a topological manifold? REPLY [14 votes]: As Jeff Strom said, the answer is no. For references to Cannon-Edwards theorem, see https://mathoverflow.net/a/316175/121665.<|endoftext|> TITLE: Is the supremum of L-definable cardinals silver-indiscernible QUESTION [8 upvotes]: Let $\kappa$ be the supremum of ordinals first order definable in L without parameters. Assume $0^\sharp$ exists. Is $\kappa$ the least silver indiscernible ordinal? REPLY [8 votes]: Updated. The answer is no. First, let me point out that in general, in ZFC we are not able to refer to the notion of first-order-definable-in-$L$, since definability is not expressible. But in your case, we have $0^\#$, from which we are able to define a truth predicate for first-order truth in $L$, and so your question can be formulated. But see the comments at the bottom concerning an ambiguity about this in nonstandard models. For each natural number $n$, the least $\Sigma_n$-correct cardinal $\kappa_n$ is definable in $L$, meaning $L_{\kappa_n}\prec_{\Sigma_n}L$, since we can express this property using a $\Sigma_n$ truth predicate. Further, any $\Sigma_n$ definable ordinal will be bounded by $\kappa_n$. Thus, $\kappa=\sup_n\kappa_n$. The union of an increasingly elementary chain is elementary, so $L_\kappa\prec L$. Thus, the supremum of the definable ordinals is the same as the first fully correct cardinal. But this is never a Silver indiscernible, since (as Monroe points out in the comments below) Silver indiscernibles $\xi$ are inaccessible in $L$ and therefore have many smaller $L_\alpha\prec L_\xi$. So the least Silver indiscernible is strictly larger than the least fully correct cardinal in $L$. This answers your question, but let me augment my answer with the following observation for the general ZFC case, which I find interesting. Theorem. For any model $M$ of ZFC, not necessarily well-founded, let $W$ be the collection of all $x\in M$ for which $x\in (V_\alpha)^M$ where $\alpha$ is an ordinal definable in $M$ without parameters. This is called the definable cut of $M$. Then $W\prec M$. Proof. Notice that the definable cut may not necessarily have a least upper bound in $M$. It could be that $W=M$ or that the supremum of $W$ is not realized in $M$. But we can nevertheless prove that $W$ is an elementary substructure of $M$ by verifying the Tarski-Vaught criterion. If $M$ has a witness for an extensional statement with parameters in $W$, then the least rank of such a witness is definable from those parameters, and by considering all possible parameters up to a definable rank, we can get a definable bound on the rank of the witnesses. And so the witness is in $W$. So we have fulfilled the Tarski-Vaught criterion, and thus $W\prec M$. $\quad\Box$ Note that in this theorem, we are using the external notion of definability, coming from outside the model, whereas in your question, we were using the internal notion of definability provided by the truth predicate defined from $0^\#$. These are not necessarily the same, even when $0^\#$ exists, since there could be $\omega$-nonstandard models of ZFC with $0^\#$. This issue can be seen as a possible ambiguity in your question, as to whether you intend to use the internal notion of definability or the meta-theoretic notion of definability. Observation. If ZFC+$\exists 0^\#$ is consistent, then there is a model of this theory in which the supremum of the definable-$L$-without-parameters ordinals (understood using meta-theoretic definitions) does not exist. Proof. Take any $\omega$-nonstandard model of the theory. For standard $n$, the first $\Sigma_n$-correct cardinal of $L$ is definable in $L$, using a standard-finite definition. The supremum of these would be the definable cut of the model, and this is the supremum of the meta-theoretically definable ordinals in $L$. This is bounded in the ordinals of $L$, because by overspill there must be some $\Sigma_n$ correct ordinals for nonstandard $n$. The definable cut can have no supremum in the model, since from it we could define the standard cut in $\omega$, which is possible. $\quad\Box$<|endoftext|> TITLE: A power of a sum in a non-commutative algebra QUESTION [6 upvotes]: Let $ A := \mathbb{C}\langle x, y \rangle / ( xy-2 ) $ where $ \mathbb{C}\langle x, y \rangle $ is the free (non-commuative) $ \mathbb{C} $-algebra which is generated by $ x $ and $ y $ and $ ( xy-2 ) $ is the two-sided ideal which generated by the element $ xy-2 $. Let us also write $ x $ resp. $ y $ for the classes of $ x $ resp. $ y $ in $ A $. Clearly, the elements $ y^k x^l $ form a basis of $ A $ for $ k, l \in \mathbb{N}_0 $. Therefore, there is a presentation $$ ( x + y )^n = \sum_{k + l = 0, \dots, n} c_{k,l}^{(n)} y^k x^l $$ for any $ n \in \mathbb N_0 $. Questions: I am interested in the sum of these coefficients, i.e. $$ S_n = \sum_{k, l} c_{k,l}^{(n)}. $$ Is there a general method to compute powers of sums in such algebras? What can be more said about the asymptotic behaviour of $ S_n $ except for the obvious bounds in $(*)$? Does anybody know a context in which this algebra (or more generally the algebra $ \mathbb C\langle x, y \rangle / ( xy - a ) $ for $ a \in \mathbb N $) plays a role? Known and maybe helpful: As we have two choices in each of the $ n $ factors in $ ( x + y )^n $, (without using the relation $ xy = 2 $) we already count at least $ 2^n $ $ ( x, y ) $-monomials. Moreover, applying the relation $ xy = 2 $ only doubles the coefficient of any of these $ 2^n $ monomials. This provides the lower bound $ 2^n $ for $ S_n $. One the other hand, mapping $ ( x, y ) \mapsto (\sqrt 2, \sqrt 2) $ yields a morphism $ A \to \mathbb{C} $ of $ \mathbb C $-algebras and gives the upper bound $ 2^{3n/2} $. So, we have the bounds \begin{align} 2^n \le S_n \le 2^{3n/2}. (*) \end{align} Furthermore, with some more effort one can show $ S_n \ll 2^{3n/2} / \sqrt n $ asymptotically as $ n \to \infty $. Anyway, I have the feeling that $ 2^{3n/2} $ is not far off, in the sense that $ (2^{3/2}-\varepsilon)^n / S_n \to 0 $ for any $ \varepsilon > 0 $. Approach: Let $ \mathbb C^{ \mathbb N } $ be the vector space of sequences with elements in $ \mathbb C $ and $ L, R \in \operatorname{End}_{\mathbb C}( \mathbb C^{ \mathbb N } ) $ be the backward and forward shifts. Then the algebra $ A $ is isomorphic to the sub algebra $ B := \mathbb C\langle L, R \rangle $ of $ \operatorname{End}_{\mathbb C}( \mathbb C^{ \mathbb N } ) $ (consider the map $ ( x, y ) \mapsto ( 2 L, R ) $). I seems to be possible to get $ S_n $ by the action of $ B $ on $ \mathbb C^{\mathbb N } $. But it is computationally involved and unclear whether the so obtained presentation of the $ S_n $ is useful. Edit As suggested in the comments, here are some additional information which were computed by using MAGMA: For $ n = 0, \dots, 15 $, the values of $ S_n $ are $$ [ 1, 2, 5, 12, 31, 78, 205, 528, 1403, 3666, 9817, 25908, 69783, 185526, 502005, 1342296 ] $$ and, for $ n = 0, \dots, 5 $ the desired presentations of $ ( x + y )^n $ are $ [ 1, \\ x + y, \\ x^2 + y*x + y^2 + 2, \\ x^3 + y*x^2 + y^2*x + y^3 + 4*x + 4*y, \\ x^4 + y*x^3 + y^2*x^2 + y^3*x + y^4 + 6*x^2 + 6*y*x + 6*y^2 + 8, \\ x^5 + y*x^4 + y^2*x^3 + y^3*x^2 + y^4*x + y^5 + 8*x^3 + 8*y*x^2 + 8*y^2*x + 8*y^3 + 20*x + 20*y ] .$ REPLY [4 votes]: Consider the sum of the coefficients of the degree-$(n-2k)$ "reduced" monomials $y^ix^j$ in $(x+y)^n$. The degree of a reduced monomial always differs from $n$ by an even number, so this will capture everything. This is clearly of the form $2^k\cdot T(n,k)$ for some unknown function $T$, since each time you reduce the product $xy$ you pick up a factor of $2$. So your sum $S_n$ equals $\sum_{k=0}^{\lfloor n/2 \rfloor} 2^k\cdot T(n,k)$. Now, I claim that $T(n,k)$ is given by A298637 on the OEIS. This is basically immediate from the definition. From the explicit formula given there, we have $$T(n,k) = \frac{(n+1-2k)^2}{n+1}\binom{n+1}{k}$$ so that $$S_n = \sum_{k=0}^{\lfloor n/2 \rfloor} 2^k\cdot \frac{(n+1-2k)^2}{n+1}\binom{n+1}{k}$$ By considering the contribution to the sum for $k=\lfloor n/2 \rfloor$ we see that this equals $2^{3n/2 - o(1)}$, confirming your suspicion.<|endoftext|> TITLE: Do escaping sets "uniformly" cover dominating sets under determinacy? QUESTION [5 upvotes]: For $\mathbb{A},\mathbb{B}\subseteq\mathcal{P}(\omega^\omega)$, say $\mathbb{A}$ spreads onto $\mathbb{B}$ iff there is some $F:\omega^\omega\rightarrow\omega^\omega$ such that for all $X\in\mathbb{A}$ the set $F[X]=\{F(r):r\in X\}$ is in $\mathbb{B}$. Let $\mathbb{D},\mathbb{E}$ be the sets of dominating, escaping families of reals respectively. Trivially $\mathbb{D}$ spreads onto $\mathbb{E}$ via the identity function, since every dominating family is escaping. The converse however is more complicated, even in $\mathsf{ZFC}$: $\mathbb{E}$ obviously can't spread onto $\mathbb{D}$ if $\mathfrak{b}<\mathfrak{d}$, but $\mathsf{CH}$ implies that $\mathbb{E}$ does spread onto $\mathbb{D}$ per vzoltan's answer to an old question of mine. I'm curious what happens in $\mathsf{ZF+DC+AD}$: Under $\mathsf{ZF+DC+AD}$, does $\mathbb{E}$ spread onto $\mathbb{D}$? REPLY [4 votes]: Here's something that seems to work. I can add more details if needed. Suppose that every set of reals has the property of Baire. Then every function from $\omega^{\omega} \to \omega^{\omega}$ is continuous on a comeager set. Note that every comeager set is unbounded. Fix $F \colon \omega^{\omega} \to \omega^{\omega}$ and let $C$ be a comeager set on which $F$ is continuous. We will find a dense open sets $D_{i}$ ($i \in \omega$) and $y \in \omega^{\omega}$ such that $F(x)$ fails to dominate $y$ whenever $x \in C \cap \bigcap_{i \in \omega}D_{i}$. Let $\langle (i_{n}, \sigma_{n}) : n \in \omega \rangle$ be an enumeration of $\omega \times \omega^{<\omega}$. In the $n$th stage of the construction choose $y \upharpoonright k_{n}$ for some $k_{n} \in \omega$ and also some $\tau_{n} \in \omega^{<\omega}$ extending $\sigma_{n}$ such that $F(x) \upharpoonright k_{n}$ is the same sequence $\rho_{n}$ for all $x \in C$ extending $\tau_{n}$, and $\{ j < k_{n} : \rho_{n}(j) < y(j)\}$ has size at least $i_n$. Then we can let $D_{i}$ be the set of $x \in \omega^{\omega}$ extending any of the sets $\tau_{n}$ where $i_{n} = i$. The $\mathbb{P}_{\mathrm{max}}$ claim has to be modified somewhat, since $\mathbb{P}_{\mathrm{max}}$ needs AD$^{+}$ to work, which is not known to follow from AD + DC. Forcing over a model of AD$^{+}$, $\mathbb{P}_{\mathrm{max}}$ produces a model of ZFC which is $\Pi_{2}$ maximal for the powerset of $\omega_{1}$ by (very roughly speaking) forming a generic direct limit of all countable models, subject to agreement about stationarity for subsets of $\omega_{1}$. There is a natural $\mathbb{P}_{\mathrm{max}}$ variation for each $\Sigma_{2}$ sentence, where each model fixes a witness to the sentence, and a model in the direct limit passes its witness on to stronger conditions. This doesn't always succeed, but for the statement $\mathfrak{b} = \aleph_{1}$ it does succeed, producing (without adding reals) a model of $\mathfrak{b} = \aleph_{1}$ in which every $\Pi_{2}$ sentence for $\mathcal{P}(\omega_{1})$ holds which can be provably forced to hold along with $\mathfrak{b} = \aleph_{1}$. For instance, the statement $\mathfrak{d} > \aleph_{1}$. This is discussed at a general level in the last section of my article from the Handbook of Set Theory, and in more detail in the Shelah-Zapletal paper "Canonical models for $\aleph_{1}$-combinatorics". Having produced such a model, one can then take an unbounded family $B$ of size $\aleph_{1}$, find a real $y$ which is not dominated by $F[B]$, and then find the unbounded set of $x$ such that $F(x)$ does not dominate $y$ back in the ground model. There is an intermediate argument one could run using the fact that, under AD$^{+}$, true $\Sigma^{2}_{1}$ sentences have Suslin, co-Suslin witnesses. This is at the heart of the $\mathbb{P}_{\mathrm{max}}$ argument. For the current problem, continuity on a comeager set works as a substitute for being Suslin and co-Suslin. Morally, anything that is consistent with large cardinals should be forceable over a determinacy model without adding reals.<|endoftext|> TITLE: For which pairs $k$ and $n$, $n\mid{{n-2} \choose {k}}$ QUESTION [7 upvotes]: The question This question that arose in a discussion with Ron Adin is quite simple: For which pairs $k$ and $n$ does $n$ divide ${{n-2} \choose {k}}$? Simple observations It is easy to see that for every $k$ there are finitely many possible values of $n$. One way to see it is to note that if $n$ divides ${{n-2} \choose {k}}$ it also divides $(n-2)(n-3)\dotsb (n-k-1)$ and therefore also (if $n \ge k+2$) it must divide $(k+1)!$. For $n=2k+2$, $\frac {1}{n} {{n-2} \choose {k}}=\frac {1}{2(k+1)}{{2k} \choose {k}}=\frac {1}{2}C_k$, where $C_k$ is the $k$-th Catalan number. So the question is when the $k$-th Catalan number is even. As $C_k={{2k+1} \choose {k}}-2 {{2k}\choose {k+1}}$, this is if and only if ${{2k+1} \choose {k}}$ is even and by Kummer's theorem this always happens, unless $k$ and $k+1$ have no common $1$ digits in base $2$, namely iff $k+1$ is a power of $2$. Maybe a similar analysis can be done in other cases as well, and perhaps a complete description is possible. Experimenting a little, it seems that given $n$ there are few values of $k$ such that $n \mid{{n-2} \choose {k}}$. Motivation These are the cases where vertex-regular $n$-vertex $k$-dimensional $\mathbb Q$-acyclic complexes with complete $(k-1)$-dimensional skeletons exist. We know (albeit by an indirect argument and not by an explicit construction) that whenever $d_1,d_2,\dotsc d_n$ are non-negative integers that sum up to ${{n-2} \choose {d}}$ then a $\mathbb Q$-acyclic complex with complete $(k-1)$-dimensional skeletons exists such that the degree of vertex $i$ (namely, the number of $k$-faces containing it) is $d_i + {{n-2} \choose {k-1}}$. Related MO question: Seeking very regular $\mathbb Q$-acyclic complexes REPLY [2 votes]: I doubt there exists a simple description for the general solution, however it's possible to give a characterization in some cases. For example, in the case of square-free odd $n$, Lucas' theorem implies that $(n,k)$ is a solution iff for every prime $p\mid n$, there exists a base-$p$ digit in $k$ that is greater than the corresponding base-$p$ digit of $n-2$. For the last digit, it means that $p\mid k+1$. From here we can construct a particular series of solutions $(n,k)$ with semi-prime $n=pq$, assuming, say, that $p\mid (k+1)$, i.e. the last base-$p$ digit of $k$ is $p-1$, and $p\equiv 1\pmod{q}$, i.e. the second but last base-$q$ digit of $n-2$ is $0$. Taking primes $q$ and $p=qs+1$ and integer $k=pt-1$ for some integers $s>0$ and $t\in[0,q]$, where to satisfy the condition on the second but last base-$q$ digit of $k=pt-1=qst+t-1$, it's enough to have $q\nmid st$. That is, we need to have $q\nmid s$ and $1\leq t\leq q-1$. Example. Take $q=23$ and $p=2\cdot 23+1=47$ giving $n=pq=1081$. Then any $k\in\{47t-1\ :\ t=1,2,\dotsc,22\}$ will do the job.<|endoftext|> TITLE: Peter Freyd on path Integral? QUESTION [15 upvotes]: In the issue Electronic Notes in Theoretical Computer Science Volume 29, 1999, Page 79 there is a very intriguing abstract by Peter Freyd. Path Integrals, Bayesian Vision, and Is Gaussian Quadrature Really Good? Physicists know how to integrate over all possible paths, computer-vision experts want to assign probabilities to arbitrary scenes, and numerical analysts act as if some continuous functions are more typical than others. In these three disparate cases, a more flexible notion of integration is being invoked than is possible in the traditional foundations for mathematics. If allowed to enter a highly speculative mode, such as the intersection of category theory and computer science, we may bump into some solutions to the problem. This was a special issue after the conference CTCS '99, Conference on Category Theory and Computer Science. Unfortunately, it seems to me that there is no additional material explaining what Freyd spoke about at the conference. Q. Do we have any clue about what Freyd spoke about that day? REPLY [15 votes]: In his late paper Algebraic real analysis (published on Theory and Applications of Categories, Vol. 20, No. 10, 2008, pp. 215–306) he writes in the very last page what follows. In September 1999 at an invited talk at the annual ctcs meeting (held that year in Edinburgh) I even characterized the mean value of real-valued continuous functions on the closed interval as an order-preserving linear operation that did the right thing to constants and had the property that the mean value on the entire interval equaled the midpoint of the mean-values on the two half intervals. I described it with a diagram that used (twice) the canonical equivalence between I and I ∨ I. But one equivalence, even used twice, doesn’t bring forth the general notion: it doesn’t prompt one to invent ordered wedges; without ordered wedges one doesn’t define zoom operators nor discover the theorem on the existence of standard models. One doesn’t learn how remarkably algebraic real analysis can become. What I needed was someone to kick me into coalgebra mode. Three months later two guys did just that and on December 22 I wrote to the category list: There’s a nice paper by Dusko Pavlovic and Vaughan Pratt. It’s entitled On Coalgebra of Real Numbers [115] and it has turned me on. A solution, alas, for the three motivating problems still awaits; but, at least, now I like analysis. Very much "We must (not) know, we will (not) know" vibes, but at least mystery solved.<|endoftext|> TITLE: Is this compactness property for "satisfiability on $\mathbb{R}$" consistent? QUESTION [7 upvotes]: This was originally part of this older question of mine, but in retrospect that question should have been broken into two parts - this is the still-unanswered part. Let $\Sigma$ be the language of ordered fields and let $\mathcal{R}$ be the field of real numbers. Say that a theory $T$ in a language $\Sigma'$ gotten by adding constant symbols to $\Sigma$ is $\mathcal{R}$-satisfiable iff $T$ has a model whose $\Sigma$-reduct is $\mathcal{R}$ itself. For cardinals $\kappa<\lambda$, say that $\mathcal{R}$-satisfiability is $(\kappa,\lambda)$-compact iff every theory of cardinality $<\lambda$ all of whose size-$<\kappa$-subsets are $\mathcal{R}$-satisfiable is itself $\mathcal{R}$-satisfiable. There are a couple easy observations about the possible extent of compactness for $\mathcal{R}$-satisfiability (see the discussion at the above-linked question): $\mathcal{R}$-satisfiability is provably not $(\omega_1,\omega_2)$-compact or $(2^\omega, (2^\omega)^{+})$-compact. If $\kappa$ is measurable then $\mathcal{R}$-satisfiability is $(\kappa,\kappa^+)$-compact. However, there is still one natural "low-level" question that remains open: Is $\mathcal{R}$-satisfiability consistently $(\omega_2,\omega_3)$-compact? Here are a couple comments: First, by the above observations $\mathcal{R}$-satisfiability can only be $(\omega_2,\omega_3)$-compact if $2^\omega\ge\omega_3$ (this was pointed out by Joel), so neither $\mathsf{CH}$ nor forcing axioms can help us. A bit more complicatedly, there are subtleties which pose potential obstacles to any "coarse" argument. Specifically, suppose we shift from $\mathcal{R}$ to its expansion $\mathcal{R}_\mathbb{Z}$ by a predicate naming the integers. In $\mathcal{R}_\mathbb{Z}$ we can talk about reals coding countable well-orderings and comparisons between the ordertypes of coded well-orderings. This gives us right off the bat a $\mathsf{ZFC}$-provable counterexample to $(\omega_2,\omega_3)$-compactness of $\mathcal{R}_\mathbb{Z}$-satisfiability: let $T$ be the theory using constant symbols $(c_\eta)_{\eta<\omega_2}$ saying that the $c_\eta$s code well-orderings of distinct ordertypes. This $T$ also demonstrates that we cannot obviously focus on complete theories WLOG: while every subtheory of $T$ of size $\omega_1$ is $\mathcal{R}_\mathbb{Z}$-satisfiable, every completion $S$ of $T$ has some subtheory of size $\omega_1$ which is not $\mathcal{R}_\mathbb{Z}$-satisfiable since every linear order of size $\omega_2$ has a suborder of size $\omega_1$ not embedding into $\omega_1$. Of course, the above reasoning breaks down for $\mathcal{R}$, but this does still suggest that there may be subtleties (and ultimately makes me very skeptical of a positive answer). REPLY [6 votes]: It looks to me like under ZFC, $\mathbb{R}$-satisfiability is not (consistently) $(\omega_2,\omega_3)$-compact. To see this, we'll emulate your argument above for $\mathbb{R}_{\mathbb{Z}}$. So basically, we want a theory with constants $c_\eta$ for $\eta<\omega_2$, which says: "$c_\eta$ codes a wellorder of $\omega$ (in ordertype $\geq\omega$)", "if $\eta_0\neq\eta_1$ then $c_{\eta_0},c_{\eta_1}$ have distinct ordertypes". It suffices to express these things in an appropriate fashion. We will in fact use a bunch more constants to help with this. We start with constants $\left_{i<\omega}$, along with the statements: "$n_0=0$" "$n_1=1$" "$n_{k+1}=n_k+1$", for each $k<\omega$ So whenever all these formulas are included, $n_k$ must be interpreted by $k$. Thus, we can talk about the integers by referencing these constants. But we cannot quantify directly over the integers. But we can do this enough for our purposes, indirectly. For each tuple $\vec{a}=(a_0,\ldots,a_{k-1})$ of constants that we add, we add further new constants $t_{\vec{a}}$ and $t^{\mathrm{wit}}_{\vec{a}}$; $t_{\vec{a}}$ will code the $\Sigma_1^{\mathbb{N},\vec{a}}$ theory (by which I really mean that $\Sigma_1^{\mathbb{N},A_{\vec{a}}}$ theory, where $A_{\vec{a}}=(A_{a_0},\ldots,A_{a_{k-1}})$ denotes the tuple of sets of integers coded by $\vec{a}$, and the theory is the collection of all $\Sigma_1$ truths in integer parameters over the structure $(\mathbb{N},A_{a_0},\ldots,A_{a_{k-1}})$), and the (interpretation of the) constant $t^{\mathrm{wit}}_{\vec{a}}$ will code witnesses to the $\Sigma_1$ assertions in this theory. Because our constants are closed under this, we end up with constants which will code the $\Sigma_k^{\mathbb{N},\vec{a}}$ theory, for each $k<\omega$, via which we can make arithmetical statements, which will be very helpful. So, for specificity, let's say that the real $x$ codes the set of rationals ${_{i<\omega}$ of all $\Sigma_0$ formulas in the language of $\mathbb{N}$ plus a predicate (implicit below, interperted as $A_\vec{a}$). We include the following statements in our theory: "If the formula `$\exists k\psi_i(k,\vec{n})$' is in $A_{t_{\vec{a}}}$, and $f_{t_{\vec{a}}^{\mathrm{wit}}}(i)=k'$, as witnessed by calculation $c$, then $\psi_i(k',\vec{n})$" (here $i,\vec{n},k',c$ are arbitrary (tuples of) integers, and we have some fixed coding of "calculations" by integers; note that "$\psi_i(k',\vec{n})$", for these particular integers $k',\vec{n}$, is expressible as noted above). "If the formula `$\exists k\psi_i(k,\vec{n})$' is not in $A_{t_{\vec{a}}}$, then $\neg\psi_i(k',\vec{n})$" (here $i,k',\vec{n}$ are arbitrary integers). If we have all these formulas present in the theory, then $t_{\vec{a}}$ must be interpreted by (some real coding) the true $\Sigma_1^{\mathbb{N},\vec{a}}$-theory. Thus (and as mentioned above) we can make arithmetical statements about the (sets coded by the) constants. We now want to express "$c_\eta$ codes a wellorder of $\omega$ (of ordertype $\geq\omega$)". We can do this by saying it codes a linear order which is comparable with all countable ordinals. Toward this, fix surjections $f:\omega\to\alpha$, for each $\alpha\in[\omega,\omega_1)$. We introduce new constants $d_\alpha$ for $\alpha\in[\omega,\omega_1)$, and, along with $c_\eta$, new constants $\pi_{\eta\alpha}$ for $\alpha\in[\omega,\omega_1)$. We want $d_\alpha$ to code a wellorder of ordertype $\alpha$, via $f_\alpha$, and $\pi_{\eta\alpha}$ to be an isomorphism comparing $c_\eta$ with $d_\alpha$. So add the following statements which achieve this: "$(m,n)\in A_{d_\alpha}$" (for integers $m,n$ and $\omega\leq\alpha<\omega_1$ such that $f_\alpha(m)\omega_1$. By ($\dagger$), $\{\kappa_x\bigm|x\in\mathbb{R}\}$ is cofinal in $\omega_2$, so there are exactly $\omega_2$-many equivalence classes. So let $\left_{\eta<\omega_3}$ be some constants. It suffices to see we can express the statements "$c_\eta\neq^* c_\beta$", for each $\eta<\beta<\omega_3$ (with the help of further constants). Given any constant $x$ we introduce, we will also introduce $s(x)$, which will be interpreted as $x^\#$. For all constants $s,t$, we will also introduce a constant $p(s,t)$ for $s\oplus t$. As before, we also introduce the $n_k$'s and close under the $\Sigma_1$-theories, so that we can make arithmetic statements. Now the main issue is to ensure that the interpretation of $s(x)$ is really $x^\#$: given this, the assertion "$c_\eta\neq^* c_\beta$" is equivalent to "$L[s(p(c_\eta,c_\beta))]\models c_\eta\neq^* c_\beta$" (where we define $=^*$ in this "model" $M$ as in $V$, except that we use $\omega_1^M$ instead of $\omega_1^V$), which will say that $L[(c_\eta\oplus c_\beta)^\#]\models$"$c_\eta\neq^* c_\beta$"; note that we can check this by just checking that $s(s(p(c_\eta\oplus c_\beta)))$ contains the right statement. So we want to ensure that $s_x$ really gets interpreted as $x^\#$. We add further constants: Add $d_\alpha$ for $\omega\leq\alpha<\omega_1$ as before (coding a wellorder $<_\alpha$ of $\omega$ of ordertype $\alpha$), and $s_{x,\alpha}$ (to code the model of the form "$L_\gamma[x]$" (as of yet possibly illfounded) one gets generated from an $\alpha$-sequence of "$x$-indiscernibles" as determined by $s_x$), for $\alpha<\omega_1$, and $\pi_{x,\alpha,\beta}$, to code an isomorphism which compares the ordinals of $s_{x,\alpha}$ with $\beta$, for each $\alpha,\beta<\omega_1$. We add formulas asserting: "$s_{x}$ is a pre-$x$-sharp" (meaning that $s_x$ has the right syntactic properties for a sharp, but ignoring iterability / wellfoundedness of the models generated as above), "$(m,n)\in A_{d_\alpha}$" or "$(m,n)\notin A_{d_\alpha}$" (as before), "$s_{x,\alpha}$ codes the term model generated by an $\alpha$-sequence of indiscernibles modelling the theory given by $s_x$ (and using $<_\alpha$ to determine the indiscernible sequence)", "$\pi_{x,\alpha,\beta}$ codes either (i) an isomorphism from the ordinals of $s_{x,\alpha}$ onto an initial segment of $<_\beta$, or (ii) an isomorphism from an initial segment of the ordinals of $s_{x,\alpha}$ onto $<_\beta$". These are all arithmetic statements about these constants, so we can make them. And they together ensure the correctness of $s_x$, because it ensures that the model $L_\gamma[x]$ mentioned above (which is countable) is wellfounded. As mentioned above, since there are $\omega_2$ distinct classes with respect to $=^*$, but not $\omega_3$, this gives a failure of $(\omega_2,\omega_3)$-compactness for $\mathbb{R}$-satisfiability. Thus, what if we work in $V=L[G]$ where $G$ adds $\geq\omega_4^L$ Cohen reals to $L$? Does $L[G]\models$"$\mathbb{R}$-satisfiability is $(\omega_3,\omega_4)$-compact"? And however, I think $(\omega_4,\omega_5)$ will be more subtle. Re $(\kappa,\kappa^{++})$-compactness for $\mathbb{R}$-satisfiability, don't we get that from $\kappa^+$-supercompactness of $\kappa$, like with the measurability giving $(\kappa,\kappa^+)$? (And likewise for $(\kappa,\lambda)$-compactness, if $\kappa$ is $<\lambda$-supercompact.) I.e. if $T$ is a theory of size $\gamma\in[\omega,\lambda)$, and all subtheories of size $<\kappa$ are $\mathbb{R}$-satisfiable, and $j:V\to M$ is elementary with $\gamma\kappa$. (On the other hand, in connection with the foregoing argument, and this question https://math.stackexchange.com/questions/4166734/does-mathsfzfc-prove-that-the-field-of-real-numbers-has-one-of-these-compac/4172264#4172264 and its answers, a reasonable question was whether every uncountable cardinal $\kappa$ such that $\mathcal{R}$-satisfiability is $(\kappa,\kappa^+)$-compact, has to be weakly compact.) It turns out that the most obvious candidate for a (counter)example works: Suppose $\kappa$ is weakly compact in $L$. Let $G$ be $L$-generic for adding a $\kappa^+$-sequence of Cohen reals with finite support. Then in $L[G]$, $\mathcal{R}$-satisfiability is $(\kappa,\kappa^+)$-compact, but $2^{\aleph_0}=\kappa^+$, so $\kappa$ is not weakly compact. Proof: Work in $V=L[G]$. Let $T$ be a relevant theory of size $\kappa$, all of whose size ${<\kappa}$ sub-theories are $\mathcal{R}$-satisfiable. We want to see $T$ is also. Note that there is $\alpha<\kappa^+$ such that $T\in L_{\alpha}[G\upharpoonright\alpha]$, and so by rearranging $G$ a little, we may assume that $T\in L_\alpha[G\upharpoonright\kappa]$ for some $\alpha<\kappa^+$. Let $X\preccurlyeq L_{\kappa^{++}}$ with $\beta=X\cap\kappa^+$ transitive and $\alpha<\beta<\kappa^+$ and $|X|=\kappa$. Let $L_\gamma$ be the transitive collapse of $X$. Using the weak compactness of $\kappa$ in $L$, let $\xi<\kappa^+$ and $j:L_{\gamma}\to L_\xi$ be elementary with $\mathrm{crit}(j)=\kappa$. So $\beta TITLE: Higher Airy functions: an exponentially decreasing solution of $f^{(n)}(x) = (-1)^n \, x^a \, f(x)$ QUESTION [9 upvotes]: Consider the differential equation $$ f^{(n)}(x) = (-1)^n\, x^a \, f(x), $$ where $a > 0$ is a real number. My numerical experiments suggest that this equation has a unique solution on $\mathbb{R}_+$ satisfying $$ f(x) \sim_{x \to \infty} x^{-\beta} e^{-x^\alpha/\alpha}, $$ where $\alpha = 1+\frac{a}{n}$, $\beta = \frac{a}{2n}$. Is this known? The solution would be a generalization of the Airy function, which is the unique bounded solution of $f''(x) = x f(x)$ (that is, $n=2$, $a=1$). The case $a=1$ is known: analogously to the Airy function, one constructs the solution using the integral $$ f(x) = \int_{\mathbb{R}} \exp\left[i \left(\frac{t^{n+1}}{n+1} + tx \right)\right] \, dt. $$ With some analysis one checks that this function satisfies the differential equation and has the right asymptotic as $x \to \infty$. If $n=2$ and $a$ is an integer, the required solution can be constructed using a much more complicated integral found in "A generalization of the Airy integral" by Gundersen and Steinbart. If $a$ is an integer, the existence of such a solution probably follows from "The possible orders of solutions for linear differential equations with polynomial coefficients" by Gundersen, Steinbart, and Wang (but I still have to check). However, the statement seems to be just as true for any real $a$. Here is a slightly more general conjecture. For any complex number $c \not=0$ the differential equation $$ f^{(n)}(x) = c^n \, x^a \, f(x) $$ has a solution satisfying $$ f(x) \sim_{x \to \infty} x^\beta e^{c \, x^\alpha/\alpha}, $$ for $\alpha = 1+\frac{a}n$ and $\beta = \frac{ac}{2n}$. It is easy to write a basis of solutions of this differential equation using (sort of) hypergeometric functions. The first solution is $$ f_0(x) = 1 + \frac{c^n \, x^{a+n}}{(a+n)(a+n-1) \dots (a+1)} + \frac{c^{2n} \, x^{2(a+n)}}{(a+n)(a+n-1) \dots (a+1) \; \cdot \; (2a+2n)(2a+2n-1)\dots(2a+n+1)} + \dots. $$ It is straightforward to check that the series converges for any positive $x$ and that $f_0$ is, indeed, a solution of the equation. Similarly one can construct a solution $f_q(x) = x^q + \dots$ for any integer $q 0 $ a real number, the possible behaviors at infinity are known, see Example 5 Section 3.4 in [1]. We get the following possible behaviors at infinity $$ f(x) \sim \gamma \, x^{\beta} \exp\left(\omega \, c \, \frac{x^\alpha}{\alpha}\right), \qquad \text{as } x \to +\infty, $$ with $ \alpha = 1 + \frac{a}{n} $, $ \beta = a \frac{1-n}{2n} $, $ \omega^n = 1 $ a $n$-th root of the unity, and $\gamma$ a constant. From this, if $ \min\{\Re(c \, \omega) \mid \omega^n = 1\} $ is reach for an unique $ \omega_c $ then there exists an unique solution $g$ such that $$ g(x) \sim x^{\beta} \exp\left(\omega_c \, c \, \frac{x^\alpha}{\alpha}\right), \qquad \text{as } x \to +\infty. $$ If we return to your first example with $c = -1$, we see that $ \min\{\Re(-\omega) \mid \omega^n = 1\} $ is uniquely reach for $ \omega = 1 $ therefore there exits an unique solution $g$ such that $$ g(x) \sim x^{\beta} \exp\left(-\frac{x^\alpha}{\alpha}\right), \qquad \text{as } x \to +\infty. $$ [1] C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer-Verlag, 1999.<|endoftext|> TITLE: Are $\log(\sigma(A(z))$ subharmonic functions? QUESTION [7 upvotes]: Let $A$ be a matrix-valued entire function. It is then well-known that $\log \Vert A(z)\Vert$ is subharmonic. In particular, the operator norm is just the largest singular value of $A$. Is it therefore also true that for any singular value $\sigma$ of $A$, in a domain where they are simple, turn $\log \sigma(A(z))$ into a subharmonic function? REPLY [9 votes]: First a positive result, which answers a different but related question. This paper of Aupetit On log sub-harmonicity of singular values of matrices, Studia Mathematica 122 (1997), 195-200; DOI 10.4064/sm-122-2-195-200 has the following abstract: Let $F$ be an analytic function from an open subset $\Omega$ of the complex plane into the algebra of $n\times n$ matrices. Denoting by $s_1,\dots,s_n$ the decreasing sequence of singular values of a matrix, we prove that the functions $\log s_1(F(\lambda))+\dots + \log s_k(F(\lambda))$ and $\log^+ s_1(F(\lambda)) + \dots+\log^+s_k(F(\lambda))$ are subharmonic on $\Omega$ for $1\leq k\leq n$. (Note: this result had been obtained independently and earlier in the 1989 PhD thesis of M. C. White, but was not published at the time.) In the same Studia article, an example is given just before Theorem 1 which shows that the answer to your stated question is negative. The example is quite easy to describe: take the entire matrix-valued function $$ F(z) = \begin{pmatrix} 1 & 1 \\ 0 & z \end{pmatrix} $$ Then the smallest singular value satisfies $s_2(z)=\min(1,|z|)$, which is not subharmonic. Consequently $\log s_2(z)$ also cannot be subharmonic.<|endoftext|> TITLE: Extending continuous functions from $\mathbb Q$ to $\mathbb R$ QUESTION [6 upvotes]: Definitions: Let $E$ be a subset of $X$. By an extension of a function $f: E \to \mathbb R$, I mean a function $\bar f: X \to \mathbb R$ such that $f = \bar f$ on $E$. Question: For every continuous function $f: \mathbb Q \to \mathbb R$, does there exist an extension $\bar f: \mathbb R \to \mathbb R$ that is continuous at each $q \in \mathbb Q$? In other words, can every continuous function on the rationals be extended to a function on the reals that is continuous at every rational point? REPLY [13 votes]: You may extend by upper limit. Details. Denote $g(y)=\limsup_{x\to y, x\in \mathbb{Q}} f(x)$ for all real $y$. So, possibly $g$ takes the value $+\infty$ or $-\infty$ at some points. But we have $g(x)=f(x)$ for rational $x$, and for each rational $x$ and each $\varepsilon>0$ there exists $\delta>0$ such that $|g(y)-f(x)|<\varepsilon$ whenever $|y-x|<\delta$, in particular $g(y)\ne \pm \infty$ if $|y-x|<\delta$. Thus if we replace all infinite values of $g$ to 0, $g$ remains continuous at rational points.<|endoftext|> TITLE: On the Galois group of the compositions of polynomials QUESTION [6 upvotes]: We reprint an old math SE question here (see https://math.stackexchange.com/questions/1241224/composition-of-polynomials-and-galois-theory): " Let $f(x)$ be a polynomial of degree $n$ over $\mathbb{Q}$, with Galois group isomorphic to the symmetric group $S_n$. How do I show that $f$ cannot be expressed as a composition $g(h(x))$ of two polynomials $g$ and $h$ of degrees $> 1$. " This old question does not have an answer, but one comment refers to the article http://www.ccms.or.kr/data/pdfpaper/jcms22_3/22_3_497.pdf of Choi. Therein, in the paragraph after Lemma 3.2, it is written: "One of the important results about Gal$(f(g(x))/K)$ is that the Galois group is a wreath product of certain groups ([6])." Here, $K$ denotes any field and the reference $[6]$ points to the article "The Galois theory of iterates and composites of polynomials" by Odoni. Alas, Choi does not give a particular Lemma or Theorem of $[6]$ as a reference. The closest we could find is Lemma $4.1$ in $[6]$: $K$ is an arbitrary field. LEMMA $4.1.$ Let $f(g(X))$ be separable over $K$, and let deg$(f)= k$, deg$(g)=l$, with $k,l\geq 1$. Then $f(X)$ is also separable over $K$. Let $\mathcal{F}$ be Gal $f(X)/K$, identified with a subgroup of the permutations of its zeros in the usual way. Then there is an injective homomorphism of Gal $f(g(X))/K$ into the wreath product of $\mathcal{F}$ with the symmetric group $S_l$. The question is now: How to derive the statement in Choi's article from Lemma $4.1$ of Odoni's article? Or is there another result of Odoni's article needed? Any additional references are very welcome. EDIT: I am interested in the question, if the Galois group of two such polynomials is a wreath product in a non-trivial way. Thank you very much for the help. REPLY [7 votes]: Since $\mathcal{F} \le S_k$, the wreath product of $\mathcal{F}$ with $S_l$ is no larger than the wreath product of $S_k$ with $S_l$. This has cardinality $$ (l!)^k k! < (k\cdot l)! = |S_{k\cdot l}|.$$ (Perhaps the wreath product we need is the opposite one, with cardinality $(k!)^l l!$. The result is the same.) To see the inequality, write $$\frac{(k\cdot l)!}{(l!)^k} = \frac{(1\cdot\ldots\cdot l)}{(1\cdot\ldots\cdot l)}\cdot\frac{ ((l+1)\cdot\ldots\cdot (2l))}{(1\cdot\ldots\cdot l)}\cdot\ldots\cdot \frac{((kl-l+1)\ldots\cdot(kl))}{(1\cdot\ldots\cdot l)}$$ $$> \frac{l}{l}\cdot\frac{2l}{l}\cdot\ldots\cdot\frac{kl}{l} =k!,$$ where to pass from the first line to the second we take the last factor from the numerator and denominator of each fraction (it is $>$ and not just $\ge$ if $k>1$). Therefore $\mathrm{Gal}(f\circ g(X)/K)$ has cardinality less than $S_{k\cdot l}$. Edit: Regarding whether the Galois group is the wreath product, the answer is no. Consider the Galois group of $f\circ f(x)$ over $\mathbb{Q}$, where $f(x)=x^2 - 2$. Its order is $4$, while the wreath product of $S_2$ with itself has order $8$.<|endoftext|> TITLE: Does a random walk on a surface visit uniformly? QUESTION [16 upvotes]: Let $S$ be a smooth compact closed surface embedded in $\mathbb{R}^3$ of genus $g$. Starting from a point $p$, define a random walk as taking discrete steps in a uniformly random direction, each step a geodesic segment of the same length $\delta$. Assume $\delta$ is less than the injectivity radius and small with respect to the intrinsic diameter of $S$. Q. Is the set of footprints of the random walk evenly distributed on $S$, in the limit? By evenly distributed I mean the density of points per unit area of surface is the same everywhere on $S$. This is likely known, but I'm not finding it in the literature on random walks on manifolds. I'm especially interested in genus $0$. Thanks! Update (6JUn2021). The answer to Q is Yes, going back 38 years to Toshikazu Sunada, as recounted in @RW's answer. REPLY [19 votes]: This problem was first considered and solved by Sunada, see his 1983 paper Mean-value theorems and ergodicity of certain geodesic random walks. Alas, the authors of the quoted arxiv paper were not aware of this. Any assumptions on curvature and dimension are not necessary - it is just enough to assume that the manifold is compact. As it has been pointed out by Pierre PC, the fact that the Riemannian volume is a stationary measure is an immediate consequence of the Liouville theorem. Its ergodicity with respect to the geodesic random walk is equivalent to the absence of invariant subsets of the manifold, which would follow, for instance, if any two points can be joined by a chain of geodesic segments of length $\delta$. Actually, geodesic random walks are always mixing for sufficiently small $\delta$ - this is a consequence of the fact that the cube of the transition operator has a density which is bounded away from 0 on the diagonal. The latter also implies the uniqueness of the stationary measure. EDIT I had misattributed the term "geodesic random walk" to Sunada. Actually, it seems to be first introduced in 1975 by Jørgensen The central limit problem for geodesic random walks whose work is quoted by Sunada.<|endoftext|> TITLE: Göbel's correspondance between rooted trees and natural numbers QUESTION [10 upvotes]: In the paper On a 1-1-correspondence between rooted trees and natural numbers by F. Goebel, a correspondence between natural numbers and rooted tree was established via prime factorization. He defines: Let $T$ be a rooted tree, $r$ its root. The connected components of $T-r$ are denoted by $T_1,\dots,T_v$, where $v$ is the degree of $r$. The graphs $T_j$ ($j=1,\dots,v$) obviously are trees, which we transform into rooted trees by defining as the root of $T_j$ the vertex of $T_v$ which is adjacent to $r$ in $T$. Figure 2 (below) shows all rooted trees up to n=45. I do understand all composite numbers, however I fail to understand the structure of how the rooted trees for prime numbers are structured. Question: What are the rules to get the rooted tree for prime numbers? REPLY [14 votes]: (I see this in the comments, too, but to ensure this has an actual answer...) Here are the first fifteen natural numbers after drawing an individual line segment (edge and node) beneath the root: As you can see, I also numbered what this change does: The $n$th natural becomes the diagram for the $n$th prime, e.g., you recover the rooted trees beginning with $2, 3, 5, 7, 11, 13$ by looking at this image. Next will be $17, 19, 23$, etcetera.<|endoftext|> TITLE: A finite dimensional algebra associated to the symmetric group QUESTION [14 upvotes]: Let $S_n$ be the finite group given as $n \times n$ permutation matrices. Define for a given field $K$ the algebra $B_n$ as the subalgebra of $M_n(K)$ generated by all permutation matrices of $S_n$. (more generally we can do this for any subgroup of $S_n$ to associate to a finite group such a subalgebra. Often we will just obtain the group algebra, but not always.) Question 1: What is the subalgebra $B_n$? Has it been studied before and does it have a name? Can we determine quiver and relations of the basic algebras of the blocks of $B_n$ (over a splitting field of $K$)? Probably I should know this but I only feel really familiar with quiver algebras. Here some things that I found out: -$B_n$ has vector space dimension $(n-1)^2+1$ and is semisimple over a field of characteristic 0. The center of $B_n$ is two dimensional. This implies that the algebra is $M_{n-1}(K) \times K$ since we know two simples that are split over $K$, see the answer of Sasha. -$B_2$ over a field of characteristic two is isomorphic to $K[x]/(x^2)$ (so it is a non-semisimple Frobenius algebra in this case). -$B_3$ over a field of characteristic 3 is isomorphic to the the Nakayama with Kupisch series [2,3] (in particular it can have finite non-zero global dimension and thus is not a Frobenius algebra in general). Question 2: Is it true that any field $K$ is a splitting field for this algebra? What is the number of simple modules? Note that if any $K$ is really a splitting field, then the number of simples of the algebra $B$ is given by the vector space dimension of $B/(rad(B)+[B,B])$, which seems to be 2-dimensional(?). It would be interesting to see what the algebra for $S_p$ is for the primes $p=5$ and $p=7$ (up to Morita equivalence), but the algebra is not baisc anymore. REPLY [2 votes]: Let $K$ be a field. The literal answer has already been given by several people but let me try and get at the algebraic structure and provide a quiver with relations (see the addition). Let $J$ be the $n\times n$ all ones matrix. Then $J$ centralizes $S_n$ and the centralizer of $J$ consists of all matrices whose rows and columns all sum to some fixed field element $k\in K$. This has dimension $(n-1)^2+1$ and several people have proved the subalgebra spanned by $S_n$ has this dimension so $B_n$ is the centralizer of $J$. If the characteristic of $K$ is zero or does not divide $n$, then we can change the basis to consists of $(1,\cdots,1)$ and $e_1-e_i$ with $2\leq i\leq n$ with $e_i$ standard basis vectors. Then $J$ becomes the elementary matrix $E_{11}$ in this basis and its centralizer is clearly $K\times M_{n-1}(K)$, which can also be seen by representation theory. This is a split semisimple algebra with two simple modules and there is nothing much to say. If $K$ has characteristic $p$ dividing $n$, then $J^2=0$ and it is easy to see the Jordan canonical form of $J$ is a $2\times 2$ block and $n-2$ blocks that are $1\times 1$. In fact the basis $e_1,(1,\ldots, 1), e_i-e_1$ for $2\leq i\leq n-1$ gives the Jordan form with $e_1,(1,\ldots,1)$ giving the $2\times 2$-block. Thus we can identify $B_n$ with the centralizer of $$J'=N_2\oplus 0_{n-2}$$ where $N_2= \begin{bmatrix} 0 & 0\\ 1&0\end{bmatrix}$. A more ring theoretic view is the following. Let $R=K[x]/(x^2)$; its a self-injective local $K$-algebra and $K$ is the unique simple where $x$ acts by $0$. Then the centralizer of $J'$ is $\mathrm{End}_R(R\times K^{n-2})$. I claim the the semisimple quotient here is $K\times M_{n-2}(K)$ and hence there are two simple modules and the algebra $B_n$ is split over $K$. This is not to difficult to check directly since $R$ is a projective indecomposable $R$-module with simple quotient $K$ and simple socle $K$. Also note that the radical of $R$ is its socle. So the radical of $\mathrm{End}_R(R\times K^{n-2})$ is the direct sum of the $1$-dimensional radical of $R$ (viewed as endomorphisms of the first summand $R$), the $n-2$-dimensional space of $R$-module homomorphisms $R\to K^{n-2}$ and the $n-2$-dimensional space of $R$-module homomorphisms $K^{n-2}\to R$ (which all land in the one-dimensional socle=radical). From this description it is easy to see that the radical squares to $0$ which means that as soon as you know the quiver, you know the basic algebra. The semismple quotient is a copy of $K$ coming from $R/\mathrm{rad}(R)$ and the endomorphisms of the semisimple module $K^{n-2}$, which is $M_{n-2}(K)$. Note that any composition $K^{n-2}\to R\to K^{n-2}$ is zero since the socle of $R$ is its radical. But the compositions $R\to K^{n-2}\to R$ give the elements of $\mathrm{rad}(R)$ and so the radical cubed is zero. If we view $B_n$ as the centralizer of $J$, then the radical (which has dimension $2n-3$) is spanned by those matrices where each row is a constant vector and each column sums to $0$ and those matrices where each column is a constant vector and each row sums to $0$. Hopefully a representation theorist can compute the basic algebra from this description. Let me add that when $n=2=p$, this description recovers that $B_2\cong K[x]/(x^2)$ and when $n=3=p$, it recovers that $B_3$ is a basic algebra with semisimple quotient $K\times K$. I would guess in general that the basic algebra is $\mathrm{End}_R(R\times K)$ but I am not 100% sure. Added and edited based on corrections by the OP and @JeremyRickard. If I am not mistaken, the quiver of $B_n$ when $p\mid n$ and $n>2$ has two vertices $v,w$. There is one edge from $v$ to $w$, one edge from $w$ to $v$.and one loop at $v$. Since its a radical-squared zero split algebra, the quiver presentation for the basic algebra then says that all paths of length $2$ are $0$. But an expert should check this. There is a relation saying the path of length two from $v$ to $v$ is zero. Indeed view $B_n$ as $A=\mathrm{End}_R(R\times K^{n-2})$. A complete set of orthogonal primitive idempotents are the projection $e$ to $R$ and the $n-2$ projections to $K$, but the latter $n-2$ all give isomorphic projective indecomposables so we just need one of them, say the projection $f$ to the first factor, for the quiver. Then since the radical squared is zero, to compute the quiver we have that $f\mathrm{rad}(A)e=fAe$ is one dimensional and so is $e\mathrm{rad}(A)f=eAf$ since there is a one-dimensional space of homomorphisms $R$ to $K$ and $K$ to $R$. Also $eAe\cong R$ but $e\mathrm{rad}(R$ is the radical squared as noted above and so $e(\mathrm{rad}(A)/\mathrm{rad}^2(A))e\cong 0$. On the other hand, $fAf\cong K$ and so $f\mathrm{rad}(A)f=0$. This gives my description of the quiver. That the compositions $K\to R\to K$ are zero but $R\to K\to R$ are not all zero gives the relation.<|endoftext|> TITLE: Are there cases in which the Weyl group _does_ act on the flag variety/springer fiber? QUESTION [6 upvotes]: In nearly every reference on the classical springer correspondence (for example Chriss/Ginzburg's book on Complex Geometry) it is stated that the action of the Weyl Group on the homology of the springer fiber is not induced by an action of the Weyl group on the fiber itself. But I couldn't find any reference or general statement that would make this precise (and I'm just learning about geometric representation theory so I don't have any good intuition/working knowledge). For concreteness (and I think this should be a simpler case), consider the springer fiber over zero -- then $\mathcal{B}_0 = \mathcal{B} = G/B$ is the flag varitey (where $B$ is a Borel subalgebra). Now there is an action of the Weyl group $W$ on the quotient $G/T$ (where $T\subseteq B$ is a maximal torus, which allows to write $W = N_G(T)/T$). Now in Yun's notes, (1.5.4), he writes that the map $G/T\to G/B$ is an "affine space bundle", that this implies that the their homologies are isomorphic and that under this isomorphism, the action of $W$ on $H(G/T)$ gives the springer action of $W$ on $H(\mathcal{B})$ (I couldn't find a reference for that though). But I don't see why there should be no way to use this projection to define an action of $W$ on $\mathcal{B}$. Also, in "Schubert cells and cohomology of the spaces $G/P$" (1973), Bernstein Gelfand and Gelfand construct in chapter 5 a correspondence that gives rise to the springer action. I was hoping that this very explicit construction in the paper would shed some more light on what's going on, but I haven't been able to do that myself. Any help, and also references, are really appreciated. REPLY [2 votes]: Here is a general fact: Given a group $G$ and a subgroup $H$, there is an isomorphism between The space $\text{End}(G/H)$, consisting of $G$-equivariant maps $G/H\to G/H$. The group $N/H$, where $N$ denoted the normalizer of $H$ in $G$. This isomorphism is induced by the surjective homomorphism $N\to \text{End}(G/H)$, taking $n\in N$ to $gH \mapsto gnH$, which kernel is $H TITLE: Pushforward of semi-small maps QUESTION [10 upvotes]: Let $f : X \rightarrow Y$ be a semi-small map of complex projective varieties with $\dim X = n$. Apparently if we assume that $X$ is smooth, then $f_*\underline{\mathbb Q}[n]$ is perverse. (It would be much appreciated if you can provide a reference for this.) Is it still true if we only assume that $X$ is locally complete intersection? I know that $\underline{\mathbb Q}[n]$ is perverse, so I think that there is a chance. REPLY [3 votes]: It seems this example https://mathoverflow.net/a/72916/89514 can be adapted to our case to produce a counterexample. Suppose $Y=\mathbb{C}^2$, $X_1$ the blow up of $\mathbb{C}^2$ at the origin, $X_2$ a copy of $X_1$, and $X=X_1\cup_Z X_2$ glued along the exceptional divisor $Z$. Then $\pi:X\to Y$ is semismall, $X$ is a locally a complete intersection because it locally looks like union of two coordinate hyperplanes in $\mathbb{C}^3$. Then $\mathbb{C}_X$ is the extension of $\mathbb{C}_{X_1}\oplus \mathbb{C}_{X_1}$ by $\mathbb{C}_{Z}[-1]$. If $R\pi_* \mathbb{C}_X$ was perverse then $R\pi_* \mathbb{C}_Z[-1]$ would only have $H^0=ker(R\pi_* \mathbb{C}_X\to R\pi_* \mathbb{C}_{X_1}\oplus R \pi_*\mathbb{C}_{X_2})$ and $H^1=coker(R\pi_* \mathbb{C}_X\to R\pi_* \mathbb{C}_{X_1}\oplus R\pi_* \mathbb{C}_{X_2})$, whatever these are in the category of perverse sheaves, but we know that $R\pi_* \mathbb{C}_Z$ is the cohomology of $\mathbb{P}^1$, so its cohomological "width" is 2.<|endoftext|> TITLE: Prehistory of Gromov-hyperbolic spaces/groups QUESTION [13 upvotes]: When speaking about hyperbolic groups/spaces, one usually refers to Gromov's monograph Hyperbolic groups for their introduction. However, coarse notions of hyperbolicity can be found in some of his earlier texts, such that Infinite groups as geometric objects and Hyperbolic manifolds, groups and action. Also, among the (many) equivalent definitions of $\delta$-hyperbolic spaces, one of them refers to Rips' condition, suggesting that E. Rips played some role here. Hence my question: What is the prehistory of Gromov-hyperbolic spaces/groups? Here, "prehistory" refers to "before Gromov's monograph". Edit: I should specify that I am not looking for the historical motivations of the definition, but the origins of the definition itself. How did the definition emerge? REPLY [9 votes]: The notion of coarse curvature was very popular in the "Leningrad math school" where Gromov is from before Gromov, for example, see the works of D. Alexandrov and Toponogov. The small cancelation idea which goes back to Dehn was very popular in other parts of the USSR, especially in Moscow (Grindlinger, Novikov, Adian, Olshanskii and others) but did not influence Gromov's work very much, although groups with thin geodesic triangles are studied already in the work by Novikov and Adian. So in some sense Alexandrov, Toponogov, Novikov and Adian predated Gromov.<|endoftext|> TITLE: Tensor representations of the quantum algebra $U_q(\mathfrak{sl}(2))$ at the roots of unity QUESTION [6 upvotes]: I'm trying to understand how the representation theory of $U_q(\mathfrak{sl}(2))$ works and I had a look to some books and lecture notes available on the internet. The case of $q^m\neq1$ is discussed everywhere, but I'm having a hard time finding reliable (and clear) sources for representation theory for $q$ being a root of unity. I've understood that when $q$ is a root of unity, some of the spin $j\in\frac{1}{2}\mathbb{Z}$ representations become reducible while remaining indecomposable. But how do their tensor product representations behave? I've read in some references about the $q$-deformed Clebsch–Gordan coefficients for $q^m\neq1$, but I was unable to find a section dedicated to tensor product representations for $q^m=1$. Can we still define some sort of Clebsch–Gordan coefficients also when $q^m=1$? I've read something about “removing” representations and defining truncated tensor products, but the statements were confusing... what happens there? I would appreciate some reference suggestions. REPLY [5 votes]: The reason you're having trouble finding similarly simple explanations in the root of unity case is that the root of unity case is a lot more complicated. So rather than trying to answer your question directly let me instead give you some key signposts for what to look out for to see why things are complicated. First, the first definition you look at for a quantum group probably works over the base field $\mathbb{Q}(q)$. This actually means it doesn't make sense to specialize $q$ to any number whatsoever! In order to talk about specializing, you have to pick an "integral form", i.e. you want to work with a ring over say $\mathbb{Q}[q,q^{-1},q+q^{-1}]$ which has the property that it gives you the old quantum group when you base extend. The good news is that so long as $q$ is not a root of unity you can choose any reasonable integral form and you'll end up with the same theory, so you can basically ignore this whole issue. But if you're working at a root of unity you can't ignore this issue. Morever, the integral form that you probably want to use (the Lusztig form) is not the obvious one (in particular it's no longer generated by $E$, $F$, and $K$!). Second, looking at all representations of the quantum group at a root of unity is too complicated and doesn't look much like what happens at generic $q$. Instead you only want to look at the "nice" representations. The name for these nice representations are "tilting modules." There's an easy elementary definition, namely they're the direct sums of direct summands of tensor products of the defining 2-dimensional representation, however in order to actually prove results about them you need a less elementary definition (in terms of certain filtrations) and you need to see that these two definitions are equivalent. Third, as you mention, for many applications you don't actually want the category of tilting modules, you actually want to take a quotient of this category (by the "negligible morphisms"). In order to show that this quotient has the properties you want you need to prove some results about the combinatorics of negligible tilting modules. Furthermore, from a concrete viewpoint this quotient is somewhat confusing because the resulting category is no longer the category of representations of any Hopf algebra. The most elementary way to get at this category is to just ignore quantum groups entirely and directly construct the category you want via the Temperley-Lieb-Jones categories. That is, you just define a category whose objects are strings of dots, whose morphisms are linear combinations of planar arc diagrams modulo planar isotopy and a circle relation, where compositon is stacking and tensor is horizontal disjoint union. Then semismplifying is just taking a quotient setting a certain Jones-Wenzl projection to zero. But then because you don't see the quantum groups anywhere, and so it's much more difficult to generalize beyond $\mathfrak{sl}_2$.<|endoftext|> TITLE: Representation theory of higher homotopy groups QUESTION [7 upvotes]: I've seen some works on the representation of fundamental groups, which are (at least for me) quite important topic in mathematics. For example, Riemann-Hilbert correspondence relates representation of a fundamental group of complex algebraic variety with differential equations on it. (There's also a result in positive characteristic by Bhatt-Lurie). Also, etale fundamental group and its representation is also important in number theory because it is closely related to Galois groups. (I don't know much about this though). However, I can't find any works that studies about representation of higher homotopy groups $\pi_{n}$ (or etale homotopy groups, whatever it is). Although $\pi_n$ is abelian for $n\geq 2$, there might exist some nontrivial stuff that $\pi_n$ can act naturally so that we can study about its representations (which would be 1-dimensional if irreducible, though). Could you give any examples if there's any interesting works about such things? REPLY [11 votes]: There are many results that generalize the Riemann–Hilbert correspondence from the fundamental groupoid to the fundamental ∞-groupoid, for example: Jonathan Block, Aaron Smith. A Riemann–Hilbert correspondence for infinity local systems. Joseph Chuang, Julian Holstein, Andrey Lazarev. Maurer-Cartan moduli and theorems of Riemann-Hilbert type. Manuel Rivera, Mahmoud Zeinalian. The colimit of an ∞-local system as a twisted tensor product. Camilo Arias Abad, Florian Schaetz. Flat Z-graded connections and loop spaces. Julian V. S. Holstein. Morita cohomology. Camilo Arias Abad, Florian Schaetz. The A_infty de Rham theorem and integration of representations up to homotopy. Vincent Braunack-Mayer. Strict algebraic models for rational parametrised spectra II. Camilo Arias Abad, Sebastian Velez Vasquez. The higher Riemann-Hilbert correspondence and principal 2-bundles. For the higher Galois theory, see Marc Hoyois. Higher Galois theory.<|endoftext|> TITLE: Maximal sublattice index in Minkowski's Second Theorem QUESTION [6 upvotes]: Let $B$ be a (small) convex compact set in $\mathbb{R}^n$, symmetric around the origin. Let $\Gamma$ be a lattice in $\mathbb{R}^n$ of dimension $n$ (I'm almost sure we can just take $\mathbb{Z}^n$, but it may not always be the most natural example). Now, we dilate $B$ by simply multiplying it by a growing $\lambda$. We define the successive minima $\lambda_1,...,\lambda_n$ as the smallest $\lambda$ such that $\lambda\cdot B$ contains $i$ linearly independent non-zero elements of $\Gamma$. Define by $v_i$ the corresponding elements of $\Gamma$ (if there are multiple, we can choose any). $v_1,...,v_n$ generate a sublattice of $\Gamma$, but not necessarily the whole lattice. In fact, Minkowski's Second Theorem states that the maximal index of that sublattice is $n!$. It is easy to give an example with index $2^n$ - simply take a cube and $\Gamma=\mathbb{Z}^{n-1}$ and choose the vertices of the cube with side 2 as $v_i$s. Another natural example is the 5-dimensional sphere with $\Gamma = \mathbb{Z}^5\cup(\mathbb{Z}^5+(0.5,...,0.5))$ - the "half-points" are further than the standard $\mathbb{Z}^5$ points, giving index 2. My question is: what is an example of set $B$ (and lattice $\Gamma$) that has the index equal to $n!$, even for a specific $n>2$? Of course it would be best if the example was easily generalized to any dimension. Or if there are no such examples, what is the largest possible index? REPLY [3 votes]: For the case of $B$ a Euclidean sphere, the paper On the Index System of Well-Rounded Lattices is probably state-of-the-art (the well-rounded assumption is known to be WLOG). It investigates bounding the exact index you are interested in (it actually describes the set of isomorphism classes of the quotients $\Lambda / \Lambda'$, so somewhat more), and: Has exact bounds for dimensions $\leq 9$ Has exact characterizations for root lattices, of which the maximal index (parametrized as a function of $n$) is given by $\mathbb{D}_n$, of index $\leq 2^{\lfloor (n-1)/2\rfloor}$ (and it achieves this index). One can generically bound the index by $\gamma_n^{n/2}$ for $\gamma_n$ the Hermite constant, which is known to satisfy the bound $\gamma_n < 1 + \frac{n}{4}$, so $\gamma_n^{n/2} = O(n^n) = O(n!)$, as you mention. The appendix of On Classifying Minkowskian Sublattices additionally mentions that the Leech lattice has a quotient of the form you are interested in of index $2^{24}$. This is all to say that for $B$ a Euclidean sphere, there appears to be a gap. A somewhat similar question is, for an $n$-dimensional lattice $L$, comparing the products: $$\frac{\prod_i\lVert \vec e_i\rVert_2^2}{\det L}$$ over independent vectors $\vec e_i$ (denoted $M(L)$) and basis vectors $\vec e_i$ (denoted $H_b(L)$), where $\det L$ is the determinant of the Gram matrix of $L$. In Hermite vs. Minkowski, Martinet shows that: $$Q_b(L) = \frac{H_b(L)}{M(L)} \leq (5/4)^{n-4}$$ for $n\geq 4$. I don't see how this might immediately imply an index bound, but it does show that on average one only has to inflate the size of each minimal independent vector by a constant factor to get a basis, which is perhaps interesting to you.<|endoftext|> TITLE: For ideals, does normal imply countably complete? QUESTION [14 upvotes]: The following little question has bugged me for a while. Suppose $Z \subseteq \mathcal P(X)$. We say an ideal $I$ on $Z$ is normal when it is closed under diagonal unions, which means that if $\{ A_x : x \in X \} \subseteq I$, then $\nabla_x A_x := \{ z \in Z : \exists x \in z ( z \in A_x) \} \in I$. We say that $I$ is fine when for all $x \in X$, $\{ z \in Z : x \notin z \} \in I$. We say that $I$ is countably complete if it is closed under countable unions. Fact 1: Normal + fine implies countably complete. See Proposition 1.5 here. Fact 2: Fine does not imply countably complete. Let $Z$ be the collection of all finite subsets of an infinite set $X$, and let $I$ be the smallest fine ideal on $Z$. Fact 3: Normal + countably complete does not imply fine. Let $Z$ be the set of all Dedekind cuts of the rationals, and let $I$ be a countably complete ideal on the reals like the Lebesgue null ideal. Then $I$ is normal by the regressive function characterization, since any regressive function on a non-null set takes some rational value on a non-null subset. But $I$ is not fine, since the set of cuts not containing a given rational is just the reals less than that rational, which is not null. Fact 4: Fine + countably complete does not imply normal. It is not hard to show that the ideal of countable subsets of $\omega_1$ is not normal. Question: Can we strengthen Fact 1? Does normality imply countable completeness? REPLY [2 votes]: Throughout, let $\kappa$ be a cardinal and let $X=\kappa$. Here is a way to cover the case left open by Paul Larson. Let $\{S_{n}\}_{n<\omega}\subseteq I$ be an arbitrary countable collection from $I$. Let $T_{n}=\{z\in S_{n}\, :\, |z|<\aleph_0\}$. Since $T_{n}\subseteq S_{n}\in I$ and $I$ is an ideal, we have $T_{n}\in I$. If we knew that $T_{\infty}=\bigcup_{n<\omega}T_n\in I$, then by taking the union of $T_{\infty}$ with the set in Paul's answer we would be done. We will prove the following (stronger) claim about possibly uncountable unions. Claim: Let $\{T_{\alpha}\}_{\alpha<\kappa}\subseteq I$. If $z\in T_{\alpha}$ implies $z\in \mathscr{P}^{<\omega}(X)$ (i.e., $z$ is finite) for each $\alpha$, then $\bigcup_{\alpha<\kappa}T_{\alpha}\in I$. Proof. The claim is trivial if $\kappa$ is finite, so hereafter we assume $\kappa$ is infinite. Working by induction on $\kappa$, suppose the claim is true for all (normal ideals on power sets of) smaller cardinals. For each ordinal $\beta<\kappa$, let $T_{\alpha}(\beta)=\{z\in T_{\alpha}\, :\, z\in \mathscr{P}^{<\omega}(\beta)\}$. Since $\lambda=|\mathscr{P}^{<\omega}(\beta)|<\kappa$, we can fix an map $f\colon \lambda\to \kappa$ such that if $z\in \bigcup_{\alpha<\kappa}T_{\alpha}(\beta)$ then $z\in T_{f(\gamma)}(\beta)$ for some $\gamma<\lambda$. Now, by our inductive hypothesis, the normal ideal generated by $\{T_{f(\gamma)}(\beta)\}_{\gamma<\lambda}$ contains $$T_{\infty}(\beta)=\bigcup_{\gamma<\lambda}T_{f(\gamma)}(\beta)=\bigcup_{\alpha<\kappa}T_{\alpha}(\beta).$$ Hence, $T_{\infty}(\beta)\in I$ too. The diagonal union of these sets (as indexed by $\beta$, possibly shifted by $1$) is $\bigcup_{\alpha<\kappa}T_{\alpha}$ (possibly after throwing in $\{\emptyset\}$, as needed). $\square$<|endoftext|> TITLE: Fermions, their path integrals and effective actions QUESTION [6 upvotes]: I just read the nice exposition Fermionic Path Integral on nLab and began to wonder about some details to which references appear to be lacking. Suppose we live on Euclidean space as in the Osterwalder-Schrader approach to QFT: Is there a deeper analogy between fermionic and bosonic integration? For bosons, we should consider e.g the space $\mathcal{S}'$ of Schwartz distributions as "path space". What is the corresponding - presumably non-commutative - fermionic space $\mathcal{F}'$? If such an analogy exists, what is the corresponding one between probability measures $\mu$ on $\mathcal{S}'$ and Berezin integrals $\nu$ on $\mathcal{F}'$? Is there an analogy to the spaces physicists would like to work on? i.e The space $\mathcal{S}$ of Schwartz functions would be nice and obliterate the need for regularisation/renormalisation, but unfortunately we have to work on $\mathcal{S}'$ instead. If such an analogy exists, what is $\mathcal{F}$? Finally, what features should an interaction $S^{\mathrm{int}}$ have in order to make \begin{equation} S^{\mathrm{eff}} \left( \phi \right) = - \ln \int_{\mathcal{F}'} \exp \left[ - S^{\mathrm{int}} \left( \phi, \psi \right) \right] \mathrm{d} \nu_{\mathrm{Berezin}} \left( \psi \right) \end{equation} well-defined? i.e How to integrate out a fermion? EDIT: For the analogies, I was expecting something along the lines: For bosons, we encode $\mathbb{R}^4$ by a commutative ring of test functions on $\mathbb{R}^4$, e.g $\mathcal{S}$ For fermions, I would expect something similar to give a non-commutative ring $\mathcal{F}$ For bosons, consider the cylindrical measure of a free theory on $\mathcal{S}$ which extends to a measure on $\mathcal{S}'$ For fermions, consider some weird (in some sense positive) linear functional on $C_b( \mathcal{F} )$ that is somehow compatible with the non-commutativity of $\mathcal{F}$, expose a failure of regularity and show that the failure disappears upon prolonging to $C_b( \mathcal{F}' )$ where $\mathcal{F}'$ is something even more untangible than $\mathcal{F}$ - but has a concrete description REPLY [2 votes]: The fermionic path integral is not an integral in the analytic sense, and that's good so because therefore we can evaluate it rigorously. So, this is quite unrelated to measures or distributions, and I think that there is no analogy like the one you are looking for. Your last point can be answered completely, more or less along the lines of the nlab page you cite. The setting has be so the bosons $\phi$ parameterize a family of Dirac operators $D_\phi$ acting usually on a space of $L^2$-sections of a spinor bundle. Associated to this family of Dirac operators must be a Pfaffian line bundle $Pfaff(D)$ over the space of all bosons $\phi$. It is not so that you would have any choice for the fermionic action functional, it only works for the standard functional $$ S^{int}(\phi,\psi) := \int \langle \psi,D_\phi \psi \rangle dvol_g, $$ where the integral is w.r.t. a volume form and orientation on the worldvolumes of fields. The reason is that the - a priori not well-defined - expression $$ \int_\psi e^{S^{int}(\phi,\psi)} d\psi $$ can then be interpreted as an element in $Pfaff(D)$ in the fibre over $\phi$. This has to do with the fact that $\langle -,D_\phi -\rangle$ is skew-hermitian (provided the setting is correctly set up), and Paffians are concerned with skew-hermitian operators. Under this interpretation, the map $$ \phi \mapsto \int_\psi e^{S^{int}(\phi,\psi)} d\psi $$ a well defined section (probably with zeros) of $Pfaff(D)$. Further discussion ("anomaly cancellation") is then concerned with the question how the Paffian line bundle can be trivialized, so that this section becomes a complex-valued function on the space of bosons. Once this is achieved, one may multiply this function with any bosonic action functional; this gives the full integrand under the bosonic path integral. For example, a spin structure on spacetime trivializes the Pfaffian line bundle in worldvolume dimension one, while a (geometric) string structure trivializes the Pfaffian line bundle in worldvolume dimension two. Some reference for this are: Freed, Daniel S.; Moore, Gregory W., Setting the quantum integrand of M-theory, Commun. Math. Phys. 263, No. 1, 89-132 (2006). ZBL1124.58011. Freed, Daniel S., Determinants, torsion, and strings, Commun. Math. Phys. 107, 483-513 (1986). ZBL0606.58013. Bunke, Ulrich, String structures and trivialisations of a Pfaffian line bundle, Commun. Math. Phys. 307, No. 3, 675-712 (2011). ZBL1238.58022. I have also talked a lot about this, and have slides on my webpage, e.g. these.<|endoftext|> TITLE: Sunflowers in maximal almost disjoint families QUESTION [8 upvotes]: Let $[\omega]^\omega$ denote the collection of infinite subsets of $\omega$. We say ${\cal A}\subseteq [\omega]^\omega$ is almost disjoint if $A \cap B$ is finite whenever $A\neq B \in {\cal A}$. Zorn's Lemma implies that every almost disjoint family is contained in a maximal one. Moreover, a diagonalization argument shows that every maximal almost disjoint (MAD) family is uncountable. A sunflower is a set ${\cal X}$ of sets such that ${\cal X} \neq \emptyset$ and there is $K\subseteq \bigcup{\cal X}$ such that $X\cap Y = K$ whenever $X\neq Y\in {\cal X}$. (We allow for $K=\emptyset$.) Does every MAD family ${\cal A}\subseteq [\omega]^\omega$ contain an infinite sunflower? If not, is it true that given $n\in \omega$, every MAD family contains a sunflower of cardinality $n$? REPLY [4 votes]: One remark on KP Hart's answer: if one does the same to a finitely branching tree with unbounded sizes of the splitting sets then one gets a MAD family with no infinite delta-system but delta systems of all finite sizes.<|endoftext|> TITLE: Is there an inscribed cube for an arbitrary compact closed surface? QUESTION [10 upvotes]: Given a compact closed surface $M$ (2-dim topological manifold) isometrically embedded in $\mathbb{R}^3$, are there 8 points $x_i\in M(i=1,\dots,8)$ such that they are the vertices of a cube $C\subset\mathbb{R}^3$? We may assume that (1)$M$ is smooth and homeomorphic to the 2-sphere $S^2$; (2)$M$ is piecewise-smooth; (3)$M$ is $C^2$-manifold. The case (1) is actually what I'm mostly curious about. REPLY [7 votes]: One can't inscribe cubes in generic surfaces by dimension reason. Indeed the space of cubes in $\mathbb R^3$ is $7=3+3+1$-dimensional, while a cube has $8$ vertices, and so a surface imposes $8$ conditions on the vertices of the cube. To make this dimension reasoning rigorous one can do the following. Take the space of polynomials of degree $\le d$ on $\mathbb R^3$ that vanish at $8$ vertices of a (non-zero) cube. For $n$ large enough (probably $d\ge 3$ will suffice) this space has codimension $8$ in the space of all polynomials of degree $\le d$. So if we take the poly $x^2+y^2+z^2-1$ and add to it $\varepsilon F$, where $F$ is a generic poly of degree $d$ then the surface $\Sigma:=\{x^2+y^2+z^2-1+\varepsilon F=0\}$ doesn't contain a cube. And $\Sigma$ has a connected component diffeomorphic to a sphere. However it is not so easy to construct a concrete example of such surface by hands, because it should be quite asymmetric.<|endoftext|> TITLE: Pfaffian systems that do not satisfy their integrability conditions QUESTION [7 upvotes]: Remark: In this question I am first and foremost interested in a local problem and local solutions therefore I assume all functions are defined on open sets of real coordinate spaces and I will not bother with explictly considering domains. To simplify notation some maps will actually be partial maps defined on an open subset. Summation convention on repeated indices is understood. All functions are taken to be $C^\infty$. Background: A Pfaffian system is a set $\theta^\alpha=\theta^\alpha_i dx^i$ of pointwise linearly independent $1$-forms ($i=1,...,m$, $\alpha=1,...,n$, $n\le m$, i.e. we have $n$ such $1$-forms in $m$ variables). A Pfaffian equation $$ \theta^\alpha\approx 0 $$ is a partial differential equation for a submanifold $\phi:\mathbb R^{m-n}\rightarrow\mathbb R^{m}$ which is a solution iff $$ 0=\phi^\ast\theta^\alpha=\theta^\alpha_i(\phi(u))\frac{\partial\phi^i}{\partial u^a}du^a. $$ If $\Lambda^\alpha_{\ \beta}$ is an invertible matrix whose elements are functions of the $x^i$, the system $$ \bar\theta^\alpha =\Lambda^\alpha_{\ \beta}\theta^\beta$$ is equivalent to the system $\theta^\beta$. Per the standard terminology, the Pfaffian system is integrable if for each $x_0$ there is a solution (integral submanifold) whose image contains the point $x_0$. The Pfaffian system is completely integrable if there is an equivalent system whose generators are exact, i.e. there exists an invertible matrix $\Lambda^\alpha_{\ \beta}$ such that $$ \Lambda^\alpha_{\ \beta}\theta^\beta=dF^\alpha. $$The system is closed if there are $1$-forms $\xi^\alpha_\beta$ such that $$ d\theta^\alpha=\xi^\alpha_\beta\wedge\theta^\beta, $$ or equivalently if $$ d\theta^\alpha\wedge\theta^1\wedge...\wedge\theta^n=0. $$ The Frobenius integrability theorem essentially states that a Pfaffian system being closed is a sufficient condition for complete integrability. (The other direction of implications completely integrable $\Rightarrow$ integrable $\Rightarrow$ closed is easily seen.) Assuming now $m$ "base" variables $x^i$ and $n$ "fibre" variables $y^\alpha$ and $n+m$ "total variables, a differential equation $$ \frac{\partial\phi^\alpha}{\partial x^i}(x)+\Pi^\alpha_i(x,\phi(x))=0 $$ which is often called a total differential equation has its existence conditions determined by the Frobenius theorem. The Frobenius condition in this case is $$ R^\alpha_{ij}=\frac{\partial\Pi^\alpha_j}{\partial x^i}-\frac{\partial\Pi^\alpha_i}{\partial x^j}+\frac{\partial\Pi^\alpha_i}{\partial y^\beta}\Pi^\beta_j-\frac{\partial\Pi^\alpha_j}{\partial y^\beta}\Pi^\beta_i=0. $$ If this equation is satisfies then for any initial condition $(x_0,y_0)$ the total differential equation has a (essentially unique) solution with $\phi^\alpha(x_0)=y_0^\alpha$. It also follows (modulo some analytical muckery regarding the smooth dependence on initial conditions) that then there exists a function $$ \Phi^\alpha(x,y_0) $$ of $m+n$ variables such that for fixed $y_0$ the function (as a function of the remaining varibles) is the unique solution corresponding to the initial values $(x_0,y_0)$, that is the general solution of the total differential equation is parametrized by $n$ constants/parameters. Questions: I am essentially looking for a modern (more explanation later on what I mean modern) treatment of Pfaffian systems which also take into account what happens when the integrability conditions are not satisfied identically. For the case of a Pfaffian system $\theta^\alpha\approx 0$ and from a geometric point of view I expect that integral submanifolds of maximal ($m-n$) dimension do not exist, but lower dimensional integral manifolds might. For total differential equations of the form $\phi^\alpha_{,i}(x)+\Pi^\alpha_{i}(x,\phi(x))=0$ I expect that a general point $(x_0,y_0)\in \mathbb R^{m+n}$ has no solution passing through it. However solutions that satisfy the algebraic constraint $$ R^\alpha_{ij}(x,\phi(x))=0 $$can still exist. I am interested in eg. how can we describe the "general solution" of the differential equation. Maybe instead of $n$ parameters, the general solution depends on $< n$ constants? I give some further context here. I know that this problem is quite classical and its solution was even known in the 19th century. But for some reason I have extreme difficulties in finding a good and concise reference that treats this and contains the proofs. I have searched in differential geometry books, but those usually only contain a geometric discussion of Frobenius' theorem but do not consider solutions of PDEs when the integrability conditions are not satisfied. I am not particularly familiar with the PDE literature but looked in a few books, which only contained treatments of Laplacian/Poissonian/wave equations, boundary conditions, etc. not Pfaffians. As far as I understand, Pfaffians belong more to differential geometry than PDE theory especially that the structure of Pfaffians can be understood via ODEs. I have found extensive discussions in very old books like those of Schouten/Eisenhart/Forsyth, however I have mainly three issues with these books, They use very archaic terminology and notations that makes reading them extremely difficult. It seems to me that pre-Bourbaki mathematicians had a very different idea what constitutes a proof than the post-Bourbaki mathematicians. They frequently assume analyticity and have often seen power series proofs in these books. I don't like that. As far as I am aware exterior differential systems more complicated than Pfaffians can only be treated in the analytic category. But the Frobenius theorem only requires $C^\infty$ (in fact $C^{\text{sufficiently high}}$) functions and thus the integrability of Pfaffian systems is treatable in the $C^\infty$ category. This is something I absolutely insist on, I have no interest in any treatment that uses analyticity. Sooooo, I'd really like to get my hands on a reasonably modern and rigorous reference on integrability conditions of Pfaffian PDEs which consider the case when the integrability conditions are not satisfied and only use $C^\infty$ arguments at most. I also note that I am not necessarily interested in constructive solution techniques that can aid with explicitly finding the solutions/integral manifolds. I am only interested in "theoretical" characterization of integral submanifolds and/or solutions of the Pfaffian system. REPLY [6 votes]: Since everything is local and $C^\infty$, it is not hard to derive sufficient conditions for there to exist solutions. Analyticity is not actually needed, but some assumption of regularity is necessary. Here is one such result: Let the Pfaffian system $\mathcal{I}$ be generated by $\theta^1,\ldots,\theta^s$ and choose $1$-forms $\omega^1,\ldots,\omega^m$ to complete to a basis of $1$-forms. (If desired, one can choose functions $x^1,\ldots, x^m$ such that $\theta^1\wedge\cdots\theta^s\wedge\mathrm{d}x^1\wedge\cdots\mathrm{d}x^m\not=0$ and let $\omega^i=\mathrm{d}x^i$, but this is not necessary.) Now compute the functions $R^\sigma_{ij}=-R^\sigma_{ji}$ such that $$ \mathrm{d}\theta^\sigma \equiv \tfrac12 R^\sigma_{ij}\,\omega^i\wedge\omega^j\ \mod \theta^1,\ldots,\theta^s. $$ Next, compute the functions $R^\sigma_{ijk}$ such that $$ \mathrm{d}R^\sigma_{ij} \equiv R^\sigma_{ijk}\,\omega^k\ \mod \theta^1,\ldots,\theta^s, $$ and, inductively, define $R^\sigma_{i_1i_2\cdots i_{k+1}}$ for $k\ge3$ by $$ \mathrm{d}R^\sigma_{i_1i_2\cdots i_k} \equiv R^\sigma_{i_1i_2\cdots i_ki_{k+1}}\,\omega^{i_{k+1}}\ \mod \theta^1,\ldots,\theta^s. $$ Let $\mathcal{R^k}$ for $k\ge 2$ denote the $C^\infty$ ideal spanned by the functions $R^\sigma_{i_1i_2\cdots i_j}$ for $j\le k$. Obviously, this is an increasing sequence of ideals, and any $m$-dimensional integral manifold of $\mathcal{I}$ will lie in the zero locus of $\mathcal{R^k}$ for all $k\ge 2$. If, at any level of $k$, $\mathcal{R}^k$ becomes equal to the ring of all $C^\infty$ functions on the domain, then there is no integral manifold of the Pfaffian system on which $\omega^1\wedge\omega^2\wedge\cdots \wedge\omega^m$ is nonvanishing. If, on the other hand, $\mathcal{R}^{k+1}=\mathcal{R}^k$ for some $k\ge 2$ and the zero locus $Z_k$ of $\mathcal{R}^k$ consists of $\mathcal{R}^k$-ordinary zeroes, then the smooth manifold $Z_k$ is smoothly foliated by $m$-dimensional integral manifolds of $\mathcal{I}$. (Note: Given a set of smooth functions $\mathcal{R}$ on a domain, a zero of $\mathcal{R}$, i.e., a point $p$ at which all of the functions in $\mathcal{R}$ vanish, is said to be an ordinary zero of $\mathcal{R}$ if there exist functions $r_1,\ldots,r_q\in \mathcal{R}$ with independent differentials at $p$ such that the zero locus of $\mathcal{R}$ is defined by $r_1=\cdots=r_q=0$ on some open neighborhood of $p$.) The proof of the above result is simple: It's a local statment, so let $p\in Z_k$ be an ordinary zero of $\mathcal{R}^k$, let $r_1,\ldots,r_q\in \mathcal{R}^k$ be functions with linearly independent differentials at $p$, and let $U$ be an open neighborhood of $p$ on which the $\mathrm{d}r_\ell$ are linearly independent and $Z_k\cap U$ is the locus in $U$ on which all of the $r_\ell$ vanish. Then, by the implicit function theorem, $Z_k\cap U$ is an embedded submanifold of $U$ of codimension $q$ and, by construction, $\mathcal{R}^k(U)= \langle r_1,\ldots,r_q\rangle$. Consequently, since $\mathcal{R}^{k+1}(U)=\mathcal{R}^k(U)$, it follows that $$ \mathrm{d}r_\ell = G_{\ell\sigma}\theta^\sigma + F_{\ell i}\,\omega^i $$ on $U$ where $F_{\ell i}\in \langle r_1,\ldots,r_q\rangle$. Since the $F_{\ell i}$ vanish on $Z_k\cap U$, it follows that the matrix $(G_{\ell\sigma})$ must have rank $q$ on $Z_k\cap U$. Consequently, when these relations are pulled back to $Z_k\cap U$, they become $q$ linearly indepdent relations among the $\theta^\sigma$, i.e., $0 = G_{\ell\sigma}\theta^\sigma$. Meanwhile, since $R^\sigma_{ij}\in \mathcal{R}^k(U)= \langle r_1,\ldots,r_q\rangle$, it follows that the pullback of the Pfaffian system $\mathcal{I}$ to $Z_k\cap U$ (which has rank $s-q$) satisfies the Frobenius integrability condition. Now apply the Frobenius Theorem to this system on $Z_k\cap U$. References: I don't know exactly where this is written in the 'modern' literature. I think we didn't include it in our book Exterior Differential Systems (B—, Chern, Gardner, Goldschmidt, and Griffiths). I know that I learned it from a paper of Élie Cartan, Les problèmes d'équivalence (Séminaire de Math., exposé D, 11 janvier 1937) [Republished in Partie II of Cartan's Œeuvres Complètes]. [Of course, it's written in Cartan's style, so you won't see anything mention of an open set $U$, ideals of smooth functions, or a formal definition of 'ordinary zero', but the concept is implicit in his argument. I just translated his proof into our 'modern' idiom.] Now, the above sufficient condition is by no means necessary. Here is an example to show why: On $xyz$-space consider the $1$-form $$ \theta = \mathrm{d}z + z^m x\,\mathrm{d}y, $$ for some integer $m>1$. Then $\mathrm{d}\theta \equiv z^m\,\mathrm{d}x\wedge\mathrm{d}y\mod\theta$ and $\mathrm{d}(z^m)\equiv - mz^{2m-1} x\,\mathrm{d}y\mod\theta$, so $\mathcal{R}^k = \langle z^m\rangle$ for all $k\ge 2$, so the locus $z=0$ (which is the unique $2$-dimensional integral manifold of $\theta$) does not consist of ordinary zeros of $\mathcal{R}^k$ for any $k\ge 2$. One could, instead, replace $\mathcal{R}^2$ by $\widehat{\mathcal{R}^2}$, the set of all $C^\infty$ functions on the domain that vanish on the zeroes of $\mathcal{R}^2$, and similarly, at each stage, replace $\mathcal{R}^k$ by $\widehat{\mathcal{R}^k}$, the `$C^\infty$ radical' of $\mathcal{R}^k$. There would be a similar theorem to the above for the ordinary zeroes of the resulting limit ideal. However, in most cases, such a computationally intensive procedure is not needed. The upshot is that one doesn't need to assume anlyticity, per se, but, since arbitrary $C^\infty$ ideals can have almost arbitrarily bad zero loci, and since one can construct $C^\infty$ Pfaffian ideals whose $\mathcal{R}^k$ can be just about any $C^\infty$ ideal, unless one makes some kind of constant rank assumptions, almost anything can happen. What is true is that there are enough constant rank assumptions to cover most cases of interest in the smooth category.<|endoftext|> TITLE: Why does elliptic cohomology fail to be unique up to contractible choice? QUESTION [22 upvotes]: It is often stated that the derived moduli stack of oriented elliptic curves $\mathsf{M}^\mathrm{or}_\mathrm{ell}$ is the unique lift of the classical moduli stack of elliptic curves satisfying some conditions, meaning the moduli space $Z$ of all such lifts is connected. This is mentioned in Theorem 1.1 of Lurie's "A Survey of Elliptic Cohomology" [Surv], for example. In Remark 7.0.2 of Lurie's "Elliptic Cohomology II: Orientations" ([ECII]), Lurie says "...beware, however, that $Z$ is not contractible". In other words, $\mathsf{M}^\mathrm{or}_\mathrm{ell}$ is not the unique lift up to contractible choice (the gold standard of uniqueness in homotopy theory). (Side note: in [ECII] and [Surv], Lurie is talking about the moduli stack of smooth elliptic curves, but the uniqueness up to homotopy of a derived stack $\overline{\mathsf{M}}_\mathrm{ell}^\mathrm{or}$ lifting the compactification of the moduli stack of smooth elliptic curves is also stated in the literature; for example, in Theorem 1.2 of Goerss' "Topological Modular Forms [after Hopkins, Miller, and Lurie]". I am interested in the compactified situation mostly, but both are related.) Although I do not hope that $\mathsf{M}^\mathrm{or}_\mathrm{ell}$ does possess this much stronger form of uniqueness, I would like to understand the reason for this failure: Why is the moduli space $Z$ not contractible? and Does a similar statement apply in the compactified case? To be a little more precise, let $\mathcal{O}^\mathrm{top}$ be the Goerss--Hopkins--Miller--Lurie sheaf of $\mathbf{E}_\infty$-rings on the small affine site of the moduli stack of elliptic curves $\mathsf{M}_\mathrm{ell}$. Denote this site by $\mathcal{U}$. The moduli space $Z$ can then be defined as the (homotopy) fibre product $$Z=\mathrm{Fun}(\mathcal{U}^{op}, \mathrm{CAlg})\underset{\mathrm{Fun}(\mathcal{U}^{op}, \mathrm{CAlg}(\mathrm{hSp}))}{\times}\{\mathrm{h}\mathcal{O}^\mathrm{top}\},$$ where $\mathrm{CAlg}$ is the $\infty$-category of $\mathbf{E}_\infty$-rings, and $\mathrm{CAlg(hSp)}$ is the 1-category of commutative monoid objects in the stable homotopy category. The presheaf $\mathrm{h}\mathcal{O}^\mathrm{top}$ can be defined using the Landweber exact functor theorem (at least on elliptic curves whose formal group admits a coordinate), and hence $Z$ can be seen as the moduli space of presheaves of $\mathbf{E}_\infty$-rings recognising the classical Landweber exact elliptic cohomology theories. To prove uniqueness up to homotopy, I am aware one should use some arithmetic and chromatic fracture squares to break down the problem into rational, $p$-complete, $K(1)$-local, and $K(2)$-local parts. The $K(2)$-local part of $\mathcal{O}^\mathrm{top}$ is unique up to contractible choice by the Goerss--Hopkins--Miller theorems surrounding Lubin--Tate spectra (see Chapter 5 of [ECII] for a reference which you might already have open). The $K(1)$-local part also seems to be unique up to contratible choice, as all of the groups occuring in the Goerss--Hopkins obstruction theory vanish (this is discussed at length in Behrens' "The construction of $tmf$" chapter in the "TMF book" by Douglas et al). Similarly, the rational case also has vanishing obstruction groups; see ibid. Edit: As pointed out by Tyler below, these claims about the function of Goerss--Hopkins obstruction theory above are wrong! I'm then lead to believe that is something interesting (being a pseudonym for "I don't know what's") going on in the chromatic/arithmetic fracture squares gluing all this stuff together. Are their calculable obstructions/invariants to see this? Or otherwise known examples that contradict the contractibility of $Z$? Any thoughts or suggestions are appreciated! REPLY [18 votes]: So the issue is with this: all of the groups occuring in the Goerss--Hopkins obstruction theory vanish In "generic terms", for the obstruction theory that you're running in either the $K(1)$-local case or the rational case you care about some bigraded obstruction groups $$ \mathfrak{E}xt^{s,t}(R;S) $$ where these are some nonabelian Ext-groups occuring in some algebraic category ($K(1)$-locally it is Andre-Quillen cohomology for derived $p$-complete $\psi$-$\theta$-algebras, and rationally it is Andre-Quillen cohomology for graded-commutative algebras over $\Bbb Q$, maybe relative to some fixed base algebra, maybe over some other algebra, maybe on a category of diagrams with some sheaf condition, etc etc etc). The obstructions to existence of some object typically occur when $t-s = -1$, and to uniqueness (up to homotopy equivalence) occur when $t-s = 0$. These are the groups that vanish when constructing $Tmf$ by obstruction theory. Those two groups, however, tell you only about path components. In general, there is some obstruction-theoretic yoga that tells you $\mathfrak{E}xt^{s,t}(R;S)$ is part of some kind of fringed spectral sequence, computing $\pi_{t-s} \mathcal{M}$ of some moduli space of realizations. As a result, for contractibility of this moduli space we would need to know about what happens to $\mathfrak{E}xt^{s,t}$ for $t-s > 0$, and those groups don't vanish. (Aside, the Goerss--Hopkins--Miller theorem about Lubin-Tate theories is so strong precisely because, in that situation, all of these higher groups vanish, and so these moduli spaces are homotopy discrete.) The rational case makes it easiest to see this. For the rational case, roughly (ignoring the sheaf stuff), we have constructed $p$-complete $Tmf_p$ and we need to construct a map $H\Bbb Q[c_4, c_6] \to (Tmf_p)_{\Bbb Q}$ to the rationalization that has a particular effect on $\pi_*$. That map is unique up to homotopy. However, the space of maps $H\Bbb Q[c_4, c_6] \to (Tmf_p)_{\Bbb Q}$ does not have contractible components: $H\Bbb Q[c_4, c_6]$ is a free rational commutative algebra on generators in degrees $8$ and $12$, and so the space of maps to $(Tmf_p)_{\Bbb Q}$ is homotopy equivalent to $$ Map(S^8, (Tmf_p)_{\Bbb Q}) \times Map(S^{12},(Tmf_p)_{\Bbb Q}). $$ Its higher homotopy give many contributions from the rational higher homotopy of $Tmf_p$. In this case, freeness actually gives us vanishing of Ext groups when $s > 0$; it's $t$ that gives us trouble.<|endoftext|> TITLE: What is the Gelfand dual of an open surjection? QUESTION [6 upvotes]: Does anyone have a handy characterisation of open continuous surjections $X \to Y$ in terms of the corresponding injective $*$-homomorphism $C(Y) \to C(X)$? (I'm only interested in the case where $X$ and $Y$ are compact Hausdorff spaces, but the question could be suitably modified for locally compact Hausdorff spaces.) Cf. Reference request for translating from Top to C*-alg REPLY [6 votes]: After more thought, I think the correct statement is the following: Theorem: Let $\pi : X \to Y$ be a continuous map between compact Hausdorff spaces. Then the following condition are equivalents: (a) $\pi$ is an open map. (b) The map $\pi^* : C(Y)^+ \to C(X)^+$ has a left adjoint which is $C(Y)^+$-linear, i.e. there is a (automatically unique) map $\pi_! : C(X)^+ \to C(Y)^+$ such that for all $f \in C(X)^+$ and $h \in C(Y)^+$ one has $\pi_!(f) \leqslant h \Leftrightarrow f \leqslant \pi^*(h)$ and $\pi_!(f \pi^*(h)) = \pi_!(f) h$. Note: By $C(X)^+$ I mean the subset of selfadjoint positive elements, i.e. continuous function with values in non-negative real numbers. Before proving the theorem, let me a recall: Lemma : Given a continuous map $\pi:X \to Y$ between comapct Hausdorff space, if $y \in Y$ and $U$ is an open containing the fiber $\pi^{-1}(y)$, then there exists a neighborhood $V$ of $y$ such that $\pi^{-1}(V) \subset U$. Proof: It feels like it is a form of openness, but it is actually related to properness. Indeed, let $K$ be the closed complement of $U$, $K$ is compact, so $f(K)$ is a compact of $Y$ not containing $y$, hence the complement $V$ of $f(K)$ is an open neighborhood of $y$, and by construction, its preimage is included in $U$. Proof: (a) $\Rightarrow$ (b) For $f$ a positive functions on $X$, We define $\pi_!(f)$ as a function $Y \to \mathbb{R}$ by: $$ \pi_!(f)(y) = \sup_{x \in \pi^{-1}(y)} f(x) $$ The sup is taken in the poset of positive real numbers, so it is $0$ if the fiber is empty, it is always finite by compactness of the fiber, so that $\pi_!(f)(y)$ is indeed a positive real number for all $y$. It is immediate that $\pi_!(f \pi^*(h))= \pi_!(f) h$ and $\pi_!(f) \leqslant h$ if and only if $f \leqslant \pi^*(h)$, what is not so clear is whether $\pi_!(f)$ is indeed an element of $C(X)^+$, i.e. is continuous. The map $\pi_!(f)$ is always an upper semi-continuous functions. It only rely on the lemma above: If $\pi_!(f)(y)q$, it means there is an element $x_0 \in \pi^{-1}(y)$ such that $f(x_0)>q$, hence there is a neighborhood of $x_0$ such that $f(x)>q$, the image of this neighborhood is a neighborhood of $y$ on which $\pi_!(f)>q$. Together they implies that $\pi_!(f)$ is continuous which concludes the proof. (b) $\Rightarrow$ (a) Assume we have such a map $\pi_!$. For $f$ a continuous positive function, I write $\{f>0\}$ for the open subset $\{ x | f(x) >0 \}$, these forms a basis of the topology, so it is enough to check that the direct image of these by $\pi$ are open subsets. We will show that $\pi_!$ has to be of the form defined in the previous part of the proof. It then imediately follows that $\pi\{f>0\} = \{ \pi_! f >0 \}$ and this concludes the proof. First one show that $\pi_!(f)(y) \geqslant \sup_{x \in \pi^{-1} y} f(x)$. Indeed, assume that $\pi_!(f)(y) \leqslant a$. Then $\forall \epsilon>0$ there exists a positive function $\chi$ such that $\chi(y)=1$ and $\chi \pi_!(f) \leqslant (a+\epsilon)\chi$ (take $\chi$ to be $1$ in a small neighborhood of $x$, and $0$ away from $x$). The adjunction formula give you that $\pi^*(\chi) f \leqslant (a+ \epsilon)\pi^*(\chi)$. Now for any $x \in \pi^{-1}(y)$, evaluating the previous inequality at $x$ gives $f(x) \leqslant a+ \epsilon$, hence $\sup_{x \in \pi^{-1}(y)} f(x) \leqslant a$. Conversely, assume that $\sup_{x \in \pi^{-1} y} f(x) \leqslant a$, i.e. $f(x) \leqslant a$ everywhere on the fiber of $a$, for all $\epsilon>0$ there is an open neighborhood $U$ of $\pi^{-1}(y)$ on which $f(x) < a + \epsilon$. Applying the lemma again, we get an open neighborhood $V$ of $y$ such that $\pi^{-1}(V)$ is included in $U$. Take $\chi$ such that $\chi(y)=1$ and $\chi$ is $0$ outside of $V$ and run the same argument as above, you have that $ \pi^*(\chi) f \leqslant a+\epsilon$, hence $\pi^*(\chi) f \leqslant \pi^*(a+ \epsilon)$, hence $\pi_!(\pi^*(\chi) f) \leqslant a+ \epsilon$, and finally $\chi \pi_!(f) \leqslant a + \epsilon$ evaluating at $y$ gives $\pi_!(f)(y) \leqslant a+ \epsilon$, which conclude the proof.<|endoftext|> TITLE: Subvarieties of Lagrangian Grassmannians QUESTION [5 upvotes]: Let $LG(n,2n)$ be the Lagrangian Grassmannian parametrizing Lagrangian subspaces (so of dimension $n$) of $\mathbb{C}^{2n}$. Then $LG(n,2n)\subset G(n,2n)$, where $G(n,2n)$ is the Grassmannian of subspaces of dimension $n$ of $\mathbb{C}^{2n}$. Fix a subspace $H\subset \mathbb{C}^{2n}$ of dimension $n+2$, and denote by $G(n,H)\subset G(n,2n)$ the Grassmannian of subspaces of dimension $n$ that are contained in $H$. The intersection $X_n := LG(n,2n)\cap G(n,H)\subset G(n,2n)$ parametrizes Lagrangian subspaces of $\mathbb{C}^{2n}$ that are contained in $H$. For instance, for $n = 2$ we have $G(2,H) = G(2,4)$ and hence $X_2 = LG(2,4)$. In general, is the variety $X_n$ smooth and irreducible? By any chance is $X_n$ a well-known variety appearing under some name in the literature? Addition: Is the subvariety $Y_n\subset G(n+2,2n)$, parametrizing $(n+2)$-dimensional subspaces $H\subset \mathbb{C}^{2n}$ that are co-isotropic, homogeneous as well? Is there a formula for the dimension of $Y_n$? Thank you very much. REPLY [2 votes]: Co-isotropic subspaces of dimension $n+k$ are in bijection with isotropic subspaces of dimension $n-k$. The variety of isotropic subspaces of dimension $n-k$ of a symplectic vector space of dimension $2n$ is the symplectic Grassmannian $SG(n-k,2n)$. Its dimension is given by $$\dim(SG(n-k,2n)) = 2n(n-k)-\frac{3(n-k)^2-(n-k)}{2}.$$ In particular, when $k = 0$ you get the Lagrangian Grassmannian $LG(n,2n)$ parameterizing maximal isotropic subspaces, and its dimension is given by $$\dim(LG(n,2n)) = \frac{n(n+1)}{2}.$$<|endoftext|> TITLE: Is there Z_n graded supersymmetry? QUESTION [6 upvotes]: I have tried searching for something similar to what is described below, but to no avail. It would be great if somebody could show some right references, where this has been done, or explain why such approach is bound to fail. Motivation Let's consider algebra $gl_n$ with ''bosonic'' generators $B_{ij}$: $$[B_{ij},B_{kl}]=\delta_{jk}B_{il}-\delta_{li}B_{jk}$$ They can be realized as bosonic differential operators $B_{ij}=x_i\partial_j$ in commuting variables $x_i x_j = x_j x_i$ There is a known way to extend this construction to the so-called supersymmetric ($\mathbb{Z}_2$) case by adding ''fermionic'' generators $F_{rs}$ with grading $\exp(\frac{2\pi i}{2})$, which can be realized with the help of Grassmanian variables $\psi_i$, $\psi_i \psi_j = - \psi_j \psi_i$. These generators satisfy similar relations, but commutation law $[,]$ is turned into anticommutation law $\{,\}$. The pattern is $[B,B]\sim B$, $\{F,F\}\sim B$, $[B,F]\sim F$. Now I want to generalize it to $\mathbb{Z}_3$ case. My approach consists in introducing another set of generators $C$ and reassigning the gradings: $|B| = 1$, $|F| = \exp(\frac{2\pi i}{3})$, $|C| = \exp(\frac{4\pi i}{3})$ so that the following pattern will take place: $[B,B]\sim B$, $[B,F]\sim F$, $[B,C]\sim C$, $[F,F]\sim C$, $[F,C]\sim B$, $[C,C]\sim F$. Here instead of usual commutators I mean suitably redefined $\mathbb{Z}_3$-graded commutators. The problem is that it doesn't work. Indeed, in order to define commutators in this way I need for elements to have commutation relations with cubic roots of $1$. But it isn't possible because if all my variables are similar to each other with the only exception of grading, then, for example, $\psi c = f(\psi,c) c \psi = f(\psi,c) f(c,\psi) \psi c$. Hence, since $f(\psi,c) = f(c,\psi)$ this factor must be a quadratic root of $1$. The only way out that I could think of was introducing more (I started with two) sets variables which kind of live on a quantum plane: $x_i^{1} x_j^{2} = q x_j^{2} x_i^{1}$. I could write some algebra, but it looked very messy and is probably wrong. The question is whether similar constructions of $\mathbb{Z}_n$ supersymmetry exist in literature or are easy to exclude on some general grounds. References where people do something else There's something called "color Lie algebras", but I haven't been able to fully digest their ideas. Richard Kerner et. al. have considered a similar problem and solved it by embracing the realm of ternary structures. Is there any hope to stay binary in this brave new world? Edit concerning explicit expressions for commutators I was trying to make use of the following realization of $q(2)$-algebra (belonging to the strange series q(n)) consisting of bosonic $B_{ij}=x_i \partial_{x_j}+\psi_i \partial_{\psi_j}$ and fermionic $F_{ij} = x_i \partial_{\psi_j}+\psi_i \partial_{x_j}$, $i,j=1,2$. One can check that these generators indeed form q(2) with signs in commutators naturally appearing according to the degree of variables. Now, for $\mathbb{Z}_n$ case I'm going to change notation to make the motivation more visible and set $G_{ij}^k \equiv x_i^l \partial_{x_j^{l-k}}$ be generators of degree $k$ in $\mathbb{Z}_n$ commuting variables $x_i^l$ of degree $l$. I use summation over repeating indices so that, for example, $\mathbb{Z}_3$ case looks like $B_{ij} = x_i \partial_{x_j}+\psi_i \partial_{\psi_j}+c_i \partial_{c_j}$, $F_{ij} = \psi_i \partial_{x_j}+c_i \partial_{\psi_j}+x_i \partial_{c_j}$, $C_{ij} = c_i \partial_{x_j}+x_i \partial_{\psi_j}+\psi_i \partial_{c_j}$. Setting the commutation rules between variables $x_i^l x_j^k = \tilde{g}(i,l;j,k) x_j^k x_i^l$ I immediately descend to the "non-quantum case" (in the sense described above) and set $\tilde{g}(i,l;j,k)\equiv g(l,k)$. Then I use the following commutation relations $$ [G_{ij}^r, G_{kl}^p] \equiv G_{ij}^r G_{kl}^p - \alpha(r,p) G_{kl}^p G_{ij}^r$$ with the goal of determining $\alpha(r,s)$ from the condition of absence of second derivatives: $$ [G_{ij}^r, G_{kl}^p] = x_i^s \partial_{x_j^{s-r}} x_k^q \partial_{x_l^{q-p}} - \alpha(r,p) x_k^q \partial_{x_l^{q-p}} x_i^s \partial_{x_j^{s-r}} = g(r-s,q) g(s,q) g(r-s, p-q) x_k^q x_i^s \partial_{x_l^{q-p}}\partial_{x_j^{s-r}} - \alpha(r,p) g(p-q,s)x_k^q x_i^s \partial_{x_l^{q-p}}\partial_{x_j^{s-r}}$$ so that $\alpha(r,p) = \frac{g(r-s,q) g(s,q) g(r-s, p-q)}{g(p-q,s)}$ for every $s$ and $q$ (which is itself a condition on $g(s,q)$). However the solution I had in mind while writing this, namely, $g(r,s) = \exp(\frac{2\pi i rs}{n})$ doesn't work here, since $\alpha(r,p) = \exp(\frac{2\pi i}{n}(rp+sq-sp-ps-qs))$ depends on $s$ and $q$. This problem originates in the fact that $x_i^r x_j^s =g(r,s) x_j^s x_i^r$ with $g(r,s)$ taking values in the $n$-th root of $1$ is incompatible with abelian $g(r,s)$, since $g(r,s)g(s,r)=1$, as follows from $x_i^r x_j^s =g(r,s) x_j^s x_i^r =g(r,s) g(s,r) x_i^r x_j^s$. It's probably not very comprehensible right now, but I'm not sure how to say it the right way. REPLY [2 votes]: The realizations (of an algebra through another algebra) you are speaking about are actually homomorphisms. And as such they should map between algebraic structures of the same kind: that is from algebras to algebras, from Lie alg to Lie alg, from graded algebras to graded algebras (graded by the same group), etc Since you are considering Lie superalgebras and realize them through boson-fermion operators, what you are actually doing is to consider the algebra mixing the bosonic/fermionic degrees of freedom as a superlagebra: this means that you consider it eqquiped with a $\mathbb{Z}_2$-grading and you pick the unique color function available (unique because there is ony a single bicharacter of the $\mathbb{Z}_2$ group). This is the function: $\theta:\mathbb{Z}_2\times\mathbb{Z}_2\rightarrow\mathbb{C}^*$ explicitly given by $\theta(a,b)=(-1)^{\deg a\deg b}$, where $a,b\in\mathbb{Z}_2$ and $\deg=0,1$ depending on whether the corresponding element is even or odd. This bicharacter actually determines the exact form of the brackets in the LS; that is whether they are commutators ($\theta(a,b)=1$) or anticommutators ($\theta(a,b)=-1$). If you go to more general gradings (that is $\mathbb{Z}_3$ which interests you -but it could even be an arbitrary finite, abelian group $\mathbb{G}$) then the "bracket" will generally be defined by $[a,b]=ab-\theta(a,b)ba$, where $\theta:\mathbb{G}\times\mathbb{G}\rightarrow\mathbb{C}^*$ so essentially the color function (take it as a synonym for a bicharacter in this setting) is what determines the form of the bracket (in higher grading groups with more bicharacters available, the bracket need not be an (anti)commutator, it can have a more general form). $\theta$ also determines the behaviour of the multiplication in the tensor product algebras, so we speak about braided, graded tensor products but this is possibly another story. If you are interested in these points of view, you coud take a look (excuse me in advance for the self-citation but i think it is relevant here) at: Super-Hopf realizations of Lie Superalgebras: Braided Paraparticle extensions of the Jordan-Schwinger map Gradings, Braidings, Representations, Paraparticles: some open problems (see expecially the discussion of p. 80-81 and section 5.1) The first paper mostly reviews the relevant notions while the second one utilizes realizations in a way which -i think- is close to what you are doing. The difference is that i am using (in this references) "bigger" algebras than the supersymmetric mixture of bosons and fermions you are considering. I am utilizing algebras which mix parabosonic and/or parafermionic generators (your algebras can be recovered as quotients of the later). This gives the opportunity to consider other grading groups (i am mainly focusing to the case of a $\mathbb{Z}_2\times\mathbb{Z}_2$ grading group and its colors or bicharacters if you prefer). It seems that such a line of investigation presents interest from both the mathematics and the physics point of view. In: $\mathbb{Z}_2\times\mathbb{Z}_2$-graded parastatistics in multiparticle quantum Hamiltonians, F. Topan the author seems to pursue further similar ideas and methods (focusing more on the physics point of view) while i (together with a couple of colleagues) have some older work on further applications of the $\mathbb{Z}_2\times\mathbb{Z}_2$-graded, color Lie algebras (focusing more on the representation-theoretic point of view: that is on the construction of representations of Lie superalgebras through paraparticle realisations). I can send you the relevant references if you are interested in these. I am not sure if the above shed some light in your questions and your research but i hope you'll find something of interest in these. Edit: Some more references which revolve around $\mathbb{Z}_n$-graded symmetries (generalizing thus the supersymmetric case mentioned in the OP) and focusing on the physics side, are: $\mathbb{Z}_n$-Graded Topological Generalizations of Supersymmetry and Orthofermion Algebra, A Mostafazadeh On a $\mathbb{Z}_2^n$-graded version of supesymmetry, A J Bruce REPLY [2 votes]: Another possibility is to consider group elements rather than commutators. If you take the matrices $A=\mathrm{diag}(1,\xi,\ldots,\xi^{n-1})$ and $B=\begin{pmatrix}0&1&0&\cdots&0&0\\ 0&0&1&\cdots&0&0\\ 0&0&0&1&\cdots&0\\ \vdots&\vdots&\ddots&\ddots&\ddots&\vdots\\ 0&0&0&\cdots&0&1\\ 1&0&0&\cdots&0&0 \end{pmatrix}$ in $gl_n$ (here $\xi=\exp(2\pi i/n)$, then $A^n=B^n=I_n$, $BA=\xi AB$, and the elements $A^iB^j$ form a basis of $gl_n$ which has a nice multiplication table. Moreover, this way the algebra of matrices is $\mathbb{Z}_n\times\mathbb{Z}_n$-graded (put $\deg(A^iB^j)=(i,j)$, it works well). Considering various homomorphisms $\mathbb{Z}_n\times\mathbb{Z}_n\to\mathbb{Z}_n$, you will get all sorts of $\mathbb{Z}_n$-gradings.<|endoftext|> TITLE: Can we construct a dessin of any genus with a cyclic automorphism group of any order? QUESTION [6 upvotes]: We consider a dessin d'enfant $D$ as a bipartite graph $D$ on a complex oriented surface $S$, such that the complement $S \backslash D$ is homotopic to a collection of disks. Definition: Let an automorphism of $D$ be an orientation preserving automorphism of $S$ taking $D$ to itself. Here is my general question: Given an arbitrary genus surface $S$, and a positive integer $k$, can we construct a a dessin $D$ on that surface whose automorphisms are a cyclic group, i.e., $\text{Aut}(D) \simeq C_{k}$? Remark: We do not require that the dessin be regular, in fact, we don't expect it to be. For example, the genus 1 example below is not a regular dessin (i.e. isomorphic to a complete bipartite graph). [See page 6 of this paper of Jones which addresses genuses possible in the regular case.] Motivating Examples: Let's stick to the examples of cyclic group $C_3$ and the dihedral group $D_3$ for simplicity. In the genus 0 and 1 cases, the $C_3$ and $D_3$ cases immediately generalize to all $C_k$ and $D_k$. In the genus 0 case, they look like this, note that the $C_2$ action in the dihedral group comes from switching the inner and outer faces (indicated by the red arrow). In the genus 1 case, they look like the following picture. [Remark: the $C_2$ automorphism in $D_3$ comes by moving the purple disk out and down and then in and up to where it was, and rotating the pink loops inwards back to where it was, this swaps the faces. (The teal lines prevent this map, thus giving us $C_3$.)] In the genus 2 case, I can only manage to create with automorphism group $C_2$ and $C_2 \times C_2$ respectively. They look like the following. No matter how many loops I add, I haven't managed to get an automorphism group larger than this on a dessin on a surface of genus 2. Similarly, I don't see render a dessin with automorphism group $C_{k+i}$ on a surface of genus $k$ in general. Hopefully this is due to my own incompetance because it would be really cool to see what it would look like! I am quite stuck. Any and all help is deeply appreciated. REPLY [2 votes]: This is possible for any $g$ and $k$. The idea is as follows: (1) Start with a dessin embedded in the sphere $S^2$ with symmetry group $C_k$ which has a lot of faces and a lot of triple points. (2) Make a surgery on the surface at a collection of triple points that are far away. When you do (2) there faces adjacent to one vertex will become one face, so the Euler characteristic of your surface will drop by $2$ and genus will increase by $1$. It a bit hard to describe this surgery in words, but the idea is the following: to each dessin embedded in a surface you can associate its thinking, that is called a ribbon graph, or a fat graph. And a ribbon graph defines the surface. Now, in order to associate the structure of a ribbon graph to a abstract graph one has to choose a cyclic order on the edges at each vertex. And the surgery in (2) amounts just to changing the cyclic order at exactly one vertex. I attach a picture which describes the above procedure. There is a honeycomb graph there with one surgery at one point. Here is one more picture, with a slightly different construction, if one glues two "disk-like" surfaces in the picture one gets a surface of genus $2$ with a dessin that has a $C_3$ symmetry (i.e. the example you mentioned). Here you start with a dessin of $S^2$ that has a $C_3$. This symmetry and then flip two vertex. The sphere is glued from two disks and you flip on vertex in each disk (it is red on the picture). Flipping one point increases genus by $2$. It changes the structure of ribbon graph, but doesn't change the dessin. This picture generalises to any $k, g$. .<|endoftext|> TITLE: On almost-(multiplicative)-morphisms between infinite groups and the complex numbers QUESTION [10 upvotes]: [cross-posting] https://math.stackexchange.com/q/4153692/522463 Let $G$ be an infinite group and $\varphi\colon G\longrightarrow\mathbb{C}$ such that \begin{eqnarray*} \exists\,\delta>0, \forall\,a,b\in G,\, |\varphi(ab)-\varphi(a)\varphi(b)|\leq\delta\quad(\mathcal{P}) \end{eqnarray*} What can be said about $\varphi$? Precisely, is it true that $(1)\quad\varphi$ is bounded, or $(2)\quad\varphi(ab)=\varphi(a)\varphi(b)$ for all $a,b\in G$? Let $e$ be the neutral element of $G$. Clearly, if $\varphi(e)\neq 1$, then $|\varphi(a)|\leq\delta|1-\varphi(e)|^{-1}$ for all $a\in G$; hence $\varphi$ is bounded. However, if $\varphi(e)=1$, it’s not clear to me if $\varphi$ remains bounded. I also think that when it’s not bounded, $\varphi$ may end up satisfying $\varphi(ab)=\varphi(a)\varphi(b)$ for all $a,b\in G$ (my attempts to build counterexamples failed, which makes me feel that the claim might be true) Note: $\mathcal{P}$ does not define a quasimorphism (the landing set is the complex numbers and the property is multiplicative instead of the usual additive one for quasimorphisms) REPLY [3 votes]: If $\varphi$ is not bounded, there is a sequence $(x_n)_{n\in\mathbb{N}}\in G^{\mathbb{N}}$ such that $|\varphi(x_n)|>2^n$ for all $n\in\mathbb{N}$. Let $x,y\in G$ and $n\in\mathbb{N}$. \begin{eqnarray*} |\varphi(x)\varphi(y)-\varphi(xy)||\varphi(x_n)|&=&|\varphi(x)\varphi(y)\varphi(x_n)-\varphi(xy)\varphi(x_n)|\\ &\leq&|\varphi(x)\varphi(y)\varphi(x_n)-\varphi(x)\varphi(yx_n)|+\\ &&|\varphi(x)\varphi(yx_n)-\varphi(xyx_n)|+\\ &&|\varphi(xyx_n)-\varphi(xy)\varphi(x_n)|\\ &\leq&\delta(|\varphi(x)|+2) \end{eqnarray*} Hence, $|\varphi(x)\varphi(y)-\varphi(xy)|\leq\delta(|\varphi(x)|+2)2^{-n}$. Take the limit as $n\longrightarrow\infty$, and you're done.<|endoftext|> TITLE: Planes in Lagrangian Grassmannians QUESTION [12 upvotes]: Let $LG(h,2h)$ be the Lagrangian Grassmannian of subspaces of dimension $h$ of a complex vector space of dimension $2h$. For instace, $LG(1,2)=\mathbb{P}^1$, and $LG(2,4)\subset\mathbb{P}^4$ is a smooth quadratic $3$-fold. So $LG(2,4)$ contains no plane (linearly embbeded $\mathbb{P}^2$). Furthermore, by Lemma 2.5.1 here https://arxiv.org/pdf/math/0209169.pdf $LG(3,6)$ containes no plane as well. Is is true that $LG(h,2h)$ containes no plane for all $h\geq 1$. REPLY [13 votes]: This is, indeed, true. To prove this, assume we have an embedding $\mathbb{P}^2 \to \operatorname{LGr}(V)$ (where $V$ is a symplectic vector space). Let $U \subset V \otimes \mathcal{O}$ be the tautological subbundle on $\operatorname{LGr}(V)$. Note that it extends to an exact sequence $$ 0 \to U \to V \otimes \mathcal{O} \to U^\vee \to 0 $$ on $\operatorname{LGr}(V)$ and then, via pullback, also on $\mathbb{P}^2$. So, we consider this sequence on $\mathbb{P}^2$. The condition that the plane is embedded linearly means that $$ c_1(U) = -h, $$ where $h$ is the line class on $\mathbb{P}^2$. Therfore, the Chern class of $U$ takes the form $$ c(U) = 1 - h + xh^2 $$ for some $x \in \mathbb{Z}$. By duality, $$ c(U^\vee) = 1 + h + xh^2. $$ Now by Whitney formula we obtain $$ 1 = c(V \otimes \mathcal{O}) = c(U)\cdot c(U^\vee) = (1 - h + xh^2)(1 + h + xh^2) = 1 + (2x - 1)h^2 $$ which implies $2x = 1$. But since $x$ is integer, this is impossible. This contradiction shows that $\operatorname{LGr}(V)$ has no planes.<|endoftext|> TITLE: Volumes of $\mathrm{SL}_n(K_\mathbb{R})/\mathrm{SL}_n(\mathcal{O}_K)$ QUESTION [9 upvotes]: $\DeclareMathOperator\SL{SL}$The volume of $\SL_n(\mathbb{R})/\SL_n(\mathbb{Z})$ can be computed under the natural measure that it inherits from $GL_n(\mathbb{R})$. Two formulae seem to be known. $$\operatorname{vol}(\SL_n(\mathbb{R})/\SL_n(\mathbb{Z}))= \zeta(2)\zeta(3) \dotsb \zeta(n).$$ This is available in notes by Paul Garrett [2]. In [1], the volume is given as $$\operatorname{vol}(\SL_n(\mathbb{R})/\SL_n(\mathbb{Z}))= \frac{\sqrt{2}}{\prod_{i=1}^{n-1}2^{i-1}i!}\zeta(2)\zeta(3) \dotsb \zeta(n).$$ Main question. 1.) Are there any such formulæ known for the quotient of $\SL_n(K\otimes_{\mathbb{Q}} \mathbb{R})/\SL_n(\mathcal{O}_K)$? I expect that it would be an expression involving Dedekind zeta functions. Here $K$ is a number field and $\mathcal{O}_K$ is the ring of integers. Another question. 2.) Why are the two expressions different? Is one of them wrong or are they under different scaling of the Haar measures? Which of them is the volume with respect to the natural measure on $\SL_n(\mathbb{R})$? [1] Paula - Comparison of volumes of Siegel sets and fundamental domains of $\operatorname{SL}_n(\mathbb Z)$ [2] Garrett - Volume of $\operatorname{SL}_n(\mathbb Z)\backslash\operatorname{SL}_n(\mathbb R)$ and $\operatorname{Sp}_n(\mathbb Z)\backslash\operatorname{Sp}_n(\mathbb R)$ REPLY [17 votes]: $\DeclareMathOperator\SL{SL}$The same argument (due to Siegel, in a classical form, of course), adelized, gives the analogous computation for any number field, and, yes, the corresponding Dedekind zeta appears. (To know that the adelic analogue computes the same thing, rather than the adelic quotient for $\SL_n$ being larger than the classical one, probably we invoke Strong Approximation, just to be safe.) Certainly there is the issue of describing a reasonable normalization of the Haar measures. Of course, in a silly way, we could normalize them to get any result we wanted. For that matter, from a contemporary viewpoint, it is more stylish to define "Tamagawa measures" on semisimple groups $G$, designed so that the measure of $G_k\backslash G_{\mathbb A}$ is (often) $1$, or some other simple integer. With hindsight, Siegel's computation gives hints about how to do this. (See Weil's book "Adeles and algebraic groups".) Operationally, from a neo-classical viewpoint, the important thing is to make natural choices of measures on $P_v$ and $K_v$, where $P_v$ is the $v$-adic minimal parabolic and $K_v$ is the $v$-adic maximal compact, which assemble to give a measure on $G_v$. It is natural to give $K_v$ measure $1$ at non-archimedean places. The measure on $P_v$ is made from a local multiplicative measure on $k_v^\times$, which has the natural normalization that $\mathfrak{o}_v^\times$ has measure $1$. At places $v$ not ramified over $\mathbb Q$, the additive measure naturally gives $\mathfrak{o}_v$ measure $1$. At absolutely ramified places, there is also a natural normalization as in Iwasawa–Tate theory, and this introduces a power of the discriminant, unsurprisingly. At archimedean places $v$, thinking in terms of Iwasawa–Tate theory gives a natural measure on $P_v$. The chief ambiguity here is about $K_v$. One choice is to relate the measure to the natural measure of spheres. Another is to give $K_v$ measure $1$. And other choices that adapt to induction over the size $n$. These lead to different constants, attributible to these choices at archimedean places. EDIT: in response to a question about why computing "the" volume of $\SL_n(\mathbb{Q})\backslash\SL_n({\mathbb A})$ should be essentially the same thing as that of $\SL_n(\mathbb Z)\backslash\SL_n(\mathbb R)$, and so on… The most typical (?) thing that is said is that (by Strong Approximation) $\SL_n(\mathbb A)=\SL_n(\mathbb Q)\SL_n(\mathbb{R})\prod_{p<\infty}\SL_n(\mathbb Z_p)$, from which \begin{gather*} \SL_n(\mathbb Q)\backslash\SL_n(\mathbb A)/\prod_{p<\infty}\SL_n(\mathbb Z_p) \approx (\SL_n(\mathbb Q)\cap \SL_n(\mathbb R)\prod_{p<\infty}\SL_n(\mathbb Z_p))\backslash\SL_n(\mathbb R) \\ \;=\; \SL_n(\mathbb Z)\backslash\SL_n(\mathbb R). \end{gather*} In particular, a right $\SL_n(\mathbb R)$-invariant measure on one gives the same on the other. The measures on the various $X_N$ can be normalized to make everything consonant… A possibly disappointing aspect of this viewpoint is that it gives no origin story, despite being an orthodox viewpoint. Another viewpoint, which in the long term may be more explanatory, is in terms of projective limits of classical quotients $X_N=\Gamma_N\backslash\SL_n(\mathbb R)$, where $\Gamma_N$ is the principal congruence subgroup of level $N$. When $M$ divides $N$, there is the natural surjection $X_N\to X_M$. Ordering positive integers by divisibility, the projective limit (a type of non-abelian solenoid) is naturally an object of interest. Already we could look at the proj lim of circles $\mathbb R/N\mathbb Z$, or even the proj lim of circles $\mathbb R/2^n\mathbb Z$. These solenoids were studied from a topological viewpoint in the early 1940s by Eilenberg and MacLane. The latter case gives $(\mathbb R\times \mathbb Q_2)/\mathbb Z[{1\over 2}]^\Delta$, where $\mathbb Q_2$ is the $2$-adic numbers. An interesting operational point is that $\mathbb Q_2$ acts on the family of circles $\mathbb R/2^n\mathbb Z$, although it does not act on the individuals. Similarly the proj lim of $\mathbb R/N\mathbb Z$ is $\mathbb A/\mathbb Q$. That is, in fact every $\mathbb Q_p$ acts on that family of circles, though not on individuals. Similarly, $\SL_n(\mathbb Q_p)$ acts on the projective family of classical arithmetic quotients $X_N$, though not on individuals. (The emergence of adelic things in some areas of physics also seems to be not just by fiat of definition, but because one takes a projective limit of "charge lattices", whatever those are. :) I don't know anything beyond those words…)<|endoftext|> TITLE: Integral isomorphism between $K_0(X)$ and $A(X)$ for toric varieties QUESTION [13 upvotes]: Let $X$ be a smooth projective toric variety. The Chern character gives an isomorphism of rings: $$\operatorname{Ch}:K_{0}(X)\otimes\mathbb{Q} \to A(X)\otimes \mathbb{Q} $$ where $K_{0}(X)$ is the Grothendieck group of vector bundles on $X$ and $A(X)$ is the Chow ring of $X$. This map seems only well-defined over $\mathbb{Q}$, but I was wondering (likely naively) if there is possibly an integral isomorphism (i.e. without tensor with $\mathbb{Q}$)? Why might we hope for such a map? Fulton and Strumfels showed that there exists an isomorphism $\mathcal{D}_{X}:A^{k}(X)\to \operatorname{Hom}(A_{K}(X),\mathbb{Z})$ where $A^{k}(X)$ and $A_{k}(X)$ are the Chow cohomology and homology groups respectively. In particular, this means that the Chow ring $A(X)$ of a smooth toric variety is torsion free. In the couple very (very) simple examples I've done $K_{0}(X)$ also seems torsion free, although I am unsure whether this is true generally. Of course, even if both $K_{0}(X)$ and $A(X)$ are torsion free there need not be an isomorphism between them, but one can hope. REPLY [5 votes]: For the permutahedral toric variety associated to the type A hyperplane arrangement, there is such an isomorphism -- see Section 10 of Berget-Eur-Spink-Tseng https://arxiv.org/abs/2103.08021 . I believe this isomorphism also descends to toric varieties with coarser fans, and many small examples (like e.g. projective space) are special cases of this.<|endoftext|> TITLE: Countable choice for countable subfamily QUESTION [5 upvotes]: Axiom of Countable Choice (CC) states that for every countable family $\left\{A_i\right\}_{i=1}^\infty$ of nonempty sets there exists choice function $f \colon \mathbb{N} \to \bigcup_{i=1}^\infty A_i$ such that $f(i) \in A_i$ for all $i \in \mathbb{N}$. Consider the following axiom strongly related to the above axiom: For every countably infinite family $\left\{A_i\right\}_{i=1}^\infty$ of nonempty sets there exists countably infinite subfamily $\left\{A_{i_k}\right\}_{k=1}^\infty$ such that $\left\{A_{i_k}\right\}_{k=1}^\infty$ has choice function i.e. there exists $f \colon \mathbb{N} \to \bigcup_{k=1}^\infty A_{i_k}$ such that $f(k) \in A_{i_k}$ for all $k \in \mathbb{N}$. Let's call it Axiom of Countable Choice for Countable Subfamily (CCCS). Of course CC implies CCCS. But I wonder if CCCS is strongly weaker than CC or equivalent to CC. I can't imagine being the first person to come up with the idea of CCCS, so probably there is literature which consider this kind of choice axiom or maybe it's unnaturally to consider such an axiom. I'm amateur in that sort of set theory. What do you think? Maybe someone knows proper literature? REPLY [8 votes]: They are equivalent. Given a countable family $B_i$ of nonempty sets, let $A_i=\prod_{j=1}^iB_j$. That these are nonempty is provable in ZF (we only use finite products). Now for any choice function $f$ with $f(k)\in A_{i_k}$, since $i_k\geq k$, projecting $f(k)$ onto the $k$-th coordinate of $A_{i_k}=\prod_{j=1}^{i_k}B_j$ gives a choice function for the $B_i$. I don't know a reference for this result, but I have seen similar arguments when establishing various equivalents of countable choice for subsets of $\mathbb R$, so I can't take full credit for the idea.<|endoftext|> TITLE: Average distance of the mean of $n$ random complex numbers in a unit disc QUESTION [6 upvotes]: Let $z_1,z_2,\dots,z_n$ be $n$ complex numbers distributed uniformly and randomly over the unit disc $x^2+y^2 \leq 1$. Let $z$ be the complex number defined by the mean of the of these numbers,that is, $$ z=\frac1n \sum_{i=1}^{n} z_i.$$ Numerical simulations reveal that $E(|z|)$ exists. I want to know if there is some analytical way of finding a closed form expression for the expected radial distance of $z$ in terms of $n$? Any hints/responses will be greatly appreciated . The graph shows the variation of $E(|z|)$ with $n$ REPLY [5 votes]: Here is another method. Since the uniform distribution on the unit disk is rotationally symmetric (invariant under orthogonal transformations), the problem can be reduced to a random walk problem in $\mathbb{R}$. Let $Z_1,Z_2,\ldots$ be independent, and identically uniformly distributed on the unit disk, with $Z_i=(X_i,Y_i)$. Clearly $X_1,X_2\ldots$ are i.i..d with density $p(x)=\frac{2}{\pi}\sqrt{1-x^2}1_{[-1,1]}(x)$. For any rotationally symmetric random vector $V=(V_1,V_2)$ with finite expectation of $|V|:=\sqrt{V_1^2+V_2^2}$ it holds that $$\mathbb{E}(|V|)=\frac{\pi}{2}\mathbb{E}(|V_1|)$$ Applying this to $Z^{(n)}:= |Z_1+\ldots+Z_n|$ we get that \begin{align*} \mathbb{E}\big(\frac{Z^{(n)}}{n}\big)&=\frac{1}{n}\frac{\pi}{2}\mathbb{E}|X_1+\ldots + X_n|\\ &=\frac{1}{n}\big(\frac{2}{\pi}\big)^{n-1}\,\int_{-1}^1\ldots\int_{-1}^1|x_1+\ldots+x_n|\prod_{i=1}^n\sqrt{1-x_i^2}\,dx_1\,\ldots dx_n\\ \end{align*} The integrals can be solved explicitly for $n=1$ and $n=2$, but that seems to be as far as it goes. But (using the central limit theorem and uniform integrability) it is easy to see that $$\mathbb{E}\big(\frac{Z^{(n)}}{\sqrt{n}}\big)\longrightarrow \sqrt{\frac{\pi}{8}}$$ Remarks: (1) a related problem appeared here An interesting triple integral (2) for similar considerations see Feller II (1971), p. 30 ff<|endoftext|> TITLE: 3-dimensional h-cobordisms QUESTION [7 upvotes]: Let $W$ be a $3$-dimensional $h$-cobordism of closed surfaces $M_0$ and $M_1$. Can we prove that $W$ is trivial? That is, $W$ is homeomorphic to $M_0 \times [0,1]$. REPLY [9 votes]: For $M_0=M_1=S^2$, this follows from the 3-dimensional Poincaré conjecture: glueing in 3-balls, you get a simply connected $3$-manifold, that has to be $S^3$, so $W=S^2\times\left[0,1\right]$. For surfaces of higher genus, the result follows from Waldhausens rigidity theorem, which says that homotopy-equivalent Haken manifolds with homeomorphic boundaries are homeomorphic. (An irreducible $3$-manifold with non-spherical boundary is necessarily a Haken manifold.)<|endoftext|> TITLE: Is cohomology always related to topology? QUESTION [15 upvotes]: I am trying to understand whether there is a sense in which cohomology always relates to topology or whether this is the case only in particular examples. According to the Wikipedia page, a cochain complex is defined as:  “… a sequence of abelian groups or modules ..., $A_0$, $A_1$, $A_2$, $A_3$, $A_4$, ... connected by homomorphisms $d^n$ : $A^n$ → $A^{n+1}$ satisfying $d^{n+1} \cdot d^n$ = 0. “ Based on this one defines the nth cohomology group $H^n$ as the group $H^n := ker \, d^n / \, im \, d^{n-1}$ . As a physicist, I am most familiar with examples of cochains and cohomology appearing in the context of topology. For instance, de Rham cohomology in which $d$ is identified with the exterior derivative and $A^n$ with the space of differential forms of degree $n$ on a smooth manifold $M$. In this case cohomology groups capture topological invariants of the manifold $M$.   Although the general definition above makes no explicit reference to a topological space, I often see the claim that cohomology is a branch of topology. In fact, from the mathoverflow page cohomology is defined as: "A branch of algebraic topology concerning the study of cocycles and coboundaries." Thus, my question is the following: Q: Is there a sense in which, given a general cochain complex, its cohomology captures topological invariants of some underlying topological space? Q': If so, what defines the underlying topological space?  Q'': If not, is there still a sense in which cohomology groups are some sort of “invariants?” REPLY [23 votes]: While it is always possible to introduce topology, it is not always the obvious or most useful thing to do. So at least in this sense, there are notions of cohomology that do not immediately connect to topology. As an example group (co)homology, Lie algebra (co)homology, Hochschild (co)homology, ... all appear in algebraic contexts without necessarily having topological interpretations all too closely connected to them. There are of course topological interpretations, e.g. group cohomology is the topology of the classifying space of the group, but that's not necessarily the most useful way to view them. More useful often is the connection to representation theory (says the guy with a degree in representation theory) which comes from the view point that they're all derived functors on some module category. EDIT: Because negatively-graded complexes were mentioned in the comments, let me just point you towards Tate cohomology which is a generalization of group cohomology to negative degrees.<|endoftext|> TITLE: Do $r(a) \leq^\oplus R$ and $r(a) = r(a^2)$ imply $r(a) = eR$ and $r(1-a) \subseteq (1-e)R$ for some idempotent $e$? QUESTION [5 upvotes]: Let $R$ be a (commutative or non-commutative) unital ring, fix $a \in R$, and denote by $r(\cdot)$ the right annihilator of an element. Question. If $r(a)$ is a (right) direct summand of $R$ and $r(a) = r(a^2)$, does there exist an idempotent $e \in R$ such that $r(a) = eR$ and $r(1-a) \subseteq (1-e)R$? It is obvious that $r(a) \cap r(1-a) = \{0\}$ and hence the sum $r(a) + r(1-a)$ is direct. Moreover, it is not difficult to check that $r(a) \oplus r(1-a) = r(a-a^2)$. So, we see that the answer to the question is yes when $R$ is a right Rickart ring, that is, a ring where every right annihilator is a direct summand: This is, in some sense, a trivial consequence of the definitions (let alone that it has nothing to do with the "if" clause of the question); however, it rules out a whole variety of potential counterexamples. The answer is still yes for (commutative or non-commutative) local rings. In a way, the reason is once again trivial: If $R$ is local, then either $a$ is a unit and then $r(a) = \{0\}$ (so we can take $e = 0$); or $a = 0$ and then $r(a) = R$ and $r(1-a) = \{0\}$ (so we can take $e = 1$); or $a$ is a non-zero non-unit and then $r(a)$ is not a direct summand of $R$, or else $R$ would have an idempotent that is neither $0$ nor $1$ (which, in a local ring, is just impossible). It is perhaps worth noting that $r(1-a) \subseteq aR$ and, if $r(a) = r(a^2)$, then $r(a) \cap aR = \{0\}$. REPLY [3 votes]: The answer is yes. Further, you don't need the condition $r(a)=r(a^2)$. Given an idempotent $e\in R$ such that $r(a)=eR$, let $e'=e(1-a)$. It is easy to check that, since $ae=0$, we have $e'^2=e'$. Also $eR=e'R$ since $e'=ee'$ and $e=e'e$. Finally, if $(1-a)x=0$ then $(1-e')x=[1-e(1-a)]x=x$, so $r(1-a)$ is contained in $(1-e')R$.<|endoftext|> TITLE: $\operatorname{STop}_{n,n-2}\simeq S^1$? QUESTION [5 upvotes]: $\DeclareMathOperator\STop{STop}$I am interested in any information about the homotopy type of the groups $\STop_{n,j}$ of homeomorphisms of $R^n$ preserving orientation and pointwise $R^j\subset R^n$. It is easy to see that $\STop_{n,n-1}$ is contractible, being homeomorphic to the square of the group of relative-to-the-boundary homeomorphisms of a disc. But what about $\STop_{n,n-2}$? How does it compare to $S^1$? At least for $n=3$ the answer is probably known. REPLY [7 votes]: The general result in this direction is due to Kirby-Siebenmann, Theorem B of Normal bundles for codimension 2 locally flat imbeddings: the map $$\mathrm{SO}(2) \longrightarrow \mathrm{STop}_{n,n-2}$$ is $(n-2)$-connected. (They state this for $n \neq 4$ but it is true in this case as well by the corresponding result for codimension 2 normal bundles due to Freedman-Quinn.)<|endoftext|> TITLE: Computing the homotopy groups $\pi_n(G,G_q)$ QUESTION [7 upvotes]: Let $G_q$ be the space of degree one maps $S^{q-1}\rightarrow S^{q-1}$ and $G=lim G_q$ under suspensions $G_q\subset G_{q+1}$. What are the known computations of $\pi_n(G,G_q)$? I found that $\pi_3(G,G_3)=\mathbb{Z}_2$ and $\pi_4(G,G_3)=\mathbb{Z}$ in Haefliger's https://doi.org/10.2307/1970475 section 5.16. REPLY [8 votes]: By Theorem A of Milgram's On the Haefliger knot groups, there is a $(2k-3)$-connected map $$\mathrm{SO}/\mathrm{SO}(k) \longrightarrow \mathrm{G}/\mathrm{G}_k,$$ and the left side is well-studied in the metastable range (see the work of Mahowald).<|endoftext|> TITLE: Revisiting the unreasonable effectiveness of mathematics QUESTION [16 upvotes]: Question: On balance, with theoretical advances in algorithmic information theory and Quantum Computation it appears that the remarkable effectiveness of mathematics in the natural sciences is quite reasonable. By effectiveness, I am generally referring to Wigner's observation that mathematical laws have remarkable generalisation power. Might there be a modern review paper on the subject for mathematicians where the original question is re-evaluated in light of modern mathematical sciences? An information-theoretic perspective: In order to motivate an information-theoretic analysis, it is worth observing that Occam's razor is an essential tool in the development of mathematical theories. From an information-theoretic perspective, a Universe where Occam's razor is generally applicable is one where information is conserved. The conservation of information would imply that fundamental physical laws are generally time-reversible. Moreover, given that Occam's razor has an appropriate formulation within the context of algorithmic information theory as the Minimum Description Length principle this information-theoretic analysis generally presumes that the Universe itself may be simulated by a Universal Turing Machine. David Deutsch and others have done significant work demonstrating the plausibility of the Physical Church-Turing thesis(which is consistent with the original Church-Turing thesis) and this would explain why mathematical methods are so effective in the natural sciences. This brief analysis has emerged from informal discussions with a handful of algorithmic information theorists(Hector Zenil, Marcus Hutter, and others) and it makes me wonder whether complementary theories from mathematical physics might help mathematicians account for the remarkable effectiveness of mathematics in the natural sciences. Clarification of particular terms: Minimum Description Length principle: Given data in the form of a binary string $x \in \{0,1\}^*$ the Minimum Description Length of $x$ is given by the Kolmogorov Complexity of $x$: \begin{equation} K_U(x) = \min_{p} \{|p|: U(p) = x\} \end{equation} where $U$ is a reference Universal Turing Machine and $p$ is the shortest program that takes as input the empty string $\epsilon$ and outputs $x$. The Law of Conservation of Information: The Law of Conservation of information which dates back to Von Neumann essentially states that the Von Neumann entropy is invariant to Unitary transformations. This is meaningful within the framework of Everettian quantum mechanics as a density matrix may be assigned to the state of the Universe. This way information is conserved as we run a simulation of the Universe forwards or backwards in time. The Physical Church-Turing thesis: The Law of Conservation of information is consistent with the observation that all fundamental physical laws are time-reversible and computable. The research of David Deutsch(and others) on the Physical Church-Turing thesis explains how a Universal Quantum computer may simulate these laws. Michael Nielsen wrote a good introductory blog post on the subject [7]. The Physical Church-Turing thesis is a key point in this discussion as it provides us with a credible explanation for the remarkable effectiveness of mathematics in the natural sciences. A remark on effectiveness: What I have retained from my discussions with physicists and other natural scientists is that the same mathematical laws with remarkable generalisation power are also constrained by Occam's razor. In fact, from an information-theoretic perspective the remarkable effectiveness of mathematics is a direct consequence of the effectiveness of Occam's razor. This may be partly understood from a historical perspective if one surveys the evolution of ideas in physics [10]. Given two compatible theories, Einstein generally argued that one should choose the simplest theory that yields negligible experimental error. To be precise, he stated: It can scarcely be denied that the supreme goal of all theory is to make the irreducible basic elements as simple and as few as possible without having to surrender the adequate representation of a single datum of experience.-Einstein(1933) As the application of Occam's razor generally requires a space of computable models, and algorithmic information theory carefully explains why simpler theories generalise better [8] it is fair to say that the notion of effectiveness may be made precise. However, the theory of algorithmic information was developed in the mid 1960s by Chaitin, Kolmogorov and Solomonoff which was after Wigner wrote his article in 1960. What is remarkable: If we view the scientific method as an algorithmic search procedure then there is no reason, a priori, to suspect that a particular inductive bias should be particularly powerful. This much was established by David Wolpert via his No Free Lunch Theorems [11]. On the other hand, the history of natural science indicates that Occam's razor is remarkably effective. The effectiveness of this inductive bias has more recently been explored within the context of deep learning [12]. References: Eugene Wigner. The Unreasonable Effectiveness of Mathematics in the Natural Sciences. 1960. David Deutsch. Quantum theory, the Church–Turing principle and the universal quantum computer. 1985. Peter D. Grünwald. The Minimum Description Length Principle . MIT Press. 2007. A. N. Kolmogorov Three approaches to the quantitative definition of information. Problems of Information and Transmission, 1(1):1--7, 1965 G. J. Chaitin On the length of programs for computing finite binary sequences: Statistical considerations. Journal of the ACM, 16(1):145--159, 1969. R. J. Solomonoff A formal theory of inductive inference: Parts 1 and 2. Information and Control, 7:1--22 and 224--254, 1964. Michael Nielsen. Interesting problems: The Church-Turing-Deutsch Principle. 2004. https://michaelnielsen.org/blog/interesting-problems-the-church-turing-deutsch-principle/ Marcus Hutter et al. (2007) Algorithmic probability. Scholarpedia, 2(8):2572. Andrew Robinson. Did Einstein really say that? Nature. 2018. The Evolution of Physics, Albert Einstein & Leopold Infeld, 1938, Edited by C.P. Snow, Cambridge University Press Wolpert, D.H., Macready, W.G. (1997), "No Free Lunch Theorems for Optimization", IEEE Transactions on Evolutionary Computation 1, 67. Guillermo Valle Pérez, Chico Camargo, Ard Louis. Deep Learning generalizes because the parameter-function map is biased towards simple functions. 2019. REPLY [2 votes]: The French logician Jean-Louis Krivine gives an interesting perspective in his essay Wigner, Curry et Howard. My simplified summary of his explanation of the "unreasonable" effectiveness is that rather than simply "discovering" theorems, mathematicians "reverse-engineer" them from the human brain, where they were written by evolution. The second-to-last paragraph summarizes a core part of Krivine's argument: Nous venons de nous apercevoir que les lois mathématiques sont, en fait, des programmes écrits dans la mémoire morte de notre cerveau. Dire que le monde physique se conforme à de telles lois revient donc à dire qu’il se conforme à des programmes écrits dans notre cerveau. L’anthropocentrisme ridicule de cette affirmation saute aux yeux, nous voilà revenus au temps où les planètes, le soleil et les étoiles tournaient autour de la terre et où les oranges étaient divisées en quartiers pour que nous puissions les consommer plus facilement. Wigner a tout à fait raison de trouver cela « déraisonnable ». In English: We just observed that the mathematical laws are in fact programs that are hard-coded in our brains. To say that the physical world conforms to these laws thus amounts to saying that it conforms to programs written in our brain. The ridiculous anthropocentrism of this statement jumps to the eyes, we have returned to the time where the planets, the sun and the stars turned around the earth and where the oranges where divided in quarters so that we could consume them more easily. Wigner is indeed right to find this "unreasonable". Thus Krivine dismisses the question, perhaps in a way similar to Lützen as quoted in Carlo Beenakker's answer.<|endoftext|> TITLE: Real rootedness of a polynomial with binomial coefficients QUESTION [10 upvotes]: It is possible to show using diverse techniques that the following polynomial: $$P_n(x)=1 + \binom{n}{2} x + \binom{n}{4} x^2 + \binom{n}{6} x^3 + \binom{n}{8} x^4 +\ldots + \binom{n}{2\lfloor\tfrac{n}{2}\rfloor} x^{\lfloor \frac{n}{2}\rfloor},$$ is real-rooted. For instance, it is an $s$-Eulerian polynomial, for which Savage and and Visontai in [1] have proved real-rootedness. Here I ask for a very similar polynomial, which I think might be solved with some other technique that perhaps I am overlooking. $$Q_n(x)=1 + \left(\binom{n}{2}-n\right) x + \binom{n}{4} x^2 + \binom{n}{6} x^3 + \binom{n}{8} x^4 +\ldots + \binom{n}{2\lfloor\tfrac{n}{2}\rfloor} x^{\lfloor \frac{n}{2}\rfloor},$$ This polynomial $Q_n(x) = P_n(x) - nx$ is the Ehrhart $h^*$-polynomial of the hypersimplex $\Delta_{2,n}$. There are several conjectures regarding unimodality/log-concavity/real-rootedness for the $h^*$-polynomial of an arbitrary hypersimplex $\Delta_{k,n}$, but for $k > 2$ the formulas are much more complicated. REPLY [12 votes]: Let's consider $n\geqslant 5$ case only ($n=1,2,3,4$ are straightforward). Then $Q_n$ has non-negative coefficients and we care on the number of negative roots of $Q_n$. We have $2P_n(-x)=(1+i\sqrt{x})^n+(1-i\sqrt{x})^n$. So, we should prove that $$2Q_n(-x)=(1+i\sqrt{x})^n+(1-i\sqrt{x})^n+2nx$$ has $\lfloor n/2\rfloor$ positive roots. Denote $x=\tan^2 t$, $0 TITLE: Zeroes of the Alexander polynomial for achiral knots QUESTION [13 upvotes]: Are there some known properties about the position (on the complex plane) of roots of the Alexander polynomial of achiral knots? They are shown as blue points in the following picture of roots for knots up to 16 crossings. Specifically what about those sitting in the middle of the holes around chiral ones. Added the zoom-out picture: Zeros (in the half unit disc) of the Alexander polynomial for knots up to 16 crossings (red - chiral,alternating; green - chiral,nonalternating; blue - achiral) blue on top of red on top of green Edit: A fragment of the roots for the positive amphichiral knots (red dots): ------------------- Edit: In the picture below we indicate (by orange dots) the roots (of multiplicity more than 1) of Alexander polynomials of prime knots up to 15 crossings (313.230 knots, regardless of their chirality). The number shows the maximal multiplicity among all polynomials. As suggested by this picture, the holes and their centers have much to do with the multiplicity of roots after all. As in the suggested paper by Hartley, for positive amphicheiral knots, the roots are more frequently have higher (than one) multiplicity because they "potentially" factors through the other polynomials (at least to the second power). The isolated roots in the interior are from Alexander polynomials like: $(t^4-3t^3+5t^2-3t+1)^2$, $(t^6-2t^5+4t^4-5t^3+4t^2-2t+1)^2$, $(t^6-5t^5+12t^4-15t^3+12t^2-5t+1)^2$, $(t^4-5t^3+9t^2-5t+1)^2$, $(2t^4-6t^3+9t^2-6t+2)^2$, $(2t^4-7t^3+11t^2-7t+2)^2$, $(t^6-3t^5+5t^4-5t^3+5t^2-3t+1)^2$ ... so they are squares of Alexander polynomials for some knots. REPLY [5 votes]: This is an extended comment, and a response to one of my earlier comments. The image below depicts the roots of all integer-coefficient Laurent polynomials $$p(t) = \sum_i a_i t^i$$ that satisfy $p(t^{-1}) = p(t)$, $p(1)=1$ that are of degree at most 6 with coefficients at most 6. i.e. the sum is over $i=-6 \cdots 6$ and the coefficients $a_i \in \{-6, \cdots, 6\}$. These latter limitations were to help make the computation reasonably-long. As in your figure, I only show the roots inside the unit disc with non-negative imaginary part. I think the main difference between your image and this one must come down to the manner in which the polynomials are generated. I believe you are getting Alexander polynomials of much higher degree, but also your coefficients are not uniformly distributed like in the way I generate the polynomials. So I think your plot is very much a reflection of the "shape" of Alexander polynomials, parametrized by the crossing number. As you have observed, it appears certain regions in the space of possible Alexander polynomials are more rapidly filled-in by the amphichiral knots. Perhaps there is a construction that can produce low-crossing amphichiral knots that can also produce a sequence of "approximating" chiral knots, where "approximating" is in terms of roots of the Alexander polynomial. If you choose the polynomials to be at most degree 26, with coefficients between -2 and 2, uniformly and randomly you get something even further from your plot. I tried a less even-handed search through Alexander polynomials, biased towards polynomials that are weakly alternating, like the typical polynomials one sees in knot tables. Interestingly (maybe only to me), the answer appears to be even further from what you are getting. Clearly the phenomena I see in the tables isn't enough to characterize what is happening.<|endoftext|> TITLE: Why is 1331 the only cube of the form $x^2 + x − 1$? QUESTION [22 upvotes]: The Wikipedia (https://en.wikipedia.org/wiki/1000_(number)#1300_to_1399) mentions that 1331 is the only cube of the form $x^2 + x − 1$, for $x = 36$. What is the proof? REPLY [17 votes]: Here is the more elementary way that one would have solved this before LLL. I will still use the computer for some calculations, but technically one could do all this by hand. Worse, I won't even do all of the steps, but I hope the part I show explains the method, for those that may be interested. Maybe someone wants to complete it or finds a shorter way. First, setting $K=\mathbb{Q}(\theta)$ with $\theta^2+\theta-1$, a.k.a the Golden ratio, we will rewrite our equation as $$N_{K/\mathbb{Q}}(y-\theta) = x^3$$ to be solved in $X$ and $y$ in $\mathbb{Z}$. The field $K$ has class number 1 and the units are generated by $\theta$. From this, using a little argument about the $\sqrt{5}$-adic valuation, one can show that, there must be a $i\in \{0,1,2\}$ such that $$y-\theta = \theta^i\cdot \alpha^3$$ for some $\alpha\in\mathbb{Z}[\theta]$, the ring of integers of $K$. Write $\alpha = a+b\,\theta$ for $a,b\in\mathbb{Z}$ and split into the three cases according to $i$. Case $i=0$ : Spanning out $\alpha^3$ and equating the coefficients in front of $\theta$, we get $$\begin{align*} y &= a^3+3ab^2-b^3 \\ -1&=b\cdot(3a^2-3ab+2b^2)\end{align*}$$ Now the second line tells us that $b=\pm 1$, but neither of the choices allows us to find $a$ as an integer, so that case does not occur. Case $i=1$ : This time the equations are $$\begin{align*} y &= 3a^2b-3ab^2+2b^3\\-1 &=a^3-3a^2b+6ab^2-3b^3\end{align*}$$ One can note here aside that the second line is a model of the curve $3$-isogenous to the original curve over $\mathbb{Q}$. This second (Thue) equation, we rewrite as $$-1 = N_{L/\mathbb{Q}}(a-b\xi)$$ where $L=\mathbb{Q}(\xi)$ and $\xi^3-3\xi^2+6\xi-3=0$. This $L$ has also trivial class number and the units are generated by $1-\xi$ of norm $1$. We are now looking for units of norm $-1$ in $L$ with no $\xi^2$ term. Once more we split into three cases $j\in\{0,1,2\}$. We want to solve $$a+b\xi = - (1-\xi)^j\cdot(1-\xi)^{3k} = -(1-\xi)^j\cdot(-2+3\xi)^k.$$ Subcase $j=0$ : We may rewrite it as $$a-b\xi = - (1+3(\xi-1))^k = -\bigl(1+3(\xi-1)k+9(\xi-1)^2k(k-1)/2 + 27\cdots \bigr)$$ and consider this as a power series in $k$ over $\mathbb{Q}_3(\xi)$, which is a totally ramified extension of $\mathbb{Q}_3$. We can compare the power series coefficient in front of $\xi^2$, to get an equation of the form $$0=9 k (k-1)\cdot \bigl(1/2 -9/24 (k-2)(k-3)+\cdots\bigr).$$ By Strassmann's theorem, the power series has at most two solutions in $\mathbb{Q}_3$. But it is easy to spot that there are two such solutions, namely $(a=-1,b=0,k=0)$ and $(a=2,b=3,k=1)$. These correspond to $(x,y)=(-1,0)$ and $(11,36)$ repsectively. Subcase $j=1$ : leads to a single solution $a-b\xi=-1+\xi$ which corresponds to $(x,y)=(1,-2)$. Subcase $j=2$ : produces a unit power series and has therefore no solutions. Case $i=2$ : I believe one just gets the $-P$ for all the $P$ found in the case $i=1$. But I admit that I have not checked. Finally, I should say that for questions on integral points on elliptic curve I often turn to Smart's "The algorithmic Resolution of Diophantine Equations" that explains things well. Like the $3$-adic method used in the subcases in chapter III.